= Solution
Use the displayed square's notation $f:A\to B$, $h:A\to C$, $g:B\to D$, $k:C\to D$, with $gf=kh$ and $g$ epic. Form the <biproduct> and the morphisms
$$
q=[g,-k]:B\oplus C\to D,\qquad j=\binom f h:A\to B\oplus C.
$$
The map $q$ is an <epimorphism>, since its restriction to $B$ is $g$: equality after $q$ implies equality after $g$. The <pullback in a category> property says exactly that $j$ is a <categorical kernel> of $q$. Indeed $qj=gf-kh=0$, and a map into $B\oplus C$ killed by $q$ is a pair $(x,y)$ with $gx=ky$, which factors uniquely through $(f,h)$.
Use the standard <abelian category> property that every <epimorphism> is the <categorical cokernel> of its <categorical kernel>. If $u:B\to X$ and $v:C\to X$ satisfy $uf=vh$, then $[u,-v]j=0$. Hence there is a unique $w:D\to X$ with $wq=[u,-v]$. Restriction to the two summands gives $wg=u$ and $wk=v$. This is the <pushout in a category> universal property. Thus \b[a <pullback of an epimorphism is a pushout in an abelian category>].
In this pushout, $g$ is the pushout of $h$ along $f$. The reflection result of part (b) therefore makes $h$ epic. This proves <pullback stability of epimorphisms in an abelian category>.
Finally, let $e:E\to B$ be an <epimorphism> and let $p_1,p_2:R\rightrightarrows E$ be its <kernel pair>. Their pullback square is a pushout by the result just proved. If $u:E\to X$ satisfies $up_1=up_2$, the two copies of $u$ form a pushout cocone. There is a unique $v:B\to X$ with $ve=u$. Hence \b[<epimorphisms in an abelian category are coequalizers of their kernel pairs>], so they are <regular epimorphisms>.
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