← Subtoposes of actions of free monoids
main.tex
% !TEX TS-program = pdflatex
% !TEX encoding = UTF-8 Unicode
\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
%%% PAGE DIMENSIONS
\usepackage{geometry} % to change the page dimensions
\geometry{a4paper}
\usepackage{graphicx} % support the \includegraphics command and options
%%% PACKAGES
\usepackage{booktabs}
\usepackage{array}
\usepackage{paralist}
\usepackage{verbatim}
\usepackage{subfig}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathrsfs}
\usepackage{stmaryrd}
\usepackage{bbm}
\usepackage{mathtools}
\usepackage{textcomp}
\usepackage{xcolor}
\usepackage{listings}
\usepackage{float}
\usepackage{centernot}
\usepackage{showkeys}
\usepackage[mathscr]{euscript}
\usepackage{tikz-cd}
\usepackage{scalerel}
\usepackage{stackengine}
\usepackage{quiver}
%%% refs
\usepackage[shortlabels]{enumitem}
\renewcommand{\labelitemi}{$-$}
\usepackage{hyperref}
\hypersetup{
colorlinks = true,
linkbordercolor = {red},
linkcolor ={teal},
anchorcolor = {pink},
citecolor = {orange},
filecolor = {teal},
menucolor = {teal},
runcolor = {teal},
urlcolor = {teal},
}
\usepackage{cleveref}
\setcounter{tocdepth}{2}
%%% Macros
\newcommand{\mr}[1]{\textcolor{orange}{#1}}
\newcommand{\hr}[1]{\textcolor{blue}{#1}}
\newcommand{\hi}[1]{\textcolor{red}{#1}}
\newcommand{\Acal}{\mathcal{A}}
\newcommand{\Bcal}{\mathcal{B}}
\newcommand{\Ccal}{\mathcal{C}}
\newcommand{\Dcal}{\mathcal{D}}
\newcommand{\Ecal}{\mathcal{E}}
\newcommand{\Fcal}{\mathcal{F}}
\newcommand{\Gcal}{\mathcal{G}}
\newcommand{\Hcal}{\mathcal{H}}
\newcommand{\Ical}{\mathcal{I}}
\newcommand{\Jcal}{\mathcal{J}}
\newcommand{\Kcal}{\mathcal{K}}
\newcommand{\Lcal}{\mathcal{L}}
\newcommand{\Mcal}{\mathcal{M}}
\newcommand{\Ocal}{\mathcal{O}}
\newcommand{\Pcal}{\mathcal{P}}
\newcommand{\Qcal}{\mathcal{Q}}
\newcommand{\Rcal}{\mathcal{R}}
\newcommand{\Scal}{\mathcal{S}}
\newcommand{\Tcal}{\mathcal{T}}
\newcommand{\Ucal}{\mathcal{U}}
\newcommand{\Wcal}{\mathcal{W}}
\newcommand{\Zcal}{\mathcal{Z}}
\newcommand{\Ibb}{\mathbb{I}}
\newcommand{\Jbb}{\mathbb{J}}
\newcommand{\Xbb}{\mathbb{X}}
\newcommand{\Cbb}{\mathbb{C}}
\newcommand{\Dbb}{\mathbb{D}}
\newcommand{\Fbb}{\mathbb{F}}
\newcommand{\Tbb}{\mathbb{T}}
\newcommand{\Nbb}{\mathbb{N}}
\newcommand{\Zbb}{\mathbb{Z}}
\newcommand{\yon}{\mathbf{y}}
\newcommand{\Set}{\mathbf{Set}}
\newcommand{\Pos}{\mathbf{Pos}}
\newcommand{\Frm}{\mathbf{Frm}}
\newcommand{\Loc}{\mathbf{Loc}}
\newcommand{\Top}{\mathbf{Top}}
\newcommand{\PFrm}{\mathbf{PFrm}}
\newcommand{\CLat}{\mathbf{CLat}}
\newcommand{\CmSLat}{\mathbf{CmSLat}}
\newcommand{\MSL}{\mathbf{MSL}}
\newcommand{\Rel}{\mathbf{Rel}}
\newcommand{\Hom}{\mathrm{Hom}}
\newcommand{\End}{\mathrm{End}}
\newcommand{\Cat}{\mathbf{Cat}}
\newcommand{\CwP}{\mathbf{CwP}}
\newcommand{\CAT}{\mathbf{CAT}}
\newcommand{\op}{^{\mathrm{op}}}
\newcommand{\fibtop}[1]{\Lcal_{#1}}
\newcommand{\fibtopp}[1]{\Rcal_{#1}}
\newcommand{\fibtoq}[1]{\Wcal_{#1}}
\newcommand{\Chat}{[\Ccal\op{,}\Set]}
\newcommand{\Dhat}{[\Dcal\op{,}\Set]}
\newcommand{\lcorner}{\rotatebox{45}{$\lrcorner$}}
\newcommand{\rcorner}{\rotatebox{45}{$\ulcorner$}}
\newcommand{\ucorner}{\rotatebox{45}{$\urcorner$}}
\newcommand{\dcorner}{\rotatebox{45}{$\llcorner$}}
\newcommand{\too}{\twoheadrightarrow}
\newcommand{\inc}[1]{\mathbf{i}_{#1}}
\newcommand{\shf}[1]{\mathbf{a}_{#1}}
\newcommand{\vsim}{\rotatebox[origin=cc] {90} {$\sim$}}
\newcommand{\pf}{\trianglelefteq}
\newcommand{\loc}{\mathrm{Loc}}
\DeclareMathOperator{\sq}{Tr}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\ob}{ob}
\DeclareMathOperator{\mor}{mor}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\dom}{dom}
\DeclareMathOperator{\cod}{cod}
\DeclareMathOperator{\PSh}{PSh}
\DeclareMathOperator{\Sh}{Sh}
\DeclareMathOperator{\sh}{sh}
\DeclareMathOperator{\Sep}{Sep}
\DeclareMathOperator{\Geom}{Geom}
\DeclareMathOperator{\Cont}{Cont}
\DeclareMathOperator{\pt}{pt}
\DeclareMathOperator{\inter}{int}
\DeclareMathOperator{\Map}{Map}
\DeclareMathOperator{\Flat}{Flat}
\DeclareMathOperator{\Max}{Max}
\DeclareMathOperator{\Sub}{Sub}
\DeclareMathOperator{\dSub}{dSub}
\DeclareMathOperator{\Reln}{Rel}
\DeclareMathOperator*{\colim}{colim}
\DeclareMathOperator{\Ran}{Ran}
\DeclareMathOperator{\Lan}{Lan}
\DeclareMathOperator{\Ind}{Ind-}
\DeclareMathOperator{\Pow}{P}
\DeclareMathOperator{\Ideal}{Ideal}
\DeclareMathOperator{\RadIdeal}{RadIdeal}
\DeclareMathOperator{\Closed}{Closed}
\DeclareMathOperator{\rad}{rad}
\DeclareMathOperator{\supp}{su}
\DeclareMathOperator{\bang}{!}
%%%THEOREM FORMATTING
\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{crly}[thm]{Corollary}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{schl}[thm]{Scholium}
\theoremstyle{definition}
\newtheorem{dfn}[thm]{Definition}
\newtheorem{xmpl}[thm]{Example}
\theoremstyle{remark}
\newtheorem{rmk}[thm]{Remark}
\newtheorem{rmks}[thm]{Remarks}
\theoremstyle{definition}
\newtheorem{question}[thm]{Question}
\theoremstyle{definition}
\newtheorem{construction}[thm]{Construction}
\theoremstyle{definition}
\newtheorem{warning}[thm]{Warning}
%%% HEADERS & FOOTERS
\usepackage{fancyhdr}
\pagestyle{fancy}
\renewcommand{\headrulewidth}{0pt}
\lhead{}\chead{}\rhead{}
\lfoot{}\cfoot{\thepage}\rfoot{}
\date{}
\title{Subtoposes of actions of free monoids}
\author{Ryuya Hora, Morgan Rogers}
\date{}
\begin{document}
\maketitle
\tableofcontents
\section{Introduction}
The authors of the present paper have extensively studied toposes whose objects are actions of topological monoids on sets, both in general \cite{TTMA,Endomodels} % could also mention MPaTP, FitzGerald, EssHype.
and in special cases \cite{horakamio}.
Such actions are subject to a continuity condition, and the resulting categories are certain (full, left exact) \textit{coreflective} subcategories of presheaf toposes.
However, a natural question that one may ask when confronted with a topos of monoid actions, as with any topos, is: \textbf{`What are its subtoposes?'} That is, what properties characterize the actions belonging to the (full, left exact) \textit{reflective} subcategories? These correspond to Grothendieck coverages\footnote{These are sometimes called Grothendieck topologies, but we avoid that terminology here because it clashes with the ordinary notion of topology.} on the monoid, or equivalently Lawvere-Tierney operators\footnote{We avoid the name `topologies' for this too.} in the category of actions.
The primary goal of the present paper is to give a classification of the subtoposes of categories of actions of \textit{free} monoids, and then to exploit that classification to understand subtoposes for more general monoids.
The majority of the paper is spent developing this classification by exploiting the topos-theoretic machinery of étendues. This was not our original approach: originally, we attempted to find combinatorial descriptions of Grothendieck coverages. Since we feel that the original approach is of independent interest from a combinatorial point of view and has potential pedagogical value, we include it in the Appendix.
In \Cref{sec:prelims} we recall the locale and topos theory required for the main results. We have tried to make these sections self-contained in the interest of less experienced readers. %Some complementary background material is included in the Appendix. ...
%While much of our classification was achieved through an approach from first principles, the final version of the paper applies some topos-theoretic machinery to more elegantly reach the same conclusions. As such, we spend much of \Cref{sec:prelims} recalling some structures naturally associated to free monoids and then the requisite topos-theoretic machinery before putting these pieces together in \Cref{sec:coverages}. %We have attempted to make this study as comprehensive as possible by examining all of the standard forms of data that determine subtoposes. We have also made an effort to include relevant definitions to the benefit of readers coming from semigroup theory.
[Automata theory etc]
\mr{To include: / To do:}
\begin{itemize}
\item More Hora papers.
\item (Andrew Pitts spoke about) Sam Staton's work on certain toposes of topological monoid actions
\item Incorporate self-similar sublocales into earlier discussion.
\item Figure out whether internal sublocales of Cantor space in $\PSh(\Sigma^*)$ really do correspond to self-similar sublocales (and hence to subtoposes we are constructing).
\item \mr{Consider whether characterization from thesis can complete observations of which toposes arise.}
\end{itemize}
Note that throughout we use classical reasoning, although we try to point out where we use non-constructive reasoning (\Cref{rmk:fan}, \Cref{rmk:notnotrank}) to make it easier for interested readers to determine when these results may be extended to an intuitionistic setting.
\section{Preliminaries}
\label{sec:prelims}
We recommend that a reader with experience in topos theory read Sections \ref{ssec:Cayley} and \ref{ssec:prefix} to understand our chosen notation and then skip to \Cref{ssec:more} to see the essential topos-theoretic tools we employ.
\subsection{Actions and their Cayley graphs}
\label{ssec:Cayley}
Throughout, let $\Sigma$ be a set and $\Sigma^*$ the free monoid on this set. We call the elements of $\Sigma^*$ \textit{words} and write $\varepsilon$ for the empty word.
Given a monoid $M$, we write $\PSh(M)$ for its category of right acts (also known as right $M$-sets or right $M$-actions). This notation is derived from the fact that when $M$ is viewed as a one-object category, the category of presheaves on $M$ (that is, functors $M\op \to \Set$ and natural transformations between these) is equivalent to the category of right acts of $M$.
\begin{dfn}
\label{dfn:Cayley}
Given a right $M$-act $X$ and a set $\Sigma$ of generators of $M$, we can construct its \textbf{Cayley graph} as follows: the vertices of the graph are the elements of $X$. Each vertex $x$ has an outgoing edge for each element $a \in \Sigma$ whose target is $x \cdot a$.
\end{dfn}
\begin{xmpl}
\label{exa:Cayley}
Consider the special case where $M = \Sigma^*$ is a free monoid and $X = \Sigma^*$ with its canonical right action. The Cayley graph (with respect to the generating set $\Sigma$) is a $|\Sigma|$-regular tree, meaning that each vertex has a set of children indexed by $\Sigma$. For brevity, we will refer to this tree throughout as the \textbf{Cayley graph of $\Sigma^*$}, and denote it $G_{\Sigma}$.
\end{xmpl}
%We will primarily use the Cayley graph in the combinatorial version in the Appendix.
\subsection{Sequence spaces and prefixes}
\label{ssec:prefix}
While the Cayley graph is a useful combinatorial structure to have to hand for conceptual reasons, there are further derived structures that will naturally arise in our attempt to classify subtoposes of $\PSh(M)$. The first of these are \textit{sequence spaces} which we define now, and which we will eventually examine using the theory of \textit{locales}, recalled in the next subsection.
Write $\Sigma^{\omega}$ for the set of countably infinite sequences of elements of $\Sigma$. We call the elements of $\Sigma^{\omega}$ \textit{infinite words}. $\Sigma^*$ has a left action on $\Sigma^{\omega}$ by concatenation of a finite word at the start of an infinite word. We write $\Sigma^{\leq \omega} := \Sigma^* \cup \Sigma^{\omega}$ for the set of finite and infinite words (without a monoid structure).
Endowing $\Sigma^{\omega}$ with the product topology, we obtain the \textbf{sequence space} on $\Sigma$. It will behoove us to give a more explicit presentation of the product topology on $\Sigma^{\omega}$, and for this we need to consider prefixes.
\begin{dfn}
\label{dfn:prefix}
Let $u \in \Sigma^*$ and $v \in \Sigma^{\leq \omega}$. We say $u$ is a \textbf{prefix} of $v$, denoted $u \pf v$ if there exists $w \in \Sigma^{\leq \omega}$ with $v = uw$; we write $u \not\pf v$ when $u$ is not a prefix of $v$.
\end{dfn}
A prebase for the product topology on $\Sigma^{\omega}$ is given by the collection of subsets of the form
\begin{equation}
\label{eq:U(u)}
U(u) := \{v \in \Sigma^{\omega} \mid u \pf v\},
\end{equation}
for $u \in \Sigma^*$. %Observe that one can identify an open set in this topology with a downward-closed subset of $G_{\Sigma}$.
A related topological space which we will need to consider is the \textbf{extended sequence space}, having $\Sigma^{\leq \omega}$ as its set of points. The topology is defined analogously, via basic opens
\begin{equation}
\label{eq:hatU(u)}
\hat{U}(u) := \{v \in \Sigma^{\leq \omega} \mid u \pf v\}.
\end{equation}
The essential distinction between sequence space and extended sequence space is that,
\[U(u) = \bigcup_{a \in \Sigma} U(ua)
\hspace{10pt} \text{ whereas } \hspace{10pt}
\hat{U}(u) \supsetneq \bigcup_{a \in \Sigma} \hat{U}(ua),\]
since in the latter case only the left-hand side contains $u$.
\begin{rmk}
\label{rmk:CantorBaire}
When $|\Sigma| = 2$, $\Sigma^\omega$ is conventional \textbf{Cantor space}. Homeomorphic sequence spaces are obtained when $2 \leq |\Sigma| < \aleph_0$. When $|\Sigma| = \aleph_0$, the sequence space $\Sigma^\omega$ coincides with \textbf{Baire space}.
\end{rmk}
\begin{rmk}
\label{rmk:fan}
One arguably non-constructive assumption we employ throughout is (an infinitary version of) the \textit{fan theorem}, which is equivalent to the assertion that the locale constructed as the infinite product $\Sigma^{\omega}$ is spatial--that there are `enough' infinite sequences to distinguish the opens sets as they are defined above. However, the consequences of refuting this result would be mostly linguistic, since we never rely on points to distinguish opens in $\Sigma^{\omega}$. \mr{I think, but need to check before submitting.}
\end{rmk}
In what follows, we will simply use $\Sigma^{\omega}$ and $\Sigma^{\leq \omega}$ to denote the respective topological spaces and write $\Ocal(\Sigma^{\omega})$ and $\Ocal(\Sigma^{\leq \omega})$ to denote their respective frames of open sets. Note that since the left action of $\Sigma^*$ is continuous, it induces a \textit{right} action on the topologies, via:
\begin{equation}
\label{eq:opensetaction}
U \cdot w := \{u \in \Sigma^{\omega} \mid wu \in U\}
\end{equation}
and similarly for $\Ocal(\Sigma^{\leq \omega})$. By inspection, this action commutes with the operations of intersection and union. In fact, these frames satisfy the Beck-Chevalley conditions required for them to constitute internal locales in $\PSh(\Sigma^*)$. In what follows, we shall need to extend this right action to arbitrary subsets, with an identical definition:
\begin{equation}
\label{eq:closedsetaction}
C \cdot w := \{u \in \Sigma^{\omega} \mid wu \in C\}.
\end{equation}
\begin{rmk}
\label{rmk:pfspec}
Observe that the specialization ordering on $\Sigma^{\leq \omega}$ is exactly the prefix ordering. Since points in any sublocale are closed under directed colimits with respect to this ordering, a sublocale contains an infinite word $w$ as soon as it contains all finite prefixes of $w$.
\end{rmk}
\subsection{Locales}
\label{ssec:locale}
While we have presented sequence spaces from a classical, topological point of view, we will actually need to consider these spaces as \textit{locales}, the objects of point-free topology. The localic approach takes the opens as primary and the points as secondary. While we refer to Mac Lane and Moerdijk's text \cite{MLM} for elementary results, a more comprehensive reference is that of Picado and Pultr \cite{PicadoPultr}, especially Chapters III and VI on sublocales.
\begin{dfn}
\label{def:locale}
Recall that a \textbf{frame} is a completely distributive lattice: a lattice having finite meets ${\wedge}$ and all joins ${\bigvee}$ such that for any element $a$ and set of elements $\{b_i \mid i \in I\}$ we have,
\[a \wedge \bigvee_{i \in I} b_i = \bigvee_{i \in I} (a \wedge b_i). \]
In particular, the collection $\Ocal(X)$ of open subsets of a topological space $X$ is always a frame. A frame homomorphism is a function preserving finite meets and all joins. We denote the category of frames and homomorphisms by $\Frm$.
A \textbf{locale} is an object of $\Loc := \Frm\op$. To distinguish locales from frames, we write a letter $L$ for the locale and $\Ocal(L)$ for the corresponding frame, so that a locale map $L \to L'$ is no more than a frame homomorphism $\Ocal(L') \to \Ocal(L)$.
\end{dfn}
This convention is chosen because the opens of a topological space $X$ form a frame $\Ocal(X)$ (beware the overloading of notation), and a continuous map between topological spaces $X \to X'$ induces a frame homomorphism $\Ocal(X') \to \Ocal(X)$, so we have a functor $\loc:\Top \to \Loc$ from the category $\Top$ of topological spaces and the category $\Loc$ of locales.
This functor is part of a canonical adjunction (cf. \cite[Theorem IX.3.1]{MLM}):
% https://q.uiver.app/#q=WzAsMixbMiwwLCJUb3AiXSxbMCwwLCJMb2MiXSxbMSwwLCJcXG1hdGhybXtwdH0iLDIseyJvZmZzZXQiOjJ9XSxbMCwxLCJcXG1hdGhybXtMb2N9IiwyLHsib2Zmc2V0IjoyfV0sWzMsMiwiIiwyLHsibGV2ZWwiOjEsInN0eWxlIjp7Im5hbWUiOiJhZGp1bmN0aW9uIn19XV0=
\[\begin{tikzcd}[ampersand replacement=\&]
\Loc \&\& \Top
\arrow[""{name=0, anchor=center, inner sep=0}, "{\mathrm{pt}}"', shift right=2, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{\mathrm{Loc}}"', shift right=2, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}\]
We can view the left adjoint as simply `forgetting points'. The right adjoint sends a locale to its space of points, with underlying set the collection of locale maps $1 \to L$, where $1$ is the terminal locale (having $\Ocal(1) = \{\bot \leq \top\}$). The fixed points of this adjunction are respectively called \textit{spatial locales} and \textit{sober spaces}.
A potentially jarring aspect of locale theory for the uninitiated is that neither functor is conservative. On one hand, the $\Loc$ functor is unable to keep track of topologically indistinguishable points; on the other, there are non-trivial locales without points and distinct locales may have the same space of points. We shall encounter a number of examples of the latter phenomenon in the present work.
Our classification of subtoposes of $\PSh(\Sigma^*)$ will lead us to consider \textit{sublocales} of the sequence space $\Sigma^{\omega}$. As Mac Lane and Moerdijk explain towards the end of \cite[\S IX.4]{MLM}, the $\mathrm{Loc}$ functor preserves regular monomorphisms. As such, if $L$ is a spatial locale, so that $L \cong \mathrm{Loc}(\pt(L))$ is fixed by the adjunction above, we have a Galois connection between its lattices of sublocales and subspaces:
% https://q.uiver.app/#q=WzAsMyxbMiwwLCJcXFN1Yl97VG9wfShcXHB0KEwpKSJdLFswLDAsIlxcU3ViX3tMb2N9KEwpIl0sWzMsMCwiXFxQY2FsKFxccHQoTCkpIl0sWzEsMCwiXFxtYXRocm17cHR9IiwyLHsib2Zmc2V0IjoyfV0sWzAsMSwiXFxtYXRocm17TG9jfSIsMix7Im9mZnNldCI6Mn1dLFswLDIsIlxcY29uZyIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6Im5vbmUifSwiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs0LDMsIiIsMix7ImxldmVsIjoxLCJzdHlsZSI6eyJuYW1lIjoiYWRqdW5jdGlvbiJ9fV1d
\begin{equation}
\label{eq:sub}
\begin{tikzcd}[ampersand replacement=\&]
{\Sub_{\Loc}(L)} \&\& {\Sub_{\Top}(\pt(L)) \cong \Pcal(\pt(L)),}
\arrow[""{name=0, anchor=center, inner sep=0}, "{\mathrm{pt}}"', shift right=2, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{\mathrm{Loc}}"', shift right=2, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}
\end{equation}
where by `subspace' on the right-hand side we always mean the a subset of points equipped with the subspace topology. The left adjoint sends a subspace to the smallest sublocale containing those points.
The properties of this Galois connection, especially in extreme cases where one or other of the maps involved is surjective, have been of historical interest in locale theory; see Suarez \cite{Suarez} for some novel proofs of those results. Several of the properties that we exploit, notably the notions of $T_D$ spaces and scattered spaces, appear there.
To exploit this adjunction, we must first refine it by leveraging properties of sequence spaces. As topological spaces, they are (relatively) easily seen to be Hausdorff, compact and zero-dimensional with no isolated points. These imply the properties appearing as hypotheses in the following subsections. Note that we could have proved localic versions of these results using the corresponding separation properties from \cite[Chapter V]{PicadoPultr}; we leave those mild generalizations to the motivated reader.
At various times we will need to exploit the alternative presentation of a sublocale of $L$ (equivalently, a quotient frame of $\Ocal(L)$) in terms of a \textit{nucleus}. Let $L' \subseteq L$. The corresponding frame homomorphism $q^*:\Ocal(L) \to \Ocal(L')$ preserves all joins, so as a map of partially ordered sets it has a right adjoint $q_*$; composing these, we obtain an idempotent, non-decreasing, meet-preserving map $j_{L'}:\Ocal(L) \to \Ocal(L)$, the \textit{nucleus on $L$ corresponding to $L'$}. Indeed, nuclei on $L$ and sublocales are in 1-to-1 correspondence: we can recover $\Ocal(L')$ from $j_{L'}$ as the frame of fixed points.
\subsubsection{$T_D$-spatial locales}
For the remainder of this section, \textit{$L$ is assumed to be a spatial locale} so we may make reference to the adjunction of \Cref{eq:sub}.
Recall that a topological space \textit{$X$ satisfies the $T_D$ separation axiom} if every $x \in X$ has an open neighbourhood $U\ni x$ such that $U \backslash \{x\}$ is open as well; see \cite[Proposition I.4.2]{PicadoPultr} for some equivalent characterizations and VI.1.2 in \textit{loc. cit.} for the fact that this is equivalent to $\loc$ being full and faithful on subspaces in \Cref{eq:sub}. We say a locale $L$ is \textbf{$T_D$-spatial} if it is spatial and $\pt(L)$ is $T_D$. A consequence of interest to us is that any singleton subspace $\{x\}$ of a $T_D$-spatial locale is complemented as a sublocale (being the intersection of an open and a closed sublocale). We write $\{x\}^c$ for the complement, which coincides with $\loc(\pt(L) \backslash \{x\})$.
\begin{lemma}
\label{lem:TD}
If $\pt(L)$ is $T_D$, the functor $\pt$ on sublocales in \Cref{eq:sub} also has a right adjoint, sending a subset of points $S$ to the intersection of their complementary sublocales,
\[\Max(S) := \bigcap_{x \in \pt(L) \backslash S} \{x\}^c. \]
\end{lemma}
\begin{proof}
The description defines an order-preserving map, so we must show that this is the maximal sublocale with the given set of points. As a sublocale, $\{x\}$ is an atom, since we are working over $\Set$, where the one-point locale has no non-trivial sublocales\footnote{See the proof of \cite[Lemma C1.2.13]{Ele} for a related argument. To generalize this result, one could extend the notion of subspace to allow parts of points.}. Thus every sublocale either contains $\{x\}$ or is disjoint from it. In the latter case, the sublocale must be contained in $\{x\}^c$ by the definition of this sublocale as a complement.
\end{proof}
In summary, we have the following adjoint triple between the respective lattices of subobjects of a $T_D$-spatial locale in $\Top$ and $\Loc$.
\begin{equation}
\label{eq:maxloc}
% https://q.uiver.app/#q=WzAsMixbMiwwLCJcXFN1Yl97XFxUb3B9KFxccHQoTCkpIl0sWzAsMCwiXFxTdWJfe1xcTG9jfShMKSJdLFsxLDAsIlxcbWF0aHJte3B0fSIsMSx7ImxhYmVsX3Bvc2l0aW9uIjo3MH1dLFswLDEsIlxcbWF0aHJte0xvY30iLDIseyJvZmZzZXQiOjR9XSxbMCwxLCJcXG1hdGhybXtNYXh9IiwwLHsib2Zmc2V0IjotNH1dLFszLDIsIiIsMix7ImxldmVsIjoxLCJzdHlsZSI6eyJuYW1lIjoiYWRqdW5jdGlvbiJ9fV0sWzIsNCwiIiwyLHsibGV2ZWwiOjEsInN0eWxlIjp7Im5hbWUiOiJhZGp1bmN0aW9uIn19XV0=
\begin{tikzcd}[ampersand replacement=\&]
{\Sub_{\Loc}(L)} \&\& {\Sub_{\Top}(\pt(L))}
\arrow[""{name=0, anchor=center, inner sep=0}, "{\mathrm{pt}}"{description, pos=0.7}, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{\mathrm{Loc}}"', shift right=4, from=1-3, to=1-1]
\arrow[""{name=2, anchor=center, inner sep=0}, "{\mathrm{Max}}", shift left=4, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=0, to=2]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}
\end{equation}
\begin{rmk}
\label{rem:TD}
The functor $\Max$ in \Cref{lem:TD} is full and faithful since $\loc$ is, by a general fact about adjoint triples. Moreover, it is easy to see that $\Max$ coincides with $\loc$ on cofinite subsets. On the other hand, it is important to note that $\Max(\pt(L')) \not\cong L' \subseteq L$ in general, since $\pt$ discards any structure `outside' $\loc(\pt(L'))$. For instance, the dense sublocale of \Cref{lem:noisolated} below is $\Max(\emptyset)$, whereas $\loc(\emptyset)$ is the trivial sublocale.
\end{rmk}
\begin{lemma}
\label{lem:maxplicit}
For $L$ a $T_D$-spatial locale and $S \subseteq \pt(L)$, the nucleus corresponding to $\Max(S)$ sends $U$ to the transfinite iteration of the operation,
\[U \mapsto \bigvee_{x \notin S} \inter(U \cup \{x\}).\]
\end{lemma}
\begin{proof}
The formula $U \mapsto \inter(U \cup \{x\})$ is the nucleus for $\{x\}^c$; the nucleus for an intersection of sublocales is given by the transfinite iteration of the union of their values, whence the given formula.
\end{proof}
A result attributed to Dana Scott in \cite[Theorem III.3.2]{PicadoPultr} is that the lattice of sublocales of any locale $L$ forms a \textit{coframe}, meaning that it becomes a frame when the ordering is reversed. We will see in the next section that this is true for subtoposes of a topos too. Independently, observe that any interval $[a,b] := \{x \mid a \leq x \leq b\}$ in a frame $\Ocal(L)$ is also a frame. This can be seen either by observing that the inherited operations are well-defined and distributive or by identifying $[a,b]$ with the sublocale of $L$ obtained as the intersection of the open sublocale corresponding to $b$ with the closed sublocale corresponding to $a$ (see \cite[Chapter III.6]{PicadoPultr} for the notions of open and closed sublocale). Putting these facts together, we conclude:
\begin{prop}
\label{prop:coframe}
Let $L$ be a $T_D$-spatial locale. For each subset $S \subseteq \pt(L)$, the collection of sublocales of $L$ having exactly $S$ as its set of points forms a coframe with bottom element $\loc(S)$ and top element $\Max(S)$.
\end{prop}
\subsubsection{Coscattered sublocales}
\label{sssec:coscat}
Our next task will be to understand when the sub-coframe of sublocales with a given set of points from \Cref{prop:coframe} is trivial. That is, we wish to extract criteria on subspaces $S \subseteq \pt(L)$ (for $L$ a $T_D$-spatial locale) which allow us to conclude that $\loc(S) = \Max(S)$.
\begin{dfn}
Let $X$ be a space and $S \subseteq X$ a subspace. Recall that a point $x \in S$ is \textbf{isolated} if there is an open $U \subseteq X$ such that $U \cap S = \{x\}$.
\end{dfn}
The relevance of isolated points to computing $\Max$ is as follows.
\begin{crly}
\label{crly:maxplicit2}
For a $T_D$-spatial locale $L$ and $S \subseteq \pt(L)$, the frame $\Ocal(\Max(S))$ can be identified with the collection of opens of $L$ whose complements' isolated points lie in $S$.
\end{crly}
\begin{proof}
If an open $U$ of is fixed by $j_{\Max(S)}$ (given explicitly in \Cref{lem:maxplicit}), we must have $U = \inter(U \cup \{x\})$ for every $x \notin S$. That is, either $x \in U$ or $x$ is not an interior point of $U \cup \{x\}$. The latter is equivalent to every neighbourhood of $x$ intersecting some other element of $U^c$, meaning $x$ is not an isolated point of $U^c$. Contraposing, we obtain the given characterisation.
\end{proof}
%When $X$ is a $T_2$ space, singleton subspaces are closed, so the notions of weakly isolated and isolated coincide. However, since we will prove at least one novel result in this section which applies to $T_D$ spaces, we will phrase our definitions in terms of weakly isolated points.
\begin{dfn}[Cantor-Bendixson derivative]
\label{dfn:Cantor-Bendixson}
Let $X$ be a $T_D$ space and let $S \subseteq X$ be a subspace. We can decompose $S$ as $S_{(1)} \cup S^{(1)}$, where $S_{(1)}$ is the set of isolated points and $S^{(1)}$ is its complement, the collection of \textit{limit points} of $S$: the collection of $x \in S$ such that every open neighbourhood $U \ni x$ has $U \cap S \backslash \{x\}$ inhabited. $S^{(1)}$ is the \textbf{Cantor-Bendixson derivative} of $S$.
\end{dfn}
Note that $S^{(1)}$ is closed in $S$, since for each isolated point $x \in S$ we have an open $U_x$ with $U_x \cap S = \{x\}$. Thus $S^{(1)} = X \backslash \bigcup_{x \in S_{(1)}} U_x$.
We can iterate this construction transfinitely:
\begin{itemize}
\item for a successor ordinal $\alpha + 1$ we let $S_{(\alpha+1)} = (S^{(\alpha)})_{(1)}$ and $S^{(\alpha+1)} = (S^{(\alpha)})^{(1)}$
\item for a limit ordinal $\gamma$, we let $S_{(\gamma)} = \emptyset$
and $S^{(\gamma)} = \bigcap_{\alpha<\gamma} S^{(\alpha)}$.
\end{itemize}
\begin{dfn}[Scattered and perfect]
\label{dfn:scattered}
We say a subspace $S$ of a $T_D$ space $X$ is \textbf{scattered} if $S = \bigcup_{\alpha < \gamma} S_{(\alpha)}$ for some ordinal $\gamma$. Complementarily, we say $S$ is \textbf{perfect} if $S$ is \textit{closed} and $S = S^{(1)}$. In particular, any closed set decomposes as a scattered set and a perfect set, since we the intersection $\bigcap_{\alpha} S^{(\alpha)}$ must stabilize.
