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Consider the following commutative diagram in a semiabelian category enter image description here

where $f,f',\alpha$ and $\beta$ are regular epimorphisms. Is it true that the induced dotted map is a regular epimorphism too? If so, how can I prove it? If not, is the previous statement true when we add the hypothesis that the right-hand square is a pushout?

Any help would be appreciated!

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It is not necessarily true that the induced map will always be a regular epi. In fact, in a semi-abelian category this map being a regular epimorphism is equivalent to the square being a pushout.

Indeed, if the square is a pushout, then the canonical arrow $( \alpha,f) :A\to A'\times_{B'}B$ given by the universal property of the pullback is a regular epimorphism (and the converse is true as well); this is a consequence Theorem 5.2 in this paper and the fact that every semi-abelian category is exact and Mal'tsev. Then you have a diagram $\require{AMScd}$ \begin{CD} Ker(f) @>{\ker(f)}>> A @>{f}>> B \\ @V{\gamma_1}VV @V{(\alpha,f)}VV @VV{1_B}V \\ Ker(\psi_2) @>{\ker(\psi_2)}>> A'\times_{B'}B @>{\psi_2}>> B \\ @V{\gamma_2}VV @V{\psi_1}VV @VV{\beta}V \\ Ker(f') @>>{\ker(f')}> A' @>>{f'}> B'. \end{CD} Now the bottom right square is pullback by definition, hence the induced map $\gamma_2$ is an isomorphism, and the map $1_B$ is a monomorphism, hence the top left square is a pullback (these two properties hold in any pointed category). In particular, since regular epimorphisms are stable, $\gamma_1$ is a regular epimorphism, and thus so is the composition $\gamma_2\gamma_1$. This is exactly your dotted arrow.

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  • $\begingroup$ The problem is that I don't see how to show the equivalence between the right-hand square being a pushout and the map $(\alpha,f)$ being regular epimorphism $\endgroup$ – I'm D. Feb 21 '18 at 13:08
  • $\begingroup$ @I'mD. It's definitely non-trivial. I've added a reference, sorry for not doing this earlier. $\endgroup$ – Arnaud D. Feb 21 '18 at 16:19
  • $\begingroup$ Thanks! A further (and probably trivial) question: is at least the induced map always epi (even without the right-hand square being a pushout)? If not, is there an easy condition to ask in order to obtain that? $\endgroup$ – I'm D. Mar 7 '18 at 11:04
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    $\begingroup$ @I'mD. The induced map being an epi is already enough to prove that the right-hand square is a pushout. So in fact, it is an epi if and only if it is a regular epi! $\endgroup$ – Arnaud D. Mar 7 '18 at 11:12

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