# pullback square of regular epimorphisms is a pushout

I'm looking for a proof of the fact that in a regular category a commutative square of regular epimorphisms which is a pullback is also a pushfoward.

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I've not seen any lemma like that, but there is one involving a pullback–pushout square of monomorphisms. –  Zhen Lin Apr 21 '13 at 20:03
I find this statement on arxiv.org/pdf/math/0607100v1.pdf at page 10, but there isn't a proof. –  Fabio Lucchini Apr 21 '13 at 23:46
What state the lemma involving a pullback–pushout square of monomorphisms? Thank'you –  Fabio Lucchini Apr 24 '13 at 12:11
It's just the statement that if $A = B \cup C$, then $A$ is the pushout of $B \leftarrow B \cap C \rightarrow C$ as well. –  Zhen Lin Apr 24 '13 at 12:54
Where I can find a proof of this lemma? Thank'you –  Fabio Lucchini Apr 26 '13 at 20:43

Let $f' : A' \to B'$, $v : A' \to A$, $w : B' \to B$, $f : A \to B$ form a pullback square, and let $K (f')$, $K (v)$, $K (w)$, $K (f)$ be the respective kernel pairs. It is not hard to show that $K (v)$ is the pullback along $f'$ of $K (w)$, and $K (f')$ is the pullback along $v$ of $K (f)$; thus we have regular epimorphisms $K (v) \to K (w)$ and $K (f') \to K (f)$. Suppose we have morphisms $h : A \to C$ and $u : B' \to C$ making the evident square commute. Then, by considering an appropriate commutative diagram, we see that $h : A \to C$ must factor through $f : A \to B$, and $u : B' \to C$ must factor through $w : B' \to B$. Since $v : A' \to A$ and $f : A \to B$ are epimorphisms, the two morphisms $B \to C$ we obtain are in fact equal. It is clearly the unique one making the obvious diagram commute, and thus the pullback square we started with is also a pushout square.

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