From 56aaf125b8bcd406142b44d78079a7529441cb28 Mon Sep 17 00:00:00 2001 From: partowp Date: Wed, 10 Jun 2026 16:33:33 +0100 Subject: [PATCH] minor --- draft/draft.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/draft/draft.tex b/draft/draft.tex index 59c763c..0d76044 100644 --- a/draft/draft.tex +++ b/draft/draft.tex @@ -2537,7 +2537,7 @@ The following example justifies why the $g$ in~\autoref{def:cogood-ord} should b \begin{cor} By~\autoref{prop:all-rel-compa}.(1), if $\appr$ is good, then the mid-lax Barr relator is a normal relation connector, and thus a sound and complete relator as well. \end{cor} -Perhaps if we can relax the definition of good by allowing $g$ to be a relation rather than a function, we can have a lax Barr relator that its symmetrization is a normal lax extension.\todo{Investigate!} +Perhaps if we can relax the definition of good by allowing $g$ to be a relation rather than a function, we can have a lax Barr relator that its symmetrization is a normal lax extension. Well, this idea actually does not work because if $g$ is supposed to be a relation, then $F$ can not be a set-functor, but it should be a functor on $\rel$, and it is already a big assumption. \begin{prop} For a natural order structure $\appr$ on a set-functor $F$, if for every $f\in\Hom(X,FY)$ we have $Ff\comp\appr=\appr\comp Ff$, then $\appr$ is cogood. \end{prop}