 Solution

In an abelian category, the image factorization in an abelian category of $f:A\to B$ iswhere 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 $\mathrm{coker}j=\mathrm{coker}f$. Since epimorphisms are cokernels of their kernels, its middle object is canonically isomorphic to $\mathrm{im}f$, uniquely compatibly with the two factors.

For a square $vf=f^{\prime }u$, define $I(u,v):\mathrm{im}f\to \mathrm{im}{f}^{\prime }$ byThe first arrow exists because $u\ker f$ factors through $\ker f^{\prime }$, so $p^{\prime }u$ annihilates $\ker f$. Its composite with $i^{\prime }$ 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, $i^{\prime }$ is therefore monic and $p^{\prime }$ is epic, while the composite $i^{\prime }{p}^{\prime }$ 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.