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

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?

• Apart from the formulation, this is a very reasonable question, in particular, in view of the excellent answer and the interesting comments. Commented Sep 6, 2023 at 9:08
• The catch is that poor posted questions should not be answered at all , no matter how excellent the answer is. Commented Sep 6, 2023 at 12:42
• Perhaps some of the users who voted to close could clarify what sort of additional context they think this question needs? The context seems very clear to me: they found a statement without proof in a reference source, and are looking for a proof of it. Commented Sep 6, 2023 at 13:46
• @Peter Would you rather the OP struggle for hours more, not realising the statement is in fact false? This is not the kind of post that clogs up the site for reasons of laziness or disrespect, it is a perfectly reasonable question. Sometimes questions are just short… Commented Sep 6, 2023 at 15:55
• Claiming that one tried it for hours is no context. This question has no context , so it should be closed. Commented Sep 6, 2023 at 15:57

$$\newcommand{\C}{\mathsf{C}}$$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.

• Yeah, I tried to prove without $f$ epic to no avail, as well. Commented Aug 26, 2023 at 21:29
• I also doubted the statement, and e-mailed its author on nLab, Zoran Škoda to ask him to enlighten us. Commented Aug 26, 2023 at 21:31
• He promptly replied: "I do remember putting that sentence from some canonical source which I was using at the time; it is likely indeed that the source had some running assumption on the type of ambient category, say of the sort of assumptions like having an initial object or alike. I do not remember now the argument. After your email, I looked through Borceux-Bourn as one of the likely sources (they usually have a zero object in their setup) and could not find it there. [...] you can simply take the statement as being uncarefully written into the nLab." Commented Aug 27, 2023 at 12:38
• @AnneBauval Thank you for following this up Commented Aug 27, 2023 at 12:42
• I fixed the wrong assertion. (diff: ncatlab.org/nlab/revision/diff/strict+epimorphism/14) Commented Aug 27, 2023 at 18:21