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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.

A preadditive category has an abelian group of morphisms between each pair of objects, and composition is additive in both variables. An additive category is a preadditive category with a zero object and finite products in a category; equivalently one may require finite coproducts. The equivalence, and the stronger assertion that these constructions agree canonically, follow from the argument below. An abelian category is an additive category with kernels in a category and cokernels in a category, in which every monomorphism is a categorical kernel and every epimorphism is a categorical cokernel. A monomorphism with the first property is a normal monomorphism; an epimorphism with the second is a conormal epimorphism.

Let $P={\prod }_{i=1}^n{A}_i$, with projections $p_i$. There is a unique $i_j:A_j\to P$ with $p_i{i}_j={\delta }_{ij}{1}_{A_j}$, interpreting the other components as zero. Since every $p_j$ sends ${\sum }_i{i}_i{p}_i$ to $p_j$, the product's universal property givesGiven arrows $f_i:A_i\to B$, put $f={\sum }_i{f}_i{p}_i$. Then $f{i}_j=f_j$. Conversely any arrow with these restrictions equals $f{1}_P={\sum }_{i}f{i}_i{p}_i$, so it is uniquely this sum. Thus $(P,i_i)$ is a coproduct in a category. The empty product is the zero object and also the empty coproduct. The dual argument starts from coproducts and constructs projections; it proves the equivalent definition. Hence finite products and coproducts are the same biproducts, with the displayed identities giving the canonical identification.

For the image factorization in an abelian category, let $h:A\to B$, let $c:B\to Q$ be its cokernel, and let $i:I\to B$ be the kernel of $c$. Since $ch=0$, write $h=ie$. We prove that $e:A\to I$ is epic. Suppose $t:I\to Z$ satisfies $te=0$ and let $l:L\to I$ be its categorical kernel. Then $e=lr$ for some $r:A\to L$. The composite $il$ is monic and hence is a normal monomorphism, say $il=\ker z$ for $z:B\to W$. Since $zh=zilr=0$, the categorical cokernel property makes $z$ factor through $c$, and therefore $zi=0$. The categorical kernel property of $il$ gives $s:I\to L$ with $ils=i$. Cancellation of $i$ yields $ls=1_I$; since $l$ is monic this also gives $sl=1_L$. Thus $l$ is invertible and $t=0$. In a preadditive category, $ue=ve$ now implies $(u-v)e=0$, hence $u=v$. This proves $e$ is an epimorphism and supplies the required factorization.

For uniqueness, an epimorphism that is a categorical cokernel is the categorical cokernel of its own categorical kernel. Indeed, if $e=\mathrm{coker}r$ and $k=\ker e$, then $r$ factors through $k$; an arrow killing $k$ therefore kills $r$ and factors uniquely through $e$. Dually, a normal monomorphism is the categorical kernel of its own categorical cokernel. If $h=mp$ is any epi–mono factorization, monicity of $m$ gives $\ker p=\ker h$. Both $p$ and $e$ are consequently categorical cokernels of this same categorical kernel. Their universal properties give a unique isomorphism $\theta$ between the intermediate objects with $\theta p=e$, and cancellation of $p$ gives $i\theta =m$. Therefore the epi–mono factorization is unique up to the unique compatible isomorphism.

Here is an additive category with truncated primary torsion, with the prime fixed as $2$. Its objects are finitely generated abelian groups having no element of order $4$, and its morphisms are all group homomorphisms. Equivalently their $2$-primary torsion subgroup has exponent at most $2$. It is a full subcategory of abelian groups, closed under finite direct sums and subgroups. Its hom groups and bilinear composition are inherited, and the zero group and finite direct sums belong to it, so it is additive. Categorical kernels are ordinary subgroup categorical kernels, which are still finitely generated and have no element of order $4$.

To construct categorical cokernels, for an arbitrary finitely generated abelian group $G$ let $t_2(G)$ be its $2$-primary torsion subgroup, and defineThe Fundamental theorem of finitely generated abelian groups shows that this quotient lies in the subcategory: it leaves the free and odd-primary parts unchanged and replaces each cyclic $2$-primary summand by either $0$ or $\mathbb{Z}/2\mathbb{Z}$. Every homomorphism $G\to H$ to an allowed group kills $2t_2(G)$, because it sends $2$-primary torsion into a subgroup killed by $2$. Thus $G\to R(G)$ has the universal property of a reflection. For $f:A\to B$, its categorical cokernel in the subcategory isThis has exactly the categorical cokernel universal property by the preceding factorization. Finite categorical limits exist because equalizers are categorical kernels of differences and finite products exist. Finite colimits exist dually because coequalizers are categorical cokernels of differences and finite coproducts exist.

The epimorphisms in this example are precisely the surjective homomorphisms. A surjection is epic already in abelian groups. Conversely, if $f$ is epic in the subcategory, its categorical cokernel is zero, so $R(B/f(A))=0$. A nonzero finitely generated abelian group has nonzero reflection: a nonzero free or odd-primary summand survives, and a nonzero finite $2$-primary summand has a nonzero quotient modulo twice itself. Hence $B/f(A)=0$ and $f$ is surjective. Its ordinary categorical kernel belongs to the subcategory, and its quotient by that categorical kernel is its codomain, which also belongs to the subcategory. It is therefore the categorical cokernel of its categorical kernel. This proves that every epimorphism is conormal.

The monomorphismis not normal. Its categorical cokernel in this category is reduction modulo $2$, since $R(\mathbb{Z}/4\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$. The categorical kernel of that categorical cokernel is the inclusion $2\mathbb{Z}\hookrightarrow \mathbb{Z}$, whereas the image of $m$ is $4\mathbb{Z}$. A normal monomorphism is the categorical kernel of its own categorical cokernel, as proved above, so $m$ cannot be normal. In fact multiplication by $2$ is itself the categorical kernel of reduction modulo $2$, up to the identification $\mathbb{Z}\cong 2\mathbb{Z}$. Its composite with itself is multiplication by $4$. Thus this example already exhibits two normal monomorphisms whose composite is not normal.

For the general assertion, let $m:A\to B$ be a nonnormal monomorphism in any additive category of the specified kind. Put $c=\mathrm{coker}m$ and $k:K\to B$ be its categorical kernel, and factor $m=kg$. Both $m$ and $k$ are monic, so $g$ is monic. Set $d=\mathrm{coker}g$ and $l:L\to K$ be its categorical kernel. Then $k,l$ are normal, and $g=lr$ for some $r$.

Suppose their composite were normal, say $kl=\ker t$. Since $tm=tklr=0$, $t$ factors through $c$, whence $tk=0$. The categorical kernel property of $kl$ supplies $s:K\to L$ with $kls=k$; cancellation shows $ls=1_K$, and monicity makes $l$ an isomorphism. Because $dl=0$, this forces $d=0$. In a preadditive category, vanishing of the categorical cokernel of $g$ implies that $g$ is epic: any $u,v$ with $ug=vg$ have $u-v$ factor through the zero categorical cokernel. By hypothesis $g$ is therefore a categorical cokernel, and it is also monic. If $g=\mathrm{coker}a$, monicity and $ga=0$ force $a=0$; a categorical cokernel of a zero arrow is an isomorphism. Thus $g$ is invertible, making $m=kg$ normal, a contradiction. Consequently normal monomorphisms cannot be closed under composition in any such category; the explicit pair $k,l$ witnesses the failure.

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.