A relational obstruction to transport through a quotient

Calculation in relations

Let $A=\{a\}$, $B=\{u,v\}$, and $C=\{c\}$, with $u\ne v$. Take

$$R=\{(a,u)\}\subseteq A\times B,\qquad S=\{(v,c)\}\subseteq B\times C.$$

Then $S\circ R=\varnothing$.

If $q:B\to\{*\}$ identifies $u$ and $v$, the transported relations are

$$R'=\{(a,*)\},\qquad S'=\{(*,c)\},$$

and $S'\circ R'=\{(a,c)\}$. Thus changing the middle object by a quotient can create new relational composites.

Scope

This is an elementary counterexample for arbitrary binary relations. It is not yet a counterexample for a specified class of legal total deterministic game strategies. Such a claim must show that $R,S$ represent allowed strategies, that $q$ is an allowed map, and that the stated transport agrees with strategy composition.

Note 13 discusses the legality issue. The double-category question remains open at that level.

@misc{sumire:2610.00003:23:v4,
  author = {hora},
  title = {{A relational obstruction to transport through a quotient}},
  year = {2026},
  note = {SUMIRE:2610.00003/23, v4},
  url = {https://sumire.aiformath.jp/p/2610.00003/n/23/v/4}
}