\end{dfn}
\begin{rmk}
The Cantor-Bendixson theorem states that for any Polish space (in particular, for the sequences spaces $\Sigma^\omega$ with $\Sigma$ countable), any closed subset is a disjoint union of a perfect set and a countable scattered set. Moreover, any perfect set is either empty or has cardinality $2^{\aleph_0}$, so \textit{the continuum hypothesis holds for closed subsets of Polish spaces}.
\end{rmk}
For the main results in this section, it will be more convenient to speak in terms of definitions complementary to those above.
\begin{dfn}[Coscattered and coperfect]
\label{dfn:coscattered}
We say a subspace $S$ of a $T_D$ space $X$ is \textbf{coscattered} if its complement is scattered. We say an open subspace $U$ is \textbf{coperfect} if its complement is perfect.
\end{dfn}
Specialising \Cref{crly:maxplicit2} to the case of the empty subspace, we obtain the following:
\begin{crly}
\label{crly:maxplicit3}
For any fit $T_D$-spatial locale $L$ with no isolated points, $\Ocal(\Max(\emptyset))$ is precisely the collection of coperfect open sets.
\end{crly}
\mr{This is not a directly useful result for us because self-similarity equivalence classes in $\Sigma^\omega$ have no isolated points!}
\begin{prop}
\label{prop:coscattered}
Let $L$ be a $T_D$-spatial locale. Let $S$ be a subspace of $\pt(L)$ and $U$ an open of $L$ such that $U \cup S$ is coscattered. Then $j_{\loc(S)}(U) = j_{\Max(S)}(U)$. In particular, when $S$ is coscattered, $\loc(S) = \Max(S)$.
\end{prop}
\begin{proof}
The nucleus $j_{\loc(S)}$ sends $U$ to $\inter(U \cup S^c)$. Meanwhile, the $j_{\Max(S)}$ sends $U$ to the least fixed point of the operation $U \mapsto \bigvee_{x \notin S} \inter(U \cup \{x\})$. We can construct this inductively as an ordinal-indexed sequence:
\begin{itemize}
\item let $U_0 = U$
\item for a successor ordinal $\alpha^+$, define $U_{\alpha^+} = \bigvee_{x \notin S} \inter(U_{\alpha} \cup \{x\})$,
\item for a limit ordinal $\gamma$, define $U_{\gamma} = \bigvee_{\alpha < \gamma} U_{\alpha}$.
\end{itemize}
This sequence eventually stabilizes at $j_{\Max(S)}(U)$.
Let $C = (U\cup S)^c$ be the hypothesised scattered subspace. To prove the claim, we must show that for any $x \in \inter(U \cup S^c)$, we have $x \in U_{\alpha}$ for some ordinal $\alpha$. We may assign each such $x$ an ordinal rank $r(x)$ as follows: if $x \in U$, let $r(x) = 0$; otherwise, $x \in C$, so $x \in C_{(\alpha^+)}$ for a unique ordinal $\alpha$ and we define $r(x) = \alpha^+$. Note that $r(x)$ will always be either $0$ or a successor ordinal.
We prove by induction that $x \in U_{r(x)}$.
The base case $r(x) = 0$ is trivial, since $x \in U$.
Now suppose that we have shown that for all $y \in \inter(U \cup S^c)$ with $r(y) < r(x)$ we have $y \in U_{r(y)}$, and let $r(x) = \alpha^+$ so that $x$ is isolated in $C^{(\alpha)}$. This means that there is some open $V \subseteq X$ such that $V \cap C^{(\alpha)} = \{x\}$. By the construction of the subspaces $C^{(\beta)}$ as a decreasing sequence, it follows that $V \cap C \subseteq \{x\} \cup \bigcup_{\beta \leq \alpha} C_{(\beta)}$ and hence that
\[V \cap \inter(U \cup S^c) \subseteq \inter\left(U \cup \{x\} \cup \bigcup_{\beta \leq \alpha} C_{(\beta)}\right).\]
By the induction hypothesis, $\inter(U \cup \bigcup_{\beta \leq \alpha} C_{(\beta)})$ is contained in $U_{\alpha}$, ensuring that $x \in \inter(U_{\alpha} \cup \{x\}) \subseteq U_{r(x)}$, as claimed. This completes the proof.
\end{proof}
\begin{rmk}
As a consequence of these results, we recover the result of Simmons \mr{TODO: lift reference from \cite{Suarez}, find a way to access.} that \textit{every sublocale of $L$ is spatial if and only if $L$ is scattered}. Indeed, the latter is equivalent to every subspace being coscattered, so we can apply \Cref{prop:coscattered}. On the other hand, if $L$ is \textit{not} scattered then it contains some non-trivial coperfect open, so $\loc(\emptyset) \subsetneq \Max(\emptyset)$ by \Cref{crly:maxplicit3}.
\end{rmk}
\subsubsection{Dense sublocales}
\label{sssec:dense}
Of particular interest for our purposes will be \textit{dense sublocales} which, just as for topological spaces, are those whose closure is the maximal sublocale. A result originally due to Isbell \cite[Theorem 1.5]{Isbellatomless} is that any locale $L$ has a smallest dense sublocale, often denoted $L_{\neg\neg}$ by later authors due to its relationship with the double-negation modality in categorical logic.\footnote{Beware that some of the terminology in the papers of Isbell we cite is now outdated.}
\begin{lemma}
\label{lem:noisolated}
If $\pt(L)$ is $T_D$ and has no isolated points, the minimal dense sublocale $L_{\neg\neg}$ of $L$ contains no points. In particular, by taking the join with this sublocale, we may lift the adjunction in \eqref{eq:sub} as well as the right adjoint constructed in Lemma \ref{lem:TD}, to adjunctions between \textit{dense} sublocales of $L$ and subspaces of $\pt(L)$.
\end{lemma}
\begin{proof}
To prove the first statement, we show that $\{x\}^c$ must be dense for each $x \in \pt(L)$; this will imply that $L_{\neg \neg}$, being the smallest dense sublocale, is contained in $\Max(\emptyset)$ (an intersection of dense sublocales) and thus has no points.
Since $x$ is not isolated, any open neighbourhood $U \ni x$ must have $U \backslash \{x\}$ inhabited, which is to say having non-trivial intersection with $\{x\}^c$, as required. A sketch of this argument is given by Johnstone in \cite[Example C1.2.6(c)]{Ele}.
The lifting of the adjunctions is immediate; the above also shows that $\Max$ already lands in dense sublocales in this situation, so it need not be modified.
\end{proof}
We can similarly lift \Cref{prop:coframe}.
\begin{crly}
\label{crly:coframe}
Let $L$ be a spatial locale satisfying the conditions of \Cref{lem:noisolated}. For each subset $S \subseteq \pt(L)$, the collection of \emph{dense} sublocales of $L$ having exactly $S$ as its set of points forms a coframe with bottom element $\loc(S) \vee L_{\neg\neg}$ and top element $\Max(S)$.
\end{crly}
\begin{rmk}
\label{rmk:maxdense}
In particular, $\Max(S)$ is dense for any $S$. More generally, $\Max(S)$ is dense as soon as $S$ contains all isolated points of $\pt(L)$.
\end{rmk}
Recall also that $\Ocal(L_{\neg\neg})$ can be reconstructed as the Boolean algebra of regular opens of $L$: those which are equal to the interior of their closure. As such, it is easy to describe the sublocales lying between $\bot = \loc(\emptyset)$ and $L_{\neg\neg}$: they too can be identified with the regular opens of $L$, since any sublocale of a Boolean locale is open. In order to describe sublocales between $L_{\neg\neg}$ and $\Max(\emptyset)$, we will need to add hypotheses about our space.
%We can at least count them. Isbell \cite[Proof of 2.9]{Isbellfirststeps} proves that there are $2^{2^{\aleph_0}}$-many pointless sublocales of Cantor space (cf.\ \Cref{rmk:CantorBaire}) by exploiting a homeomorphism $2^{\omega} \cong 2^\omega \times 2^\omega$ and showing that for distinct $S,S' \subseteq 2^\omega$, the maximal pointless parts of $S \times 2^\omega$ and $S' \times 2^\omega$ yield distinct sublocales. Emulating that proof, we observe that since there is a countable dense subspace $D \subseteq 2^\omega$, we can simply consider subsets $S$ containing $D$, and there are still uncountably many of these.
\subsubsection{Perfect vs regular}
Not only does every closed set in sequence space decompose as the union of a scattered subspace and a perfect subspace, we can decompose the perfect subspace into a regular part and a nowhere-dense part. Expressing this observation in terms of opens, we obtain the following result.
\begin{prop}
Let $U$ be an open in a $T_D$-spatial locale $L$ with no isolated points. Then:
\begin{enumerate}
\item There is a smallest coperfect open $U^p$ containing $U$.
\item The smallest regular open containing $U$, $\neg\neg U$, has $U^p \subseteq \neg\neg U$.
\item There is a largest coperfect open $U^{n}$ such that $U^p = \neg\neg U \cap U^{n}$. This open satisfies $\neg U^{n} = \emptyset$.
\item $U = U^p \cap \bigcap_{x \in U^p \backslash U} \overline{\{x\}}^c$.
\end{enumerate}
\end{prop}
\begin{proof}
The first statement is a rephrasing of \Cref{crly:maxplicit3} and the second is a consequence of the inclusion $L_{\neg\neg} \subseteq \Max(\emptyset)$.
For the third, we define $U^{n} := \inter(U^p \cup (\neg\neg U)^c)$. This has the required property:
\[U^p \subseteq \inter(U^p \cup (\neg\neg U)^c) \cap \neg\neg U,\]
since $U^p$ is contained in both parts of the right-hand side, while
\[\inter(U^p \cup (\neg\neg U)^c) \cap \neg\neg U \subseteq U^p\]
because any point in the intersection must have a neighbourhood lying entirely in $U^p$.
It is the largest such, since given an open $U'$ with the same properties, the inclusion
\[U' \subseteq U^p \cup (\neg\neg U)^c\]
is necessary for $U' \cap \neg\neg U \subseteq U^p$. Finally,
\[\neg U^{n} \subseteq \neg U \cap \neg\neg U = \emptyset. \qedhere\]
\end{proof}
Our reason for identifying this decomposition is that nuclei respect intersections of open, so $j(U^p) = j(\neg\neg U) \cap j(U^{n})$; we can treat regular opens and coperfect sets with nowhere-dense complement separately.
\begin{lemma}
Let $L$ be a $T_D$-spatial locale $L$ with no isolated points and $S \subseteq \pt(L)$ a dense subspace. Suppose $U$ is a regular open. Then the $U$ is a fixed point for the nuclei corresponding to both $\loc(S)$ and $\Max(S)$.
\end{lemma}
\begin{proof}
Since $S$ is dense, $L_{\neg\neg} \subseteq \loc(S) \subseteq \Max(S)$, so the closure with respect to the nuclei corresponding to the latter two sublocales is contained in $\neg\neg U = U$.
\end{proof}
More generally, a regular open $U$ containing the complement of $\overline{S}$ is fixed by the nucleus of $\loc(S)$ and $\Max(S)$.
%For any $S$, $\loc(\overline{S})_{\neg\neg} \subseteq \loc(S) \subseteq \Max(S)$, since $\loc(\overline{S})_{\neg\neg}$ is the smallest subtopos whose closure is $\loc(\overline{S})$, so the closure with respect to the nuclei corresponding to the latter two sublocales is contained in $((U \vee \overline{S}^c) \rightarrow \overline{S}^c) \rightarrow \overline{S}^c = U$.
Similarly, we can deduce from \Cref{prop:coscattered} that when $U$ is coscattered we have $j_{\loc(S)}(U) = j_{\Max(S)}(U)$ for any $S$. \mr{However, since the closure of a scattered subspace need not be scattered, we can't eliminate coscattered opens, can we?}
$j_{\loc(S)}(U)^p = j_{\loc(S)}(U^p)$ and $j_{\Max(S)}(U)^p = j_{\Max(S)}(U^p)$?
Want to say that the difference depends only on the perfect part. So $j_{\loc(S)}(U) = j_{\loc(S)}(U^n)$
\mr{Working hypothesis: the gap between $\loc(S)$ and $\Max(S)$ for (dense,) self-similar $S$ is determined by the largest nowhere-dense perfect subspace outside $S$. The intermediate sublocales are determined by filters in the collection of nowhere-dense perfect subspaces.}
-----
What can be said about the (co)perfect sets from an algebraic point of view?
Let's write $P := \Max(\emptyset)$, so $\Ocal(P)$ is the collection of coperfect open sets in $L := \Sigma^\omega$.
Being closed, an intersection of two perfect sets has a canonical largest perfect part. An infinite union of perfect sets need not be perfect: consider a sequence of intervals, whose union need not be closed. However, the closure of such a union is surely perfect since only limit points are added. In particular, this means that the intersection in the frame of coperfect sets coincides with that in the original frame. (This is just a direct description of the right adjoint to the quotient frame homomorphism! Infinite meets are preserved but finite joins are not.)
What about the Heyting operation? -- automatically preserved, since we have shown that coperfect sets form a quotient frame.
%Let $U,V$ be open sets. $U \rightarrow V$ is the union of open sets $W$ with $W \cap U \subseteq V$, which is to say the interior of $V \cup U^c$. Correspondingly, its complement is the closure of $V^c \cap U$. This is necessarily perfect when $U,V$ are coperfect: indeed, the intersection of a basic open set and a perfect set is perfect, so this is the closure of a union of perfect sets. Thus the inclusion of $\Ocal(P)$ into $\Ocal(L)$ preserves the Heyting operation.
What intrinsic properties does this Heyting algebra have? Could it be that for any $x$ in the complement of $V$ there is a sparser (nowhere-dense) perfect set containing $x$, so that $\bigcap \{U \mid U \rightarrow V = V\} = \emptyset$? (Note that any such $U$ must have $U \not\subseteq V$)
We already checked that this frame is not at all de Morgan (since the de Morganisation coincides with the dense Boolean subtopos).
Consider $U^c \cap \neg\neg U$. This is the `irregular' part of the perfect set $U^c$, in the sense that $\neg\neg U$ is the smallest regular open containing $U$. Can we characterize this in some way?
$\inter(U^c \cap \neg\neg U) \subseteq \inter(U^c) \cap \neg\neg U = \neg U \cap \neg\neg U = \bot$, so $U^c \cap \neg\neg U$ is a nowhere-dense part of $U^c$. The remainder of $U^c$ is $(\neg\neg U)^c$, which has interior $\neg U$. Thus we have a canonical decomposition of the perfect set into a regular closed set and a nowhere-dense locally closed set. Can we take the closure of the latter? Yes!
\[\inter(\overline{U^c \cap \neg\neg U}) \subseteq \inter(U^c \cap \overline{\neg\neg U})
\subseteq \neg U \cap \neg\neg U = \bot.\]
So the perfect set $U^c$ further decomposes into a union of a nowhere-dense perfect set and a regular closed set.
\mr{Perhaps this decomposition can take us somewhere? }
Dually, any coperfect open set decomposes into a regular open and an irregular coperfect open (one whose complement is a nowhere-dense closed set).
Algebraic characterization: an element $w$ of a lattice is \textbf{meet-irreducible} if for all elements $u,v$, $u \wedge v \leq w$ implies $u \leq w$ or $v \leq w$. A meet-irreducible $w$ is either the top element or is \textbf{prime}, meaning that it contains a member of any finite meet of elements below it. A lattice is \textbf{prime-free} if the only meet-irreducible element is the top element.
Using the Heyting operation we can eliminate one of the quantified variables above: $w$ is meet-irreducible if and only if for all elements $v$, either $v \leq w$ or $v \rightarrow w \leq w$. With this presentation, we can present the axiom of prime free-ness as:
\[\left(\forall v.\; (v \leq w) \vee ((v \rightarrow w) \leq w) \right) \Rightarrow w = \top.\]
Observe that eliminating isolated points amounts to making $U \subseteq U'$ dense (for the nucleus) whenever $U \rightarrow U'$ is prime, and hence $\Ocal(P)$ is the largest prime-free quotient frame of $\Ocal(L)$. This is because of a happy coincidence (that the join-prime closed sets are just the singletons in this case) that isn't necessarily true in less separated (i.e. non-$T_D$) locales/frames.
Unlike the Boolean case, there is no hope of extracting a purely equational presentation directly equivalent to `prime-free', since the `or' condition is (provably) not representable by an element of the Heyting algebra. However, there is a non-trivial algebraic consequence of this:
\[\top = v \rightarrow w \vee (v\rightarrow w) \rightarrow w\]
when $w$ is meet-irreducible. Clearly this is strictly weaker: either of the summands in the right-hand side could fail to be equal to top. Asking for $w$ satisfying this condition to be dense is clearly much stronger, since as soon as the Heyting algebra is de Morgan, the bottom element satisfies this identity...
\subsubsection{Baire locales}
%From $T_D$, we skip several levels of separation axiom to arrive at regularity. Recall that a locale is \textbf{regular} if every open $U$ is the union opens $V \subseteq U$ which are \textit{well inside} $U$, meaning that there exists an open set $V'$ with $V \cap V'= \bot$ but $U \cup V' = \top$. Any compact Hausdorff space is regular, and any regular spatial locale is $T_3$ (and hence $T_D$).
The final property of interest is not quite a separation property, and we shall need to put some extra work into proving that sequence spaces have it.
\begin{dfn}\label{dfn:Baire}
We say a locale is \textbf{Baire} if a countable intersection of open sublocales is complemented.
\end{dfn}
The Baire property for topological spaces, which demands that a countable intersection of open sets is dense, is trivial for locales. \Cref{dfn:Baire} thus provides a non-trivial property that holds for Baire spaces as locales.
\begin{lemma}
\label{lem:Baire2}
Let $\{U_i\}_{i = 1}^{\infty}$ be a countable collection of open subsets in sequence space $\Sigma^{\omega}$. Let $C$ be a closed subset such that each $U_i \cap C$ is dense in $C$. Then the intersection of subspaces $\bigcap_{i=1}^\infty U_i$ is dense in $C$.
\end{lemma}
\begin{proof}
This proof is based on that of \cite[Proposition VII.7.1]{PicadoPultr}.
Let $V$ be any open with $C \cap V \neq \emptyset$. Then there is some finite word $u_i$ such that $C \cap U(u_i) \subseteq C \cap V \cap U_1$, since $U_1$ is dense in $C$ and $V \cap U_1$ is open. Suppose we have already found $u_i$, $i = 1,\dotsc,n$, such that
\[ \emptyset \neq C \cap U(u_i) \subseteq C \cap U(u_{i-1}) \cap U_{i-1} \text{ and } V \supseteq U(u_1) \supseteq U(u_2) \supseteq U(u_3) \supseteq \cdots,\]
which is to say $u_1 \pf u_2 \pf u_3 \pf \cdots$. Then choose a $u_{n+1}$ such that $\emptyset \neq C \cap U(u_{n+1}) \subseteq C \cap U(u_n) \cap U_n$.
Either the sequence $(u_i)_{i=1}^\infty$ stabilizes or it converges to some word $w$. Either way, we obtain an inhabitant of $C \cap V \cap \bigcap_{i=1}^\infty U_i$: in the latter case, the inhabitants of $C \cap U(u_n) \cap U_n$ converge to $w$ and so $w$ also lies in $C$, since $C$ is closed.
\end{proof}
\begin{prop}
\label{prop:Baire3}
Let $\{U_i\}_{i = 1}^{\infty}$ be a countable collection of open subsets in $\Sigma^{\omega}$. Then the intersection of sublocales $\bigwedge_{i=1}^\infty U_i$ is spatial (and hence is equal to $\loc(\bigcap_{i=1}^\infty U_i)$). In particular, sequence space is Baire.
\end{prop}
\begin{proof}
This proof is based on that of \cite[Proposition VII.7.2]{PicadoPultr}.
The nucleus corresponding to $\loc\left(\bigcap_{i=1}^\infty U_i\right)$ sends $U$ to $\inter(U \cup (\Sigma^\omega \backslash \bigcap_{i=1}^\infty U_i))$. Thus $U$ is fixed for this nucleus if and only if, for open $V$, $V \cap \bigcap_{i=1}^\infty U_i \subseteq U$ implies $V \subseteq U$.
By comparison, the nucleus corresponding to the sublocale $\bigwedge_{i=1}^\infty U_i$ sends $U$ to the least fixed point of $\bigcup \inter(U \cup (\Sigma^\omega \backslash U_i))$. Thus $U$ is fixed by the latter if and only if, for open $V$:
\begin{equation}
\label{eq:countableintersect}
\left(\exists n, V \cap U_n \subseteq U \right) \Rightarrow V \subseteq U.
\end{equation}
Suppose $U$ is a fixed point for the latter nucleus, and let $C = \Sigma^\omega \backslash U$. For every $V$ such that $V \cap C \neq \emptyset$ (that is, $V \not\subseteq U$) we have, by contraposition of \eqref{eq:countableintersect},
\[\forall n, V \cap U_n \not\subseteq U.\]
That is, $V \cap U_n \cap C \neq \emptyset$. Thus each $C \cap U_n$ is dense in $C$, whence $\bigcap_{i=1}^\infty C \cap U_i$ is dense in $C$ by \Cref{lem:Baire2}. Thus if $V \cap \bigcap_{i=1}^\infty U_i \subseteq U$ then $V \cap \bigcap_{i=1}^\infty U_i \cap C = \emptyset$, so $V \subseteq U$, which shows that $U$ is fixed for the former nucleus, as required.
\end{proof}
---
\begin{rmk}
Following results of Isbell \cite[1.3, 1.4]{Isbellfirststeps}, we observe that the intersections of \Cref{prop:Baire3} are complemented sublocales as well as subspaces. Indeed, a union of closed sublocales is always spatial, so the complement is simply the union of the complementary (nowhere dense) closed subspaces. A complemented sublocale of a locale is (classically) always spatial, by an argument related to to that used in \Cref{lem:noisolated}.
\end{rmk}
\begin{crly}\label{crly:Bairelyworkin}
For $L$ a $T_D$-spatial Baire locale, $\Max(S) = \loc(S)$ whenever $S \subseteq \pt(L)$ has countable complement.
\end{crly}
A further consequence of \Cref{prop:Baire3} is that we can partially answer the question posed at the end of \Cref{sssec:fit}. Indeed, if an open $U$ is a countable intersection of opens belonging to a filter $\Fcal$ then adding $U$ to $\Fcal$ will not change the induced coating map $c_{\Fcal}$. We can thus restrict attention to countably complete filters without missing any sublocales.
Conversely, recall that any closed set in sequence space is the union of a (closed) perfect set and a countable set. As such, any open $U$ is the intersection of a coperfect set and a countable collection of cofinite sets. In particular, it suffices to establish what happens over coperfect sets. Given a perfect set $C$, how many ways can it be presented as a union of a countably complete ideal of closed sets?
Max: just countable subsets.
Inbetween: include some sub-perfect-sets.
Max: include the perfect set itself.
i.e. \textbf{after its intersection, a filter is determined by the perfect sets which are present}.
In summary, the abstract properties of sequence spaces are already enough to enable us to organize their sublocales according to the points they contain, and \Cref{prop:Baire3} shows that when the collection of points is `co-countable', they uniquely determine a sublocale. We will see later on that there are surprisingly strict foundational restrictions on how much we can understand about the coframe of sublocales associated to a set of points $S$ whose complement is uncountable.
\subsection{Subtoposes}
$\PSh(M)$ is a presheaf topos, meaning it is a special case of a Grothendieck topos%\footnote{When not accompanied by a qualifier, `topos' will always mean `Grothendieck topos' in the remainder.}
, which in turn are always \emph{elementary toposes}. Let us recall some of the essential categorical features that will be of interest to us.
Any elementary topos has a \emph{subobject classifier}, usually denoted $\Omega$, equipped with a \emph{universal subobject} $\top : 1 \to \Omega$, where $1$ is the terminal object, so-called because any subobject $s: S \hookrightarrow X$ in the topos has a unique \emph{classifying map} $\chi_s:X \to \Omega$ such that $s$ is the pullback of $\top$ along $\chi_s$. The subobject classifier has the structure of an internal frame; for instance, there is a morphisms $\wedge : \Omega \times \Omega \to \Omega$ which represents the intersection operation on subobjects.
For $\PSh(M)$, $\Omega$ is the set of right ideals of $M$ equipped with the \textit{inverse image action} defined for an ideal $I$ and $m \in M$ by:
% \[I \cdot m := \{x \in M \mid mx \in I\}.\]
\begin{equation}
\label{eq:IdealAction}
I \cdot m := \{x \in M \mid mx \in I\};
\end{equation}
the universal subobject $\top$ picks out the maximal ideal. The classifying map of a sub-$M$-set $S \subseteq X$ sends $x \in X$ to the ideal $\{m \in M \mid x \cdot m \in S\}$. The frame operations are the set-theoretic operations one would expect: $\wedge$ is just intersection of ideals, for instance.
A \textbf{subtopos} of a topos is a full, left exact reflective subcategory. For a general elementary topos, the data required to specify a subtopos can be presented as follows.
\begin{dfn}\label{def:ltoperator}
Let $\Ecal$ be an elementary topos with subobject classifier $\Omega$. A \textbf{Lawvere-Tierney operator} in $\Ecal$ is an endomorphism $j: \Omega \to \Omega$ which:
\begin{itemize}
\item is idempotent, $j \circ j = j$,
\item preserves the universal subobject, $j \circ \top = \top$,
\item preserves meets, $j \circ \wedge = \wedge \circ j \times j$.
\end{itemize}
A monomorphism $m$ is \textbf{$j$-dense} if $j \circ \chi_m = \chi_m$. An object $X$ of $\Ecal$ is a \textbf{$j$-sheaf} if for every $j$-dense monomorphism $m:S \rightarrowtail T$, each morphism $t: S \to X$ extends along $m$:
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJTIl0sWzAsMSwiVCJdLFsxLDEsIlgiXSxbMCwyLCJ0Il0sWzAsMSwibSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1vbm8ifX19XSxbMSwyLCJcXGV4aXN0cyIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==
\[\begin{tikzcd}[ampersand replacement=\&]
S \\
T \& X
\arrow["m"', tail, from=1-1, to=2-1]
\arrow["t", from=1-1, to=2-2]
\arrow["\exists"', dashed, from=2-1, to=2-2]
\end{tikzcd}\]
We write $\sh_j(\Ecal)$ for the (full, left exact) subcategory of $\Ecal$ on the $j$-sheaves. Every subtopos is of this form, meaning we have an adjunction:
% https://q.uiver.app/#q=WzAsMixbMCwwLCJcXHNoX2ooXFxFY2FsKSJdLFsyLDAsIlxcRWNhbCJdLFswLDEsIiIsMix7Im9mZnNldCI6Miwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwwLCIiLDAseyJvZmZzZXQiOjJ9XSxbMywyLCIiLDAseyJsZXZlbCI6MSwic3R5bGUiOnsibmFtZSI6ImFkanVuY3Rpb24ifX1dXQ==
\begin{equation}
\label{eq:jshf}
\begin{tikzcd}[ampersand replacement=\&]
{\sh_j(\Ecal)} \&\& \Ecal
\arrow[""{name=0, anchor=center, inner sep=0}, shift right=2, hook, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, shift right=2, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}
\end{equation}
\end{dfn}
Our overarching goal is to find a classification of Lawvere-Tierney operators in $\PSh(M)$. \mr{(ref weak LT topologies thesis).}
Consider the topos $\Sh(L)$ of sheaves on a locale $L$, defined as a subtopos of $\PSh(\Ocal(L))$ just as for sheaves on a space, see \cite[Chapter II]{MLM}. We can recover $\Ocal(L)$ as the subterminal objects of $\Sh(L)$. A Lawvere-Tierney operator induces a \textit{nucleus} on $L$ and vice versa. In other words, we have the following correspondence.
\begin{lemma}[{\cite[Corollary IX.6.6]{MLM}}]
\label{lem:subloc}
Subtoposes of $\Sh(L)$ correspond to sublocales of $L$.
\end{lemma}
\subsection{Geometric morphisms}
\label{ssec:more}
The morphisms typically considered between toposes are geometric morphisms.
\begin{dfn}
Let $\Ecal, \Fcal$ be toposes. A \textbf{geometric morphism} $f:\Fcal \to \Ecal$ consists of an adjunction:
% https://q.uiver.app/#q=WzAsMixbMCwwLCJcXEVjYWwiXSxbMiwwLCJcXEZjYWwiXSxbMCwxLCJmXyoiLDIseyJvZmZzZXQiOjJ9XSxbMSwwLCJmXioiLDIseyJvZmZzZXQiOjJ9XSxbMywyLCIiLDIseyJsZXZlbCI6MSwic3R5bGUiOnsibmFtZSI6ImFkanVuY3Rpb24ifX1dXQ==
\[\begin{tikzcd}
\Fcal && \Ecal
\arrow[""{name=0, anchor=center, inner sep=0}, "{f_*}"', shift right=2, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{f^*}"', shift right=2, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}\]
such that $f^*$ preserves finite limits. The functors $f^*$ and $f_*$ are called the \textbf{inverse image} and \textbf{direct image} functor of $f$, respectively.
\end{dfn}
\begin{xmpl}
\label{xmpl:globalsections}
Every Grothendieck topos $\Ecal$ admits a unique geometric morphism to $\Set$, the \textbf{global sections morphism}, whose direct image functor maps an object $X$ to its set $\Ecal(1,X)$ of global elements. We will provide further examples below.
\end{xmpl}
%To streamline the definition of some properties of geometric morphisms,
Recall that a \textbf{subquotient} of an object $X$ in a topos is a subobject of a quotient of $X$ (or equivalently a quotient of a subobject of $X$).
\begin{dfn}
\label{dfn:fourtypesdef}
Let $f: \Fcal \to \Ecal$ be a geometric morphism. We will primarily be concerned with the following five properties in the present article.
\begin{enumerate}
\item $f$ is an \textbf{inclusion} if the direct image functor $f_*$ is fully faithful. We denote these with hooked arrows, $f:\Fcal \hookrightarrow \Ecal$.
\item Almost dually, $f$ is a \textbf{hyperconnected} morphism if the inverse image functor $f^*$ is fully faithful \textit{and closed under subquotients}. We use double-headed arrows $f:\Fcal \twoheadrightarrow \Ecal$ for these.
\item $f$ is a \textbf{surjection} if $f^*$ is faithful, or equivalently if $f^*$ is comonadic.
\item $f$ is \textbf{localic} if every object of $\Ecal$ is a subquotient of one in the essential image of $f^*$.
\item $f$ is an \textbf{\'{e}tale morphism}\footnote{Johnstone calls these \textit{local homeomorphisms} in \cite{Ele}.} if it takes the form of the canonical morphism $\bang:\Ecal/X \to \Ecal$ (up to an equivalence $\Fcal \simeq \Ecal/X$) for some object $X$ of $\Ecal$. The canonical morphism $\bang$ has $\bang^*:Y \mapsto (\pi_1: X \times Y \to X)$. We denote these with slashed arrows $f:\Fcal \nrightarrow \Ecal$.
\end{enumerate}
\end{dfn}
Implicit in the final definition is the so-called `fundamental theorem of topos theory': that $\Ecal/X$ is an elementary topos (resp. Grothendieck topos, presheaf topos) as soon as $\Ecal$ is. Each of these classes of geometric morphisms is stable under composition, and all but surjections are stable under pullback in the $2$-category of toposes.
In this section, we collect together some examples and general results regarding the interactions between these classes of geometric morphisms, which we will apply to our toposes of interest in the next section.
\begin{xmpl}
\label{xmpl:fourtypeexa}
\begin{enumerate}
\item \label{ex:inc} Every inclusion has the form $\sh_j(\Ecal) \hookrightarrow \Ecal$ of \cref{eq:jshf}, by \cite[Theorem VII.4.6]{MLM}. That is, an inclusion identifies its domain as a \textit{subtopos} of its codomain.
\item \label{ex:monoidhype} Let $M$ be a monoid.
\begin{enumerate}
\item Any hyperconnected morphism, including those to follow, is a surjection. The forgetful functor $\PSh(M) \to \Set$ is the inverse image functor of a surjective geometric morphism which is not hyperconnected.
\item If $M$ is a quotient of the free monoid $\Sigma^*$, the surjective monoid homomorphism $\Sigma^* \to M$ induces a hyperconnected morphism $\PSh(\Sigma^*) \twoheadrightarrow \PSh(M)$. See \cite[Example A4.6.9]{Ele} for this specific fact and \cite{EDMA} for more general relationships between properties of monoid homomorphisms and their induced geometric morphisms.
