In a finitely complete and cocomplete category. Does it always hold that the composition of two regular epimorphisms is regular? And if it's not the case, what kind of additional constraints can make it true (say, a pre-abelian category)?

What I already knew is it holds for categories where regular epimorphisms and strong epimorphisms conincide.

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    $\begingroup$ Well, strong epimorphisms are always composable, so the hypothesis that regular epi = strong epi just transfers that property to regular epimorphisms. I do not believe that regular epimorphisms are composable in general. $\endgroup$ – Zhen Lin Feb 12 '16 at 15:55
  • $\begingroup$ @ZhenLin Do you think it holds for pre-abelian categories? I tried to prove it in this special case by mimicking diagram chasing trick, but failed. $\endgroup$ – Censi LI Feb 12 '16 at 17:08
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    $\begingroup$ No, it is false even there. See exercise 5(c) here. $\endgroup$ – Zhen Lin Feb 12 '16 at 19:01
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    $\begingroup$ A sufficient condition is that regular epimorphisms are pullback stable epimorphisms. $\endgroup$ – Nex Feb 12 '16 at 19:03

In a category with all kernel pairs and coequalisers of kernel pairs, the following conditions are equivalent:

  1. regular epimorphisms are stable under composition;
  2. regular epimorphisms coincide with strong epimorphisms;
  3. for any morphism $f$, if $m_f \circ e_f$ is its factorisation through the coequaliser of its kernel pair, $m_f$ is a monomorphism;
  4. regular epimorphisms and monomorphisms form a factorisation system.

This is proved in Monomorphisms, Epimorphisms, and Pull-backs by Kelly (Propositions 2.7 and 3.8). Note: Kelly takes as definition of regular epimorphism what is called elsewhere strict epimorphism, but these notions coincide when kernel pairs exist. (And Kelly doesn’t give the “factorisation system” version.)

Kelly gives an example of a pre-abelian category in which regular monomorphisms do not compose: the category of abelian groups with no elements of order 4 (last paragraph of p. 126). The dual category is a pre-abelian category in which regular epimorphisms do not compose.

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    $\begingroup$ Note: 3 of the implications are easy. 3 $\to$ 4: we have the factorisation; the orthogonality is the inclusion of regular into strong epimorphisms. 4 $\to$ 2: in a factorisation system, the “epi” morphisms are exactly the morphisms orthogonal to the “mono” morphisms. 2 $\to$ 1: strong epimorphisms compose. The proof of 1 $\to$ 3 is a bit more involved: you factorise $m_f$ as $m' \circ e'$ through the coequaliser of its own kernel pair, and use the regularity of $e' \circ e_f$ to prove that $e'$ is an isomorphism, so that the kernel pair of $m_f$ is trivial and $m_f$ is a monomorphism. $\endgroup$ – Vej Kse Feb 12 '16 at 20:05

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