= Solution
Take any <morphism> $h:A\to B$ and let $c:B\to C$ be its <categorical cokernel>. Let $i:I\to B$ be the <categorical kernel> of $c$. Since $ch=0$, kernel universality gives $e:A\to I$ with $h=ie$. The map $i$ is a <monomorphism>. We must establish the <pseudo-epimorphism> property of $e$, rather than merely name $I$ an image.
Suppose $t:I\to Z$ satisfies $te=0$. Let $k:K\to I$ be its <categorical kernel>. Then $e=ka$ for some $a:A\to K$. The composite $ik$ is a <monomorphism>, so the hypothesis that every <monomorphism> is normal provides a map $q:B\to D$ of which $ik$ is a <categorical kernel>, up to its unique compatible isomorphism. Now $qh=qika=0$. The <categorical cokernel> property of $c$ gives $q=\bar q c$, and therefore $qi=\bar q ci=0$.
Since $ik$ is the <categorical kernel> of $q$, the map $i$ factors through it: $i=ik\ell$ for some $\ell:I\to K$. Cancel the <monomorphism> $i$ to get $k\ell=1_I$. As $k$ is also a <monomorphism>, $\ell k=1_K$, so $k$ is an <isomorphism>. Finally $t=tk\ell=0$, proving that $e$ is a <pseudo-epimorphism>.
Thus
$$
\boxed{h=ie,\qquad i=\ker(\operatorname{coker}h),\qquad e\text{ pseudo-epic},\quad i\text{ monic}.}
$$
The <pseudo-epimorphism factorization through the kernel of a cokernel> proof uses a <pointed category>, existence of <categorical kernels> and <categorical cokernels>, and normality of every monic map. It does not silently replace pseudo-epic cancellation against zero by the stronger <epimorphism> condition.
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