\item If $\tau$ is a topology on $M$, the category of actions of $M$ on sets which are continuous with respect to $\tau$ (and the discrete topology on the sets) form a topos, $\Cont(M,\tau)$ (see \cite{TTMA}). There is a hyperconnected morphism $\PSh(M) \twoheadrightarrow \Cont(M,\tau)$ whose inverse image functor is the inclusion of this subcategory.
\item More generally, any homomorphism of topological monoids with dense image induces a hyperconnected morphism, \cite[Lemma 4.20]{TTMA}. In particular, the global sections morphism of $\Cont(M,\tau)$, induced by the unique homomorphism to the terminal monoid, is hyperconnected. We say a topos is \textbf{hyperconnected} if its global sections morphism is.
\end{enumerate}
\item \label{ex:localic}
\begin{enumerate}
\item Inclusions and \'{e}tale morphisms are localic.
\item Any locale map $L \to L'$ induces a localic morphism $\Sh(L) \to \Sh(L')$ and vice versa; this is where this class gets its name.
\item \label{ex:posetal} We call a Grothendieck topos \textbf{localic} if its global sections morphism is, which happens if and only if it is equivalent to $\Sh(L)$, with $\Ocal(L)$ the frame of subterminal objects. $\PSh(\Ccal)$ is localic if and only if $\Ccal$ is a preorder, and the corresponding frame is that of upward-closed subsets in $\Ccal$.
\item \label{ex:locrefl} Any Grothendieck topos $\Ecal$ has a \textbf{localic reflection}: the subterminal objects of $\Ecal$ form a frame $\Ocal(L)$, and there is a canonical hyperconnected morphism $\Ecal \to \Sh(L)$ sending $X$ to the sheaf $(U \mapsto \Ecal(U,X))$.
\end{enumerate}
\item \label{ex:openinc}
\begin{enumerate}
\item Suppose $U$ is a subterminal object of $\Ecal$. Then the canonical morphism $\bang:\Ecal/U \nrightarrow \Ecal$ is an inclusion, called an \textbf{open inclusion} because it corresponds to an open subspace when $\Ecal$ is localic.
\item If instead $X$ is a \textbf{well-supported} object of $\Ecal$, meaning the unique morphism $X \to 1$ is epic, the induced morphism $\Ecal/X \to \Ecal$ is a surjection.
\item \label{ex:facta} More generally, we define the \textbf{support} of $X$, denoted $\supp(X)$, to be the intermediate (subterminal) object in the epi-mono factorization of the unique morphism $X \to 1$. By construction, $X$ is well supported as an object of $\Ecal/\supp(X)$, so this induces a factorization of $\bang:\Ecal/X \to \Ecal$ as an \'{e}tale surjection followed by an open inclusion.
\item \label{ex:mapetale} Any morphism $f:Y \to X$ induces an \'{e}tale morphism $\Ecal/Y \to \Ecal/X$ which we also denote by $f$, via the equivalence $\Ecal/Y \simeq \Ecal/(f:Y \to X)$.
\end{enumerate}
\end{enumerate}
\end{xmpl}
The factorizations presented in \Cref{xmpl:fourtypeexa}.\ref{ex:locrefl} and \ref{xmpl:fourtypeexa}.\ref{ex:facta} are respectively special cases of the \textbf{(hyperconnected, localic)} and \textbf{(surjection, inclusion)} factorization systems on geometric morphisms. We will not go into detail on how these are constructed in general (see \cite[Theorems A4.6.5 and A4.2.10]{Ele}, respectively), but we will occasionally exploit their existence and properties.
\begin{lemma}
\label{lem:subtoposesarehype}
Any subtopos of a hyperconnected topos is either degenerate or hyperconnected.
\end{lemma}
\begin{proof}
Let $\Ecal$ be hyperconnected and $\Fcal \to \Ecal$ an inclusion. Consider the hyperconnected--localic factorization of the global sections morphism $\Fcal \to \Set$ which is equal to the composite $\Fcal \to \Ecal \to \Set$. By \cite[Proposition A4.6.10]{Ele}, this factorization coincides with the surjection--inclusion factorization, so $\Fcal$ is hyperconnected over a subtopos of $\Set$. If the subtopos is the degenerate subtopos, then $\Fcal$ is degenerate too; otherwise, the subtopos is all of $\Set$ and $\Fcal$ is a hyperconnected topos, as claimed.
\end{proof}
\begin{rmk}\label{rmk:denseinc}
Observe that the non-degenerate subtoposes in \Cref{lem:subtoposesarehype} must contain the initial object $0$ of $\Ecal$, which is one of the equivalent definition for a subtopos to be \textit{dense}.
\end{rmk}
Next, let us return to the slice topos $\Ecal/X$. Since the inverse image functor of the canonical morphism $\Ecal/X \to \Ecal$ is \textit{logical} (\cite[Lemma C3.5.4(iii)]{Ele}), it preserves the subobject classifier, meaning that the subobject classifier in $\Ecal/X$ is $\pi_1:X \times \Omega \to X$, equipped with universal subobject $X \cong X\times 1 \xrightarrow{\id \times \top} X \times \Omega$. A Lawvere-Tierney operator is in particular an endomorphism $k$ of $\pi_1: X \times \Omega \to X$. %by adjointness, we can translate this to a morphism $\tilde{k}: X \times \Omega \to \Omega$ (which amounts to composing $k$ with the projection $\pi_2:X \times \Omega \to \Omega$). In particular,
Considering a subtopos of $\Ecal$ with corresponding Lawvere-Tierney operator $j$, we have a pullback square,
% https://q.uiver.app/#q=WzAsNCxbMCwxLCJcXHNoX2ooXFxFY2FsKSJdLFsxLDEsIlxcRWNhbCJdLFsxLDAsIlxcRWNhbC9YIl0sWzAsMCwiXFxzaF9qKFxcRWNhbCkvaV4qKFgpIl0sWzAsMSwiaSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMV0sWzMsMF0sWzMsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=
\[\begin{tikzcd}[ampersand replacement=\&]
{\sh_j(\Ecal)/i^*(X)} \& {\Ecal/X} \\
{\sh_j(\Ecal)} \& \Ecal.
\arrow["i/X",hook, from=1-1, to=1-2]
\arrow["\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow["\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=1-2, to=2-2]
\arrow["i"', hook, from=2-1, to=2-2]
\end{tikzcd}\]
The upper inclusion has corresponding Lawvere-Tierney operator $\id_X \times j$; we say this operator \textbf{lifts} $j$. Pullback-stable properties of the lower inclusion will be inherited by the upper inclusion, but more can be said for the specific case of a slice topos.
\begin{crly}
\label{crly:dense}
Suppose $i:\sh_j(\Ecal) \hookrightarrow \Ecal$ is a dense subtopos. Then for any $X$, the inclusion $i/X:\sh_{\id_X \times j}(\Ecal/X) \hookrightarrow \Ecal/X$ is dense. \\
In particular, if $\Ecal$ is hyperconnected and $i$ is non-degenerate, $i/X$ is dense.
\end{crly}
\begin{proof}
An efficient proof exploits the Beck-Chevalley condition for \'{e}tale morphisms; a direct proof is as follows. The direct image functor $(i/X)_*$ sends $f:Y \to i^*(X)$ to the pullback of $i_*(f)$ along the unit $\eta_X:X \to i_*i^*(X)$. If $i_*(Y)$ is an initial object, this pullback will be initial too, when $(i/X)_*$ preserves the initial object whence $i_*$ does. The particular case follows from \Cref{rmk:denseinc}.
\end{proof}
It will serve us to characterize lifted Lawvere-Tierney operators in $\Ecal/X$ more intrinsically. We exploit the factorization of \Cref{xmpl:fourtypeexa}.\ref{ex:facta} here.
\begin{lemma}\label{lem:openlifts}
Let $U$ be a subterminal object. Then every Lawvere-Tierney operator in $\Ecal/U$ lifts some (typically not unique) such operator $j$ in $\Ecal$.
\end{lemma}
\begin{proof}
Let $k$ be a Lawvere-Tierney operator in $\Ecal/U$ and define $i$ to be the composite inclusion $\sh_k(\Ecal/U) \hookrightarrow \Ecal/U \hookrightarrow \Ecal$. The pullback of inclusions is their intersection, whence the following square is a pullback:
% https://q.uiver.app/#q=WzAsNCxbMCwxLCJcXHNoX2soXFxFY2FsL1UpIl0sWzEsMSwiXFxFY2FsIl0sWzEsMCwiXFxFY2FsL1UiXSxbMCwwLCJcXHNoX2soXFxFY2FsL1UpIl0sWzAsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIiIsMix7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=
\[\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{\sh_k(\Ecal/U)} \& {\Ecal/U} \\
{\sh_k(\Ecal/U)} \& \Ecal
\arrow[hook, from=1-1, to=1-2]
\arrow[equals, from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow["\slash"{marking,font=\fontsize{8}{2}\selectfont}, hook, from=1-2, to=2-2]
\arrow[hook, "i", from=2-1, to=2-2]
\end{tikzcd}\]
Following the argument above, letting $j$ be the operator in $\Ecal$ induced by $i$, we have $k = \id_U \times j$. Note that any Lawvere-Tierney operator $j$ in $\Ecal$ inducing a subtopos whose intersection with $\Ecal/U$ coincides with $\sh_k(\Ecal/U)$ will also satisfy $k = \id_U \times j$; this is possible because the projection map $\pi_2: U \times \Omega \to \Omega$ is not an epimorphism.
\end{proof}
\begin{prop}
\label{prop:operatorlift}
Consider the following morphism in $\Ecal/X$,
\begin{equation}\label{eq:pi13}
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJYIFxcdGltZXMgWCBcXHRpbWVzIFxcT21lZ2EiXSxbMiwwLCJYIFxcdGltZXMgXFxPbWVnYSJdLFsxLDEsIlgiXSxbMCwxLCJcXHBpX3sxM30iXSxbMCwyLCJcXHBpXzEiLDJdLFsxLDIsIlxccGlfMSJdXQ==
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times \Omega} \&\& {X \times \Omega.} \\
\& X
\arrow["{\pi_{13}}", from=1-1, to=1-3]
\arrow["{\pi_1}"', from=1-1, to=2-2]
\arrow["{\pi_1}", from=1-3, to=2-2]
\end{tikzcd}
\end{equation}
A Lawvere-Tierney operator $k$ in $\Ecal/X$ lifts one in $\Ecal$ if and only if the following square commutes in $\Ecal/X$:
\begin{equation}\label{eq:pi13comm}
% https://q.uiver.app/#q=WzAsNCxbMSwxLCJYIFxcdGltZXMgXFxPbWVnYSJdLFswLDEsIlggXFx0aW1lcyBYIFxcdGltZXMgXFxPbWVnYSJdLFsxLDAsIlggXFx0aW1lcyBcXE9tZWdhIl0sWzAsMCwiWCBcXHRpbWVzIFggXFx0aW1lcyBcXE9tZWdhIl0sWzEsMCwiXFxwaV97MTN9IiwyXSxbMiwwLCJrIl0sWzMsMiwiXFxwaV97MTN9Il0sWzMsMSwiXFxpZCBcXHRpbWVzIGsiLDJdXQ==
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times \Omega} \& {X \times \Omega} \\
{X \times X \times \Omega} \& {X \times \Omega}
\arrow["{\pi_{13}}", from=1-1, to=1-2]
\arrow["{\id \times k}"', from=1-1, to=2-1]
\arrow["k", from=1-2, to=2-2]
\arrow["{\pi_{13}}"', from=2-1, to=2-2]
\end{tikzcd}
\end{equation}
\end{prop}
\begin{proof}
When $X$ is subterminal, \eqref{eq:pi13} is an isomorphism and \eqref{eq:pi13comm} commutes for any $k$, so the condition is vacuous, which is consistent with \Cref{lem:openlifts}.
By employing the factorization of \Cref{xmpl:fourtypeexa}.\ref{ex:facta}, we may without loss of generality assume that $X$ is well-supported in $\Ecal$.
% \hr{it needs an explanation, since what we get it is not $\Omega\to \Omega$, but $\Omega \times \supp(X)\to \Omega \times \supp(X)$. Possibly, this problem might be corrected by considering (a nicer version of) the injectibvity of $\Omega$.}
When $k = \id \times j$, the square \eqref{eq:pi13comm} trivially commutes, so the condition is necessary.
To see it the square \eqref{eq:pi13} commuting as a diagram in $\Ecal/X$ is sufficient, we construct the following diagram in $\Ecal$:
\begin{equation}
\label{eq:doublecoeq}
% https://q.uiver.app/#q=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
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times \Omega} \& {X \times \Omega} \& \Omega \\
{X \times X \times \Omega} \& {X \times \Omega} \& \Omega
\arrow["{\pi_{13}}", shift left, from=1-1, to=1-2]
\arrow["{\pi_{23}}"', shift right, from=1-1, to=1-2]
\arrow["{\id \times k}"', from=1-1, to=2-1]
\arrow["{\pi_2}", from=1-2, to=1-3]
\arrow["k", from=1-2, to=2-2]
\arrow["j", dashed, from=1-3, to=2-3]
\arrow["{\pi_{13}}", shift left, from=2-1, to=2-2]
\arrow["{\pi_{23}}"', shift right, from=2-1, to=2-2]
\arrow["{\pi_2}"', from=2-2, to=2-3]
\end{tikzcd}
\end{equation}
The rows are coequalizer diagrams (this is where we use well-supportedness of $X$). The other left-hand square, featuring $\pi_{23}$ in the horizontal positions commutes for any $k$. As such, we conclude by diagram chasing that $\pi_{13};k;\pi_2 = \pi_{23};k;\pi_2$ and we obtain a morphism $j:\Omega \to \Omega$ inheriting the required properties from $k$ and such that $k = \id_X \times j$ (this follows from the above diagram plus the fact that $k$ commutes with $\pi_1$, being an endomorphism in $\Ecal/X$).
\end{proof}
\begin{rmk}
When $X$ is well-supported, the inverse image functor of the canonical map $\Ecal/X \to \Ecal$ is not only comonadic, it is also monadic. In this case, the morphism of \eqref{eq:pi13} presents the subobject classifier of $\Ecal$ as an algebra for the induced monad on $\Ecal/X$.
\end{rmk}
While \Cref{prop:operatorlift} has the advantage of generality, the construction of $\id \times k$ requires applying the functors in the canonical morphism $\Ecal/X \to \Ecal$. For this reason, we derive further necessary conditions that will be sufficient in our case of interest.
Let $X$ be an object of $\Ecal$ and $m: X \to X$ an endomorphism. Consider the corresponding étale geometric morphism $\Ecal/X \to \Ecal/X$ of \Cref{xmpl:fourtypeexa}.\ref{ex:mapetale}. Thanks to the trivial commutativity of the triangle,
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiWCJdLFsxLDEsIjFfe1xcRWNhbH0iXSxbMCwyLCIhIiwyXSxbMSwyLCIhIl0sWzAsMSwibSJdXQ==
\[\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
X \&\& X \\
\& {1_{\Ecal}}
\arrow["m", from=1-1, to=1-3]
\arrow["{!}"', from=1-1, to=2-2]
\arrow["{!}", from=1-3, to=2-2]
\end{tikzcd}\]
in $\Ecal$, any object of the form $\bang^*(Y)$ is fixed by $m^*$, and similarly for morphisms. As such, we obtain a right action of the endomorphism monoid $\End(X)$ on the set of \textit{all} morphisms $\bang^*(Y) \to \bang^*(Z)$.
\begin{rmk}\label{rmk:TheActionOfEndoMorphisms}
The $\End(X)$-action can be explicitly written as follows. First, passing across the adjunction, we have:
\[
\Ecal(X\times Y , Z) \xrightarrow{\cong} \Ecal/X(\bang^* (Y), \bang^*(Z)) : f\mapsto \langle \pi_1, f\rangle
\]
On the left-hand side, the action of $m : X \to X$ on an element $f\colon X\times Y \to Z$ is given by precomposition with $(m \times \id_Y)$. %$f*m = f\circ (m\times \id_Y)$ due to the following diagram:
% https://q.uiver.app/#q=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
% \[\begin{tikzcd}
% && {X\times Z} \\
% {X\times Y} &&&&& {X\times Z} \\
% &&& {X\times Y} \\
% & X \\
% &&&& X
% \arrow["{m\times \id_Z}", from=1-3, to=2-6]
% \arrow["{\pi_1}", from=1-3, to=4-2]
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=5-5]
% \arrow["{\langle \pi_1, f\circ(m\times \id_Y)\rangle}", from=2-1, to=1-3]
% \arrow["{m\times \id_Y}", from=2-1, to=3-4]
% \arrow["{\pi_1}"', from=2-1, to=4-2]
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-1, to=5-5]
% \arrow["{\pi_1}", from=2-6, to=5-5]
% \arrow["{\langle \pi_1, f\rangle}", from=3-4, to=2-6]
% \arrow["{\pi_1}"', from=3-4, to=5-5]
% \arrow["m"', from=4-2, to=5-5]
% \end{tikzcd}\]
\end{rmk}
\begin{dfn}\label{def:self-similar}
We say $g:\bang^*(Y) \to \bang^*(Z)$ in $\Ecal/X$ is \textbf{self-similar} if it is fixed by the action of $\End(X)$ described above.
\end{dfn}
Since every morphism of the form $\bang^*(h)$ is self-similar, we conclude the following result.
\begin{lemma}\label{lem:self-similar}
A Lawvere-Tierney operator in $\Ecal/X$ lifting one in $\Ecal$ is self-similar. In particular, the corresponding subtopos is stable under pullback along each \'{e}tale morphism $m:\Ecal/X \to \Ecal/X$ induced by an endomorphism of $X$.
\end{lemma}
When $X$ is `symmetric enough', the converse of \Cref{lem:self-similar} holds, and we may use this in place of \Cref{prop:operatorlift}.
\begin{dfn}\label{dfn:EnoughEndomorphismsSupport}
We say an object $X$ of a topos $\Ecal$ \textbf{has enough endomorphisms} if the joint coequalizer of its endomorphisms is its support, $\supp(X)$.
\end{dfn}
\begin{prop}\label{prop:self-similar}
%\hr{I don't think this is true: $X=0_\Ecal $.}
Suppose $X$ has enough endomorphisms and let $g : \bang^*(Y) \to \bang^*(Z)$ in $\Ecal/X$. The following conditions are equivalent:
\begin{enumerate}
\item $g$ is self-similar.
\item \label{pt:lifted} There exists a unique morphism $h: \supp(X) \times Y \to \supp(X) \times Z$ in $\Ecal$ such that $g=\bang^*(h)$.
\end{enumerate}
In particular, a Lawvere-Tierney operator $k$ in $\Ecal/X$ is lifts one in $\Ecal$ if and only if $k$ is self-similar.
\end{prop}
\begin{proof}
By self-similarity, the following diagram commutes (in fact, is a pullback) in $\Ecal$ for each endomorphism $m:X \to X$:
% https://q.uiver.app/#q=WzAsNCxbMSwxLCJYIFxcdGltZXMgWiJdLFswLDEsIlggXFx0aW1lcyBaIl0sWzEsMCwiWCBcXHRpbWVzIFkiXSxbMCwwLCJYIFxcdGltZXMgWSJdLFsyLDAsImciXSxbMywxLCJnIiwyXSxbMywyLCJtIFxcdGltZXMgXFxpZCJdLFsxLDAsIm0gXFx0aW1lcyBcXGlkIiwyXV0=
\[\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times Y} \& {X \times Y} \\
{X \times Z} \& {X \times Z}
\arrow["{m \times \id}", from=1-1, to=1-2]
\arrow["g"', from=1-1, to=2-1]
\arrow["g", from=1-2, to=2-2]
\arrow["{m \times \id}"', from=2-1, to=2-2]
\end{tikzcd}\]
Taking the joint coequalizer of the horizontal morphisms, similarly to \eqref{eq:doublecoeq}, we recover the required unique morphism $h:\supp(X) \times Y \to \supp(X) \times Z$. %, which is unique, although if $X$ was not well-supported, there may be several morphisms in $\Ecal$ lifting to this morphism in $\Ecal/\supp(X)$.
\end{proof}
\begin{rmk}
Unlike in \Cref{prop:operatorlift}, we cannot deduce that the morphism $h$ produced in \Cref{prop:self-similar} is of the form $\id_{\supp(X)} \times h'$ for some $h':Y \to Z$, since applying this result when $X$ is the initial object (which always has enough endomorphisms) we could deduce that there is a morphism $Y \to Z$ for \textit{any} objects $Y$ and $Z$ of $\Ecal$, which is false in general.
\end{rmk}
\subsection{\'{E}tendues}
\label{ssec:etendues}
\begin{dfn}
\label{def:etendue}
A Grothendieck topos $\Ecal$ is called an \textbf{\'{e}tendue} if there is some well-supported object $X$ such that $\Ecal/X$ is a localic topos, meaning that we have an \'{e}tale morphism $\Sh(L) \nrightarrow \Ecal$ with faithful inverse image functor.
\end{dfn}
Rosenthal \cite{rosenthal1981etendues} worked extensively with \'{e}tendues. However, much of the locale theory that we have collected in \Cref{ssec:locale} has emerged only since Rosenthal's contributions.
The reason for our interest in \'{e}tendues is that $\PSh(\Sigma^*)$ is an \'{e}tendue. Indeed, viewing $\Sigma^*$ itself as an object of $\PSh(\Sigma^*)$ (as a set with right action by concatenation), we have $\PSh(\Sigma^*)/\Sigma^* \simeq \PSh(G_{\Sigma}^*)$, where $G_{\Sigma}^*$ denotes the poset freely generated by the Cayley graph of $\Sigma^*$, and this topos is localic by \Cref{xmpl:fourtypeexa}.\ref{ex:posetal}. The corresponding locale is exactly the extended sequence space $\Sigma^{\leq \omega}$, as can be seen by identifying the basic opens of \eqref{eq:hatU(u)} with the principal upward-closed sets in $G_{\Sigma}^*$.
Given $X$ in $\Ecal$, we have that $\Ecal/X$ is localic if and only if the hyperconnected geometric morphism to its localic reflection of \Cref{xmpl:fourtypeexa}.\ref{ex:locrefl} is an equivalence. A subterminal object in $\Ecal/X$ is a subobject of $X$ (which we must in this instance take care not to identify with its domain). As such, these are the opens of the resulting locale $L$; explicitly, the morphism sends $f:Y \to X$ to the \textit{sheaf of partial sections},
\[(u:U \rightarrowtail X) \mapsto \Ecal/X(u,f),\]
where the right-hand side is the set of commuting triangles over $X$.
Even when the morphism $\Ecal/X \to \Sh(L)$ is not an equivalence, we can still ask which sublocales of $\Sh(L)$ are the images of subtoposes of $\Ecal/X$:
% https://q.uiver.app/#q=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
\[\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{\Sh(L')} \& {\Sh(L)} \\
{\sh_j(\Ecal)/i^*(X)} \& {\Ecal/X} \\
{\sh_j(\Ecal)} \& \Ecal
\arrow[hook, from=1-1, to=1-2]
\arrow[two heads, from=2-1, to=1-1]
\arrow[hook, from=2-1, to=2-2]
\arrow["\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=2-1, to=3-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-1, to=3-2]
\arrow["h"',two heads, from=2-2, to=1-2]
\arrow["\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=2-2, to=3-2]
\arrow["i"', hook, from=3-1, to=3-2]
\end{tikzcd}\]
In full generality, this would amount to applying $h_*$ to the diagram \eqref{eq:pi13comm} of \Cref{prop:operatorlift}, but since in our case of interest $X$ will always have enough endomorphisms, we directly translate the condition of \Cref{prop:self-similar}. Given an endomorphism $m$ of $X$, applying the localic reflection to the \'{e}tale morphism $m:\Ecal/X \to \Ecal/X$ produces a localic morphism $\Sh(L) \to \Sh(L)$ and hence a locale map $L \to L$. In other words, $L$ too inherits an action of $\End(X)$.
\begin{dfn}\label{def:self-similarsublocale}
Let $L$ be a locale equipped with a (left) action of a (discrete) monoid $M$. We say a sublocale $L'$ of $L$ is \textbf{self-similar} with respect to the action if it is preserved by it, in the sense that for each element $m \in M$, we have a pullback square in the category of locales:
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJMJyJdLFswLDEsIkwnIl0sWzEsMCwiTCJdLFsxLDEsIkwiXSxbMSwzLCJpIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwyLCJpIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJtIl0sWzAsMSwibXxfe0wnfSIsMl0sWzAsMywiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d
\[\begin{tikzcd}[ampersand replacement=\&]
{L'} \& L \\
{L'} \& L
\arrow["i", hook, from=1-1, to=1-2]
\arrow["{m|_{L'}}"', from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow["m", from=1-2, to=2-2]
\arrow["i"', hook, from=2-1, to=2-2]
\end{tikzcd}\]
That is, $m \circ i$ factors through $i$ and the factoring map is a pullback of $m$ along $i$.
We use the same definition for a subspace $X'$ of a topological space $X$, where it reduces to demanding that $m(p) \in X'$ \textit{if and only if} $p \in X'$ for each $m \in M$.
\end{dfn}
% We would like to translate the conditions of \Cref{prop:operatorlift} along this morphism so that we can uniquely identify subtoposes of $\Ecal$ with certain sublocales of $L$ when $\Ecal$ is an \'{e}tendue.
% Considering \eqref{eq:pi13}, observe that for any monomorphism $u:U \rightarrowtail X$ we have isomorphisms,
% \begin{align*}
% \Ecal/X(u, X \times X \times \Omega \xrightarrow{\pi_1} X) &\cong \Ecal(U,X) \times \Sub_{\Ecal}(U) \\
% \Ecal/X(u, X \times \Omega \xrightarrow{\pi_1} X) &\cong \Sub_{\Ecal}(U),
% \end{align*}
% by passing across the adjunction. Up to these isomorphisms, applying the direct image of the morphism to $\Sh(L)$ to the morphism \eqref{eq:pi13} produces the natural transformation whose component at $u$ is simply:
% \[\pi_2 : \Ecal(U,X) \times \Sub_{\Ecal}(U) \to \Sub_{\Ecal}(U).\]
% In \eqref{eq:pi13comm}, $k$ becomes the closure operator $\Sub_{\Ecal}(U) \to \Sub_{\Ecal}(U)$ induced by some nucleus on $L$. \mr{I have been trying for hours to come up with an intrinsic description of (the translation of) $\id \times k$. In $\Ecal/X$, it sends $\langle u,t,\chi_v \rangle$ to $\langle u, k(\langle t, \chi_v \rangle) \rangle$, but the argument passed to $k$ on the right-hand side is not a valid morphism of $\Ecal/X$ (this expression only formally makes sense in $\Ecal$) so I don't know how to translate it properly!}
\begin{crly}
\label{crly:inducedsublocalesaress}
Let $\Sh(L)$ be the localic reflection of $\Ecal/X$. Then any subtopos $\Sh(L')$ of $\Sh(L)$ which is the image of the lift of a subtopos of $\Ecal$ corresponds to a sublocale $L'$ which is self-similar with respect to the action of $\End(X)$ on $L$. If $X$ has enough endomorphisms, the converse holds.
\end{crly}
\begin{proof}
It suffices to check that $h$ maps a subtopos induced by a Lawvere-Tierney operator which is self-similar as a morphism in $\Ecal/X$ (\Cref{def:self-similar}) to a subtopos corresponding to a sublocale which is self-similar in the sense of \Cref{def:self-similarsublocale}. This is straightforward once we recall that the direct image functor of a hyperconnected morphism preserves the subobject classifier, \cite[Proposition A4.6.6(v)]{Ele}.
\end{proof}
\begin{lemma}
\label{lem:ssintersect}
A meet of self-similar sublocales of $L$ is self-similar.
\end{lemma}
\begin{proof}
This is an instance of limits commuting with limits. The meet of a collection of sublocales is their (wide) pullback, and the restriction of the endomorphisms of $L$ restrict along this pullback to provide a restriction to the meet. This restriction is universal (completes a pullback square) since the meet is.
\end{proof}
\begin{lemma}
\label{lem:ssintersect}
Suppose a monoid $M$ acts on $L$ by inclusions or local homeomorphisms. Then a (finite) join of sublocales is self-similar.
\end{lemma}
\begin{proof}
For inclusions, the pullback along an inclusion $m$ is the intersection with $m$ in the lattice of subtoposes. Since this lattice is distributive, this intersection distributes over finite joins.
\end{proof}
When $h$ is an equivalence (notably, when $\Ecal$ is an \'{e}tendue) conditions improve even further, since the endomorphism of $\Ecal/X \simeq \Sh(L)$ induced by an endomorphism of $X$ is \'{e}tale, thus corresponding to a \textit{local homeomorphism} $L \to L$, a very strong property. %We will only need to exploit the weaker condition that these maps are \emph{open}; see \cite[Lemma C1.5.3]{Ele} for a characterization of open maps of locales.
\begin{lemma}\label{lem:denseself-similar}
Suppose $L$ is equipped with an action of a (discrete) monoid $M$ by local homeomorphisms. Then the double-negation sublocale $L_{\neg\neg}$ is self-similar.
\end{lemma}
\begin{proof}
Considering the corresponding toposes, this amounts to observing that the double-negation topology is stable under slicing.
\end{proof}
\begin{rmk}
Note that the restriction $m|_{L_{\neg\neg}}$ exists if and only if $m$ is a \textit{skeletal} map, a much weaker condition. When $M$ acts by skeletal \textit{inclusions}, we can again deduce that $L_{\neg\neg}$ is self-similar by \cite[Proposition 3.4]{JohnstoneFactorizationII}. However, we cannot expect the squares to be pullbacks for skeletal surjections \textit{a priori}: open surjections are skeletal and stable under pullback, but the domain of an open surjection need not be Boolean (for example, $\Sh(X) \to \Set$ is open for \textit{any} topological space $X$) whereas the double-negation sublocale must be Boolean. We do not know whether the constraint that $m$ is an endomorphism of $L$ makes a difference here.
\end{rmk}
Nonetheless, the condition that $M$ should act by open maps is enough to provide very concrete constraints on the nucleus corresponding to a self-similar sublocale, as the following technical lemma demonstrates.
\begin{lemma}\label{lem:jdensepres}
Suppose $L$ is equipped with an action of a (discrete) monoid $M$ by open maps (such as local homeomorphisms). Let $j:\Ocal(L) \to \Ocal(L)$ be the nucleus corresponding to a sublocale $i:L' \to L$ which is self-similar with respect to the action. Let $V$ be a subobject of $m_!(1)$ (the image of $m$) and let $v:V' \rightarrowtail V$ be an inclusion of opens of $L$. Then $v$ is $j$-dense if and only if $m^*(v)$ is $j$-dense.
\end{lemma}
\begin{proof}
Consider the pullback square defining self-similarity of \Cref{def:self-similarsublocale}. After decomposing $m$ into an open surjection followed by an open inclusion, we can construct the following diagram:
% https://q.uiver.app/#q=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
\[\begin{tikzcd}[ampersand replacement=\&,column sep=large]
{L'/i^*m^*(V')} \& {L'/i^*m^*(V)} \& {L/m^*(V)} \& L \\
{L'/i^*(V')} \& {L'/i^*(V)} \& {L/V} \& {L/m_!(1)} \\
{L'/i^*(V')} \& {L'/i^*(V)} \& {L/V} \& L
\arrow["{L'/i^*m^*(v)}", hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=1-1, to=1-2]
\arrow[from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow["{i/m^*(V)}", hook, from=1-2, to=1-3]
\arrow[from=1-2, to=2-2]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=2-3]
\arrow[hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=1-3, to=1-4]
\arrow[from=1-3, to=2-3]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=2-4]
\arrow["{m_1}", from=1-4, to=2-4]
\arrow["{L'/i^*(v)}", hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=2-1, to=2-2]
\arrow[equals, from=2-1, to=3-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-1, to=3-2]
\arrow["{i/V}", hook, from=2-2, to=2-3]
\arrow[equals, from=2-2, to=3-2]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=3-3]
\arrow[hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=2-3, to=2-4]
\arrow[equals, from=2-3, to=3-3]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-3, to=3-4]
\arrow["{m_0}", hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=2-4, to=3-4]
\arrow["{L'/i^*(v)}", hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=3-1, to=3-2]
\arrow["{i/V}", hook, from=3-2, to=3-3]
\arrow[hook, "\slash"{marking,font=\fontsize{8}{2}\selectfont}, from=3-3, to=3-4]
\end{tikzcd}\]
where, for instance, $L/V$ denotes the open sublocale of $L$ corresponding to $V$ and we mark the arrows to indicate the corresponding properties of the induced geometric morphisms. The lower-right square being a pullback corresponds to $V$ being contained in $m_!(1)$, the middle column follows from self-similarity of $L'$ and the remainder is straightforward.
