= Solution
In an <abelian category>, the <image factorization in an abelian category> of $f:A\to B$ is
$$
A\xrightarrow{p}\operatorname{im}f\xrightarrow{i}B,
\qquad p=\operatorname{coker}(\ker f),\quad i=\ker(\operatorname{coker}f),\quad f=ip,
$$
where the abelian-category axiom identifies coimage with image. Thus $p$ is epic and $i$ is monic. Any other epi-mono factorization $f=jq$ has $\ker q=\ker f$ and $\operatorname{coker}j=\operatorname{coker}f$. Since <epimorphisms> are cokernels of their kernels, its middle object is canonically isomorphic to $\operatorname{im}f$, uniquely compatibly with the two factors.
For a square $vf=f'u$, define $I(u,v):\operatorname{im}f\to\operatorname{im}f'$ by
$$
I(u,v)p=p'u,\qquad i'I(u,v)=vi.
$$
The first arrow exists because $u\ker f$ factors through $\ker f'$, so $p'u$ annihilates $\ker f$. Its composite with $i'$ equals $vi$ after the <epimorphism> $p$, proving the second equation. Uniqueness after $p$ proves preservation of identities and composition. This gives the <functoriality of abelian image factorization> as a <functor> from the <arrow category>.
For <pullback stability of abelian image factorization>, state the standard facts that pullbacks preserve <monomorphisms>, <epimorphisms> in an <abelian category> are stable under pullback, and two adjoining pullback squares have pullback outer rectangle. In the given diagram, $i'$ is therefore monic and $p'$ is epic, while the composite $i'p'$ is the pullback of $f$. Its epi-mono factorization is an image factorization by the uniqueness just proved. Thus the top row is the image factorization of the pulled-back arrow, with its middle object canonically the pullback of the original image subobject.
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