# Image factorisation in pointed categories

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 $f = k \circ g$.

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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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.
The lemma in Mac Lane makes $g$ an epimorphism, but seems to make use also of a condition that there always exist equalizers. Without that assumption, are there cases in which $g$ is not an epimorphism but only a pseudo-epimorphism? – jdc Dec 6 '15 at 6:30
For posterity, the answer to my question is in the affirmative. In the category of groups, with basepoint the trivial group, kernels are inclusions of normal subgroups and monomorphisms are injections, so most monomorphisms aren't kernels. The cokernel of a map $\phi\colon G \to H$ is the map to the quotient $H/\langle\langle\phi(G)\rangle\rangle$ by the normal closure of the set-theoretic image, and $\im(\phi)$ is the inclusion of this normal closure in $H$, so the factorization above is $G \to \langle\langle\phi(G)\rangle\rangle \hookrightarrow H$. – jdc Dec 9 '15 at 4:11
(cont'd) The first map in the above factorization is the $g$ in the initial question, and it is a pseudo-epimorphism since its cokernel is the quotient map $\langle\langle\phi(G)\rangle\rangle \to \langle\langle\phi(G)\rangle\rangle/\langle\langle\phi(G)\rangle\rangle \cong 1$. But in the category of groups an epimorphism is just a surjection, so $g$ is an epimorphism if and only if $\phi(G)$ is normal in $H$. – jdc Dec 9 '15 at 4:16