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In an abelian category, the coimage of a morphism is its domain modulo its kernel in a category, expressed as a cokernel in a category. The canonical arrow from the coimage to the image is an isomorphism, giving the image factorization in an abelian category.

For commuting five-term exact sequences in an abelian category, the first vertical map being epic and the second invertible give an invertible induced map on the images entering the middle objects, because these images are the cokernels of the first horizontal maps. The fourth vertical map being invertible and the fifth monic similarly give an invertible map on the images leaving the middle objects, because these are the kernels of the last horizontal maps. The image factorization in an abelian category packages each middle object between these two images in a short exact sequence. The short five lemma then makes the middle vertical map invertible. The proof is categorical and does not require an embedding into a category of modules.

In an abelian category, for $f:A\to B$ set $\mathrm{coim}f=\mathrm{coker}(\ker f)$ and $\mathrm{im}f=\ker (\mathrm{coker}f)$. The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.

The short five lemma states that in a commuting diagram of short exact sequences $0\to L\to M\to R\to 0$ and $0\to L^{\prime }\to M^{\prime }\to R^{\prime }\to 0$ in an abelian category, if the vertical maps $L\to L^{\prime }$ and $R\to R^{\prime }$ are isomorphisms, then so is $M\to M^{\prime }$.

For the Five lemma, consider commuting exact rowswith vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.

Put $L=\mathrm{im}{d}_2$ and $R=\mathrm{im}{d}_3$, and define $L^{\prime },R^{\prime }$ similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.

First, $L\cong \mathrm{coker}{d}_1$ and $L^{\prime }\cong \mathrm{coker}{d}_1^{\prime }$: exactness at $A_2$ identifies $\mathrm{im}{d}_1$ with $\ker d_2$, and the image-coimage isomorphism for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q^{\prime }:B_2\to L^{\prime }$ for these categorical cokernels. To construct an inverse to the induced map $\ell :L\to L^{\prime }$, observeCancellation of the epimorphism $v_1$ gives $q{v}_2^{-1}{d}_1^{\prime }=0$, so $q{v}_2^{-1}$ factors uniquely as $\bar{\ell }{q}^{\prime }$ with $\bar{\ell }:L^{\prime }\to L$. From $\ell q=q^{\prime }{v}_2$, composing with the epic $q,q^{\prime }$ gives $\bar{\ell }\ell =1_L$ and $\ell \bar{\ell }=1_{L^{\prime }}$.

Second, $R\cong \ker d_4$ and $R^{\prime }\cong \ker d_4^{\prime }$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j^{\prime }:R^{\prime }\to B_4$. The equality $d_4^{\prime }{v}_4=v_5{d}_4$ gives the induced map $\rho :R\to R^{\prime }$. Sinceand $v_5$ is a monomorphism, $d_4{v}_4^{-1}{j}^{\prime }=0$. Kernel universality gives $\bar{\rho }:R^{\prime }\to R$ with $j\bar{\rho }=v_4^{-1}{j}^{\prime }$. Cancelling the monic $j,j^{\prime }$ proves $\bar{\rho }\rho =1_R$ and $\rho \bar{\rho }=1_{R^{\prime }}$.

Apply the short five lemma to the two short exact sequences: their outer maps $\ell ,\rho$ are invertible, so $v_3$ is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.

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.

Pull back the two arrows of an image factorization in an abelian category. Monomorphisms are preserved by pullback, and pullback stability of epimorphisms in an abelian category preserves its epic part. Pasting the squares gives the pulled-back composite. The resulting epi-mono factorization is therefore its image factorization by uniqueness.