Since open surjections are stable under pullback, the result follows by inspection of the upper-left square: the lower morphism is a homeomorphism if and only if the upper morphism is, corresponding respectively to $i^*(v)$ and $i^*m^*(v)$ being isomorphisms and hence to $v$ and $m^*(v)$ being $j$-dense.
\end{proof}
\begin{crly}
\label{crly:basisdensity}
Suppose $L$ is equipped with an action of a (discrete) monoid $M$ by open maps and $L'$ is a self-similar sublocale. Suppose further that the action of $M$ \emph{generates a basis of $L$}, in the sense that for every $U \in \Ocal(L)$, either $U = 0$ or there exists some $m \in M$ with $m^*(U) = 1$. Then $L'$ is either dense or trivial in $L$.
\end{crly}
\begin{proof}
Let $j$ be the nucleus on $L$ corresponding to $L'$ as in \Cref{lem:jdensepres}. The `exterior' $\mathrm{ext}(j)$ of $j$ is the open computed as (the domain of) the $j$-closure of $0 \rightarrowtail 1$ \cite[Lemma A4.5.9]{Ele}, and $L'$ is dense if and only if the exterior is $0$.
Otherwise, $\mathrm{ext}(j) \neq 0$ and the inclusion $0 \rightarrowtail \mathrm{ext}(j)$ is $j$-dense. By hypothesis, there exists some $m$ with $m^*(\mathrm{ext}(j)) = 1$; applying \Cref{lem:jdensepres}, we conclude that $0 = m^*(0) \rightarrowtail m^*(\mathrm{ext}(j)) = 1$ is $j$-dense, which means $j$ is trivial.
\end{proof}
Returning to our discussion of sublocales from \Cref{ssec:locale}, we can add a final refinement to the adjunction between subspaces and sublocales. Observe that when $L$ has an action of a monoid $M$, the same is true of its points. For a single point, the action is as one might expect:
% https://q.uiver.app/#q=WzAsMyxbMiwxLCJMIl0sWzIsMCwiTCJdLFswLDAsIjEiXSxbMSwwLCJtIl0sWzIsMSwicCJdLFsyLDAsIm0ocCkiLDJdXQ==
\[\begin{tikzcd}[ampersand replacement=\&]
1 \&\& L \\
\&\& L
\arrow["p", from=1-1, to=1-3]
\arrow["{m(p)}"', from=1-1, to=2-3]
\arrow["m", from=1-3, to=2-3]
\end{tikzcd}\]
\begin{lemma}\label{lem:sspts}
Let $L$ be any locale equipped with an action of a (discrete) monoid $M$. %local homeomorphisms.
For any self-similar sublocale $L'$ of $L$ with respect to the action of $M$, $\pt(L')$ is self-similar.
\end{lemma}
\begin{proof}
The $\pt$ functor preserves pullbacks, being a right adjoint, hence preserves self-similarity. More explicitly, a point $p \in L$ belongs to a sublocale $L'$ if and only if the corresponding locale map from $1$ factors through it. Inspecting the pullback diagram,
% https://q.uiver.app/#q=WzAsNSxbMiwxLCJMIl0sWzIsMCwiTCJdLFsxLDAsIkwnIl0sWzEsMSwiTCciXSxbMCwwLCIxIl0sWzEsMCwibSJdLFsyLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJtfF97TCd9IiwyXSxbNCwzLCIiLDIseyJjdXJ2ZSI6Miwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzQsMSwicCIsMCx7ImN1cnZlIjotMn1dLFs0LDIsIiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDAsIiIsMCx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==
\[\begin{tikzcd}[ampersand replacement=\&]
1 \& {L'} \& L \\
\& {L'} \& L,
\arrow[dashed, from=1-1, to=1-2]
\arrow["p", curve={height=-12pt}, from=1-1, to=1-3]
\arrow[curve={height=12pt}, dashed, from=1-1, to=2-2]
\arrow[hook, from=1-2, to=1-3]
\arrow["{m|_{L'}}"', from=1-2, to=2-2]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=2-3]
\arrow["m", from=1-3, to=2-3]
\arrow[hook, from=2-2, to=2-3]
\end{tikzcd}\]
we indeed see $m(p)$ belongs to $L'$ if and only if $p$ belongs to $L'$.
\end{proof}
With all of this theory developed, we end this section with the result we will use to structure our classification of subtoposes (via sublocales) in the sequel.
\begin{prop}[Stratification adjunctions]
\label{prop:sssubspaceadjunction}
Let $L$ be a spatial locale equipped with an action of a monoid $M$ by local homeomorphisms. Then the adjunction $(\loc \dashv \pt)$ between sublocales and subspaces of $L$ restricts to $M$-self-similar sublocales and $M$-self-similar spaces. If $L$ is moreover $T_D$-spatial, the adjoint triple \eqref{eq:maxloc} also restricts:
% https://q.uiver.app/#q=WzAsMixbMiwwLCJcXFN1Yl5NX3tcXFRvcH0oXFxwdChMKSkiXSxbMCwwLCJcXG1hdGhybXtkfVxcU3ViXk1fe1xcTG9jfShMKSJdLFsxLDAsIlxcbWF0aHJte3B0fSIsMSx7ImxhYmVsX3Bvc2l0aW9uIjo3MH1dLFswLDEsIlxcbWF0aHJte0xvY30oLSkgXFx2ZWUgTF97XFxuZWdcXG5lZ30iLDIseyJvZmZzZXQiOjR9XSxbMCwxLCJcXG1hdGhybXtNYXh9IiwwLHsib2Zmc2V0IjotNH1dLFszLDIsIiIsMix7ImxldmVsIjoxLCJzdHlsZSI6eyJuYW1lIjoiYWRqdW5jdGlvbiJ9fV0sWzIsNCwiIiwyLHsibGV2ZWwiOjEsInN0eWxlIjp7Im5hbWUiOiJhZGp1bmN0aW9uIn19XV0=
\[\begin{tikzcd}[ampersand replacement=\&,column sep=large]
{\Sub^M_{\Loc}(L)} \&\& {\Sub^M_{\Top}(\pt(L))}
\arrow[""{name=0, anchor=center, inner sep=0}, "{\mathrm{pt}}"{description, pos=0.7}, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "{\mathrm{Loc}}"', shift right=4, from=1-3, to=1-1]
\arrow[""{name=2, anchor=center, inner sep=0}, "{\mathrm{Max}}", shift left=4, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=0, to=2]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}\]
If $L$ has no isolated points, we can restrict to dense sublocales on the left by replacing $\mathrm{Loc}$ with $\mathrm{Loc}(-) \vee L_{\neg\neg}$.
\end{prop}
\begin{proof}
\Cref{lem:sspts} shows that $\pt$ restricts in this way. Similarly, $\Max$ preserves pullbacks and hence self-similarity as soon as it is defined, so it remains to show that $\mathrm{Loc}$ produces self-similar sublocales.
$\mathrm{Loc}$ need not preserve pullbacks (or even products) in general, but it does preserve pullbacks of local homeomorphisms, which is all that is needed: the domain of a local homeomorphism into a spatial locale is spatial.
Regarding the restriction to dense sublocales, we already observed in \Cref{lem:noisolated} that $\Max$ produces dense sublocales under the given hypotheses, so the lower adjunction restricts. Meanwhile, $L_{\neg\neg}$ is self-similar by \Cref{lem:denseself-similar} and joins of self-similar sublocales are self-similar thanks to compatibility of joins with pullbacks in $\Loc$, so $\loc(S) \vee L_{\neg\neg}$ is the smallest dense $M$-self-similar sublocale whose set of points is $S$, as required.
\end{proof}
% \begin{rmk}
% \label{rmk:self-similar-counter}
% In case it is not clear from the proof of \Cref{prop:sssubspaceadjunction}, $\mathrm{Loc}$ cannot be expected to produce self-similar sublocales on its own: we know from \Cref{crly:dense} that when $\Ecal$ is a hyperconnected \'{e}tendue then every subtopos of $\Ecal/X$ is either degenerate or dense... \mr{actually, I notice that whenever the set of points is not empty, it is already dense, so I don't have an example supporting this argument unless we find an example of an $\Ecal$ and (well-supported) $X$ such that $X$ doesn't have enough endomorphisms but $\Ecal/X$ is localic?}
% \end{rmk}
A strategic consequence of \Cref{prop:sssubspaceadjunction}, recalling that this restriction of $\Loc$ is a coreflective inclusion under these hypotheses, is that we can index the collection of self-similar sublocales of $L$ over the collection of self-similar sub\textit{spaces} of $L$. The latter are much easier to classify, since the action of $M$ induces an equivalence relation on the points of $L$ such that the self-similar subspaces are necessarily unions of equivalence classes of points.
\section{Subtoposes of monoid actions}
Our strategy for classifying subtoposes of $\PSh(\Sigma^*)$, then $\PSh(M)$, then $\Cont(M,\tau)$ is as follows. Consider the succession of hyperconnected morphisms which we described in \Cref{xmpl:fourtypeexa}.\ref{ex:monoidhype}:
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJcXFBTaChcXFNpZ21hXiopIl0sWzEsMCwiXFxQU2goTSkiXSxbMiwwLCJcXENvbnQoTSxcXHRhdSkiXSxbMCwxLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMSwyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XV0=
\[\begin{tikzcd}[ampersand replacement=\&]
{\PSh(\Sigma^*)} \& {\PSh(M)} \& {\Cont(M,\tau)}
\arrow[two heads, from=1-1, to=1-2]
\arrow[two heads, from=1-2, to=1-3]
\end{tikzcd}\]
The left-hand topos is an \'{e}tendue, so its subtoposes can be characterized using the topos and locale theory elaborated in the previous section, identifying them with self-similar sublocales of the extended sequence space $\Sigma^{\leq \omega}$. Given a subtopos of one of the other toposes, we can pull it back along the hyperconnected morphism to obtain a subtopos of $\PSh(\Sigma^*)$; since this operation is injective, we can identify subtoposes of $\PSh(M)$ and $\Cont(M)$ with subtoposes of $\PSh(\Sigma^*)$ verifying further conditions, which we can make explicit. %In the opposite direction, given a subtopos of $\Sh(\Sigma^{\leq \omega})$, we can compose the inclusion with the geometric morphisms to $\PSh(\Sigma^*)$ (and beyond), then take the \textbf{(surjection, inclusion) factorization} to recover a subtopos of the latter. In other words, subtoposes of $\PSh(\Sigma^*)$ can be identified with a \mr{check variance, could be co-} reflective subposet of the subtoposes of $\Sh(\Sigma^{\leq \omega})$ and our task is reduced to understanding sublocales of $\Sigma^{\leq \omega}$ followed by isolating the fixed points of the reflection.
% Any subtopos of a topos of the form $\Cont(M,\tau)$ is the image along that hyperconnected morphism of a subtopos of $\PSh(\Sigma^*)$ for some $\Sigma$, as we can see by constructing a pullback of geometric morphisms:
% https://q.uiver.app/#q=WzAsNCxbMSwxLCJcXENvbnQoTSxcXHRhdSkiXSxbMSwwLCJcXFBTaChcXFNpZ21hXiopIl0sWzAsMSwiXFxTaChcXENjYWxfcyxKKSJdLFswLDAsIlxcU2goXFxTaWdtYV4qLEopIl0sWzEsMCwiaCIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFsyLDAsImkiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsImknIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMywyLCJoJyIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFszLDAsIiIsMix7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==
% \[\begin{tikzcd}[ampersand replacement=\&]
% {\Sh(\Sigma^*,J)} \& {\PSh(\Sigma^*)} \\
% {\Sh(\Ccal_s,J)} \& {\Cont(M,\tau)}
% \arrow["{i'}", hook, from=1-1, to=1-2]
% \arrow["{h'}"', two heads, from=1-1, to=2-1]
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
% \arrow["h", two heads, from=1-2, to=2-2]
% \arrow["i"', hook, from=2-1, to=2-2]
% \end{tikzcd}\]
% \mr{May want to introduce the canonical site $(\Ccal_s,J)$ for $\Cont(M,\tau)$.}
% Since the first named author established a classification of hyperconnected quotients of presheaf toposes in \cite{hquotients}, we can thus entirely characterize subtoposes of toposes of topological actions(!)
\subsection{Subtoposes of $\PSh(\Sigma^*)$}
\subsubsection{The maximal and minimal self-similar sublocales}
To apply the results of \Cref{ssec:locale}, we must first observe that $\Sigma^*$ equipped with a right action on itself by multiplication, viewed as an object of $\PSh(\Sigma^*)$, has enough endomorphisms in the sense of \Cref{dfn:EnoughEndomorphismsSupport}. Indeed, the endomorphisms of $\Sigma^*$ also correspond to elements of $\Sigma^*$ (acting by concatenation on the left) and the generators from $\Sigma$ along with the empty word already have the terminal object as their joint coequalizer.
The inherited action of $m \in \Sigma^*$ on the space $\Sigma^{\leq \omega}$ simply sends a word $w$ to $mw$. In terms of the (basic) opens of \eqref{eq:hatU(u)}, the inverse image map sends $\hat{U}(u)$ to $\hat{U}(w)$ when $u = mw$ and to $\emptyset$ otherwise. Specializing \Cref{def:self-similarsublocale} to this situation, we obtain the following.
\begin{lemma}
\label{lem:self-similar}
A sublocale $L'$ of sequence space $\Sigma^{\leq \omega}$ is \textbf{self-similar} (in the sense of \Cref{def:self-similarsublocale}) with respect to the action of $\Sigma^*$ if and only if for each $a \in \Sigma$, we have a pullback square:
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJMJyJdLFswLDEsIkwnIl0sWzEsMCwiXFxTaWdtYV57XFxsZXEgXFxvbWVnYX0iXSxbMSwxLCJcXFNpZ21hXntcXGxlcSBcXG9tZWdhfSJdLFsxLDMsImkiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDIsImkiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDMsImEgXFxjZG90LSJdLFswLDEsImEgXFxjZG90IC18X3tMJ30iLDJdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==
\[\begin{tikzcd}[ampersand replacement=\&]
{L'} \& {\Sigma^{\leq \omega}} \\
{L'} \& {\Sigma^{\leq \omega}}
\arrow["i", hook, from=1-1, to=1-2]
\arrow["{a \cdot -|_{L'}}"', from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow["{a \cdot-}", from=1-2, to=2-2]
\arrow["i"', hook, from=2-1, to=2-2]
\end{tikzcd}\]
where $a \cdot -$ is the concatenation map.
\end{lemma}
\begin{proof}
Since we can glue pullback squares together, it suffices to check the self-similarity condition on the generators. It is straightforward to verify that the endomorphism of $\Sigma^*$ in $\PSh(\Sigma^*)$ corresponding to $a$ induces the endomorphism of $\Sigma^{\leq \omega}$ given (on points) by concatenation.
\end{proof}
From here on, we leave the action of $\Sigma^*$ on the locale $\Sigma^{\leq \omega}$ (by local homeomorphisms!) implicit, and speak only of `self-similar sublocales'. All of the results of \Cref{ssec:etendues} apply, so that in particular \Cref{crly:inducedsublocalesaress} yields:
\begin{crly}
\label{crly:geometricClassification'}
Subtoposes of $\PSh(\Sigma^*)$ are in one-to-one correspondence with self-similar sublocales of the extended sequence space $\Sigma^{\leq \omega}$.
\end{crly}
We would of course like to apply \Cref{prop:sssubspaceadjunction} to structure a classification of the self-similar sublocales of $\Sigma^{\leq \omega}$. Since $\Sigma^{\leq \omega}$ is not $T_D$ (any infinite word fails to be complemented as a sublocale) we first show that we can restrict to dense sublocales of $\Sigma^{\omega}$. To this end, we introduce the following definitions.
\begin{dfn}\label{lem:prefix-dense}
We say a subset $C \subseteq \Sigma^*$ is \textbf{prefix-complete} if every infinite word (element of $\Sigma^{\omega}$) has a prefix in $C$ (in the sense of \Cref{dfn:prefix}).
%Dually, we say $U \subseteq \Sigma^*$ is \textbf{prefix-dense} if every word in $\Sigma^*$ is a prefix of some word in $U$.
Finally, we say $P \subseteq \Sigma^*$ is \textbf{prefix-independent} if for any $u,v \in P$, $u \not \pf v$ ($u$ is not a prefix of $v$).
\end{dfn}
\begin{lemma}[Ordinal rank lemma]
\label{lem:rank}
We can assign an ordinal \textbf{rank} to any prefix-complete, prefix-independent set $C \subseteq \Sigma^*$ as follows:
\begin{itemize}
\item The rank of the minimal such set, $\{\epsilon\}$, is defined to be $0$.
\item The rank of any other $C$ is the supremum of the successors of the ranks of $C\cdot a$ across $a \in \Sigma$ (recalling \eqref{eq:closedsetaction}).
\end{itemize}
\end{lemma}
\begin{proof}
Prefix-independence here ensures that each $C \cdot a$ is non-empty as soon as $C \neq \{\epsilon\}$. The rank of $C$ is well-defined as soon as the rank of its pullbacks $C \cdot a$ are. That is, $C$ fails to have a well-defined rank if and only if there is an infinite word $w$ such that at each finite prefix $u \pf w$, the rank of $C \cdot u$ is not defined, but this would be a contradiction to prefix-completeness, since some prefix $u \pf w$ must belong to $C$, and for this $u$ we have $C \cdot u = \{\epsilon\}$ of rank $0$.
\end{proof}
\begin{rmk}
\label{rmk:notnotrank}
This argument is non-constructive in the following sense: we have actually shown that every prefix-complete, prefix-independent set can \textit{not not} be assigned an ordinal rank. We will not examine in detail whether this is an obstacle to extending the induction argument in the proofs that follow to an intuitionistic setting.
\end{rmk}
\begin{lemma}\label{lem:prefix-complete}
For each finite word $u \in \Sigma^*$, the singleton subspace $\{u\} \subseteq \Sigma^{\leq \omega}$ is complemented. Considering the sublocale $\bigcap_{u \in \Sigma^*} \{u\}^c$, an open set $U$ is a fixed point of the corresponding nucleus if and only if whenever there is a prefix-complete set $C$ and finite word $u$ with $uC := \{uv \mid v \in C\} \subseteq U$, then $u \in U$.
\end{lemma}
\begin{proof}
For the first sentence, observe that $\{u\}$ is the intersection of the basic open $\hat{U}(u)$ and the (closed) complement of $\bigcup_{a \in \Sigma}\hat{U}(ua)$
Let $U$ be any open of $\Sigma^{\leq \omega}$. The closure operation corresponding to $\{u\}^c$ sends $U$ to the interior of $U \cup \{u\}$. Thus $U$ is a fixed point for the nucleus corresponding to $\bigcap_{u \in \Sigma^*} \{u\}^c$ if and only if $u \in U$ whenever $\{ua \mid a \in \Sigma\} \subseteq U$.
To show that this is equivalent to the hypothesised condition (which implies the above, since $\Sigma \subseteq \Sigma^*$ is prefix-complete), observe first that we can assume that the prefix-complete set $C$ in question is prefix-independent. Indeed, any word in $\Sigma^{\omega}$ must have a \textit{shortest} prefix in $C$, and no pair of such shortest prefixes can be prefixes of one another.
Thus assuming that $uC \subseteq U$ for a prefix-complete, prefix-independent set $C$, we can show that $u \in U$ by induction on the rank of $C$ from \Cref{lem:rank}:
\begin{itemize}
\item If $C$ has rank $0$, $uC = \{u\} \subseteq U$.
\item Otherwise, each $C \cdot a$ has strictly lower rank than $C$, so we can by the induction hypothesis deduce that $ua \in U$ for each $a \in \Sigma$, which implies that $u \in U$ by the closure condition.
\end{itemize}
This concludes the proof.
\end{proof}
\begin{prop}\label{prop:maxproper}
$\Sigma^{\leq \omega}$ has a unique maximal self-similar sublocale homeomorphic to $\Sigma^{\omega}$.
\end{prop}
\begin{proof}
Although $\Sigma^{\leq \omega}$ is not $T_D$, we can still exploit the $(\loc \dashv \pt)$ adjunction between self-similar sublocales and self-similar spaces.
Let $L'$ be any self-similar sublocale. By inspection of \Cref{lem:self-similar}, $L'$ contains a finite word if and only if it contains all finite words. Any sublocale containing all finite words also contains all of the infinite words, by \Cref{rmk:pfspec}. But then $L'$ contains all points of $\Sigma^{\leq \omega}$, so must be equal to it! As such, any non-trivial self-similar sublocale must exclude all finite words. That is, $\pt(L') \subseteq \pt(\Sigma^{\omega})$.
It remains to show that there are no larger sublocales with the same set of points. By the argument of \Cref{lem:TD}, the sublocale $\bigcap_{u \in \Sigma^*} \{u\}^c$ defined in \Cref{lem:prefix-complete} is the largest sublocale with this set of points. %Taking the intersection over words up to a fixed length, we can conveniently present this as a codirected meet of opens $\bigcap_{n=0}^\infty \bigcup_{|w| = n} \hat{U}(w)$. \mr{It stops being obvious that this presentation is equivalent as soon as $|\Sigma|$ is infinite.}
We will show that this is equal to $\Sigma^\omega$. Indeed, suppose that $U$ satisfies the hypothesis of \Cref{lem:prefix-complete} and that we are given an open $U'$ with $U' \cap \Sigma^\omega = U \cap \Sigma^\omega$. Given a finite word $u' \in U'$, consider the standard open $U(u') \subseteq U' \cap \Sigma^\omega$. Every word in $U' \cap \Sigma^\omega$ has a prefix in $U$. In other words, there is a prefix-complete set $C$ with $u'C \subseteq U$. By \Cref{lem:prefix-complete}, it follows that $u' \in U$. Thus $U' \subseteq U$, which completes the proof, since it shows that $U$ is already the largest open of $\Sigma^{\leq \omega}$ having the required intersection with $\Sigma^\omega$.
\end{proof}
Thus we can reduce to considering self-similar sublocales of $\Sigma^{\omega}$. Since $\Sigma^{\omega}$ is a $T_D$-spatial locale with no isolated points, we can apply \Cref{prop:sssubspaceadjunction} to fibre the collection of self-similar sublocales over the collection of self-similar subspaces.
\begin{prop}
\label{prop:mindense}
The dense sublocale ${\Sigma^\omega}_{\neg\neg}$ is the minimal non-trivial self-similar sublocale.
\end{prop}
\begin{proof}
We know from \Cref{lem:denseself-similar} that ${\Sigma^\omega}_{\neg\neg}$ is self-similar and non-trivial, so it suffices to show that the action of $\Sigma^*$ satisfies the condition of Corollary \ref{crly:basisdensity} to conclude that every non-trivial self-similar sublocale is dense.
That is, we must show that $\Sigma^*$ `generates a basis' of $\Sigma^\omega$, meaning that for every $U \in \Ocal(\Sigma^\omega)$ there exists some $u \in \Sigma^*$ with $u^*(U)$ the whole space. This is immediate from the fact that the basis of the topology on $\Sigma^\omega$ has opens of the form $U(u)$: any open must contain an open of this form, and $u^*(U(u)) = 1$, as required.
\end{proof}
% Version 2:
% \begin{lemma}
% For any two-valued topos, the double-negation subtopos is the minimal non-trivial subtopos.
% \end{lemma}
% \begin{proof}
% Any non-trivial subtopos of a two-valued subtopos is dense, since it must contain both the initial and terminal objects. The double-negation subtopos is the smallest dense subtopos and hence the smallest non-trivial subtopos.
% \end{proof}
\subsubsection{Self-similar sub\textit{spaces} of sequence space}
With Propositions \ref{prop:maxproper} and \ref{prop:mindense}, we have finally arrived at the adjoint triple situation prepared in \Cref{prop:sssubspaceadjunction}.
Let us first classify the self-similar sub\textit{spaces} of $\Sigma^{\omega}$. Consider the equivalence relation on words generated by $u \sim au$ for letters $a \in \Sigma$, $u \in \Sigma^{\leq \omega}$. Two infinite words are related by this equivalence relation if and only if they have a shared suffix. The self-similar subspaces of $\Sigma^{\omega}$ are simply the unions of equivalence classes for ${\sim}$.
\begin{lemma}
Each equivalence class of $\Sigma^{\omega}$ for the equivalence relation ${\sim}$ defined above is dense and, when $|\Sigma| \geq 2$, has cardinality equal to $|\Sigma^*|$.
\end{lemma}
\begin{proof}
Consider the equivalence class containing some fixed word $w$. For density, observe that for any finite word $u$, we have $w \sim uw$ and the latter lies in $U(u)$; that is, every basic open contains an element of the equivalence class.
For the cardinality, %observe that for $\Sigma$ empty, $\Sigma^{\omega}$ is empty or a singleton, there is a single equivalence class, so the
observe that any $w' \sim w$ has a shared suffix with $w$, meaning there exists an infinite word $v$ and finite words $u,u'$ with $w=uv$ and $w'=u'v$. In other words, we can obtain any such $w'$ by specifying $|u|$ (the size of the prefix to be deleted to obtain $v$) and $u'$ (the prefix to be added to obtain $w'$). Thus the equivalence class of $w$ injects into $\Nbb \times \Sigma^*$. When $|\Sigma| \geq 2$ we also have a surjection onto $\Sigma^*$, and, since $\Sigma^*$ is infinite in this case, the cardinality of $\Nbb \times \Sigma^*$ is equal to that of $\Sigma^*$. %Classically, the above implies that the actual cardinality lies between these two and hence is equal to them both. \mr{might not bother with the cardinality argument!}
\end{proof}
Recall that a subspace of a topological space is \textbf{nowhere dense} if it has empty interior.
\begin{lemma}
If a union of equivalence classes has non-empty interior, it is all of $\Sigma^\omega$.
\end{lemma}
\begin{proof}
$S$ having non-empty interior means that it contains some $U(u)$. By self-similarity it must contain $U(\epsilon) = \Sigma^\omega$.
\end{proof}
As such, the proper self-similar sublocales of $\Sigma^\omega$ are nowhere dense, and we should try to understand nowhere-dense subsets a little better.
\begin{lemma}
\label{lem:nwdensepres}
Let $C \subseteq \Sigma^\omega$ be a closed, nowhere-dense set. Then the sets
\begin{itemize}
\item $C \cdot u := \{w \in \Sigma^\omega \mid uw \in C\}$ and
\item $\Sigma C := \{aw \mid a \in \Sigma, w \in C\}$
\end{itemize}
are closed and nowhere dense.
\end{lemma}
\begin{proof}
Note that $w \notin C \cdot u$ if and only if for $uw \notin C$. Since $C$ is closed, there is a finite prefix $uv \pf uw$ with $U(uv)$ contained in the complement of $C$. Then $U(v) \ni w$ is contained in the complement of $C \cdot u$.
Given $a \in \Sigma$ with $aw \notin \Sigma C$, we must have $w \notin C$, hence $v \pf w$ with $U(v)$ contained in the complement of $C$, whence $U(av) \ni aw$ is contained in the complement of $\Sigma^k C$.
To see that each is nowhere dense, consider the dual argument: $U(uv) \subseteq C \cdot u$ if and only if $U(v) \subseteq C$, if and only if $U(av) \subseteq \Sigma C$ for any $a$. None of these can happen since $C$ is nowhere dense.
\end{proof}
\begin{lemma}
\label{lem:countableunion}
Each equivalence class for ${\sim}$ is a union of countably many closed, nowhere dense subsets. Dually, the complement of any equivalence class is a countable intersection of dense opens.
\end{lemma}
\begin{proof}
For $\Sigma$ empty, the result is trivial, so assume not. Suppose $w = a_0a_1a_2a_3\cdots$. Let $w^{(k)} = a_ka_{k+1}a_{k+2}\cdots$, so $w = w^{(0)}$. Consider the subsets,
\[C_{i,j} := \Sigma^{i} \{w^{(j)}\},\]
meaning those elements obtained by adding a prefix of length $i$ to $w^{(j)}$. Each of these is closed and nowhere-dense by repeated application of \Cref{lem:nwdensepres} and their union is the full equivalence class $[w]$. The complement of each is dense, so the claim regarding the complement of $[w]$ also follows.
\end{proof}
Combining \Cref{lem:countableunion} with \Cref{prop:Baire3}, we immediately obtain the following:
\begin{thm}
\label{thm:countablecomplement}
Suppose $L' \subseteq \Sigma^\omega$ is a self-similar subspace such that $\pt(L')$ includes all but countably many ${\sim}$-equivalence classes. Then $L'$ is spatial.
\end{thm}
Thus we have found that sufficiently large members in the lattice of dense, self-similar sublocales of $\Sigma^\omega$ are determined entirely by their points.
% \begin{lemma}
% \label{lem:perfection?}
% Let $|\Sigma| \geq 2$. Each equivalence class for ${\sim}$ is a perfect subspace. \mr{Not true, not closed...} Dually, the complement of any equivalence class is coperfect.
% \end{lemma}
% Let $S$ be a union of equivalence classes.
\subsubsection{Self-similar sublocales of sequence space}
Consider the case that $S$ consists of the single ${\sim}$-equivalence class in $\{0,1\}^{\omega}$ generated by $(01)^\omega$ -- this is the class of sequences which eventually alternate between $0$ and $1$. Meanwhile, let $C$ be the nowhere-dense perfect set $\{00,11\}^* \subseteq \{0,1\}^*$ consisting of sequences of paired digits. We clearly have $C \cap S = \emptyset$.
By \ref{crly:maxplicit2}, the elements of $\Ocal(\Max(S))$ can be identified with the opens of $\Sigma^\omega$ whose complements' isolated points lie in $S$, while those of $\Ocal(\Loc(S))$ are those $U$ such that $U = \int(U \cup S^c)$.
Letting $U = C^c$, this trivially satisfies the former definition (since $C$ has no isolated points). However, $U \cup S^c = \{0,1\}^\omega = \int(U \cup S^c)$.
Since we can take any closed set and eliminate its isolated points, for any $S$ the separation between $\Loc(S)$ and $\Max(S)$ is controlled by the number of nowhere-dense perfect sets which avoid $S$. \Cref{thm:countablecomplement} essentially says that when $S$ is large enough, no such perfect sets exist.
Such sets are stable under finite unions. By taking the perfect part of an intersection, we also have arbitrary meets, so we have a filter of perfect sets. The shape of a nowhere-dense perfect set is also determined by the size of a cover; any such must be homeomorphic to $\Gamma^\omega$ for some set $\Gamma$ with $|\Gamma| \leq |\Sigma|$.
-----
Let $S$ be a union of equivalence classes. Is anything else required for $\Max(S)$ to be distinct from $\loc(S)$?
Note: I write $\bigcup^\infty \cdots$ for the operations iterated transfinitely.
Given an open set $U \subseteq \Sigma^\omega$, the respective closure operations are:
$\Max(S)(U) = \bigcup_{w' \sim w}^\infty \inter(U \cup \{w\})$.
$\loc(S)(U) = \bigcup \{U' \mid U' \cap S = U \cap S\}$.
Suppose that $U$ is closed for the upper operator; by \Cref{lem:maxplicit}, this means that any isolated points of $U^c$ lie in $S$, and hence that $U^c \cap S^c$ has no isolated points.
Given $U'$ with $U' \cap S = U \cap S$ and $w' \in U' \cap S^c$, %suppose $w' \notin U$
there is some basic open neighbourhood of $U(u') \subseteq U'$ with $u' \pf w'$. Now $U(u') \cap S \subseteq U$. If we have $w' \in U^c$ then $w' \in U^c \cap S^c$ (so $w' \sim w$) then there are infinitely many points in $U(u') \cap S^c$. ... This is true for any open by density of $S^c$ ...
---
\noindent
\mr{TODO: Cite Caramello (or earlier) for construction of De Morganisation.}
\begin{lemma}
The De Morganization of $\Sh(\Sigma^{\omega})$ (the largest De Morgan subtopos) coincides with the double-negation subtopos.
\end{lemma}
\begin{proof}
Note that when $\Sigma$ has at most one element, both toposes coincide with $\Sh(\Sigma^{\omega}) \simeq \Set/\Zbb$ since this is already Boolean.
When $\Sigma$ has at least two elements, we decompose $\Sigma$ as $X \sqcup Y$ with $X,Y$ non-empty (relying on excluded middle). Given a dense open set $U$, let $P_{L(U)}$ be the prefix-independent generating set for the corresponding ideal. We can construct two prefix-independent sets
\begin{align*}
Q_X &:= \{ux \mid u \in P_{L(U)}, x \in X\} \\
Q_Y &:= \{uy \mid u \in P_{L(U)}, y \in Y\}.
