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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.

A semi-additive category is a category whose hom-sets are commutative monoids and whose composition is additive in each variable, with finite products and coproducts.

Suppose first that $P=A\times B$ is a binary product in a category, with projections $p_1,p_2$. The zero morphisms and the product property define mapsbyThe two projections of $i_1{p}_1+i_2{p}_2$ equal those of $1_P$, soFor $f:A\to C$ and $g:B\to C$, the mapsatisfies $h{i}_1=f$ and $h{i}_2=g$. If $k$ has the same restrictions, thenThus $(P,i_1,i_2)$ is also the binary coproduct. The dual argument starts from a coproduct and makes it a product. Hence binary products and coproducts coincide canonically as biproducts.

Let $f,g:A\rightrightarrows B$ be a reflexive pair in an additive category, with $fr=gr=1_B$. For every object $C$, regard $x:C\to A$ as an arrow from $fx$ to $gx$ between objects of $\mathcal{C}(C,B)$. The identity at $b:C\to B$ is $rb$.

If $gx=fy$, define the composite byIts source and target areThe identities follow fromand associativity follows immediately by expanding both iterated composites and using the matching equations. The inverse of $x$ iswhose source is $gx$, whose target is $fx$, and whose two composites with $x$ are the appropriate identity arrows. These formulas are natural in $C$, so the Yoneda lemma identifies them with structure morphisms in $\mathcal{C}$. The pair is therefore an internal groupoid, proving that every reflexive pair in an additive category is an internal groupoid.

This fails for semi-additive categories. In the category of commutative monoids, letunder coordinatewise addition. The two projections $f,g:R\rightrightarrows \mathbb{N}$ have the common splitting $r(n)=(n,n)$, so they form a reflexive pair. Its underlying reflexive graph is the usual order category on $\mathbb{N}$: there is an arrow $m\to n$ exactly when $m\le n$. If it were an internal groupoid, the arrow $0\to 1$ would have an inverse $1\to 0$, but $(1,0)\notin R$. Therefore this reflexive pair is not an internal groupoid, and “additive” cannot be weakened to “semi-additive.”

A semi-additive category is enriched in commutative monoids: each hom-set has a zero and addition, and composition is additive in each variable. It also has finite products and coproducts. Every finite product is canonically a coproduct and conversely, producing finite biproducts.