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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.

Let $q:B\to Q$ be the categorical cokernel of $f$ in an abelian category. If $f$ is an epimorphism, the equality $qf=0f$ implies $q=0$. Since $q$ is also an epimorphism, $1_{Q}q=0q$ implies $1_Q=0$. An object with zero identity is a zero object: every morphism to or from it is zero. Hence the cokernel object is zero.

Conversely, suppose the cokernel object is zero. If $uf=vf$, additivity gives $(u-v)f=0$. The categorical cokernel property makes $u-v$ factor through the zero object, so $u-v=0$ and $u=v$. ThusThis is the zero-cokernel criterion for epimorphisms. In a pushout in a category, the two horizontal morphisms have isomorphic cokernels by part (a). Therefore the lower morphism is epic if and only if the upper morphism is epic. In particular pushouts reflect epimorphisms in an abelian category; the same argument also proves preservation.

Use the displayed square's notation $f:A\to B$, $h:A\to C$, $g:B\to D$, $k:C\to D$, with $gf=kh$ and $g$ epic. Form the biproduct and the morphismsThe map $q$ is an epimorphism, since its restriction to $B$ is $g$: equality after $q$ implies equality after $g$. The pullback in a category property says exactly that $j$ is a categorical kernel of $q$. Indeed $qj=gf-kh=0$, and a map into $B\oplus C$ killed by $q$ is a pair $(x,y)$ with $gx=ky$, which factors uniquely through $(f,h)$.

Use the standard abelian category property that every epimorphism is the categorical cokernel of its categorical kernel. If $u:B\to X$ and $v:C\to X$ satisfy $uf=vh$, then $[u,-v]j=0$. Hence there is a unique $w:D\to X$ with $wq=[u,-v]$. Restriction to the two summands gives $wg=u$ and $wk=v$. This is the pushout in a category universal property. Thus a pullback of an epimorphism is a pushout in an abelian category.

In this pushout, $g$ is the pushout of $h$ along $f$. The reflection result of part (b) therefore makes $h$ epic. This proves pullback stability of epimorphisms in an abelian category.

Finally, let $e:E\to B$ be an epimorphism and let $p_1,p_2:R\rightrightarrows E$ be its kernel pair. Their pullback square is a pushout by the result just proved. If $u:E\to X$ satisfies $u{p}_1=u{p}_2$, the two copies of $u$ form a pushout cocone. There is a unique $v:B\to X$ with $ve=u$. Hence epimorphisms in an abelian category are coequalizers of their kernel pairs, so they are regular epimorphisms.

An abelian category is an additive category with a kernel in a category and a cokernel in a category for every morphism, in which every monomorphism is a kernel and every epimorphism is a cokernel. In particular, every epimorphism $p$ is the cokernel of its own kernel: if $p$ is initially a cokernel of $u$, then $u$ factors through $\ker p$, and the two cokernel universal properties agree. The dual statement holds for monomorphisms. Finite biproducts and kernels construct finite limits.

For pullback stability of epimorphisms in an abelian category, let $p:B\to C$ be an epimorphism and $k:C^{\prime }\to C$ any morphism. The mapis epic, since its restriction to the $B$ summand is $p$. Its kernel in a category $i=(h,p^{\prime }):B^{\prime }\to B\oplus C^{\prime }$ exhibits the pullback in a category: the equation $d(h,p^{\prime })=0$ means $ph=k{p}^{\prime }$ and has exactly that universal property.

Suppose $v:C^{\prime }\to D$ satisfies $v{p}^{\prime }=0$. The morphism $[0,v]:B\oplus C^{\prime }\to D$ annihilates $i$, so, because $d$ is the cokernel in a category of $i$, it factors as $[0,v]=wd$ for some $w:C\to D$. Restricting to $B$ gives $wp=0$, and epicity gives $w=0$. Hence $v=0$. Applying this to the difference of any two morphisms agreeing after $p^{\prime }$ proves that $p^{\prime }$ is an epimorphism. ThereforeThis proof uses normality from the definition, rather than assuming the stability to be proved.