\end{align*}
By construction, the corresponding opens in $\Sigma^\omega$ are double-negation closed (must show this), and their union is $U$. In particular, $U$ must be covering for the De Morgan coverage. Since the De Morgan coverage is contained in the double-negation coverage, this shows that the coverages coincide, as claimed.
\end{proof}
\section{Topological point of view: Cantor space, Jonsson-Tarski algebra, and Radical ideals}
This section aims to provide a topological point of view on the infinite branch argument. For a full tree, the set of infinite branches gives a closed subset of the Cantor space. Here is the summary of the correspondence:
\begin{table}[ht]
\centering
\begin{tabular}{|c|l|c|} \hline
Word combinatorics &cf.& Topology of Cantor space\\ \hline
finite word $w$ &\cref{rmk:GeometricIntuitionofV}& function over $\Sigma^{\omega}$\\ \hline
infinite word $p$ &\cref{rmk:GeometricIntuitionofV}& point of $\Sigma^{\omega}$\\ \hline
prefix $w\pf p$ &\cref{rmk:GeometricIntuitionofV}& $w$ is non-zero at $p$\\ \hline
non-prefix $w \not \pf p$ &\cref{rmk:GeometricIntuitionofV}& $w$ vanishes at $p$\\\hline
ideal &\cref{rmk:GeometricIntuitionofV}& set of functions\\\hline
radical ideal &\cref{crly:IsomOfRadicalsAndOpen}& open set $\leftrightarrow$ closed set\\\hline
Description of radicals &\cref{thm:Nullstellensatz}&Nullstellensatz\\\hline
(Jonsson-Tarski) covering ideal&\cref{prop:coveringIdeal}&The whole space $\Sigma^{\omega}$\\\hline
dense covering &\cref{xmple:denseIdealisDense}&dense open set\\\hline
\textbf{non-trivial coverage}&\cref{thm:geometricClassification} & \textbf{Self-similar sublocale}\\\hline
\end{tabular}
\end{table}
This section also provides a full description of non-trivial coverages.
\mr{[I have extracted material from this section, moving it into Section \ref{ssec:prefix}.]}
Just like any other binary relation, the relation ${\not \pf}\subset\Sigma^* \times \Sigma^{\omega}$ defines an anti-Galois connection between their power sets. Namely, we will consider the following functions:
\[
V\colon \Pow(\Sigma^*) \to \Pow(\Sigma^{\omega})\colon S \mapsto V(S)\coloneqq \{p\in \Sigma^{\omega}\mid \forall w\in S, w\not \pf p\}
\]
\[
I\colon \Pow(\Sigma^{\omega}) \to \Pow(\Sigma^{*})\colon C \mapsto I(C)\coloneqq \{w\in \Sigma^{*}\mid \forall p\in C, w\not \pf p\}
\]
\begin{prop}
\label{prop:galois-connection}
Two functions $V, I$ define an anti-Galois connection. In other words, for any $S\subset \Sigma^*$ and $C\subset \Sigma^{\omega}$,
\[
S\subset I(C) \iff C\subset V(S).
\]
\end{prop}
\begin{proof}
Both conditions are equivalent to
$
\forall w\in S, \forall p \in C, w\not \pf p.
$
\end{proof}
\begin{rmk}
\label{rmk:VandTrees}
For an ideal $I$, the set $V(I)$ is the set of all infinite branches of the corresponding tree $T_I$.
\end{rmk}
Hereafter, we will consider the product topology on the set $\Sigma^{\omega}$. If $2\leq |\Sigma|<\omega$, this topological space is isomorphic to the Cantor space. Notice that it is non-compact if $\Sigma$ is an infinite set.
Just for later reference, we prove the compatibility with the $\Sigma^*$ action.
For a subset $C \subset \Sigma^{\omega}$ and $w\in \Sigma^*$, we define $C\cdot w\coloneqq \{p\in \Sigma^{\omega}\mid wp\in C\}$ just like \cref{eq:IdealAction}.
\begin{prop}
\label{prop:compatibilityOfActions}
The Galois connections \cref{thm:GaloisConnectionOfIdealsAndClosed}
and \cref{crly:GaloisConnectionOfIdealsAndOpen} are internal to $\PSh(\Sigma^*)$, in the sense that
\begin{itemize}
\item for any subset $C\subset \Sigma^{\omega}$ and $w\in \Sigma^*$, $I(C\cdot w)=I(C)\cdot w$.
\item for any ideal $I\in \Ideal(\Sigma^{\ast})$ and $w\in \Sigma^*$, $V(I\cdot w)=V(I)\cdot w$.
\end{itemize}
Similarly, $D$ and $I$ in \cref{crly:GaloisConnectionOfIdealsAndOpen} commutes with the action of $\Sigma^*$.
\end{prop}
\begin{proof}
Since the $\Sigma^*$ action is compatible with taking complement, it is enough to prove it for \cref{crly:GaloisConnectionOfIdealsAndOpen}.
Notice that $w\pf p \iff wv\pf wp$ for any $w,v\in \Sigma^*$ and $p\in \Sigma^{\omega}$.
For an arbitrary subset $C\subset \Sigma^{\omega}$ and $w\in \Sigma^*$,
\begin{align*}
& u \in I(C\cdot w)\\
\iff&\left( \forall q \in \Sigma^{*}, u\pf q \implies q\in C\cdot w \right) \\
\iff&\left( \forall q \in \Sigma^{*}, wu\pf wq \implies wq\in C\right) \\
\iff &\left( \forall p \in \Sigma^{*},
% w\pf p \land
wu\pf p \implies p\in C\right) \\
\iff & u\in I(C)\cdot w.
\end{align*}
For an arbitrary subset $S\subset \Sigma^*$ and $w\in \Sigma^*$,
\begin{align*}
& p\in D(S\cdot w)\\
\iff & \exists v\in (S\cdot w), v\pf p\\
\iff & \exists v\in (S\cdot w), wv\pf wp\\
\iff & \exists u\in S, (w\pf u \land u\pf wp)\\
\implies & \exists u \in S, u\pf wp\\
\iff & p\in D(S)\cdot w.
\end{align*}
If $S$ is an ideal, the above implication is equivalent.
% $p\in D(I) \iff \$
\end{proof}
This Galois connection is the bridge between the topology of the Cantor space and the word combinatorics. The natural question is: is it equivalence of posets? The answer is no. While $V(I(C)) = C$ for a closed set $C\subset \Sigma^{\ast}$, the inclusion $S \subset I(V(S))$ may not be equal in general, even for an ideal $S$. Our next aim is to describe the set $I(V(S))$, just like Nullstellensatz in algebraic geometry!
\begin{rmk}
If $\Sigma$ is finite, being covering is equivalent to being generated by prefix-complete set. If $\Sigma$ is infinite, then they are not equivalent. There exists a infinite height tree $T_I$ without infinite branches.
\end{rmk}
\begin{lemma}
\label{lem:CovIsNonTrivialCoverage}
The set $J_{\text{cov}}$ is a non-trivial Grothendieck coverage.
\end{lemma}
\begin{proof}
$J_{\text{cov}}$ is a Grothendieck coverage corresponding to the local operator $\rad$ (\cref{thm:RadAsLocalOperator}).
Since ideal $\Sigma^* \setminus \{\varepsilon\} \subsetneq \Sigma^{*}$ is a covering ideal, $J_{\text{cov}}$ is non-trivial.
\end{proof}
To prove \cref{thm:RadAsMinimalLocalOperator}, we define a technical tool for measuring the complexity of an ideal.
\begin{dfn}
We define the \textbf{rank} of an ideal $I$ by the following recursion:
\begin{itemize}
\item an ideal $I$ has rank $0$, if $I=\Sigma^{*}$.
\item an ideal $I$ has rank $\alpha$ for an ordinal $\alpha$, if $I\cdot a$ has rank (strictly) lower than $\alpha$.
\item When an ideal $I$ has a rank, the smallest rank is denoted by $r(I)$.
\end{itemize}
\end{dfn}
\begin{lemma}
\label{lem:CoveringIFFrank}
An ideal is covering if and only if it has rank.
\end{lemma}
\begin{proof}
(If-part) Suppose $I$ has a rank. We prove that $I$ is covering by transfinite induction of $r(I)$. If $r(I) =0$, then $I$ is equal to $\Sigma^*$, which is trivially covering. If $r(I)>0$, then every $I\cdot a$ for $a\in \Sigma$ has a lower rank, then is covering. This implies that $I$ is also covering.
(Only-if-part) Suppose $I$ does not have rank. This implies that there exists $a_1\in \Sigma$ such that $I\cdot a_1$ does not have rank. Repeating this process, we can take an infinite sequence $a_1, a_2, \dots$ such that $I\cdot (a_1 \dots a_n)$ does not have rank for every $n\geq 0$. Let $p\in \Sigma^{\omega}$ denote the infinite sequence $a_1 a_2 a_3 \dots$.
Since $I\cdot (a_1 \dots a_n)$ cannot be $\Sigma^{*}$, any prefix of $p$ cannot be an element of $I$. This implies that $I$ is not covering.
% If there exists $w\in I$ such that $w\pf p$, then there exists $n\geq 0$ such that $I\cdot (a_1 \dots a_n) =\Sigma^{*}$. This contradicts the fact that $I\cdot (a_1 \dots a_n)$
\end{proof}
\begin{thm}
\label{thm:TheMinimalNontrivialCoverage}
The Grothendieck coverage $J_{\text{cov}}$ is minimum among non-trivial Grothendieck coverage.
\end{thm}
\begin{proof}
Let $J$ be the Grothendieck coverage generated by the ideal $\Sigma\setminus \{\varepsilon\}$. Since every non-trivial coverage contains the ideal $\Sigma\setminus \{\varepsilon\}$, $J$ is the minimum non-trivial coverage. This observation and \cref{lem:CovIsNonTrivialCoverage} imply that $J \subset J_{\text{cov}}$.
We prove $J \supset J_{\text{cov}}$ by transfinite induction of rank. Due to \cref{lem:CoveringIFFrank}, every covering ideal has rank.
Let $I$ be a covering ideal. If $r(I)=0$, then $I=\Sigma^{*} \in J$. If $r(I)>0$, then $I\cdot a$ for every $a\in \Sigma$ belongs to $J$ by the induction hypothesis. Since $\Sigma\setminus \{\varepsilon\} \in J$, the transitivity axiom of Grothendieck coverage guarantees that $I\in J$.
% For an covering ideal $I \in J_{\text{cov}}$, we define its \textbf{complexity} $c(I)$ to be the minimum ordinal larger than $c(I\cdot a)$ for every $a\in \Sigma$:
% \[
% c(I) \coloneqq \min\{\alpha \colon \text{ordinal}\mid \alpha> c(I\cdot a) \text{ for every }a\in \Sigma\}
% \]
% This definition utilize transifinite recursion. We need to check its well-definedness.
\end{proof}
\Cref{thm:RadAsMinimalLocalOperator} is just a paraphrase of \cref{thm:TheMinimalNontrivialCoverage}
% \begin{thm}
% \label{thm:RadIsminimal}
% The local operator $\rad\colon \Ideal(\Sigma^{*})\to \Ideal(\Sigma^*)$ is minimum among non-trivial local operators.
% \end{thm}
\subsection{Geometric classification of subtopoi}
\begin{dfn}
\label{dfn:self-similar-original}
A sublocale $L$ of the locale $\Sigma^{*}$ (equipped with the product topology) is \textbf{self-similar} if the homeomorphism
\[
\coprod_{a\in \Sigma} \Sigma^{\omega} \to \Sigma^{\omega} \colon (p\text{ indexed by }a) \mapsto ap
\]
restricts to a homeomorphism
\[
\coprod_{a\in \Sigma} L \to L.
\]
\end{dfn}
\begin{thm}
\label{thm:geometricClassification}
Non-trivial subtopoi of $\PSh(\Sigma^*)$ are in one-to-one correspondence with self-similar sublocale of the generalized Cantor space $\Sigma^{\omega}$.
\end{thm}
\begin{rmk}
This theorem should be reinterpreted in terms of internal locale in $\PSh(\Sigma^*)$. See \cite[Proposition 4.6]{Wrigley} as well.
\end{rmk}
Recall that the poset of radical ideals is isomorphic to the frame of the open subsets of the generalized Cantor space (\cref{crly:IsomOfRadicalsAndOpen}):
\[\RadIdeal(\Sigma^{*})\cong \Ocal(\Sigma^{\omega})\]
\begin{lemma}
\label{lem:restrictionConstruction}
For a non-trivial local operator $j\colon \Ideal(\Sigma^*) \to \Ideal(\Sigma^*)$,
\begin{itemize}
\item $j(I)$ is a radical ideal for every ideal $I$,
\item the restriction $j\colon \RadIdeal(\Sigma^{\ast})\to \RadIdeal(\Sigma^{\ast})$ is a nucleus of the frame $\RadIdeal(\Sigma^{*})\cong \Ocal(\Sigma^{\omega})$.
\end{itemize}
\end{lemma}
\begin{proof}
The first half of the statement follows since $\rad\circ j = j$, which is a paraphrase of \cref{thm:RadAsMinimalLocalOperator}.
The latter half follows from the fact that $j$ is a local operator.
\end{proof}
For a non-trivial local operator $j$, let $L_j$ denote the corresponding sublocale of the nucleus $j\colon \RadIdeal(\Sigma^{\ast})\to \RadIdeal(\Sigma^{\ast})$.
--- \mr{results in relation to Section \ref{ssec:locale}.}
\begin{lemma}
Non-trivial self-similar sublocales of $\Sigma^\omega$ are dense.
\end{lemma}
\begin{proof}
\mr{TODO}
\end{proof}
\begin{lemma}
A sublocale $L$ of $\Sigma^\omega$ is self-similar if and only if and \mr{please complete; should be able to give a condition in terms of the expression as an intersection of opens}
\end{lemma}
\begin{proof}
\mr{TODO}
\end{proof}
\begin{lemma}
If $L$ is a self-similar sublocale of $\Sigma^\omega$ then $\pt(L)$ is a union of $\Sigma^*$-equivalence classes.
\end{lemma}
Referring back to the results in \Cref{ssec:locale}, we have thus already done most of the work required to classify subtoposes of $\PSh(M)$. Note that when $S \subseteq \Sigma^\omega$ is a union of $\Sigma^*$-equivalence classes, we automatically have that $\Max(S)$ and $\mathrm{Loc}(S)$ are self-similar sublocales, so we are reduced to considering the self-similar sublocales between these extremes. \mr{......}
---
% \begin{lemma}
% The correspondence $j \mapsto L_j$ is injective, in the sense that $L_{j}=L_{j'}$ implies $j=j'$ for non-trivial local operators $j,j'$.
% \end{lemma}
% \begin{proof}
% Since $j$ is non-trivial, we have $j\circ \rad = j$. This implies that $j$ can be reconstructed from the restriction $j\colon \RadIdeal(\Sigma^{\ast})\to \RadIdeal(\Sigma^{\ast})$ i.e., the corresponding sublocale $L_j$. Similarly, $j'$ can be reconstructed from the corresponding sublocale $L_{j'}$.
% \end{proof}
\begin{prop}
\label{prop:nontirivialIsEquivNucleus}
The restriction construction in \Cref{lem:restrictionConstruction} gives a one-to-one correspondence between
\begin{itemize}
\item non-trivial local operators $j\colon \Ideal(\Sigma^*) \to \Ideal(\Sigma^*)$
\item nuclei $k\colon \RadIdeal(\Sigma^{\omega}) \to \RadIdeal(\Sigma^{\omega})$ that commute with the $\Sigma^{\ast}$-action.
\end{itemize}
\end{prop}
\begin{proof}
For a non-trivial local operator $j\colon \Ideal(\Sigma^*) \to \Ideal(\Sigma^*)$, it is not hard to prove that its restriction $k_j\colon \RadIdeal(\Sigma^{\omega}) \to \RadIdeal(\Sigma^{\omega})$ is a nucleus that commutes with the $\Sigma^{\ast}$-action.
We prove that this correspondence is bijective.
For a non-trivial local operator $j$, since \cref{thm:RadAsMinimalLocalOperator} implies $j\circ \rad = j$, $j$ can be reconstructed from the restriction to $\RadIdeal(\Sigma^*)$. This shows the injectivity.
For proving surjectivity, let $k_j\colon \RadIdeal(\Sigma^{\omega}) \to \RadIdeal(\Sigma^{\omega})$ be a nucleus that commutes with the $\Sigma^{\ast}$-action. Then
\[
\begin{tikzcd} j_k \colon \Ideal(\Sigma^*)\ar[r,"\rad",twoheadrightarrow]&\RadIdeal(\Sigma^*)\ar[r,"k"]&\RadIdeal(\Sigma^*)\ar[r,hookrightarrow]&\Ideal(\Sigma^*)
\end{tikzcd}
\]
is a non-trivial local operator, whose restriction is $k$.
\end{proof}
\begin{lemma}
\label{lem:equivIsSelfSimilarity}
For a nucleus $j \colon \Ocal(\Sigma^{\omega})\to \Ocal(\Sigma^{\omega})$, the corresponding sublocale $L_j \rightarrowtail \Sigma^{\omega}$ is self-similar if and only if $j$ commutes with the $\Sigma^{\ast}$ action.
% (defined via the isomorphism
% \[\RadIdeal(\Sigma^{*})\cong \Ocal(\Sigma^{\omega}) \text{ (\cref{crly:IsomOfRadicalsAndOpen})}\])
\end{lemma}
\begin{proof}
The nucleus $j$ commutes with the $\Sigma^*$-action if and only if the following diagram commutes
\[
\begin{tikzcd}[column sep = 70pt]
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']\ar[d,"j"]&\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"j^{\Sigma}"]\\
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']&\Ocal(\Sigma^{\omega})^{\Sigma}
\end{tikzcd}
\]
(Notice that this diagram is not a diagram in $\Frm$, but it is just a diagram in $\Set$, because $j$ is not a frame homomorphism.)
On the other hand, the self-similarity of the sublocale $L_j \rightarrowtail \Sigma^\omega$ is equivalent to the existence of the dashed isomorphism
% in the diagram
\[
\begin{tikzcd}[column sep = 70pt]
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']\ar[d,"j^*", twoheadrightarrow]&\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"(j^*)^{\Sigma}", twoheadrightarrow]\\
\Ocal(L_j)\ar[r,"\cong", dashed]&\Ocal(L_j)^{\Sigma},
\end{tikzcd}
\]
in the category of frames $\Frm$ by the definition of the self-similarity translated via the equivalence $\Frm \cong \Loc^{\op}$.
Here, even if $\Sigma$ is infinite, the product $\Ocal(\Sigma^{\omega})^{\Sigma}$ in the category of frames coincides with the set-theoretic product, since the forgetful functor $\Frm \to \Set$, which is monadic, creates all small limits. Furthermore, the existence of the frame isomorphism is equivalent to the existence of the bijection in this case, since every bijection that makes the above diagram commutative is automatically a frame isomorphism.
% , and
% % Considering
% the frame isomorphism
% \[
% \Ocal(\Sigma^{\omega})^{\Sigma} \cong \Ocal\left(\coprod_{a\in \Sigma}\Sigma^{\omega}\right).
% \]
% the above commutative diagram is equivalent to saying that the corresponding sublocale of $j$ is self-similar. \mr{Why? (make explicit argument about limit preservation)}
If $j$ commutes with the $\Sigma^*$-action, then the bijection $\Ocal(\Sigma^{\omega}) \cong \Ocal(\Sigma^{\omega})^{\Sigma}$ induces the bijection between their fixed points $\Ocal(L_j)\cong \Ocal(L_j)^{\Sigma}$ since the naturality of taking coequalizers
\[
\begin{tikzcd}[column sep = 70pt]
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']\ar[d,"j"', shift right
] \ar[d,"\id", shift left]&\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"j^{\Sigma}"', shift right] \ar[d,"\id", shift left]\\
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']\ar[d,"j^*", twoheadrightarrow] &\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"(j^*)^{\Sigma}", twoheadrightarrow]\\
\Ocal(L_j)\ar[r,"\cong", dashed]&\Ocal(L_j)^{\Sigma}.
\end{tikzcd}
\]
This proves that $L_j$ is self-silimar.
Conversely, we assume that $L_j$ is self-similar, i.e., the diagram
\[
\begin{tikzcd}[column sep = 70pt]
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsto (U\cdot a)_{a\in \Sigma}"']\ar[d,"j^*", twoheadrightarrow]&\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"(j^*)^{\Sigma}", twoheadrightarrow]\\
\Ocal(L_j)\ar[r,"\cong", dashed]&\Ocal(L_j)^{\Sigma},
\end{tikzcd}
\]
commutes. Since every functions are frame homomorphisms, their right adjoints also commutes
\[
\begin{tikzcd}[column sep = 70pt]
\Ocal(\Sigma^{\omega})\ar[r,"\cong","U\mapsfrom (U\cdot a)_{a\in \Sigma}"', leftarrow]\ar[d,"j_*", twoheadleftarrow]&\Ocal(\Sigma^{\omega})^{\Sigma}\ar[d,"(j_*)^{\Sigma}", twoheadleftarrow]\\
\Ocal(L_j)\ar[r,"\cong", dashed, leftarrow]&\Ocal(L_j)^{\Sigma}.
\end{tikzcd}
\]
Combining the above two diagrams, we obtain the required commutativity, since $j= j_* \circ j^*$ and $j^\Sigma= (j_*)^\Sigma \circ (j^*)^\Sigma$
% Then, two frame homomorphisms
% \[
% j^*\colon \Ocal(\Sigma^\omega)\twoheadrightarrow\Ocal(L_j)
% \]
% and
% \[
% (j^*)^{\Sigma}\colon \Ocal(\Sigma^\omega)^{\Sigma}\twoheadrightarrow\Ocal(L_j)^{\Sigma}
% \]
% are isomorphic in the category of frame homomorphisms $\Frm^{\to}$, so their right adjoints
% \[
% j^*\colon \Ocal(\Sigma^\omega)\twoheadrightarrow\Ocal(L_j)
% \]
% and
% \[
% (j^*)^{\Sigma}\colon \Ocal(\Sigma^\omega)^{\Sigma}\twoheadrightarrow\Ocal(L_j)^{\Sigma}
% \]
\end{proof}
\begin{proof}[Proof of \cref{thm:geometricClassification}]
Due to the bijective correspondence between nuclei and sublocales, the theorem follows from \cref{prop:nontirivialIsEquivNucleus} and \cref{lem:equivIsSelfSimilarity}.
\end{proof}
\begin{xmpl}[$\Sigma=\emptyset$]
If $\Sigma=\emptyset$, the generalized Cantor space $\Sigma^{\omega}$ is also empty. The unique self-similar sublocale is $\emptyset$ itself. This corresponds to the fact that the only non-trivial subtopos of $\PSh(\emptyset^*)\simeq \Set$ is the degenerate one.
\end{xmpl}
\begin{xmpl}[$\Sigma$ is singleton]
If $\Sigma$ is singleton, the generalized Cantor space $\Sigma^{\omega}$ is also singleton. There are two self-similar sublocales, namely $\Sigma^{\omega}$ and $\emptyset$. These correspond to the two non-trivial subtopoi of $\PSh(\Sigma^{*})\simeq \PSh(\Nbb)$.
\end{xmpl}
\begin{rmk}[Interpretation of \hr{a modified version of} \cref{thm:finitebranches}]
The locale $\Sigma^{\omega}$ itself is the maximum self-similar sublocale. This corresponds to the minimum non-trivial coverage.
\end{rmk}
\begin{rmk}[Interpretation of \cref{thm:dense}]
If the corresponding sublocale $L$ is not dense, the self-similarity forces $L$ to be empty. Thus ${\Sigma^{\omega}}_{\lnot\lnot}$ gives the minimum self-similar sublocale. This is an interpretation of \Cref{thm:dense}.
\end{rmk}
\begin{rmk}[relationship with the infinite branch argument]
% In short, \textbf{an ideal is in $J$, iff all infinite branches are outside of the corresponding self-similar sublocale.}
For an ideal $I$, let $V(I) \subset \Sigma^{\omega}$ be the set of infinite branches of the tree $T_I$ (\Cref{thm:GaloisConnectionOfIdealsAndClosed}). This is a closed subset of $\Sigma^{\omega}$. Let $D(I)$ be its complement, which is open by \Cref{crly:GaloisConnectionOfIdealsAndOpen}.
For a non-trivial coverage $J$, let $j$ denote the corresponding nucleus on $\Ocal(\Sigma^{\omega})$. Then the following conditions are equivalent:
\begin{itemize}
\item $j(D(I))=\Sigma^{\omega}$ (``All infinite branches are outside of the self-similar sublocale.")
\item $I\in J$.
\end{itemize}
We can interpret arguments on infinite branches via \cref{thm:geometricClassification}.
\begin{itemize}
\item \Cref{lem:branchgeneration} corresponds to the self-similarity condition on the sublocale.
\item Subsets of $\Bcal$ are in one-to-one correspondence with self-similar topological subspaces. Since $\Sigma^{\omega}$ is a $T_1$ topological space, distinct subsets of $\Bcal$ define distinct self-similar sublocales of $\Sigma^{\omega}$.
\end{itemize}
\end{rmk}
\begin{crly}[Number of subtopoi]
\[
\text{The number of subtopoi of } \PSh(\Sigma^*) =
\begin{cases}
2 & |\Sigma|= 0\\
3 & |\Sigma|= 1\\
2^{2^{\aleph_0}} & 1<|\Sigma|<\aleph_0\\
2^{2^{|\Sigma|}} & \aleph_0\leq |\Sigma|
\end{cases}
\]
% The number of subtopoi of $\PSh(\Sigma^*)$ is,
% \begin{itemize}
% \item $2$ if $|\Sigma|= 0$.
% \item $3$ if $|\Sigma|= 1$.
% \item $\beth_2 = 2^{2^{\aleph_0}}$ if $1<|\Sigma|<\infty$.
% \item $2^{2^{\kappa}}$ if $\kappa=|\Sigma|\geq \aleph_0$.
% \end{itemize}
\end{crly}
\begin{proof}Supppose $|\Sigma|>1$.
Since different subsets of $\Bcal$ define a different subtopoi, there are at least $2^{|\Bcal|}= 2^{2^{\max{(\aleph_0,|\Sigma|)}}}$ subtopoi. Since the cardinality of $\Ocal(\Sigma^{\omega})\cong \RadIdeal(\Sigma^*)$ is $2^{\max{(\aleph_0,|\Sigma|)}}$, the number of its nuclei is bounded by $\left(2^{\max{(\aleph_0,|\Sigma|)}}\right)^{2^{\max{(\aleph_0,|\Sigma|)}}} = 2^{2^{\max{(\aleph_0,|\Sigma|)}}}$.
\end{proof}
There are remaining questions:
\begin{itemize}
\item Is this easily followed by the fact that the Jonsson-Tarski topos is \'{e}tendue?
\item Is this able to be written in terms of internal locale?
\item Is this result generalized to the general topos of monoid actions?
\end{itemize}
\begin{rmk}
\label{rmk:JTtopos}
The J\'{o}nsson-Tarski topos is named after the variety constructed by J\'{o}nsson and Tarski in \cite[Theorem 5]{jonsson1961two} as a counterexample. The observation that it is a topos is attributed to Freyd in \cite[Example 1.3]{johnstone1985variety}. %\cite[Exercise 6.9]{TT}
\mr{To move up:} Whereas the only coverages on a group are the trivial covering and the degenerate covering, which renders the situation uninteresting, there can be many non-trivial coverages on a monoid, as Johnstone observes \cite[Example 2.1.11(i)]{Ele}; see \Cref{rmk:JTtopos} below for many more references on the particular example of the J\'{o}nsson-Tarski Johnstone gives.
Bunge and Funk investigate \textit{distributions} on the J\'{o}nsson-Tarski topos \cite[\S8.5]{bunge2006singular}. They moreover attribute the result that it is an \'{e}tendue, covered by sheaves on Cantor space (Proposition 8.5.2 in \textit{loc.cit.}) to Freyd.
Rosenthal considers \'{e}tendues more generally in \cite{rosenthal1981etendues}; the J\'{o}nsson-Tarski topos appears as Example 1 there. Rosenthal also considers the case corresponding to $\Sigma = \Nbb$ in their Example 2. \mr{Worth noting that $\Nbb^\Nbb$ with the pointwise convergence topology is often called \textbf{Baire space}.}
To our knowledge, no one has previously explicitly exploited the fact that a topos is an étendue to study their subtoposes.
\end{rmk}
\appendix
\section{Grothendieck coverages on free monoids}
\label{sec:coverages}
$\PSh(M)$ is a special case of a Grothendieck topos\footnote{When not accompanied by a qualifier, `topos' will always mean `Grothendieck topos' in the remainder.}, and indeed of a presheaf topos. Recall that a Grothendieck topos is defined to be (any category equivalent to) a left exact reflective subcategory of a presheaf topos, which is to say a category $\Ecal$ admitting an adjunction:
% https://q.uiver.app/#q=WzAsMixbMCwwLCJcXEVjYWwiXSxbMiwwLCJcXFBTaChcXENjYWwpIl0sWzAsMSwiIiwyLHsib2Zmc2V0IjoyfV0sWzEsMCwiIiwwLHsib2Zmc2V0IjoyfV0sWzMsMiwiIiwwLHsibGV2ZWwiOjEsInN0eWxlIjp7Im5hbWUiOiJhZGp1bmN0aW9uIn19XV0=
\begin{equation}
\label{eq:subtoposofPSh}
\begin{tikzcd}[ampersand replacement=\&]
\Ecal \&\& {\PSh(\Ccal)}
\arrow[""{name=0, anchor=center, inner sep=0}, shift right=2, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, shift right=2, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}
\end{equation}
where the right adjoint is fully faithful and the left adjoint preserves finite limits. In this situation, $\Ecal$ is said to be a \textit{subtopos} of $\PSh(\Ccal)$. In this article we will primarily classify the subtoposes of $\PSh(M)$ for $M = \Sigma^*$ a free monoid (viewed as a one-object category). We shall see in the next subsection how this will enable us to classify subtoposes of $\PSh(M)$ an arbitrary monoid $M$ and indeed subtoposes of the category $\Cont(M)$ of continuous actions of a topological monoid on sets (see \Cref{xmpl:contmon}).
This abstract definition of topos is difficult to work with. Fortunately, a subtopos of $\PSh(\Ccal)$ is determined by the much more concrete structure of a \textit{Grothendieck coverage} on the category $\Ccal$. We recall here the basic definitions from \cite[Chapter III]{MLM}.
\begin{dfn}
\label{dfn:coverage}
Recall that a \textbf{presieve} over an object $C$ in a category $\Ccal$ is a collection of morphisms with codomain $C$. A \textbf{sieve} over $C$ is a presieve which is closed under composition on the right; any presieve generates a sieve.
A \textbf{Grothendieck coverage} on $\Ccal$ is a collection $J(C)$, for each object $C$ of $\Ccal$, of sieves over $C$ satisfying the following closure conditions:
\begin{itemize}
\item The \textit{maximal sieve} on $C$ (generated by the presieve $\{\id_C\}$) is a member of $J(C)$;
\item If $f:C' \to C$ is any morphism and $S \in J(C)$ then the \textit{pullback sieve}
\[f^*(S) := \{g:\dom(g) \to C' \mid f \circ g \in S\}\]
is a member of $J(C')$;
\item If $S \in J(C)$ is generated by a presieve $\{f_i:C_i \to C\}_{i \in I}$ and we are given $S_i \in J(C_i)$ for each $i\in I$, the \textit{multicomposite sieve}
\[S \circ \{S_i\}_{i \in I} := \{g \circ f_i \mid i \in I, g \in S_i\}\]
is also a member of $J(C)$.
\end{itemize}
\end{dfn}
When $\Ccal$ is a monoid, there is only one collection of sieves $J$ to consider.
\begin{rmk}
\label{rem:upwardclosed}
A consequence of the conditions in \Cref{dfn:coverage} is that Grothendieck coverages are \textit{upward closed}. That is, if $S \in J(C)$ and $S \subseteq S'$ then $S' \in J(C)$. Indeed, if we take $S' \cup \{\id_C\}$ as a presieve generating the maximal sieve, the multicomposite with the maximal sieves on $S'$ and the sieve $S$ on $\id_C$ produces precisely $S'$.
In the present article, we will typically be working with categories in which sieves admit minimal generating presieves. As a converse to the above, rather than considering arbitrary presieves generating sieves, it will suffice (and be more convenient) to demand that multicomposites exist for minimal generating presieves and to \textit{add upward closure of sieves to the axioms}.
