# Is the induced map in the following diagram a regular epimorphism?

Consider the following commutative diagram in a semiabelian category

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!

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.
• 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 – I'm D. Feb 21 '18 at 13:08