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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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.