\end{rmk}
Given a Grothendieck topology on $\Ccal$, the corresponding reflective subcategory of $\PSh(\Ccal)$ consists of the collection of presheaves satisfying corresponding \textit{sheaf conditions}.
\begin{dfn}[{\cite[\S III.4]{MLM}}]
Given a sieve $S$ over an object $C$ and a presheaf $X$ on $\Ccal$, $X$ \textit{satisfies the sheaf condition relative to $S$} if the following diagram is an equalizer diagram:
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJYKEMpIl0sWzMsMCwiXFxwcm9kX3tmIFxcaW4gU31YKFxcZG9tKGYpKSJdLFs2LDAsIlxccHJvZF97ZixnIFxcaW4gXFxzcShTKX1YKFxcZG9tKGYgXFxjaXJjIGcpKSJdLFswLDEsInggXFxtYXBzdG8gKFgoZikoeCkpX3tTfSJdLFsxLDIsIih4X2YpX3tTfVxcbWFwc3RvICh4X3tmIFxcY2lyYyBnfSkpX3tcXHNxKFMpfSIsMix7Im9mZnNldCI6Mn1dLFsxLDIsIih4X2YpX3tTfVxcbWFwc3RvIChYKGcpKHhfZikpX3tcXHNxKFMpfSIsMCx7Im9mZnNldCI6LTJ9XV0=
\[\begin{tikzcd}[ampersand replacement=\&]
{X(C)} \&\&\& {\prod_{f \in S}X(\dom(f))} \&\&\& {\prod_{f,g \in \sq(S)}X(\dom(f \circ g))}
\arrow["{x \mapsto (X(f)(x))_{S}}", from=1-1, to=1-4]
\arrow["{(x_f)_{S}\mapsto (x_{f \circ g}))_{\sq(S)}}"', shift right=2, from=1-4, to=1-7]
\arrow["{(x_f)_{S}\mapsto (X(g)(x_f))_{\sq(S)}}", shift left=2, from=1-4, to=1-7]
\end{tikzcd}\]
where the right-hand product is indexed over the set $\sq(C)$ of triangles in $S$, which is to say diagrams of the form,
% https://q.uiver.app/#q=WzAsMyxbMCwxLCJcXGJ1bGxldCJdLFsyLDEsIkMiXSxbMSwwLCJcXGJ1bGxldCJdLFsyLDEsImYiXSxbMCwyLCJnIl0sWzAsMV1d
\[\begin{tikzcd}[row sep = small, ampersand replacement=\&]
\& \bullet \\
\bullet \&\& C
\arrow["f", from=1-2, to=2-3]
\arrow["g", from=2-1, to=1-2]
\arrow[from=2-1, to=2-3]
\end{tikzcd}\]
with $f \in S$, and notation such as $(x_f)_S$ is shorthand for a tuple indexed by $f \in S$.
$X$ is said to be a \textbf{$J$-sheaf} if $X$ satisfies the sheaf condition relative to all $J$-covering sieves.
As is standard, we write $\Sh(\Ccal,J)$ for the full subcategory of $\PSh(\Ccal)$ on the $J$-sheaves. For any $J$ this is a reflective subcategory such that the left adjoint to the inclusion preserves finite limits.
\end{dfn}
For a fixed category $\Ccal$, the collection of Grothendieck coverages on $\Ccal$ forms a complete lattice under inclusion. This is contravariantly isomorphic to the lattice (hence coframe) of subtoposes of $\PSh(\Ccal)$: more covering sieves means fewer sheaves. Notably, the so-called \textit{trivial coverage} containing only the maximal sieves corresponds to the maximal subtopos, which is $\PSh(\Ccal)$ itself, and whenever $J' \supseteq J$ we have an adjunction,
% https://q.uiver.app/#q=WzAsMixbMCwwLCJcXFNoKFxcQ2NhbCxKKSJdLFsyLDAsIlxcUFNoKFxcQ2NhbCxKJykiXSxbMCwxLCIiLDIseyJjdXJ2ZSI6MX1dLFsxLDAsIiIsMCx7ImN1cnZlIjoxfV0sWzMsMiwiIiwwLHsibGV2ZWwiOjEsInN0eWxlIjp7Im5hbWUiOiJhZGp1bmN0aW9uIn19XV0=
\begin{equation}
\label{eq:subtoposSh}
\begin{tikzcd}[ampersand replacement=\&]
{\Sh(\Ccal,J')} \&\& {\Sh(\Ccal,J)}
\arrow[""{name=0, anchor=center, inner sep=0}, curve={height=6pt}, from=1-1, to=1-3]
\arrow[""{name=1, anchor=center, inner sep=0}, curve={height=6pt}, from=1-3, to=1-1]
\arrow["\dashv"{anchor=center, rotate=-90}, draw=none, from=1, to=0]
\end{tikzcd}
\end{equation}
in which the right adjoint is fully faithful and the left adjoint preserves finite limits, so we say that $\Sh(\Ccal,J')$ is a \textit{subtopos} of $\Sh(\Ccal,J)$. Conversely, we can identify subtoposes of $\Sh(\Ccal,J)$ with toposes $\Sh(\Ccal,J')$ such that $J' \supseteq J$.
Before we restrict to the one-object case, here is an example where $\Ccal$ has many objects, presenting the topos of continuous actions of a topological monoid on sets. We will use this in \mr{create section}.
\begin{xmpl}
\label{xmpl:contmon}
Let $M$ be a topological monoid, and denote by $\Cont(M)$ the category of (right) actions of $M$ on sets, $\alpha: X \times M \to X$ which are continuous with respect to the topology on $M$ and the discrete topology on $X$. The morphisms are $M$-set homomorphisms, all such being trivially continuous. Categories of this form are shown to be Grothendieck toposes in \cite{TTMA} as follows. Consider the category $\Ccal_s$ of \textit{principal actions} of $M$, which is the subcategory of $\Cont(M)$ on those actions generated by a single one of their elements. A natural Grothendieck coverage\footnote{Indeed, this is the so-called \textit{canonical} topology on $\Ccal_s$, but we shall not prove this here.} $J_s$ on this category is that whose sieves contain a surjection. With these definitions, one can show that $\Cont(M) \simeq \Sh(\Ccal_s,J_s)$, with the latter being a Grothendieck topology by definition. Explicitly, we can identify any continuous $M$-set $(X,\alpha)$ with the $J_s$-sheaf on $\Ccal_s$ whose value at a given principal action $(P,\beta)$ is the set of $M$-set homomorphisms $P \to X$.
\end{xmpl}
The correspondence between Lawvere-Tierney operators and Grothendieck coverages on a presheaf topos is straightforward: given $j$, the covering sieves in $J$ are exactly the $J$-dense subobjects of representable objects. Conversely, the pullback stability of a Grothendieck coverage $J$ means we can assemble it into a presheaf $J$; being a collection of sieves, this is a sub-presheaf of $\Omega$, and $j$ is its classifying map. Explicitly, in $\PSh(M)$ the Lawvere-Tierney operator corresponding to a Grothendieck coverage $J$ has:
\[j(I) := \{m \in M \mid I \cdot m \in J\},\]
using the action given in \Cref{eq:IdealAction}. Conversely, for a given $j$, we have
\[I \in J \iff j(I) = M.\]
\section{The structure of right ideals}
In order to understand Grothendieck coverages on $\Sigma^*$, we will identify their elements (right ideals in $\Sigma^*$) with related structures whose content is more easily described and constrained.
\subsection{Ideals correspond to prefix-independent sets}
\begin{dfn}
\label{dfn:prefixindependent}
We say a subset $U \subseteq \Sigma^*$ is \textbf{prefix-independent} if for any $u,v \in U$, $u \not \pf v$ ($u$ is not a prefix of $v$).
\end{dfn}
% Observation 0
\begin{lemma}
\label{lem:prefixindependent}
Let $I$ be a right ideal of $\Sigma^*$. Then there is a unique prefix-independent set $P_I \subseteq I$ which generates $I$. As such, there is a correspondence,
\[\{\text{right ideals of }\Sigma^*\} \longleftrightarrow \{\text{prefix-independent sets in }\Sigma^*\}\]
\end{lemma}
\begin{proof}
For each $w \in I$, consider the minimal prefix $w'$ of $w$ which lies in $I$. Let $P_I$ be the collection of such minimal prefixes. $P_I$ trivially generates $I$, since $w$ is in $I$ as soon as any prefix of $w$ is. $P_I$ is prefix-independent: if $u \in P_I$ were a prefix of $v \in P_I$ then the latter would not be minimal. Finally, any set generating $I$ must contain every element of $P_I$, since these cannot be presented as a non-trivial product (this would again contradict minimality), and conversely any further element of such a set must have a member of $P_I$ as a prefix, so can only be prefix-independent if it coincides with $P_I$, as required.
\end{proof}
Let us translate the Grothendieck coverage axioms across the correspondence of \Cref{lem:prefixindependent}.
\begin{itemize}
\item The maximal ideal corresponds to the singleton $\{\varepsilon\}$.
\item We may define
\begin{equation}
\label{eq:Pwstar}
w^*(P) :=
\begin{cases}
\{\varepsilon \} & \text{if } \exists x \in P \text{ such that } x \text{ is a prefix of } w \\
\{ x \mid wx \in P\} & \text{ otherwise.}
\end{cases}
\end{equation}
This definition is chosen such that for an ideal $I$, $P_{w^*(I)} = w^*(P_I)$.
\item Similarly, for prefix-independent sets $P$ and $\{P_w\}_{w \in P}$, the multicomposition operation corresponds to:
\begin{equation}
\label{eq:Pmulticomp}
P \circ \{P_w\} := \{ uw \mid w \in P, u \in P_w\},
\end{equation}
\item Noting that $P$ is merely the \textit{minimal} presieve generating its corresponding ideal, we can recover the full axiomatization using the condition derived in \Cref{rem:upwardclosed}. Indeed, the prefix-independent sets inherit the ordering from ideals as:
\begin{equation}
\label{eq:Porder}
P \leq P' \iff (\forall u \in P) (\exists v \in P') \; v \text{ is a prefix of } u
\end{equation}
\end{itemize}
We could have chosen to proceed with the remainder of the paper with prefix-independent sets. However, it is not easy to describe structure in prefix-independent sets. As such, we introduce a further class of structures determined by ideals.
\subsection{Ideals correspond to fully $|\Sigma|$-ary subtrees}
\label{ssec:fulltree}
Consider again the Cayley graph of $\Sigma^*$ from \Cref{exa:Cayley}.
Given an ideal $I$, the complement of $I$ determines a rooted subtree. \textit{A priori}, it would seem that this tree can take any shape, but if we adjoin to this subtree the prefix-independent generating set of $I$, we find that we can understand the structure more clearly.
\begin{figure}
\centering
% https://q.uiver.app/#q=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
\[\begin{tikzcd}[ampersand replacement=\&, column sep = small]
\textcolor{rgb,255:red,128;green,128;blue,128}{aaa} \& \textcolor{rgb,255:red,128;green,128;blue,128}{aab} \& \textcolor{rgb,255:red,128;green,128;blue,128}{aba} \& \textcolor{rgb,255:red,128;green,128;blue,128}{abb} \&\& \textcolor{rgb,255:red,128;green,128;blue,128}{baa} \& \textcolor{rgb,255:red,128;green,128;blue,128}{bab} \& \textcolor{rgb,255:red,128;green,128;blue,128}{bba} \& \textcolor{rgb,255:red,128;green,128;blue,128}{bbb} \\
\& \textcolor{rgb,255:red,128;green,128;blue,128}{aa} \& \textcolor{rgb,255:red,128;green,128;blue,128}{ab} \&\&\&\& \textcolor{rgb,255:red,255;green,51;blue,75}{ba} \& \textcolor{rgb,255:red,255;green,51;blue,75}{bb} \\
\&\& \textcolor{rgb,255:red,255;green,51;blue,75}{a} \&\&\&\& b \\
\&\&\&\& \varepsilon
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-2, to=1-1]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-2, to=1-2]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-3, to=1-3]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-3, to=1-4]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-7, to=1-6]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-7, to=1-7]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-8, to=1-8]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=2-8, to=1-9]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=3-3, to=2-2]
\arrow[color={rgb,255:red,128;green,128;blue,128}, from=3-3, to=2-3]
\arrow[color={rgb,255:red,255;green,51;blue,75}, from=3-7, to=2-7]
\arrow[color={rgb,255:red,255;green,51;blue,75}, from=3-7, to=2-8]
\arrow[color={rgb,255:red,255;green,51;blue,75}, from=4-5, to=3-3]
\arrow[from=4-5, to=3-7]
\end{tikzcd}\]
\caption{For $\Sigma = \{a,b\}$, the prefix-independent set $P = \{a,ba,bb\}$ in $\Sigma^*$ generates the ideal partially illustrated here in the Cayley graph in (red and) grey. The complement of this ideal is $\{\varepsilon,b\}$; if we include $P$, we see that we obtain a subtree in which every vertex is a leaf or has exactly two children.}
\label{fig:enter-label}
\end{figure}
\begin{dfn}
We say a rooted tree $T$ (where all edges are directed away from the root) is \textbf{fully $\alpha$-ary} (for a given cardinality $\alpha$) if every vertex is either a leaf or has exactly $\alpha$ children %(\textit{aka} successors)
in the tree.
\end{dfn}
% Observation 1 & 2
\begin{prop}
\label{prop:narytrees}
Given an ideal $I$, consider the subset $P_I \cup (\Sigma^* \backslash I) \subseteq \Sigma^*$. The corresponding rooted subtree of the Cayley graph $G_{\Sigma}$, which we denote $T_I$, is fully $|\Sigma|$-ary. Conversely, any fully $|\Sigma|$-ary subtree determines an ideal, namely that generated by its leaves.
\end{prop}
\begin{proof}
Consider the children (in $G_{\Sigma}$) of a vertex $v \in T_I$. If $v \in P_I$, then all of its children are in $I$ (but not in $P_I$), so $v$ is a leaf of $T_I$. Otherwise, $v \in \Sigma^* \backslash I$; if a child of $v$ is a member of $I$, then it must be in $P_I$ since no proper prefix of that child is in $I$. Otherwise, the child is in the complement of $I$. So all children are in $T_I$, as required to be fully $|\Sigma|$-ary. We conclude the converse by appealing to \Cref{lem:prefixindependent}.
\end{proof}
From here on we shall abbreviate `fully $|\Sigma|$-ary rooted subtree of the Cayley graph of $\Sigma^*$' to \textbf{full~tree} and denote the set of full trees by $\Tcal$. We write $\ell(T)$ for the set of leaves of a full tree $T$; this is a prefix-independent set, and we have $\ell(T_I) = P_I$ as one would expect.
%Observation 6
Let us translate the Grothendieck coverage axioms to full trees across the correspondence determined by \Cref{prop:narytrees}.
\begin{itemize}
\item The maximal ideal corresponds to the tree containing just the root $\{\varepsilon\}$.
\item Given a full tree $T$ and a word $w \in \Sigma^*$, pulling back an ideal corresponds to the operation:
\begin{equation}
\label{eq:Twstar}
w^*(T) :=
\begin{cases}
\{\varepsilon \} & \text{if } w \notin T \\
\{ x \mid wx \in T\} & \text{ otherwise,}
\end{cases}
\end{equation}
where the second line intuitively corresponds to taking the full subtree above the vertex corresponding to $w$ (and displacing it to the root).
\item Similarly, for full trees $T$ and $\{T_w\}_{w \in \ell(T)}$, the multicomposition operation corresponds to:
\begin{equation}
\label{eq:Tmulticomp}
T \circ \{T_w\}_{w \in \ell(T)} := T \cup \{ uw \mid w \in \ell(T),\ u \in T_w\}.
\end{equation}
Intuitively, the trees $T_w$ are grafted onto the leaves of $T$.
\item The ordering on full trees given by the subtree relation is dual to that on ideals, in the sense that
\[I \subseteq I' \iff T_I \supseteq T_{I'},\]
by inspection of the definition of $T_I$. Thus a collection of ideals is upward closed if and only if the corresponding collection of full trees is closed under taking rooted full subtrees.
\end{itemize}
\mr{Might consider making diagrams illustrating these operations.}
\subsection{Full trees determine closed subsets of Cantor spaces}
\label{ssec:Cantorsubsets}
Observe that a full tree may have infinite branches, which correspond to points of $\Sigma^{\omega}$, and conversely a set of points of $\Sigma^{\omega}$ generates the minimal full tree containing the corresponding branches. Thus we have a pair of functions:
% https://q.uiver.app/#q=WzAsMixbMSwwLCJcXFBjYWwoXFxTaWdtYV57XFxvbWVnYX0pIl0sWzAsMCwiXFxUY2FsIl0sWzEsMCwiQiIsMCx7Im9mZnNldCI6LTF9XSxbMCwxLCJUIiwwLHsib2Zmc2V0IjotMX1dXQ==
\[\begin{tikzcd}[ampersand replacement=\&]
\Tcal \& {\Pcal(\Sigma^{\omega})}
\arrow["B", shift left, from=1-1, to=1-2]
\arrow["\tr", shift left, from=1-2, to=1-1]
\end{tikzcd}\]
Explicitly, $\tr(C) := \{wu \mid u \in \Sigma, w \pf w' \in C\}$, while $B(T) = \{w \mid u \pf w \implies u \in T\}$. The composite $\tr \circ B$ is the operation of pruning off any finite branches a tree $T$ may have. More precisely, we obtain the following definition:
\begin{dfn}
We say a full tree $T$ is \textbf{reduced} if every vertex is either a leaf or belongs to an infinite branch.
\end{dfn}
On the other side, it is clear that $S \subseteq B(\tr(S))$, but the tree $B(S)$ will also contain the infinite branch $a_0a_1a_2\cdots$ whenever $a_0\cdots a_k$ is a prefix of some branch in $S$ for every $k$. We already have a convenient way of expressing this property.
\begin{lemma}
For $S \subseteq \Sigma^{\omega}$, $B(\tr(S))$ is the \textit{closure} of $S$ with respect to the topology on $\Sigma^{\omega}$.
\end{lemma}
As such, restricting the functions above we obtain an order-equivalence between the set of reduced trees and the set of closed subsets of $\Sigma^{\omega}$. We have a dual order-equivalence between the ideals corresponding to full trees and the open subsets complementary to the closed ones; for completeness, we derive it from first principles using the prefix relation.
Given a subset $I \subseteq \Sigma^*$ (typically an ideal), we can define the subset,
\[O(I):= \{u \in \Sigma^{\omega} \mid \exists w \in I, \, w \pf u\} \subseteq \Sigma^{\omega}.\]
This subset is, by inspection, the union of the basic open subsets $U(w)$ for $w \in I$, whence it is open. Conversely, a subset $S \subseteq \Sigma^{\omega}$ determines an ideal of $\Sigma^*$ by,
\[L(S) = \{w \in \Sigma^* \mid \forall u \in \Sigma^{\omega}, \, w \pf u \implies u \in U\}.\]
We can characterize the resulting ideals as follows.
\begin{dfn}
\label{dfn:readicalIdeal}
We call an ideal $I \subset \Sigma^{*}$ \textbf{radical} if for any $w\in \Sigma^*$,
$
\left(\forall a\in \Sigma, wa \in I\right) \implies w\in I
$.
\end{dfn}
Observing that an intersection of radical ideals is a radical ideal, we deduce that there exists a smallest radical ideal containing a given ideal $I\subset \Sigma^*$, which we denote $\rad(I)$. By inspection, an ideal is radical if and only if its corresponding full tree is reduced.
\begin{lemma}
\label{lem:radical}
For any ideal $I$, $\rad(I) = L(O(I))$.
\end{lemma}
\begin{proof}
We can easily check that for an ideal $I$, $I \subseteq L(O(I))$. To show that $\rad(I) \subseteq L(O(I))$, it therefore suffices to show that $L(U)$ is radical for any $U$ (and so in particular when $U = O(I)$). For any infinite word $u$, $w \pf u$ implies that for some $a$, $wa \pf u$. Thus if $wa \in L(U)$ for every $a$, then $w \in L(U)$, by definition of $L$.
Conversely, we show that $L(O(I)) \subseteq \rad(I)$ by contradiction. An element of $L(O(I))$ is a word $w$ such that every infinite word $u$ with $w \pf u$ has a prefix in $I$. If we had $w \in L(U(I)) \backslash \rad(I)$, there must be some $a_1 \in \Sigma$ such that $wa_1 \notin \rad(I)$ but this must still lie in $L(O(I))$ since the latter is an ideal. Repeating this argument we construct an infinite word $u = wa_1a_2 \cdots$ with no prefix in $\rad(I)$ and hence no prefix in $I$, a contradiction.
\end{proof}
\begin{rmk}
\label{rmk:GeometricIntuitionofV}
The reason for this choice of naming convention is the following intuition coming from algebraic geometry. One can associate to each finite word $w\in \Sigma^{*}$ a continuous \textbf{characteristic function} $\chi_{w}\colon \Sigma^{\omega}\to S$ (where $S$ is the Sierpinski space $\{0,1\}$ with $\{1\}$ open) via
\[
\chi_{w}(u) =
\begin{cases}
1& (w\pf u)\\
0& (w\not \pf u).
\end{cases}
\]
Observe that $\chi_{w}^{-1}(1) = U(w)$. In terms of this function, $w\pf p$ translates to `$\chi_w$ is non-zero at the point $p$,' and dually $w\not \pf p$ says `$\chi_w$ vanishes at the point $p$.' If we apply $O$ and then take the complementary closed subset, we thus send a set of words to the set of vanishing points of their characteristic functions. An ideal $I$ is radical exactly when it contains every word whose characteristic function vanishes on the points in the complement of $O(I)$. This is a kind of ``Nullsteltensatz'' of word combinatorics.
\end{rmk}
Conversely, for any $U \subseteq \Sigma^{\omega}$, it is straightforward to see that $O(L(U))$ is the interior of $U$ for the topology on $\Sigma^{\omega}$. Extracting the fixed points of these operations re-establishes the correspondence between radical ideals and open subsets of $\Sigma^{\omega}$.
In summary, we have constructed the following correspondences and mappings.
% https://q.uiver.app/#q=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
\[\begin{tikzcd}[ampersand replacement=\&]
{\text{ideals}} \&\& {\text{open in }\Sigma^{\omega}} \\
\\
\begin{array}{c} \text{prefix-independent}\\\text{sets} \end{array} \\
\\
{\text{full trees}} \&\& {\text{closed in }\Sigma^{\omega}}
\arrow["{I \mapsto\bigcup_{u \in I} U(u)}"', shift right=2, from=1-1, to=1-3]
\arrow["\bot"{marking, allow upside down}, draw=none, from=1-1, to=1-3]
\arrow["{I \mapsto P_I}"', shift right=2, from=1-1, to=3-1]
\arrow["\sim"{marking, allow upside down}, draw=none, from=1-1, to=3-1]
\arrow["{L(U) \mapsfrom U}"', shift right=3, from=1-3, to=1-1]
\arrow["{\text{complement}}"{description}, tail reversed, from=1-3, to=5-3]
\arrow["{P \mapsto {\uparrow}P}"', shift right=2, from=3-1, to=1-1]
\arrow["{P \mapsto\bigcup_{u \in P} U(u)}"'{pos=0.4}, from=3-1, to=1-3]
\arrow["{P \mapsto P \cup \Sigma^*\backslash{\uparrow}P}"', shift right=2, from=3-1, to=5-1]
\arrow["\sim"{marking, allow upside down}, draw=none, from=3-1, to=5-1]
\arrow["{T \mapsto\ell(T)}"', shift right=2, from=5-1, to=3-1]
\arrow["{T \mapsto B(T)}"', shift right=2, from=5-1, to=5-3]
\arrow["\bot"{marking, allow upside down}, draw=none, from=5-1, to=5-3]
\arrow["{\tr(C) \mapsfrom C}"', shift right=3, from=5-3, to=5-1]
\end{tikzcd}\]
where the operations in the respective columns are bijections and the horizontal adjunctions are those constructed in the present section. \mr{Should we make the notation consistent (that is, use $L(S)$ etc) and include equation numbers instead to declutter this diagram?}
\section{Subtoposes of monoid actions again}
\subsection{Extremal cases}
Any (non-empty) category $\Ccal$ admits two extreme Grothendieck coverages: the \textbf{trivial coverage}, in which only the maximal sieves cover and the \textbf{degenerate coverage} in which all sieves are covering sieves. The corresponding subtoposes of $\PSh(\Ccal)$ are $\PSh(\Ccal)$ itself and the degenerate subtopos, which consists of just the terminal object of $\PSh(\Ccal)$.
When $\Ccal = \Sigma^*$ is a free monoid, we can use the results of the previous section to identify a Grothendieck coverage with a collection of trees.
\begin{lemma}
\label{lem:extremal}
We make the following observations:
\begin{itemize}
% Observation 7
\item Given a full tree in any non-degenerate coverage, any directed path within the tree can be extended to a path ending at a leaf.
% Observation 3
\item Every non-trivial coverage must contain the smallest non-trivial full tree, whose leaves are the generators in $\Sigma$.
\end{itemize}
\end{lemma}
\begin{proof}
If a tree in a coverage contains a path with no extension to a leaf, then pulling back along the word at the end of that path produces a complete tree, which corresponds to the empty sieve, so the topology must be degenerate.
Any non-trivial full tree has the smallest non-trivial full tree as a subtree, whence the latter must be included in any Grothendieck coverage by \Cref{rem:upwardclosed}.
\end{proof}
\begin{dfn}
We say $U \subseteq \Sigma^*$ is \textbf{prefix-dense} if every word in $\Sigma^*$ is a prefix of some word in $U$. Dually, we say a subset $U \subseteq \Sigma^*$ is \textbf{prefix-complete} if every infinite word (element of $\Sigma^{\omega}$) has a prefix in $U$ (in the sense of \Cref{dfn:prefix}).
\end{dfn}
\begin{thm}
\label{thm:dense}
% Ryuya's connection
$\Sigma^*$ has a maximal non-degenerate Grothendieck coverage consisting of the family of all ideals generated by prefix-dense sets.
\end{thm}
\begin{proof}
Prefix-dense ideals exactly correspond to the trees described in the first point of \Cref{lem:extremal}, in which any directed path extends to a leaf. We must verify that these really do form a Grothendieck coverage (and hence the maximal non-degenerate such); indeed, applying the operations of \Cref{ssec:fulltree} to trees in this class produces a tree with the same property.
\end{proof}
\begin{rmk}
The coverage of \Cref{thm:dense} coincides with the so-called \textbf{dense coverage}. As a consequence, every non-degenerate subtopos of $\PSh(\Sigma^*)$ is dense, in the sense that it contains the initial object. Another way to see this is by exploiting the dense--closed factorization of geometric inclusions (cf. \cite[Corollary A4.5.20]{Ele}). Since $\PSh(\Sigma^*)$ has exactly two subterminal objects, the only closed subtoposes are the trivial subtopos and the degenerate subtopos. Thus the closure of a non-trivial subtopos must be all of $\PSh(\Sigma^*)$: they are all dense. Dually, sheaves for the dense coverage yield the unique non-degenerate Boolean subtoposes of $\PSh(\Sigma^*)$: any subtopos of a Boolean topos is closed, so if there were a larger Boolean subtopos of $\PSh(\Sigma^*)$, the dense subtopos would have to be a strictly smaller subtopos corresponding to one of the two subterminal subobjects, which is impossible.
\end{rmk}
\mr{The closed sets corresponding to ideals in the dense coverage are exactly the \textbf{nowhere-dense} subsets of $\Sigma^{\omega}$ for the topology of (ref section).}
\mr{To place appropriately:}
\begin{xmpl}
As soon as $|\Sigma| \geq 2$, there exist uncountable nowhere-dense sets in $\Sigma^\omega$. Indeed, any injective $s: \Sigma \to \Sigma^*$ whose image is prefix independent extends to an injective function $\overline{s}:\Sigma^{\leq\omega} \to \Sigma^{\leq\omega}$ by substitution. The image of $\overline{s}|_{\Sigma^\omega}$ will be nowhere-dense as soon as the image of $s$ fails to be prefix-dense, since extending any finite word in the image of $\overline{s}$ by a word with no prefix in the image of $s$ produces a word that cannot be in the image of $\overline{s}$. For instance, when $\Sigma = \{0,1\}$, we can take
\[s:\begin{cases}
0 &\mapsto 00\\
1 &\mapsto 11,
\end{cases}\]
and the image of $\overline{s}|_{\Sigma^\omega}$ is a nowhere-dense uncountable subset of $\Sigma^\omega$ on the infinite words in which each block of $0$s and $1$s has even size. \mr{Conclude that we have a gap between trees having countably many branches and those having uncountably many.}
\end{xmpl}
For the other extreme, more work is required.
\begin{prop}
\label{prop:noinfinitebranches}
The collection of trees with no infinite branches is the Grothendieck coverage generated by the smallest non-trivial full tree (appearing in the first point of \Cref{lem:extremal}).
\end{prop}
\begin{proof}
\mr{This is now incorporated into \Cref{lem:prefix-complete}.}
First, it should be clear that the set of trees with no finite branches is closed under the operations determining a Grothendieck coverage. Conversely, given such a tree, we can assign an ordinal \textbf{rank} to each vertex by setting the rank of a leaf to be $0$ and the rank of any other vertex to be the supremum of the successors of the ranks of its children.
Note that the rank of a vertex is well-defined as soon as the rank of its children are; as such, the root fails to have a well-defined rank if and only if there is an infinite word $a_0a_1a_2\cdots$ whose finite prefixes represent vertices at which the rank is undefined, which yields a contradictory infinite branch.
Observing that this definition of rank is stable under the pullback operation for trees, the tree can straightforwardly be reconstructed from the smallest non-trivial tree by induction on the rank of the root.
\begin{itemize}
\item If the root has rank $0$, the tree is trivial.
\item Otherwise, the tree is non-trivial and its children all have strictly lower rank, so we can recover the tree by multicomposition of the minimal tree with the pulled back trees, themselves reconstructible by the induction hypothesis.
\end{itemize}
This concludes the proof.
\end{proof}
In terms of ideals, this amounts to the following.
\begin{thm}
\label{thm:finitebranches}
% Observation 4
$\Sigma^*$ has a minimal non-trivial Grothendieck coverage consisting of the family of all ideals generated by prefix-complete sets (in the sense of \Cref{dfn:prefix}).
\end{thm}
\begin{proof}
It suffices to show that the prefix-complete sets correspond to the trees with no infinite branches. By definition any prospective infinite branch has a prefix in a prefix-complete set, so it cannot be a branch of the corresponding tree, and vice versa.
\end{proof}
\begin{lemma}
The pullback of the maximal non-trivial subtopos of $\PSh(\Sigma^*)$ of \Cref{thm:finitebranches} along the projection $\Sh(\Sigma^{\leq \omega}) \to \PSh(\Sigma^*)$ yields $\Sh(\Sigma^{\omega})$.
\end{lemma}
\begin{proof}
\mr{TODO}
\end{proof}
\begin{xmpl}
\label{exa:N}
When $\Sigma$ is a singleton, so $\Sigma^* \cong \Nbb$, the coverages of \Cref{thm:dense} and \Cref{thm:finitebranches} both coincide with the \textit{atomic coverage}, in which the covering sieves are generated by single morphisms. The sheaves for this coverage coincide with the actions of $\Zbb$, in that they are precisely the $\Nbb$-actions for which the generator acts by a bijection.
As such, these coverages provide a complete classification in this case: the only three Grothendieck coverages on $\Nbb$ are the trivial coverage, the atomic coverage and the degenerate coverage.
\end{xmpl}
\begin{xmpl}
\label{exa:F2}
% Ryuya's sketch
When $\Sigma = \{a,b\}$, the topos of sheaves for the coverage of \Cref{thm:finitebranches} yields the \textit{J\'{o}nsson-Tarski topos}. These are the actions of $\Sigma^*$ on a set $X$ such that $a,b$ are the product projections presenting $X$ as isomorphic to $X \times X$.
The pattern continues: the sheaves for the coverage of prefix-complete sets on $\Sigma^*$ are exactly the actions of $\Sigma^*$ on sets $X$ presenting $X$ as a $\Sigma$-indexed product of copies of itself. Moreover, since this coverage is minimal, sheaves for any finer coverage must retain this property. \mr{Might be better to explain this at the point where we make the sheaf conditions explicit.}
\end{xmpl}
\subsection{Accounting for infinite branches}
A consequence of \Cref{prop:noinfinitebranches} is that any finer Grothendieck coverage must contain trees with infinite branches.
\begin{lemma}
\label{lem:radicalreduction}
Let $J$ be a non-trivial coverage on $\Sigma^*$. Then a tree $T$ is in $J$ if and only if, in the notation of \Cref{ssec:Cantorsubsets}, $\tr(B(T)) \in J$.
