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Let $\mathbf{C}$ be a pointed category, that is to say, a category with a zero object $0$, and suppose that $\mathbf{C}$ has all kernels and cokernels, and suppose also that every monomorphism in $\mathbf{C}$ is a kernel. $\DeclareMathOperator{coker}{coker}$

Let us say that a morphism $g$ is a pseudo-epimorphism if $l \circ g = 0$ implies $l = 0$: so $g$ is a pseudo-epimorphism if and only if $\coker g = 0$. Consider a morphism $f : A \to B$; let $k = \ker (\coker f)$, the regular image of $f$. Since $\coker(f) \circ f = 0$, $f$ must factor through $k$, say $f = g \circ k$.

Question. Under these hypotheses, why is $g$ pseudo-epic?

Examples of categories satisfying these hypotheses: the category of pointed sets $\textbf{Set}_*$, the opposite category of groups $\textbf{Grp}^\textrm{op}$, and of course any abelian category. The proof in the case of abelian categories is reasonably straightforward, due to the presence of sufficient colimits and exactness conditions: indeed, if $l \circ g = 0$, we take the pushout $\tilde{k}$ of $l$ along $k$, since $\coker k = \coker f$, we find that $\tilde{k} \circ l = 0$, and $\tilde{k}$ is monic since $k$ is, so $l = 0$ as required. Unfortunately, $\mathbf{C}$ does not have pushouts in general...

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This is a special case of a lemma in Categories for the Working Mathematician [Ch. VIII, §1, Lemma 1].

Suppose $l \circ g = 0$; then $g$ factors through $s = \ker(l)$, say $g = s \circ g'$. To show $g$ is pseudo-epic, it suffices to show $s$ is an isomorphism, since in that case $l = 0$.

Let $k' = k \circ s$; then $f = k \circ g = k' \circ g'$. Since $k'$ is a composite of two monos, it is monic; thus by our hypothesis on $\mathbf{C}$, $k'$ is a kernel, and thus $k' = \ker (\coker k')$. But $\coker(k') \circ f = 0$, so $\coker k'$ factors through $\coker f$, and $\coker f = \coker k$, so $k$ factors through $k'$, say $k = k' \circ r$. But $k$ is monic, so we get $\textrm{id} = s \circ r$ by cancelling $k$. Hence, $s$ is both monic and split epic, and is therefore an isomorphism as required.

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