**Proof.**
Let $i:\mathop{\mathrm{Ker}}(g)\to y$ be the canonical injection. Let $p:x\to \mathop{\mathrm{Coim}}(f)$ be the canonical projection. Let $j:\mathop{\mathrm{Im}}(f)\to \mathop{\mathrm{Ker}}(g)$ be the canonical injection.

Suppose (1) holds. Let $h:w\to y$ with $g\circ h=0$. There exists $c:w\to \mathop{\mathrm{Ker}}(g)$ with $i\circ c=h$. Let $v=x\times _{\mathop{\mathrm{Ker}}(g)}w$ with canonical projections $k:v\to w$ and $l:v\to x$, so that $c\circ k=j\circ p\circ l$. Then, $h\circ k=i\circ c\circ k=i\circ j\circ p\circ l=f\circ l$. As $j\circ p$ is an epimorphism by hypothesis, $k$ is an epimorphism by Lemma 12.5.13. This implies (2).

Suppose (2) holds. Then, $g\circ i=0$. So, there are an object $w$, an epimorphism $k:w\to \mathop{\mathrm{Ker}}(g)$ and a morphism $l:w\to x$ with $f\circ l=i\circ k$. It follows $i\circ j\circ p\circ l=f\circ l=i\circ k$. Since $i$ is a monomorphism we see that $j\circ p\circ l=k$ is an epimorphism. So, $j$ is an epimorphisms and thus an isomorphism. This implies (1).
$\square$

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