Does always gf regular epi imply g regular epi?

Question

In a category, suppose that a composite $gf$ is a regular epimorphism. I can prove that then also $g$ is a regular epimorphism under the assumption that $f$ is an epimorphism. My question is: is it possible to show $gf$ regular epi $\to g$ regular epi, without the assumption that $f$ must be epi?

Answer 1

For statements like these, it's relatively easy to come up with small counterexamples. Consider the category generated by the following diagram $\require{AMScd}\begin{CD} \overset{h_1}{\underset{h_2}\rightrightarrows}\bullet@>f>>\bullet\\ @Vh\circ fVV \overset {h}\swarrow @VgVV\\ \bullet @<k<<\bullet \end{CD}$ subject to the relations

1. $g\circ f\circ h_1=g\circ f\circ h_2$
2. $k\circ g\circ f=h\circ f$

Then $g\circ f$ is coqualizer of $h_1$ and $h_2$ because the only other morphisms out of the codomain of $\bullet\overset{h_1}{\underset{h_2}\rightrightarrows}\bullet$ is $h\circ f$ and that factors uniquely as $k\circ g\circ f$.

But $g$ is not a coequalizer, because the only pair of morphisms it could coequalizer would be $f\circ h_1$ and $f\circ h_2$, yet we have that $h\circ f\circ h_1=h\circ f\circ h_2$ without $h$ factoring through $g$ (we only have $h\circ f=k\circ g\circ f$, not $h=k\circ g$).

Hence in general, you need to know more about the category to conclude that $g$ is a coequalizer from knowing that $g\circ f$ is a coequalizer.

Answer 2

In addition to Vladimir's counterexample, here is one that occurs in the category $\mathbf{Cat}$ of (small) categories. It is taken from Exercise 7.S. in the book "Abstract and Concrete Categories : The Joy of Cats" by Adamek, Herrlich and Strecker.

Let $A$ be the category with two objects $x,y$ and exactly one non-identity arrow $\alpha:x\to y$ (the "walking arrow", as the nLab would call it) and let $B$ be the monoid of natural numbers, seen as a category with one object. The functor $p:A\to B$ that maps $\alpha$ to $1$ is a regular epimorphism, because it is the coequalizer of the two functors from the terminal category to $A$ that map the unique point in the terminal category to $x$ and $y$ respectively. Consider now the category $A'$, which has the same objects as $A$ and has two non-identity arrows $\alpha,\beta:x\to y$. You can define a functor $p':A'\to B$ by saying that $p'(\alpha)=1$ and $p'(\beta)=2$, so that $p=p'\circ i$, where $i$ is the inclusion $A\to A'$. But unlike $p$, $p'$ is not a regular epi; indeed, any pair of functors into $A'$ that are coequalized by $p'$ would also be coequalized by the functor $p'':A'\to B$ that map $\alpha$ to $1$ and $\beta$ to $3$, but there is no functor $h:B\to B$ such $h\circ p'=p''$, because it is equivalent to asking for a monoid endomorphism of $\mathbb{N}$ such that $h(1)=1$ but $h(2)=3\neq 2=h(1)+h(1)$.