Why $g$ is strict epimorphism if $g \circ f$ is a strict epimorphism?

Question

In strict epimorphism (revision at the time when I posted this question) on nLab, it says "If the composition $g \circ f$ is a strict epimorphism then $g$ is a strict epimorphism."

I've tried to prove it for about five hours but I couldn't. Why can $g$ be a strict epimorphism?

Answer 1

$\newcommand{\C}{\mathsf{C}}$Fix a category $\C$ and arrows $a\overset{f}{\longrightarrow}b\overset{g}{\longrightarrow}c$ in $\C$; assume $gf$ is a strict epimorphism.

Unwinding definitions (colimit of the diagram ...) we would like to show the following:

If $\lambda\in\C(b,d)$ satisfies $\lambda h=\lambda h'$ for all pairs $h,h'\in\C(x,b)$ with $gh=gh'$ then $\lambda$ factors uniquely through $g$.

Firstly let's note that, as $gf$ is necessarily an epimorphism we have that $g$ is an epimorphism too, giving uniqueness of the factorisation when it exists.

Say $u,v\in\C(y,a)$ satisfy $gf\circ u=gf\circ v$. Then $g\circ(fu)=g\circ(fv)$ of course, so that $\lambda\circ(fu)=\lambda\circ(fv)$ follows, as does $(\lambda f)\circ u=(\lambda f)\circ v$. We find that $\lambda f$ coequalises all parallel pairs which $gf$ coequalises; since $gf$ is a strict epimorphism, this implies $\lambda f$ equals $\gamma\circ gf$ for a (unique) $\gamma\in\C(c.d)$.

We would like to show that $\lambda=\gamma\circ g$. In general, this seems to be a bit of a thorny problem. There does not seem to be any way to get at arrows into $b$ which don't factor through $f$ (to use the full strength of the hypothesis on $\lambda$) nor any other tool at our disposal to deduce $\lambda=\gamma\circ g$; why should $f$ be an epimorphism? It's not true that if $gf$ is epimorphic then so is $f$... But, if $f$ is an epimorphism - as an additional hypothesis - we can conclude the result.

Doing a little Googling, I find on page $31$, proposition $1.3.7$ of this the same proof as mine with the same additional hypothesis. This article references a source from Kelly, and I would assume that Kelly did not make any unnecessary assumptions; I would assume that we should suppose $f$ is epimorphic as an additional hypothesis.

In other words, I think we should treat the nLab statement as false - or "open" - without further context or hypotheses.

Indeed, we can easily make a counterexample. We will build a category that looks somewhat like this:

Where $fu\neq fv$ but $gfu=gfv$, $\lambda u=\lambda v$ and $\gamma g f=\lambda f$ and $a,b,c,d,x$ are unequal and $\gamma g\neq\lambda$. It's easy to check that $gf$ is a strict epimorphism here. The only thing to do is convince ourselves we could formalise this in such a way that the axioms of a category are satisfied, but this is also easy.