For the pullback of a short exact sequence in an abelian category, use the displayed square's notation $g,g^{\prime },h,k$. Exactness makes $f:A\to B$ the kernel in a category of $g$. Since $gf=0=k0$, the pullback in a category gives a uniqueIf $u:X\to B^{\prime }$ has $g^{\prime }u=0$, then $g(hu)=k{g}^{\prime }u=0$, so there is a unique $v:X\to A$ with $fv=hu$. Both projections of the pullback give $f^{\prime }v=u$: their composites with $h$ are equal and their composites with $g^{\prime }$ are zero. The uniqueness of $v$ follows from monicity of $f$. Thus $f^{\prime }$ is the kernel in a category of $g^{\prime }$. By the preceding stability result, $g^{\prime }$ is an epimorphism. We obtain the required exact sequence in an abelian category:

Finally form the pullback in a category $E=P{\times }_C{P}^{\prime }$ of the two projective presentations. Pulling back each of their short exact sequences in an abelian category givesBecause each of $P^{\prime }$ and $P$ is a projective object in a category, the respective final epimorphisms have sections: lift their identity morphisms through those epimorphisms. The usual additive category splitting argument identifies each middle object with the biproduct of its kernel and quotient. Explicitly, a kernel inclusion $j$ and section $s$ give the isomorphism $[j,s]$ from that biproduct to the middle object; $1-sq$ factors through $j$ and provides its inverse's kernel component. Henceand we conclude the Schanuel lemma in an abelian category:

In a pointed category, a normal monomorphism is a monomorphism that is the kernel in a category of some morphism. Suppose $m:M\hookrightarrow A$ is the kernel of $f:A\to B$, and let $q:A\to Q$ be the cokernel in a category of $m$. Since $fm=0$, there is a unique $\bar{f}:Q\to B$ with $f=\bar{f}q$. If $qx=0$, then $fx=\bar{f}qx=0$, so the universal property of $m=\ker f$ factors $x$ uniquely through $m$. Therefore $m=\ker q=\ker (\mathrm{coker}m)$. The converse is immediate: if $m$ is the kernel of its cokernel, it is the kernel of a morphism and hence normal.

An abelian category is an additive category with kernels and cokernels in which every monomorphism is normal and every epimorphism is a conormal epimorphism. Finite biproducts and kernels give finite limits. The image and coimage in an abelian category give every $f:A\to B$ the canonical factorizationand the middle map is an isomorphism. The first map is a cokernel and therefore a regular epimorphism. Every epimorphism in an abelian category is the cokernel of its kernel, and epimorphisms are stable under pullback; consequently regular epimorphisms are pullback-stable. This proves that every abelian category is regular.

Define the additive indexing category for chain complexes $\mathbf{Z}$ as follows. Its objects are the integers andLet the generator of $\mathbf{Z}(n,n)$ be $1_n$ and the generator of $\mathbf{Z}(n,n-1)$ be ${\partial }_n$. Composition is bilinear, the $1_n$ are identities, andbecause the target hom-group $\mathbf{Z}(n,n-2)$ is zero. An additive functor $C:\mathbf{Z}\to \mathcal{A}$ chooses objects $C_n=C(n)$ and differentials $d_n=C({\partial }_n)$ satisfying $d_{n-1}{d}_n=0$, hence a complex in an abelian category. Conversely every chain complex defines this unique additive functor.

For self-duality, put $Z_n=\ker d_n$, $B_n=\mathrm{im}{d}_{n+1}$, and let $q:C_n\to Q_n=\mathrm{coker}{d}_{n+1}$. Since $d_n{d}_{n+1}=0$, there is a unique ${\bar{d}}_n:Q_n\to C_{n-1}$ with ${\bar{d}}_{n}q=d_n$. The image-to-kernel factorization gives a canonical isomorphismPassing to the opposite category exchanges kernels with cokernels and images with coimages. The usual construction in ${\mathcal{A}}^{\mathrm{op}}$ is therefore the expression on the right, which is canonically the original homology object. This proves the self-duality of homology.

An abelian category is an additive category with all kernels and cokernels in which every monomorphism is a kernel and every epimorphism is a cokernel. If $m:A\rightarrowtail B$ is monic and $c=\mathrm{coker}m$, then $m$ factors through $k=\ker c$. The canonical coimage-to-image morphism is an isomorphism in an abelian category, so this factorization identifies $A$ with $\ker c$; hence $m$ is the kernel of its own cokernel. Dually, every epimorphism is the cokernel of its own kernel. The assignments $m\mapsto \mathrm{coker}m$ and $q\mapsto \ker q$ therefore give inverse bijections between subobjects and quotient objects of a fixed object. Well-poweredness is consequently equivalent to well-copoweredness.

The category ${\mathcal{C}}^{\mathcal{A}}$ of complexes has chain complexes as objects and chain maps as morphisms. Write $Z_n=\ker d_n$ and $Q_n=\mathrm{coker}{d}_{n+1}$. The equation $d_n{d}_{n+1}=0$ gives both an induced map $\mathrm{im}{d}_{n+1}\to Z_n$ and a map $Q_n\to \mathrm{im}{d}_n$. The homology object has the two canonically isomorphic descriptionsPassing to the opposite category exchanges these descriptions, proving self-duality.