\end{lemma}
\begin{proof}
Since $\tr(B(T)) \subseteq T$, the presence of the latter in $J$ implies the presence of the former. Conversely, $T$ can be recovered from $\tr(B(T))$ by multicomposition with trees containing no infinite branches, which are necessarily present in $J$ by \Cref{thm:finitebranches}.
\end{proof}
Lemma \ref{lem:radicalreduction} allows us to exploit Section \ref{ssec:Cantorsubsets}, since it is sufficient to identify the closed subsets of $\Sigma^{\omega}$ corresponding to the trees in the coverage to determine the coverage completely. As such, we should examine the consequences of the Grothendieck coverage axioms on the collection of subsets of $\Sigma^{\omega}$ corresponding to the trees in a coverage.
\begin{enumerate}
\item For any tree $T$ with no infinite branches, $B(T) = \emptyset$.
\item For a word $w$, $B(w^*(T)) = B(T) \cdot w$, where the right-hand side denotes the right action of $\Sigma^*$ on the closed sets described in \Cref{eq:closedsetaction} in \Cref{ssec:prefix}.
\item Let $T \in J$ and $\{T_w \in J \mid w \in \ell(T)\}$ be a set of trees. %For $w' \in \ell(\tr(B(T)))$, let $t_{w'} = w'^*(T)$; this is a tree with no infinite branches, and as described in \Cref{lem:radicalreduction} we have $T = \tr(B(T)) \circ \{t_{w'}\}_{w' \in \ell(\tr(B(T)))}$. Exploiting associativity of (multi)composition, we have:
\[B(T \circ \{T_w\}) = \coprod_{w \in \ell(T)} w \cdot B(T_w),\]
where $w\cdot C := \{wu \mid u \in C\}$.
\item If $T \subseteq T'$ then clearly $B(T) \subseteq B(T')$.
\end{enumerate}
To eliminate the apparent dependence on trees in the final point, we use the following lemma to characterize the prefix-independent set $\ell(T)$.
\begin{lemma}
\label{lem:leaves}
$B(T) = \Sigma^* \backslash \bigcup_{w \in \ell(T)} U(w)$, where $U(w)$ is the open defined in \Cref{eq:U(u)}.
\end{lemma}
\begin{proof}
\mr{TODO}
\end{proof}
\begin{prop}
\label{prop:closedsetsenough}
a collection $\Kcal$ of closed sets of $\Sigma^\omega$ corresponds to a collection of trees forming a Grothendieck topology if and only if:
\begin{enumerate}
\item $\emptyset \in \Kcal$;
\item $\Kcal$ is closed under the action of $\Sigma^*$;
\item For each prefix-independent set $P$ such that $\Sigma^* \backslash \bigcup_{w \in P} U(w) \in \Kcal$, given $C_w \in \Kcal$ for each $w \in P$. Then $\coprod_{w \in P} wC_w \in \Kcal$.
\item A closed subset of $C \in \Kcal$ belongs to $\Kcal$.
\end{enumerate}
closed under sup?
\end{prop}
\mr{I think at this point, I would like to derive the notion of self-similar sublocale directly before providing the abstract proof of the result with LT topologies.}
\subsection{A concrete classification of (some) self-similar sublocales}
A further consequence of these constructions is that we can put an equivalence relation on elements of $\Sigma^{\omega}$ as follows:
\[(a_0a_1a_2\cdots) \sim (b_0b_1b_2\cdots)
\iff (\exists N,N')(\forall k \in \Nbb) \; a_{N+k} = b_{N' +k}.\]
We write $[a_n]_{n\in \Nbb}$ for the equivalence class of a given sequence $a_0a_1a_2\cdots$. Let $\Bcal$ be the collection of all equivalence classes of branches.
\begin{rmk}
\label{rem:countable}
% Observation 10
Note that there are fewer than $|\Sigma^*| \times \Nbb$ elements in an equivalence class $[a_n]_{n\in \Nbb}$, since we can reconstruct any member of the class by specifying the following finite data: a word $a'_0\cdots a'_{k-1}$ plus an index $m$ such that the remainder of the sequence is $a_{m}a_{m+1}\cdots$. %this data is unique if we insist that $m$ is taken to be as small as possible.
\end{rmk}
From the preceding discussion, we can conclude that a collection of closed subsets of $\Sigma^{\omega}$ corresponding to a Grothendieck coverage must be a union of equivalence classes.
\begin{lemma}
\label{lem:branchgeneration}
% Observation 9
Let $T$ be \textit{any} full tree containing an infinite branch $(a_n)_{n\in \Nbb}$. Then the minimal Grothendieck coverage containing (the ideal corresponding to) $T$ must also contain the collection of all trees with a single infinite branch lying in the equivalence class $[a_n]_{n\in \Nbb}$.
\end{lemma}
To complete the contents of the Grothendieck coverage generated by a single infinite branch $T$, we must consider what happens when we take a multicomposite with many infinite branches. By taking finite height trees and grafting infinite branches (sharing an equivalence class with $T$) onto any subset of the leaves, we obtain the collection of trees with finitely many branches, all of whose equivalence classes coincide with that of $T$. It is not possible for any infinite branch to be created besides those being grafted on, since leaves are prefix-independent.
If we take a tree constructed as above and again graft on branches equivalent to $T$, we can have infinitely many branches equivalent to $T$, but once again prefix-independence of the leaves prevents the appearance of new equivalence classes of infinite branch. However, merely grafting on individual branches doesn't account for everything: since we can multicompose with any of the trees that we have shown to be in the Grothendieck topology so far, and since there are infinitely many leaves (when $\Sigma$ contains at least $2$ generators) onto which these trees may be grafted, the resulting trees can become arbitrarily complex. We expect it would not be possible to explicitly classify all possible branch structures. However, all trees produced in this way share the property that the only equivalence class of infinite branch present is that represented by $T$; surprisingly, this is enough to capture the class of trees we are after!
\begin{thm}
\label{thm:singlebranchtype}
Let $T$ be a full tree with a single infinite branch $(a_n)_{n\in \Nbb}$. Then the minimal Grothendieck coverage containing (the ideal corresponding to) $T$ consists of the collection of trees all of whose infinite branches lie in the equivalence class $[a_n]_{n\in \Nbb}$.
\end{thm}
\begin{proof}
Using points \ref{pt:finbranches} and \ref{pt:justbranch} above, we may without loss of generality assume that $T$ is the minimal full tree containing the given branch. Let $T^0$ be any tree whose infinite branches are in the equivalence class $[a_n]_{n\in \Nbb}$.
We will construct an ordinal-indexed sequence $T^{\alpha}$ of subtrees of $T^0$, showing inductively that $T^0$ can be recovered from each subtree in a finite number of multicomposition steps \textit{with trees in the Grothendieck topology generated by $T$}. Once we have shown that this process terminates at the trivial tree, we can conclude that $T^0$ is also in the Grothendieck topology.
\textbf{Successor step:} Given $T^{\alpha}$, consider for each of its vertices the number of infinite branches to which it belongs. Take $T^{\alpha + 1}$ to be the subtree obtained by truncating $T^{\alpha}$ at the minimal vertices for which this number is finite. $T^{\alpha}$ can be recovered from $T^{\alpha+1}$ by grafting on trees with finitely many infinite branches via multicomposition, which are in the Grothendieck topology generated by $T$ by the comments following \Cref{lem:branchgeneration}, whence $T^{0}$ can be reconstructed from $T^{\alpha+1}$ in a finite number of multicomposition steps.
\textbf{Limit step:} Let $\lambda$ be a limit ordinal. Having constructed $T^{\alpha}$ for all $\alpha < \lambda$, take $T^{\lambda}$ to be the intersection of the $T^{\alpha}$. For the reconstruction, observe that for each leaf $w$ of $T^{\lambda}$, either $w$ was already a leaf of $T^0$ or there is a last ordinal $\alpha$ for which it was an internal vertex of $T^{\alpha}$. By induction, the subtree of $T^0$ obtained by truncating at any leaf of $T^{\lambda}$ is therefore obtained from the trivial tree in a finite number of multicomposition steps from trees in the Grothendieck topology generated by $T$, whence they too are members of the Grothendieck topology. Taking the multicomposite of $T^{\lambda}$ with these trees recovers $T^0$, as claimed.
To prove that iterating this process will result in the trivial tree for any choice of $T^0$, we use the following Lemma:
\begin{lemma}
\label{lem:descentcontinues}
A full tree whose infinite branches fall into a single equivalence class such that every internal vertex belongs to infinitely many branches must be trivial.
\end{lemma}
\begin{proof}
Supposing to the contrary that there is at least one internal vertex (and hence infinitely many), we inductively construct an infinite branch in a distinct equivalence class for a contradiction. To do so, we employ Cantor's pairing function, the bijection $\pi: \Nbb \times \Nbb \to \Nbb$ defined by $\pi(x,y) = (x+y+1)(x+y)/2 + y$.
Starting with the empty word, we construct a sequence of generators in the interior of the tree $a'_0,a'_1,\dotsc$ until we reach a vertex for which several children belong to multiple branches; this must happen above every internal vertex, or else there is only one infinite branch beyond that point, a contradiction. Upon reaching the $k$th such branching point, we choose the next generator such that, if $\pi^{-1}(k) = (N,N')$, we ensure that $a_N,a_{N+1},\dotsc$ is distinct from $a'_{N'},a'_{N'+1},\dotsc$. In this way, we guarantee that there cannot exist $N,N'$ witnessing that the sequence so produced lies in the same equivalence class.
\end{proof}
From \Cref{lem:descentcontinues} it follows that each step in the induction preceeding the trivial tree produces a strict subtree, in the sense that $T^{\alpha} \supsetneq T^{\beta}$ when $\alpha < \beta$. Hence if we label each non-root vertex with the last ordinal $\alpha$ such that the vertex appears in $T^{\alpha}$, each ordinal is assigned to at least one distinct vertex. Considering an ordinal of cardinality exceeding $|\Sigma^*|$, we must therefore reach the trivial tree, or else we have more distinct ordinals than vertices to label them with, a contradiction.
\end{proof}
\begin{rmk}
\label{rem:reconstructionhazard}
When $\Sigma$ is countable, there are only countably many branches in a branch-equivalence class by \Cref{rem:countable}, and hence only countably many branches in any of the trees featuring in \Cref{thm:singlebranchtype}. One might therefore observe that any such tree can be reconstructed as a countable union of the trees obtained by finite multicomposition with a given selection of the finite branches and hope to perform a simpler reconstruction that way. However, Grothendieck topologies are not closed under countable intersections (which would correspond to such a countable union of trees), and indeed there are families of trees each of which have finitely many branches in a single branch class whose union contains arbitrarily many branch classes. Indeed, since every word is a prefix of a branch in $[a_n]_{n \in \Nbb}$, the maximal tree (corresponding to the empty ideal) is such a countable union!
\end{rmk}
\subsection{Many infinite branches}
Some of the observations above extend to cases where more than one equivalence class of branches is present. In particular, the operations under which a Grothendieck coverage must be closed cannot create branch types which were not already present, and conversely if a branch type is present in $T$ then the corresponding Grothendieck coverage of \Cref{thm:singlebranchtype} must be contained in the coverage generated by $T$.
The following is the maximal generalization of \Cref{thm:singlebranchtype}.
\begin{thm}
\label{thm:manybranchtypes}
Let $E \subseteq \Bcal$ be a countable set of equivalence classes of branches. Then the class of full trees whose branches lie exclusively in $E$ constitutes the Grothendieck topology generated by any subset of those trees whose infinite branches collectively account for all equivalence classes in $E$.
\end{thm}
Along the way to proving this, we shall need a variant of \Cref{lem:descentcontinues} which we shall prove first. Throughout, if $E$ is finite, we index over it with a countably infinite set by cycling through the elements of $E$.
\begin{lemma}
\label{lem:descentcontinues2}
Let $T'$ be a full tree whose infinite branches all lie in some countable set of equivalence classes $E \subseteq \Bcal$. Suppose that every internal vertex belongs to infinitely many branches in each equivalence class. Then $T'$ is trivial.
\end{lemma}
\begin{proof}
This time we employ a three-variable variant of Cantor's pairing function. The details are unimportant, but we could take $\varpi: \Nbb \times \Nbb \times \Nbb \to \Nbb$ given by
\[(x,y,z) \mapsto \binom{x+y+z+2}{3} + \binom{y+z+1}{2} + \binom{z}{1}.\]
The third variable will index the equivalence classes of $E$: we write $[a^z_n]_{n \in \Nbb}$ for the $z$th equivalence class. From here we proceed as before, constructing a sequence of generators $a'_1,a'_2,\dotsc$ forming a branch that is not in any of the equivalence classes. At the $k$th branching point, letting $(N,N',z) := \varpi^{-1}(k)$ we choose $a'_{i_k}$ such that $a'_{N},a'_{N+1},\cdots$ is distinct from $a^{z}_{N'},a^{z}_{N'+1},\dotsc$. This completes the proof.
\end{proof}
\begin{proof}[Proof of \Cref{thm:manybranchtypes}]
Let $T^0$ be a tree whose branches' equivalence classes lie in $E$. We can define a deconstruction process as in the proof of \Cref{thm:singlebranchtype}.
\textbf{Successor step}: If $\alpha = \lambda + z$ for $\lambda$ a limit ordinal and $z$ a finite ordinal and we have already constructed $T^{\lambda},T^{\lambda+1},\dotsc,T^{\alpha}$, let $T^{\alpha + 1}$ be the subtree of $T^{\alpha}$ obtained by truncating at the minimal vertices which are in finitely many branches in the $z$-indexed of $E$. Unlike in the proof of \Cref{thm:singlebranchtype}, it may \textit{a priori} be the case that $T^{\alpha+1}$ is equal to $T^{\alpha}$, but in any case $T^{\alpha}$ can be reconstructed by gluing on finitely many branches of type indexed by $z$ at each leaf, as required.
\textbf{Limit step}: If $T^\lambda$ is constructed, for a limit ordinal $\lambda$, by taking the intersection of subtrees $T^{\alpha}$ of $T^0$ over ordinals $\alpha < \lambda$. The reconstruction argument here is identical.
Finally, \Cref{lem:descentcontinues2} ensures that this process must terminate, this time merely guaranteeing that for limit ordinals $\lambda <\kappa$ preceding termination of the process at the trivial tree, $T^{\kappa}$ is a strict subtree of $T^\lambda$, whence we can always select an ordinal large enough to guarantee termination.
\end{proof}
To summarize, we have thus far shown that each countable subset of $\Bcal$ determines a unique Grothendieck coverage consisting of the trees whose branches's equivalence classes lie in that subset. However, that's far from the end of the story.
\begin{xmpl}
\label{exa:uncountablebranches}
Even with finitely many generators, if $\Sigma$ has at least two elements, there are uncountably many infinite branches in the Cayley graph, even though there are only countably many elements of $\Sigma^*$. More significantly, it is possible for a tree featuring in a non-degenerate coverage (satisfying the second condition of \Cref{lem:extremal}) to have uncountably many branches. For instance, when $\Sigma$ has three elements, say $a,b,c$, the tree whose leaves are words whose unique appearance of $c$ is the rightmost letter (corresponding to the right ideal of words containing at least one instance of $c$) has uncountably many branches.
%Ryuya's example
As a smaller example, if $\Sigma$ has just two elements $a$ and $b$, The ideal of $\Sigma^*$ on those words having a prefix in which $b$ appears more times than $a$ corresponds to a tree with uncountably many infinite branches.
\end{xmpl}
Before considering Grothendieck coverages containing trees with uncountably many branch types, we can at least observe the existence of extremal topologies determined by branch classes.
\begin{prop}
\label{prop:coveragegap}
Let $E \subseteq \Bcal$ be any collection of equivalence classes of branches. Then there is a maximal coverage corresponding to $E$ consisting of all trees whose branches lie in the equivalence classes of $E$ and a minimal coverage consisting of the coverage generated by the branches.
\end{prop}
\begin{proof}
Both claims follow from the observation that the tree operations under which a Grothendieck coverage must be closed can only produce trees in the same respective classes.
\end{proof}
\begin{rmk}
\label{rem:anydense}
A consequence of \Cref{prop:coveragegap} is that whenever $E$ does not include even one equivalence class, any tree whose branches lie in $E$ must have the denseness property from the second point of \Cref{lem:extremal}: every path must extend to one ending in a leaf. Indeed, we can see this directly: from a given vertex, consider the infinite path determined by any of the equivalent branches omitted from $E$. Since that branch cannot be in the tree, the path must reach a leaf.
\end{rmk}
In light of Proposition \ref{prop:coveragegap}, we can summarize the preceding results as saying that when $E$ is at most countable, the minimal and maximal coverages coincide. We can easily show that they will not coincide for larger choices of $E$, and more strongly that the there can be Grothendieck coverages between the minimal and maximal ones for a given set of branch equivalence classes.
\begin{lemma}
\label{lem:mintrees}
Let $\Sigma$ be a finite or countably infinite set and $E \subseteq \Bcal$ an uncountable collection of equivalence classes of branches. Then the class of trees having countably many branches, each of whose equivalence classes is in $E$, forms the minimal Grothendieck coverage for $E$.
\end{lemma}
\begin{proof}
It suffices to observe that the stated collection is closed under the operations determining a Grothendieck coverage. Indeed, truncating and selecting subtrees trivially produces trees in the same class. When $\Sigma$ is at most countable, any full tree has at most countably many leaves, so multicomposition (grafting) also keeps us in the same class of trees.
\end{proof}
\begin{xmpl}
\label{exa:densevsdegenerate}
The dense topology of \Cref{thm:dense} and the degenerate topology both have maximal sets of branch equivalence classes. The latter is the maximal Grothendieck topology for that collection of branches. In light of \Cref{lem:mintrees}, when $\Sigma$ is countable (and has at least two elements), the minimal topology for the maximal collection of branches is strictly smaller than the dense topology: it doesn't include the trees constructed in \Cref{exa:uncountablebranches}, for instance.
\end{xmpl}
\Cref{lem:mintrees} witnesses a divergence in behaviour depending on the cardinality of the generating set: if $\Sigma$ were uncountable, we could graft on a collection of branches of size $|\Sigma|$, so there would be trees with uncountably many branches in the minimal coverage. \mr{Trying to make precise `which uncountable branch sets' seems challenging... maybe I'll think about it some more.}
Observe that if $J$ is generated by a single tree $T$, the collection of branch equivalence classes $\Bcal$ appearing in $J$ is exactly the union of closed nowhere-dense sets,
\[\Bcal = \bigcup_{u,v \in \Sigma^*} u \cdot v^*(B(T)). \]
Conversely, if $\Bcal$ is a $(\Sigma^* \times \Sigma^*)$-indexed union of closed nowhere-dense sets, can we reconstruct a single tree $T$ whose branches include representatives of all classes in $\Bcal$?
\subsection{Suptoposes of $\PSh(M)$}
Consider a monoid $M$ equipped with a surjective homomorphism $F:\Sigma^* \to M$ (viewed as a full, surjective-on-objects functor). This induces a hyperconnected geometric morphism $\Sigma^* \to M$. For any Grothendieck coverage $J$ on $M$, we can construct the pullback:
% https://q.uiver.app/#q=WzAsNCxbMSwxLCJcXFBTaChNKSJdLFswLDEsIlxcUFNoKFxcU2lnbWFeKikiXSxbMSwwLCJcXFNoKE0sSikiXSxbMCwwLCJcXFNoKFxcU2lnbWFeKixLKSJdLFsyLDAsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMCwiIiwxLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzMsMiwiIiwxLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzMsMSwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMywwLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=
\[\begin{tikzcd}[ampersand replacement=\&]
{\Sh(\Sigma^*,K)} \& {\Sh(M,J)} \\
{\PSh(\Sigma^*)} \& {\PSh(M)}
\arrow[two heads, from=1-1, to=1-2]
\arrow[hook, from=1-1, to=2-1]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\arrow[hook, from=1-2, to=2-2]
\arrow[two heads, from=2-1, to=2-2]
\end{tikzcd}\]
where an ideal $I \subseteq \Sigma^*$ is $K$-covering if and only if $I \supseteq F^{-1}(I'):=\{w \mid F(w) \in I'\}$ for some $J$-covering ideal $I'$ in $M$, by \cite[Lemma C2.3.19]{Ele}.\footnote{In general one would need to close under multicomposition to obtain a coverage from the given sieves, but since $F$ is full and surjective on objects this closure is automatic.}
Restricting attention to $K$-covering ideals of the form $F^{-1}(I')$, any such must be a union of equivalence classes for the equivalence relation on morphisms induced by $F$. This in turn induces an equivalence relation on infinite branches in the corresponding full trees: infinite words $w,w'$ are related if for any $N$ there exist $m,n \geq N$ such that the prefix of $w$ of length $m$ has the same image under $F$ as the prefix of $w'$ of length $n$. \mr{I think this is the right equivalence...}
Converse direction:
The image of a subtopos $\Sh(\Sigma^*,K)$ of $\PSh(\Sigma^*)$ along the geometric morphism induced by $F: \Sigma^* \to M$ has $I$ covering if and only if $F^{-1}(I)=\{w \mid F(w) \in I\}$ is $K$-covering. %Since $F$ is surjective, we can identify the latter ideal with $u^*F^{-1}(I)$ for any $u$ such that $F(u) = m$, so this amounts to asking that $u^*(F^{-1}(I))$ is $K$-covering for every $u$, and hence taking $u$ to be the empty word is sufficient.
By inspection, taking the image along the geometric morphism induced by $F$ therefore preserves the degenerate subtopos and the double-negation subtopos, and these remain distinct. It also preserves the trivial subtopos.
Let us consider the maximal proper subtopos of $\PSh(\Sigma^*)$. The covering sieves for the image subtopos are those $I$ whose pullbacks' complements have no infinite branches. That is, for any infinite word $w$, there is some finite prefix $u \pf w$ such that $F(u) \in I$.
\begin{prop}
The image of the maximal non-trivial subtopos of $\PSh(\Sigma^*)$ in $\PSh(M)$ coincides with the double-negation subtopos if and only if $M$ satisfies.....
It coincides with the trivial subtopos if and only if $M$ contains an invertible element besides the identity.
\end{prop}
\begin{proof}
Consider an ideal for the double-negation coverage on $M$: for any finite word $u$, there exists $v$ with $F(v) \in I$ and $u \pf v$. Given an infinite word $w$, if $M$ satisfies the right Ore condition, ...
If the set of non-identity words forms a right ideal, then this ideal must be a non-trivial covering in $J$, so the category of sheaves is distinct from the presheaf topos. Otherwise, there is some non-identity invertible element $m \in M$. The criterion for $I$ to be a covering ideal for the non-trivial coverage implies that $m^k \in I$ for some $k \geq 1$. Since all multiples of this element $m^k$ will be present, the identity will also be present, so this sieve coincides with the maximal sieve, as claimed.
\end{proof}
\section{Scrap}
\subsection{One infinite branch}
If a tree does have infinite branches, it turns out that these dominate the structure of the Grothendieck coverage generated by the tree!
We shall reconstruct the smallest Grothendieck coverage generated by (the ideal corresponding to) $T$. To do so, we consider the effect on such a tree of applying the operations determining a Grothendieck coverage. We shall shortly see that the structure of $T$ outside of the branch is unimportant, so we do not specify further which tree we mean.
\begin{enumerate}
\item The trivial tree is automatically present. By \Cref{thm:finitebranches}, we will also have all trees of finite height.
\item \label{pt:pullback} Pulling back along a word that is not a prefix of the branch produces a tree of finite height, which was already in the coverage. Pulling back along a word which is a prefix of length $k$ produces another tree with an infinite branch corresponding to the sequence $(a_{n+k})_{n \in \Nbb}$.
\item \label{pt:finbranches} Conversely, by choosing a tree of finite height and taking a multicomposite in which we append $T$ at exactly one leaf at height $k$, we obtain a tree with an infinite branch corresponding to the sequence $(a_{n-k})_{n \in \Nbb}$ where we may choose $a_{-k},\dotsc,a_{-1}$ to be any finite sequence. Dually, by taking a multicomposite of $T$ with finite-height trees, we can add arbitrary finite side paths onto $T$. We shall consider the possibility of grafting multiple infinite branches onto a tree below.
\item \label{pt:justbranch} There is a minimal full subtree containing the infinite branch of $T$. Combined with the previous multicomposition construction, we see that any tree with the same single infinite branch of $T$ must belong to the Grothendieck coverage.
\end{enumerate}
% Observation 8
In light of these constructions, we put an equivalence relation on sequences in $\Sigma$ (and by extension on infinite branches in the Cayley graph of $\Sigma^*$) as follows:
\[(a_n)_{n \in \Nbb} \sim (b_{n'})_{n' \in \Nbb}
\iff (\exists N,N')(\forall k \in \Nbb) \; a_{N+k} = b_{N' +k}.\]
We write $[a_n]_{n\in \Nbb}$ for the equivalence class of a given sequence $(a_n)_{n \in \Nbb}$. Let $\Bcal$ be the collection of all equivalence classes of branches.
From the preceding discussion, we can conclude the following result.
\begin{lemma}
% Observation 9
Let $T$ be \textit{any} full tree containing an infinite branch $(a_n)_{n\in \Nbb}$. Then the minimal Grothendieck coverage containing (the ideal corresponding to) $T$ must also contain the collection of all trees with a single infinite branch lying in the equivalence class $[a_n]_{n\in \Nbb}$.
\end{lemma}
---
\begin{proof} of \Cref{thm:singlebranchtype}
\mr{\textit{Attempt 1.}}
Using points \ref{pt:finbranches} and \ref{pt:justbranch} above, we may without loss of generality assume that $T$ is the minimal full tree containing the given branch. Letting $T'$ be any tree whose infinite branches are in the equivalence class $[a_n]_{n\in \Nbb}$, we may similarly assume without loss of generality that $T'$ does not have any finite branches whose internal vertices are not on any infinite branch.
We first show that $T'$ must be in the minimal Grothendieck coverage generated by $T$. Let $\Bcal$ be the set of infinite branches of $T'$, viewed as full subtrees. For each $B \in \Bcal$, consider the subtree $\bigcup_{B \neq B' \in \Bcal} B \cap B'$. Define a new tree $T''$ retaining only these maximal intersections. Clearly $T'$ can be recovered from $T''$ by grafting on the removed branches via multicomposition.
We claim that iterating this process will result in a tree with finitely many infinite branches after finitely many steps. Indeed, suppose not. Let $T^{\infty}$ be the intersection of the....
\mr{Got stuck, recallibrating.}
\end{proof}
\begin{xmpl}
\label{exa:unbalancedtree}
\mr{This is a counterexample to the attempted decomposition above: the union can be obtained by gluing on trees with finitely many infinite branches to the right-most branch $T_0$, but when we prune the tree as in the attempted proof, the right-most branch is never isolated.}
In the case $\Sigma = \{a,b\}$, consider the full subtree constructed as follows.
Base case: take $T_0$ to be the tree $\{b^k \mid k \in \Nbb\} \cup \{b^k\hi{a} \mid k \in \Nbb\}$.
Induction procedure: at stage $n$, multicompose $T_{n-1}$ with itself at the leaves whose prefix starts with at least $n$ repetitions of $b$.
So $T_1$ is
\[\{b^k \mid k \geq 0\} \cup \{a\} \cup \{b^kab^{k'} \mid k \geq 1, k' \geq 0\} \cup \{b^kab^{k'}\hi{a} \mid k \geq 1, k' \geq 0\}\]
$T_2$ is
\begin{align*}
& \{b^k \mid k \geq 0\} \cup \{a\} \cup \{b^kab^{k'} \mid k \geq 1, k' \geq 0\} \cup \{ba\} \\
& \cup \{b^kab^{k'}ab^{k''} \mid k \geq 2, k' \geq 0, k'' \geq 0\} \cup \{b^kab^{k'}a\hi{a} \mid k \geq 2, k' \geq 0\} \\
& \cup \{b^kab^{k'}ab^{k''}ab^{k'''} \mid k \geq 2, k' \geq 0, k'' \geq 1,k''' \geq 0\} \cup \{b^kab^{k'}ab^{k''}ab^{k'''}\hi{a} \mid k \geq 2, k' \geq 0, k'' \geq 1,k''' \geq 0\}
\end{align*}
$T_3$ is
\begin{align*}
& \{b^k \mid k \geq 0\} \cup \{a\} \cup \{b^kab^{k'} \mid k \geq 1, k' \geq 0\} \cup \{ba\} \\
& \cup \{b^kab^{k'}ab^{k''} \mid k \geq 2, k' \geq 0, k'' \geq 0\} \cup \{b^2ab^{k'}aa \mid k' \geq 0\} \\
& \cup \{b^kab^{k'}aab^{k''} \mid k \geq 3, k' \geq 0, k'' \geq 0\} \cup \{b^kab^{k'}aa\hi{a} \mid k \geq 3, k' \geq 0\}\\
& \cup \{b^kab^{k'}aab^{k''}ab^{k'''} \mid k \geq 3, k' \geq 0, k'' \geq 1, k''' \geq 0\} \cup \{b^kab^{k'}aab\hi{a} \mid k \geq 3, k' \geq 0\}\\
& \cup \{b^kab^{k'}aab^{k''}ab^{k'''}ab^{k''''} \mid k \geq 3, k' \geq 0, k'' \geq 2, k''' \geq 0, k'''' \geq 0\} \cup \{b^kab^{k'}aab^{k''}ab^{k'''}a\hi{a} \mid k \geq 3, k' \geq 0, k'' \geq 2, k''' \geq 0\}\\
& \cup \{b^kab^{k'}aab^{k''}ab^{k'''}ab^{k''''}ab^{k'''''} \mid k \geq 3, k' \geq 0, k'' \geq 2, k''' \geq 0, k'''' \geq 1, k''''' \geq 0\}\\
& \cup \{b^kab^{k'}aab^{k''}ab^{k'''}ab^{k''''}ab^{k'''''}\hi{a} \mid k \geq 3, k' \geq 0, k'' \geq 2, k''' \geq 0, k'''' \geq 1, k''''' \geq 0\}\\
& \cup \{b^kab^{k'}ab^{k''}ab^{k'''} \mid k \geq 2, k' \geq 0, k'' \geq 1,k''' \geq 0\} \\
& \cup \{b^2ab^{k'}ab^{k''}ab^{k'''}\hi{a} \mid k' \geq 0, k'' \geq 1,k''' \geq 0\} \\
& \cup \{b^kab^{k'}ab^{k''}ab^{k'''}ab^{k''''} \mid k \geq 3, k' \geq 0, k'' \geq 1,k''' \geq 0,k'''' \geq 0\} \\
& \cup \{b^kab^{k'}ab^{k''}ab^{k'''}a\hi{a} \mid k \geq 3, k' \geq 0, k'' \geq 1,k''' \geq 0,k'''' \geq 0\} \\
& \cup \{b^kab^{k'}ab^{k''}ab^{k'''}ab^{k''''}ab^{k'''''} \mid k \geq 3, k' \geq 0, k'' \geq 1,k''' \geq 0,k'''' \geq 1,k''''' \geq 0\} \\
& \cup ...
\end{align*}
I expect that taking the union of these should produce a counterexample.
Inspecting and simplifying, we can take the tree consisting of elements of the form $b^{k_0}a^{l_1}b^{k_1}\cdots a^{l_n}b^{k_n}$ or $b^{k_0}a^{l_1}b^{k_1}\cdots a^{l_n}b^{k_n}a$, where $n \leq k_0$ and $1 \leq k_i$ and $1 \leq l_i \leq k_0$ for $i=1,\dotsc,n$. All infinite branches for this tree are in the equivalence class of the sequence which is constant at $b$, since the number of instances of $a$ in any branch is bounded by the square of the index of the position of the first $a$ (if there is one).
\end{xmpl}
\subsection{Ordering branches}
If we equip $\Sigma$ with an ordering, the collection of sequences $\Sigma^{\omega}$ inherits the ordering. Thus each tree determines a sub-poset of $\Sigma^{\omega}$ equipped with a map to $\Bcal$. We call a pair consisting of a sub-order of $\Sigma^{\omega}$ equipped with a map to $\Bcal$ an \textbf{order type}.
\mr{With various strengths of assumption on the existence of orderings on $\Sigma$, cf https://mathoverflow.net/a/37281/95129, we should be able to classify coverages by order type! Note that assuming the existence of a mere linear/total order on $\Sigma$ implies a dense linear order on $\Sigma^{\omega}$.}
\begin{itemize}
\item maximality, pullbacks and upward closure collectively mean the collection of order types should be closed under sub-orders.
\item multicomposition allows us to insert order types into one another.