Let $S_n(A)=A[n]$. A chain map $A[n]\to C_{\bullet }$ is exactly a map $A\to Z_n(C)$, while a chain map $C_{\bullet }\to A[n]$ is exactly a map $Q_n(C)\to A$. Therefore

The Snake lemma states that a commutative diagram with exact rows yields the exact sequence of the three kernels, followed by its connecting morphism and the three cokernels. Apply it degree by degree to a short exact sequence of complexes $0\to A_{\bullet }\to B_{\bullet }\to C_{\bullet }\to 0$, using the diagrams of cycles, boundaries, and degree objects. The connecting map sends a cycle of $C_n$ to a lift in $B_n$, takes its boundary in $B_{n-1}$, and identifies that boundary with a class in $H_{n-1}(A)$. The Snake lemma gives exactness and producesthe algebraic Mayer–Vietoris theorem.

In an abelian category with enough projectives, a projective object has a resolution concentrated in degree zero, so every first left derived functor vanishes on it. Conversely choose $0\to K\to P\to A\to 0$ with $P$ projective and use the additive functor $F=\mathcal{A}(-,K)$ with target the opposite of the category of abelian groups. This functor is right exact in that target. If $L_{1}F(A)=0$, the derived exact sequence makes $F(K)\to F(P)$ monic, meaning that $\mathrm{Hom}(P,K)\to \mathrm{Hom}(K,K)$ is surjective in abelian groups. Extending $1_K$ gives a retraction of $K\to P$, so the sequence splits and $A$ is a direct summand of the projective $P$.

In a pointed category with categorical kernels and categorical cokernels, where every monomorphism is normal, every $h$ factors through $i=\ker (\mathrm{coker}h)$ as $h=ie$, with $e$ a pseudo-epimorphism. If $te=0$, write $k=\ker t$. Then $e$ factors through $k$, and the monic $ik$ is a categorical kernel of some $q$ by normality. Since $qh=0$, the cokernel of $h$ forces $qi=0$. Hence $i$ factors through $ik$, forcing $k$ to be an isomorphism and therefore $t=0$. This establishes the factorization without assuming the stronger abelian category axioms.

In an abelian category, suppose a commutative square $gf=kh$ is a pullback in a category and $g$ is an epimorphism. Then $(f,h):A\to B\oplus C$ is the categorical kernel of the epimorphism $[g,-k]$. Since an epimorphism is the cokernel of its kernel, any compatible pair $u:B\to X$, $v:C\to X$ induces a unique map $D\to X$ by factoring $[u,-v]$. Hence the square is also a pushout in a category.

In an abelian category, the pullback in a category of an epimorphism $p:B\to C$ along $k:C^{\prime }\to C$ is the kernel of the epimorphism $[p,-k]:B\oplus C^{\prime }\to C$. If $v:C^{\prime }\to D$ annihilates its projection, $[0,v]$ annihilates that kernel and therefore factors through $[p,-k]$. Restriction to $B$ forces the factor to vanish, giving $v=0$ and proving the projection epic. This uses the definition's normality of epimorphisms.

An additive functor between abelian categories is right exact functor when it sends every exact sequence $X\to Y\to Z\to 0$ to an exact sequence $FX\to FY\to FZ\to 0$. Equivalently, it preserves cokernels in a category, since additivity already preserves finite biproducts. This is the convention used in constructing left derived functors. Left exactness preserves the initial part $0\to X\to Y\to Z$ instead. Contravariant Hom becomes a right exact functor covariant functor with target the opposite abelian category.

An additive functor between abelian categories is right exact when it preserves categorical cokernels, equivalently when it sends every exact sequence ending in zero to an exact sequence ending in zero. Additivity and preservation of cokernels in a category imply preservation of finite colimits. The additivity assumption is part of the ordinary left derived functor construction.

Two projective presentations of the same object in an abelian category, with kernels $K,K^{\prime }$ and middle objects $P,P^{\prime }$, satisfy $K\oplus P^{\prime }\cong K^{\prime }\oplus P$. Form their pullback in a category. Its two induced short exact sequences in an abelian category split because each quotient is a projective object in a category. The two resulting biproduct descriptions of the pullback give the isomorphism.

Let $q:C_n\to \mathrm{coker}{d}_{n+1}$. Since $d_n{d}_{n+1}=0$, there is an induced ${\bar{d}}_n:\mathrm{coker}{d}_{n+1}\to C_{n-1}$. In an abelian category,The left side is the usual homology construction and the right side is its construction in the opposite category, so the definition of homology is self-dual.

In an abelian category, a morphism is an epimorphism if and only if its categorical cokernel object is zero. For the reverse implication, any difference of maps annihilating the morphism factors through its cokernel and is therefore zero. The cokernel invariance under pushout then proves both preservation and reflection of epimorphisms by pushout.