\end{itemize}
\mr{Given an order type, are all trees with that order type necessarily present in the same Grothendieck coverage? Is ordering the right relation to consider here, or should we be using the canonical topology on $\Sigma^{\omega}$ instead?}
\subsection{Scrap locale stuff}
I think this is subsumed by \Cref{thm:countablecomplement}
\begin{prop}
\label{prop:countableprefices}
Suppose $C \subseteq \Sigma^\omega$ is a nowhere-dense closed set with the property that for each $k$, there are at most countably many words $u \in \Sigma^k$ with $u \pf w$ for some $w \in C$. Write $[C]$ for the union of the equivalence classes of words in $C$. If a sublocale $L' \subseteq \Sigma^\omega$ has $\pt(L') = \Sigma^\omega \backslash [C]$ then $L'$ is spatial.
\end{prop}
\begin{proof}
We can express $[C]$ as the countable union:
\[C_{i,u} := \Sigma^i (C \cdot u),\]
where $u$ varies over the prefixes in $C$. By \Cref{lem:nwdensepres}, each of the $C_{i,u}$ is closed and nowhere-dense, whence the complement of their union is spatial by \Cref{prop:Baire3}.
\end{proof}
\begin{lemma}
The closed sets as in \Cref{prop:countableprefices} are exactly the Lindelöf ones (see below). Equivalently, they are the \mr{length-preserving? closed?} images of maps out of $\Nbb^\omega$.
\end{lemma}
\begin{proof}
Sketch: enumerate prefixes of length $1$ to map $\Nbb$ to them, then for each such consider their extension by $1$ letter. Continue inductively. Requires dependent countable choice.
\end{proof}
\subsubsection{Fit locales}
\label{sssec:fit}
%We will shortly consider the consequences of \textit{regularity}. Before that, we consider the intermediate property of fitness.
Isbell showed \cite[Proposition 2.3]{Isbellatomless} that regular locales are \textbf{fit}, meaning that every sublocale is an intersection of open sublocales. Moreover, dense sublocales are intersections of dense open sublocales. If nothing else, this gives us a convenient and exhaustive way to \textit{describe} (dense) sublocales of sequence spaces.
As shown in \cite[V.1]{PicadoPultr}, a fit $T_D$-spatial locale is $T_1$, since the complements $\{x\}^c$ of singleton sublocales used in \Cref{lem:TD} must be open.
\mr{I realised that we can already use this formula in the $T_D$ setting, so this section was dispensable!}
\begin{lemma}
\label{lem:maxplicit}
Suppose that $L$ is a fit\footnote{Since $\Max$ is already defined as an intersection of open sublocales as soon as $\pt(L)$ is $T_1$, we don't need the full strength of fitness here: assuming that $L$ is subfit (\cite[V.1.1]{PicadoPultr}) would be enough.} $T_D$-spatial locale and let $S \subseteq \pt(L)$. Then $\Ocal(\Max(S))$ can be identified with the collection of opens of $L$ whose complements' isolated points lie in $S$.
\end{lemma}
\begin{proof}
Recall that a sublocale $L'$ of $L$ induces a nucleus on $\Ocal(L)$, so $\Ocal(L')$ can be identified with the fixed point of this operator. Explicitly, the operator corresponding to an open $V$ sends $U$ to $\inter(U \cup (L\backslash V))$, where $\inter$ denotes the interior. In particular, when $V = \{x\}^c$ is the complement of a point, $U$ is sent to $\inter(U \cup \{x\})$.
The nucleus corresponding to a meet of sublocales is the smallest nucleus containing the union of their closures. That is, recalling the definition of $\Max(S)$ from \Cref{lem:TD}, the nucleus is the transfinite iteration of the function sending $U$ to $\bigcup_{x \in \pt(L) \backslash S} \inter(U \cup \{x\})$. It is immediately apparent that $U$ is a fixed point if and only if any isolated point of the complement of $U$ lies in $S$, as claimed.
\end{proof}
By comparison, we can identify $\Ocal(\loc(S))$ with the collection of opens $U$ which are equal to $\inter(U \cup (\pt(L) \backslash S))$.
It seems appropriate to relate the above result to classical results about sequence spaces.
Fitness enables us to interpolate between the adjoints $\Max$ and $\loc$ of \eqref{eq:maxloc}.
\begin{dfn}
Let $L$ be any locale and let $\Fcal\subseteq \Ocal(L)$ be a filter (i.e. an upward closed, downward directed collection) of open sets. Then we have an order-preserving map,
\begin{align*}
c_{\Fcal} : \Sub_{\Top}(\pt(L)) &\to \Sub_{\Loc}(L) \\
S &\mapsto \bigcap \{ U \in \Fcal \mid S \subseteq \pt(U)\},
\end{align*}
where the right-hand side is a meet of sublocales. We call $c_{\Fcal}(S)$ the \textbf{$\Fcal$-coating of $S$}.
\end{dfn}
For instance, when $L$ is fit and $T_D$-spatial, $\Max(S) = c_{\Fcal_{\omega'}}(S)$, where $\Fcal_{\omega'}$ is the filter of cofinite (open) subspaces.
\begin{lemma}
Let $L$ be a fit locale. Then for any sublocale $L' \subseteq L$ there exists a filter $\Fcal$ such that $L' = c_{\Fcal}(\pt(L'))$.
\end{lemma}
\begin{proof}
By fitness, $L'$ is the intersection of the open sublocales containing it, which necessarily form a filter. If an open sublocale $U$ contains $L'$, then $\pt(U)$ contains $\pt(L')$, so the intersection recovers $L'$, as claimed.
\end{proof}
Of course, there is no guarantee that distinct filters will give distinct sublocales. If $L$ is $T_D$-spatial then whenever $\Max(S)$ coincides with $\loc(S)$, which is the case when $S = \pt(U)$ for some open $U$, any choice of $\Fcal$ containing $\Fcal_{\omega'}$ will produce the same result. This observation also applies to the subspaces derived in \Cref{lem:Baire2} below. For a fixed $\Fcal$, can we find $S$ for which the $\Fcal$-coating of $S$ lies strictly between $\loc(S)$ and $\Max(S)$?
% \subsubsection{Hausdorff spaces}
% For a $T_1$-spatial frame, being Hausdorff is equivalent to the following condition of Dowker and Strauss $T_2$
% \begin{prop}
% Let $L$ be a DS-Hausdorff spatial locale. Then any sublocale $L'$ such that $\pt(L')$ has countable complement in $\pt(L)$ is spatial.
% \end{prop}
% \begin{proof}
% Suppose that $S := \pt(L')$ has countable complement $T = \{x_i\}_{i<\omega}$. It suffices to show that $\Max(S)$ is spatial so that $\loc(S)$ and $\Max(S)$ coincide, since $L'$ lies between these. %Indeed,
% %\[\Max(S) = \bigcap_{i<\omega} \{x_i\}^c\]
% Let $U$ be an open set with complement $C$, so the isolated points of $C$ lie in $S$. We will show that $U = \inter(U \cup T)$.
% Suppose $x_i \in \inter(U \cup T)$ but $x_i \notin U$. Then $x_i$ is not an isolated point of $C$, so any neighbourhood of $x_i$, notably those contained in $\inter(U \cup T)$, must contain infinitely many points outside $U$. That is, $C \cap T \cap V$ is infinite for each open neighbourhood $V$ of $x_i$.
% \end{proof}
-----
\textbf{Lindelöf}
Complementing the above, we consider a class of (sub)locales which are stable under countable unions.
\begin{dfn}
A locale is \textbf{Lindelöf} if every countably directed open cover of the maximal element $\top \in \Ocal(L)$ includes $\top$.
\end{dfn}
\begin{xmpl}
The sequence space $\Sigma^\omega$ is Lindelöf whenever $\Sigma$ is countable.
\end{xmpl}
\begin{lemma}
Any closed sublocale of a Lindelöf locale is Lindelöf.
\end{lemma}
\begin{lemma}
Let $\{L_i\}_{i=1}^\infty$ be a countable collection of Lindelöf sublocales of a locale $L$. Then the sublocale $\bigcup_{i=1}^\infty$ is Lindelöf.
\end{lemma}
\mr{The following is true if the converse of the above holds: is every Lindelöf sublocale a countable union of closed Lindelöf sublocales?}
\begin{crly}
For $L$ a fit, $T_D$-spatial Baire locale, let $S \subseteq \pt(L)$ have Lindelöf complement. Then... \mr{originally had $\loc(S) = \Max(S)$, but that isn't right!}
\end{crly}
\begin{proof}
% Observe that $\loc(S) = \bigwedge \{U \mid S \subseteq U\}$.
\end{proof}
\section{Scrap: by hora}
\mr{This sounds like a make-up brand haha}
A pleasant implication of the previous subsection is that we can reduce our problem to radical ideals, and topological settings.
\begin{lemma}
\label{lem:radicalPrinciple}
For an ideal $I$ and a non-trivial coverage $J$, $I \in J \iff \rad(I) \in J$.
\end{lemma}
\begin{proof}
Since $I \subset \rad(I)$, it suffices to prove $\rad(I) \in J \implies I \in J$. Suppose $\rad(I)\in J$. For any $w\in \rad(I)$, \cref{lem:RadIsNatural} implies that $\rad(I\cdot w) = \rad(I)\cdot w = \Sigma^{\ast}$. This implies $I\cdot w$ is covering, and \cref{thm:TheMinimalNontrivialCoverage} implies that $I\cdot w \in J$. By the transitivity of coverage, $I \in J$.
\end{proof}
\begin{dfn}
For a non-trivial coverage $J$, let $D(J)$ denote the image of $J$ by $D$ (see \cref{crly:GaloisConnectionOfIdealsAndOpen}), i.e.,
\[
D(J) \coloneqq \{D(I) \mid I\in J\} .
\]
Similarly, let $V(J)$ denote the image of $J$ by $V$ (see \cref{thm:GaloisConnectionOfIdealsAndClosed}), i.e.,
\[
V(J) \coloneqq \{V(I) \mid I\in J\} .
\]
\end{dfn}
The next proposition is what makes topological classification possible:
\begin{prop}
For non-trivial coverages $J_1,J_2$,
\[
D(J_1) = D(J_2) \iff V(J_1) = V(J_2) \implies J_1 = J_2.
\]
\end{prop}
\begin{proof}
The left equivalence is trivial. We will prove $J_1 = J_2$ assuming $D(J_1)= D(J_2)$. Take an element $I_1 \in J_1$. By the assumption, $D(I_1) \in D(J_1) = D(J_2)$, which means there exists an element $I_2 \in J_2$ such that $D(I_1)=D(I_2)$. By the Nullstellensatz (\cref{thm:Nullstellensatz}), we have $\rad(I_1)= \rad(I_2)$. \Cref{lem:radicalPrinciple} implies $I_1 \in J_2$. The inclusion $J_1 \supset J_2$ is similarly proven.
\end{proof}
\begin{xmpl}
For the minimum non-trivial coverage $J_{\text{cov}}$, the corresponding filter of open subsets is the sinhleton $\{\Sigma^{\ast}\}$.
\end{xmpl}
\begin{xmpl}
\label{xmple:denseIdealisDense}
For the maximal non-degenerate coverage \cref{thm:dense} $J_{\text{dense}}$, the corresponding filter of open subsets is the set of open dense subsets of $\Sigma^{\omega}$.
\end{xmpl}
\hr{I will write the rest of this section later. It will be the ``self-similar (= $\Sigma^*$-equivaliant ) sublocale of the generalized cantor spaces"}
\subsection{Radical ideals and Nullstellensatz}
The aim of this subsection is to prove a kind of "Nullsteltensatz" of word combinatorics \cref{thm:Nullstellensatz}.
% The foundation of the following argument is the analogy of Nullstellensatz.
\begin{thm} We have the analogy of Nullstellensatz, as follows:
\label{thm:Nullstellensatz}
\begin{itemize}
\item For an ideal $I$, $I(V(I)) = \rad(I)$ (cf. \cref{thm:GaloisConnectionOfIdealsAndClosed}).
\begin{itemize}
\item More concretely, $
\rad(I) = \{w\in \Sigma^*\mid \forall p \in \Sigma^{\omega}, \left(\forall w_0 \in I, w_0\not \pf p \right)\implies w \not\pf p\}.
$
\end{itemize}
\item Equivalently, for an ideal $I$, $I(D(I)) = \rad(I)$ (cf. \cref{crly:GaloisConnectionOfIdealsAndOpen}).
\begin{itemize}
\item More concretely, $
\rad(I) = \{w\in \Sigma^*\mid \forall p \in \Sigma^{\omega}, w \pf p \implies \left(\exists w_0 \in I, w_0\pf p \right)\}.
$
\end{itemize}
\end{itemize}
\end{thm}
\begin{rmk}
\label{rmk:geometricIntuitionOfNullstellensatz}
Not only the abstract formulations with $V,I,D$, but the concrete description $\rad(I) = \{w\in \Sigma^*\mid \forall p \in \Sigma^{\omega}, \left(\forall w_0 \in I, w_0\not \pf p \right)\implies w \not\pf p\}
$ also emphasizes the analogy with algebraic geometry. It says: the radical $\rad(I)$ is the set of functions that vanishes at the common zeros of $I$.
\end{rmk}
\begin{lemma}
\label{lem:VandIareClosedAndIdeal}
We have the following properties about $V,I$.
\begin{itemize}
\item For any subset $S\subset \Sigma^{*}$, $V(I)$ is a closed subset of the product space $\Sigma^{\omega}$.
\item For any subset $C\subset \Sigma^{\omega}$, $I(C)$ is an (right) ideal of the monoid $\Sigma^{*}$.
\end{itemize}
\end{lemma}
\begin{proof}
If $S$ (resp. $C$) is a singleton, then $V(S)$ (resp. $I(C)$) is a closed set (resp. an ideal). Since closed sets (resp. ideals) are closed under arbitrary intersection, an arbitrary image of $V$ (resp. $I$) is a closed set (resp. an ideal).
\end{proof}
\begin{thm}
\label{thm:GaloisConnectionOfIdealsAndClosed}
The anti-Galois connection of \cref{prop:galois-connection} is restricted to an anti-Galois connection
\[
\begin{tikzcd}
\Ideal(\Sigma^{*})\ar[r,shift left,"V"]&\Closed(\Sigma^{\omega})\ar[l, shift left, "I"]
\end{tikzcd}
\]
\end{thm}
Since the frame of open subsets $\Ocal(\Sigma^{\omega})$ is contravariantly isomorphic to $\Closed(\Sigma^{\omega})$, we obtain a (covariant) Galois connection as well.
\begin{crly}
\label{crly:GaloisConnectionOfIdealsAndOpen}
There is a (covariant) Galois connection
\[
\begin{tikzcd}
\Ideal(\Sigma^{*})\ar[r,shift left,"D"]&\Ocal(\Sigma^{\omega})\ar[l, shift left, "I"],
\end{tikzcd}
\]
where $D(I)= \{p \in \Sigma^{\omega}\mid \exists w\in I, w\pf p\}$, $I(U)= \{w\in \Sigma^{*}\mid \forall p\in \Sigma^{\omega}, w\pf p \implies p\in U\}$
\end{crly}
% $\rad(I)$ is the set of function $w$, such that if $w$} doesn't vanish at a point $p$, then there is a function $w_0$ in the original ideal that does not vanish at the point $p$.
As a preparation, we prove a refinement of \cref{lem:VandIareClosedAndIdeal}.
\begin{lemma}
\label{lem:RadicalIdeals}
For a subset $C\subset \Sigma^{\omega}$, $I(C) \subset \Sigma^{*}$ is a radical ideal.
\end{lemma}
\begin{proof}
We have already proven that it is an ideal \cref{lem:VandIareClosedAndIdeal}. We will prove that it is radical.
Take an arbitrary $w\in \Sigma^*$ and suppose $wa\in I(C)$ for any $a\in \Sigma$. Then, for any infinite word $p\in C$, if $w\pf p$, then there exists $a\in \Sigma$ such that $wa \pf p$. Its contraposition proves $w\in I(C)$.
\end{proof}
\begin{proof}[Proof of \cref{thm:Nullstellensatz}]
The equivalence between the four formulations is easy to prove. We will prove
\[
\rad(I) = I(V(I)).
% =\{w\in \Sigma^*\mid \forall p \in \Sigma^{\omega}, w \pf p \implies \left(\exists w_0 \in I, w_0\pf p \right)\}.
\]
Since the right-hand side $I(V(I))$ is a radical ideal containing $I$ (\cref{lem:RadicalIdeals}), the inclusion $\rad(I) \subset I(V(I))$ follows. We will prove the inclusion of their complements $\left(\Sigma^* \setminus \rad(I)\right) \subset \left(\Sigma^* \setminus I(V(I))\right)$. Take an arbitrary element $w\in \Sigma^* \setminus \rad(I)$. Since $w\notin \rad(I)$ and $\rad(I)$ is radical, we can take an element $a_1 \in \Sigma$ such that $w a_1 \notin \rad(I)$. Recursively, we can take an infinite sequence $a_1, a_2, \dots$ such that $w a_1 \dots a_n \notin \rad(I)$ for any $n\geq 0$. Let $p$ denote the infinite word $p\coloneqq w a_1 a_2 \dots \in \Sigma^{\omega}$. By the construction, while $w \pf p$, any prefix of $p$ is not in $\rad(I)$, and a fortiori, it is not in $I$. This proves that $w$ does not belongs to $I(V(I))$, since
\[
I(V(I))= \{w\in \Sigma^*\mid \forall p \in \Sigma^{\omega}, \left(\forall w_0 \in I, w_0\not \pf p \right)\implies w \not\pf p\}.
\]
\end{proof}
There are a lot of pleasant corollaries of the Nullstellensatz. One is the following ``algebraic" description of (the locale of) generalized Cantor space $\Sigma^{\omega}$.
\begin{crly}
\label{crly:IsomOfRadicalsAndOpen}
(For a possibly infinite set $\Sigma$), the Galois connection in \cref{crly:GaloisConnectionOfIdealsAndOpen} is restricted to the following isomorphism between the poset of radical ideals and the poset (or the frame) of open subsets of the space $\Sigma^{\omega}$:
\[
% \begin{tikzcd}
% \RIdeal(\Sigma^{*})\ar[r,shift left,"D"]&\Ocal(\Sigma^{\omega})\ar[l, shift left, "I"],
% \end{tikzcd}
\RadIdeal(\Sigma^{*})\cong \Ocal(\Sigma^{\omega})
\]
\end{crly}
\begin{proof}
It is enough to prove the surjectivity of $D$ and $I$. The surjectivity of $D$ follows from the fact that $\{D(w)\}_{w\in \Sigma^{\omega}}$ form a basis of the topology. The surjectivity of $I$ is due to the Nullstellensatz (\cref{thm:Nullstellensatz}).
\end{proof}
\hr{Is there a generalization for arbitrary monoid?}
\subsection{Radical ideals and the maximum non-trivial subtopoi}
We've established the analogy of algebraic geometry so far. But how is it related to the classification of subtopoi? The first answer is that the function $\rad$ is a local operator corresponding to \hr{a modified version of} \cref{thm:finitebranches}!
Recall that the set $\Ideal (\Sigma^*)$ with the inverse image action (\cref{eq:IdealAction}) is the subobject classifier of the topos $\PSh(\Sigma^*)$.
% \begin{lemma}
% The set $\Ideal(\Sigma^*)$ equipped with the action $I* w \coloneqq $
% \end{lemma}
The aim of this subsection is to prove the following theorem:
\begin{thm}
\label{thm:RadAsLocalOperator}
The operation $\rad\colon \Ideal(\Sigma^{*})\to \Ideal(\Sigma^*)$ is a local operator.
% \footnote{Local operators are also called Lawvere-Tierney operator. However, we avoid using the confusing word `topology,' since there is the other notion of topology here.}.
% Furthermore, it is minimum among all non-trivial local operators.
\end{thm}
\begin{thm}
\label{thm:RadAsMinimalLocalOperator}
The local operator $\rad\colon \Ideal(\Sigma^{*})\to \Ideal(\Sigma^*)$ is
% a local operator\footnote{Local operators are also called Lawvere-Tierney operator. However, we avoid using the confusing word `topology,' since there is the other notion of topology here.}.
% Furthermore, it is
minimum among all non-trivial local operators.
\end{thm}
For the theorem making sense, it needs to prove that $\rad$ is a morphism of the topos $\PSh(\Sigma^*)$:
\begin{lemma}
\label{lem:RadIsNatural}
The operation $\rad$ is natural, in the sense that
for an ideal $I$ and a word $w\in \Sigma^*$, we have $\rad(I)\cdot w = \rad(I\cdot w)$.
\end{lemma}
\begin{proof}
This follows from \cref{prop:compatibilityOfActions} and the Nullstellensatz (\cref{thm:Nullstellensatz}).
\end{proof}
\begin{proof}[Proof of \cref{thm:RadAsLocalOperator}]
\Cref{lem:RadIsNatural} implies that $\rad$ is a morphism of the topos $\PSh(\Sigma^*)$.
% Since $\rad(I) \supset I$, $\rad$ preserves the tops element $\Sigma^* \in \Ideal(\Sigma^*)$.
By the definition of $\rad$, it is easy to prove $\rad(\Sigma^*) = \Sigma^*$ and $\rad(\rad(I))=\rad(I)$. We will prove $\rad(I_1 \cap I_2)=\rad(I_1)\cap \rad(I_2)$. By the Nullstellensatz (\cref{thm:Nullstellensatz}), for $w\in \Sigma^{*}$,
\begin{align*}
& w \in \rad(I_1 \cap I_2)\\
\iff & \forall p \in \Sigma^{\omega}, \left(w \pf p \implies \exists w_0 \in I_1 \cap I_2, w_0\pf p \right)\\
\iff & \forall p \in \Sigma^{\omega}, \left(w \pf p \implies \exists w_1 \in I_1, w_1\pf p \right) \land \left(w \pf p \implies \exists w_2 \in I_2, w_2\pf p \right)\\
\iff & w\in \rad(I_1) \cap \rad(I_2).
\end{align*}
This proves that $\rad$ is a local operator.
\end{proof}
We will prove \cref{thm:RadAsMinimalLocalOperator} in terms of coverage. As a preparation, we describe the corresponding coverage.
\begin{prop}
\label{prop:coveringIdeal}
For an ideal $I$, the following conditions are equivalent:
\begin{itemize}
\item $\rad(I) = \Sigma^{\ast}$
\item $\forall p \in \Sigma^{\omega}, \exists w \in I, w\pf p$
\item $V(I) = \emptyset$ ($T_I$ does not have infinite branches.)
\item $D(I) = \Sigma^{\omega}$
\end{itemize}
\end{prop}
\begin{proof}
Due to the Nullstellensatz (\cref{thm:Nullstellensatz}).
\end{proof}
\begin{dfn}
\label{dfn:coveringIdeal}
An ideal $I$ is \textbf{covering} if $I$ satisfies the equivalent conditions of \cref{prop:coveringIdeal}.
Let $J_{\text{cov}}$ denote the set of covering ideals.
\end{dfn}
\hr{I know this terminology is terrible. I think this should be called ``prefix-complete".} \mr{This part is redundant now; \Cref{ssec:Cantorsubsets} subsumes the Nullstellensatz discussion.}
\subsection{A self-similarity observation by Hora}
\begin{dfn}
An object $X$ of a topos $\Ecal$ is said to have enough endomorphisms if the colimit of the canonical left-action of the endomorphisms monoid $\End(X)$ on $X$ is the terminal object of $\Ecal$.
\end{dfn}
\begin{rmk}
If an object $X$ has enough endomorphisms, then $X$ is well-supported.
\end{rmk}
Take an étale geometric morphism $f\colon \Fcal \to \Ecal$, which correspond to an object $X$ of $\Ecal$. Then the monoid $\End(X)$ acts on the topos $\Fcal$. Due to the trivial commutativity
\[
\begin{tikzcd}
X \ar[rd]\ar[dd]&\\
&1_\Ecal\\
X\ar[ru]
\end{tikzcd}
\]
in $\Ecal$, the étale morphism $f\colon \Fcal \to \Ecal$ is stable under the monoid action.
\begin{lemma}
For any two objects $A, B$ in $\Ecal$ and a morphism $k\colon f^*A \to f^*B$, the following conditions are equivalent:
\begin{itemize}
\item There exists a map $j\colon A \to B$ such that $k=f^*j$.
\item There exists a unique map $j\colon A \to B$ such that $k=f^*j$.
\item $k$ is stable under the inverse image part of the action of $\End(X)$.
\end{itemize}
\end{lemma}
\begin{proof}
If $k=f^*j$ for some $j\colon A \to B$, then it is easy to prove that it's stable.
We prove the converse. Let us identify $k\colon f^* A\to f^*B$ with $k:X\times A\to X\times B$.
If $k\colon f^*A \to f^*B$ is stable under the monoid action, then the stability condition states that the following diagram
\[
\begin{tikzcd}
X\times A \ar[r, "k"]\ar[d, "l\times \id_A"]&X\times B\ar[d, "l\times \id_B"]\\
X\times A \ar[r, "k"]&X\times B
\end{tikzcd}
\]
commutes for any endomorphism $l\colon X \to X$. Then, by taking the colimit of this $\End(X)$ action, we obtain the unique morphism $j\colon A\to B$ such that
\[
\begin{tikzcd}
X\times A \ar[r, "k"]\ar[d, "\pi_2", twoheadrightarrow]&X\times B\ar[d, "\pi_2", twoheadrightarrow]\\
A \ar[r, "j"]&B.
\end{tikzcd}
\]
Here, we use the assumption and the fact that a topos is cartesian closed and the functors $A\times{-}, B\times{-}$ preserve small colimits.
Since $k$ is a morphism in $\Ecal/X\simeq \Fcal$
\[
\begin{tikzcd}
X\times A \ar[r, "k"]\ar[d, "\pi_1", twoheadrightarrow]&X\times B\ar[d, "\pi_1", twoheadrightarrow]\\
X \ar[r, "\id_X"]&X,
\end{tikzcd}
\]
we have $k=\id_X\times j = f^*j$.
\end{proof}
\begin{prop}
For a topos $\Ecal$ and an object $X$ that has enough endomorphisms, a Lawvere-Tierney topology $k$ in $\Ecal/X$ is a lift of a Lawvere-Tierney topology of $\Ecal$ if and only if $k$ is stable under the $\End(X)$ action on $\Ecal/X$.
\end{prop}
\begin{crly}
For a topos $\Ecal$ and an object $X$, if $X$ has enough endomorphisms and $\Ecal/X \simeq \Sh(L)$ for a locale $L$, then there is a bijective correspondence between
\begin{itemize}
\item $\End(X)$-action stable sublocales of $L$, and
\item Subtopoi of $\Ecal$.
\end{itemize}
\end{crly}
We can apply this to $\PSh(\Sigma^*)$. It should be also possible for the Jonsson-Tarski topos.
\subsubsection{Comparison}
\begin{lemma}\label{lem:ComparisonBetweenThreeInvarianceCOnditions}
For any objects $X,Y,Z$ and any morphism $f\colon X\times Y\to Z$, the following conditions are equivalent, where $k\colon X\times Y \to X \times Z$ denotes the map $\langle \pi_1, f \rangle \colon X\times Y \to X \times Z$. (Notice that every map $k\colon !^*Y \to !^*Z$ in $\Ecal/X$ is of this form.)
\begin{itemize}
\item (Morgan's commutativity condition) The diagram
\begin{equation}\label{eq:pi13commDash}
% https://q.uiver.app/#q=WzAsNCxbMSwxLCJYIFxcdGltZXMgXFxPbWVnYSJdLFswLDEsIlggXFx0aW1lcyBYIFxcdGltZXMgXFxPbWVnYSJdLFsxLDAsIlggXFx0aW1lcyBcXE9tZWdhIl0sWzAsMCwiWCBcXHRpbWVzIFggXFx0aW1lcyBcXE9tZWdhIl0sWzEsMCwiXFxwaV97MTN9IiwyXSxbMiwwLCJrIl0sWzMsMiwiXFxwaV97MTN9Il0sWzMsMSwiXFxpZCBcXHRpbWVzIGsiLDJdXQ== %But eddited by Hora.
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times Y} \& {X \times Y} \\
{X \times X \times Z} \& {X \times Z}
\arrow["{\pi_{13}}", from=1-1, to=1-2]
\arrow["{\id \times k}"', from=1-1, to=2-1]
\arrow["k", from=1-2, to=2-2]
\arrow["{\pi_{13}}"', from=2-1, to=2-2]
\end{tikzcd}
\end{equation}
commutes.
\item (Lifting condition) There exists a lift in the following diagram
\begin{equation}\label{eq:liftingpi13}
% \begin{tikzcd}
% X\times A \ar[r, "f"] \ar[d, twoheadrightarrow, "\pi\times \id_A"']&A\\
% \supp(X) \times A \ar[ru, dashed, "\exists"']&
% \end{tikzcd}
\begin{tikzcd}
X\times Y \ar[d, "f"'] \ar[r, twoheadrightarrow, "!\times \id_Y"]& \supp(X) \times Y \ar[dl, dashed, "\exists"]\\
Z&
\end{tikzcd}
\end{equation}
\end{itemize}
Furthermore, if $X$ has enough endomorphisms in the sense of \Cref{dfn:EnoughEndomorphismsSupport}, these conditions are also equivalent to the condition:
\begin{itemize}
\item (Self-similarity condition) $k$ is self-similar in the sense of \Cref{def:self-similar}.
\end{itemize}
\end{lemma}
\begin{proof}
The commutativity of \Cref{eq:pi13commDash} is decomposed into the commutativity at the first coordinate
\[
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times Y} \& {X \times Y} \\
{X \times X \times Z} \& {X}
\arrow["{\pi_{13}}", from=1-1, to=1-2]
\arrow["{\id \times k}"', from=1-1, to=2-1]
\arrow["\pi_1", from=1-2, to=2-2]
\arrow["{\pi_{1}}"', from=2-1, to=2-2]
\end{tikzcd}
\]
and the commutativity at the second coordinate
\[
\begin{tikzcd}[ampersand replacement=\&,row sep=scriptsize]
{X \times X \times Y} \& {X \times Y} \\
{X \times X \times Z} \& {Z.}
\arrow["{\pi_{13}}", from=1-1, to=1-2]
\arrow["{\id \times k}"', from=1-1, to=2-1]
\arrow["f", from=1-2, to=2-2]
\arrow["{\pi_{3}}"', from=2-1, to=2-2]
\end{tikzcd}
\]
The first commutativity is trivial, and the second is
equivalent to the commutativity of
\[
\begin{tikzcd}
X\times X\times Y\ar[r, shift left, "\pi_{13}"]\ar[r, shift right, "\pi_{23}"']& X \times Y \ar[r, "f"]& Z.
\end{tikzcd}
\]
On the other hand, since the coequializer diagram
\[
\begin{tikzcd}
X\times X\ar[r, shift left, "\pi_{1}"]\ar[r, shift right, "\pi_{2}"']& X \ar[r, twoheadrightarrow, "!"]& \supp(X).
\end{tikzcd}
\]
is preserved by the left adjoint functor $Y\times {-}\colon \Ecal \to \Ecal$, we have the coequalizer diagram
\[
\begin{tikzcd}
X\times X\times Y\ar[r, shift left, "\pi_{1}\times \id_Y"]\ar[r, shift right, "\pi_{2} \times \id_Y"']& X\times Y \ar[r, twoheadrightarrow, "!\times \id_Y"]& \supp(X)\times Y.
\end{tikzcd}
\]
The universality of this coequalizer proves the equivalence between the first two conditions.
Asuuming the self-similarity of $X$, we know that the colimit of the $\End(X)$-action on $X$ is given by $!\colon X \twoheadrightarrow \supp(X)$. By sending it by the left-adjoint functor $Y\times {-}$, we observe that the map $!\times \id_Y \colon X\times Y \twoheadrightarrow \supp(X) \times Y$ is the colimit cocone of the $\End(X)$-action on the object $X\times Y$. The action of a morphism $m\colon X \to X$ is given by $m\times\id_Y\colon X\times Y \to X\times Y$. By \Cref{rmk:TheActionOfEndoMorphisms}, the self-similarity of $k$ is equivalent to the condition that $f\colon X \times Y \to Z$ is a cocone under the $\End(X)$-action on $X\times Y$. Then, the universality of the counit proves the equivalence between the second condition and the last condition.
\end{proof}
\begin{crly}
Assume the self-similarity of $X$.
\Cref{lem:ComparisonBetweenThreeInvarianceCOnditions} implies that
a map $k\colon \bang^* \Omega \to \bang^* \Omega$ satisfies the commutativity condition \Cref{eq:pi13comm} if and only if it is self-similar. In particular, \Cref{prop:operatorlift} implies that a Lawvere-Tierney topology $k\colon !^*\Omega \to !^*\Omega$ in $\Ecal/X$ is a lift of one in $\Ecal$ if and only if $k$ is self-similar.
\end{crly}
\bibliography{montop}
\bibliographystyle{alpha}
\end{document}