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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.

A preadditive category has an abelian-group structure on every hom-set and bilinear composition. One equivalent definition of an additive category is a preadditive category with a zero object and finite products in a category; the following argument proves that these products in a category are automatically coproducts in a category. An abelian category is an additive category with kernels and cokernels in which every monomorphism is a kernel and every epimorphism is a cokernel. Equivalently, for every arrow the canonical map from its coimage to its image is invertible.

Let $P={\prod }_{i=1}^n{A}_i$, with projections $p_i$. Define ${\iota }_i:A_i\to P$ by $p_j{\iota }_i={\delta }_{ji}$, where the diagonal entry is the identity and the others are zero. Since its composites with all projections are identities,Given $f_i:A_i\to B$, the map $f={\sum }_i{f}_i{p}_i$ satisfies $f{\iota }_i=f_i$. If another map has those composites, multiplying the displayed identity by that map forces it to equal $f$. Thus $P$ is also a coproduct in a category. The empty case is the zero object. Reversing arrows proves that a finite coproduct in a category is likewise a product in a category. Hence finite products in a category and coproducts in a category are canonically biproducts in an additive category.

For the category of abelian groups, let $(A_i,D_{ij})$ be a directed diagram, with no injectivity assumption on its transition maps. Its set colimit consists of classes $[i,a]$, withDirectedness proves transitivity of this relation. Define addition by moving representatives to a common upper index:Moving all indices involved to one further upper bound proves independence of representatives and of $k$. Define $-[i,a]=[i,-a]$ and use the common class of the zeros as zero. The abelian-group laws hold at a common index, so descend to the quotient. Every canonical map $A_i\to C$ is a homomorphism, and a compatible cocone induces the homomorphism $[i,a]\mapsto f_i(a)$. This is the unique extension. Moreover the formula for addition is forced by those canonical homomorphisms, proving that the underlying-set functor creates directed colimits. This establishes directed colimits created by the abelian-group forgetful functor explicitly.

If a cocone ${\nu }_i:A_i\to B$ has injective legs, the induced map $C\to B$ is injective. Indeed, equal images of $[i,a]$ and $[j,b]$ give, at a common index $k$,Injectivity of ${\nu }_k$ gives equal representatives at $k$. Since monomorphisms of abelian groups are exactly injective homomorphisms, we have proved that the category of abelian groups is a finitary abelian category in the directed-union sense of the question.

Now let $\mathcal{A}$ be complete, cocomplete and finitary. Index by finite subsets $J\subseteq I$ and put $S_J={⨁}_{j\in J}{A}_j$. For $J\subseteq K$, insert the additional zero coordinates. A compatible cocone on these finite biproducts is exactly a family of maps from every $A_i$, so their directed colimit is $S={\coprod }_{i\in I}{A}_i$.

Each finite biproduct has a map $k_J:S_J\to P={\prod }_{i\in I}{A}_i$ using its coordinates in $J$ and zeros outside $J$. Projection to the finite set of coordinates in $J$ is a retraction, so $k_J$ is a split monomorphism. These maps form a cocone of monomorphisms. The finitary property makes its induced map a monic arrow, and that induced map is exactly the canonical infinite identity matrix:This proves the coproduct-to-product comparison in a finitary abelian category. The PDF's use of finite products in a category in the hint is valid because these are the same finite biproducts.

If $\mathcal{A}$ is also a cofinitary abelian category, apply this argument in its opposite category. The same canonical comparison $j$ is then an epic arrow in $\mathcal{A}$. In an abelian category an arrow that is both a monomorphism and an epimorphism is invertible, so $j$ is an isomorphism.

For countably many copies of $A$, put $S={\coprod }_{n\ge 0}A$ and let $q=j^{-1}\Delta :A\to S$, where $\Delta$ is the product in a category diagonal. Let ${\iota }_n$ denote the coproduct in a category injections, let $s:S\to S$ shift them by $s{\iota }_n={\iota }_{n+1}$, and let $\nabla :S\to A$ be the codiagonal. Comparing every product in a category coordinate givesSince $j$ is a monic arrow, $q={\iota }_0+sq$. Also $\nabla {\iota }_0=1_A$ and $\nabla s=\nabla$. Therefore, for $z=\nabla q$,Hom-sets are abelian groups, so cancellation gives $1_A=0$. Any arrow to or from $A$ is consequently zero by composition with its identity. Thus every object is both initial and terminal, andThis is the countable shift-and-fold argument behind countable biproducts in an additive category force triviality.

A semi-additive category has commutative-monoid enrichment: each hom-set is a commutative monoid, and composition in a category preserves addition and zero in each variable. In the finite-biproduct convention, which we use here, it also has finite products in a category and coproducts in a category. A preadditive category has abelian groups as its hom-sets, with composition additive in each variable; this definition alone does not require finite products in a category or a zero object.

Let $P={\prod }_{i=1}^n{A}_i$ with projections $p_i$. Define $i_j:A_j\to P$ by $p_i{i}_j=1_{A_j}$ for $i=j$ and $p_i{i}_j=0$ otherwise. Bilinearity givesso product uniqueness gives ${\sum }_j{i}_j{p}_j=1_P$. For a family $f_j:A_j\to X$, set $f={\sum }_j{f}_j{p}_j$. Then $f{i}_j=f_j$. If $h:P\to X$ has the same restrictions, $h=h{\sum }_j{i}_j{p}_j={\sum }_j{f}_j{p}_j=f$. Thus the product in a category is also the coproduct in a category, with its canonical injections. Conversely, the analogous construction from a finite coproduct in a category gives projections and the same identities, proving that it is a product in a category.

For the empty family, a terminal object $T$ has a singleton endomorphism monoid, so $1_T=0$. Every map $T\to X$ equals its composite with $1_T=0$, hence equals the zero map; this map exists by enrichment. Therefore $T$ is also an initial object, a zero object. The dual argument applies to an initial object. Consequently finite products in a category and coproducts in a category coincide canonically as biproducts, including the nullary case.

Fix an object $C$. For $x:C\to B$, use $rx:C\to A$ as its identity arrow. For $a:C\to A$, its source is $fa$ and target is $ga$. If $ga=fb$, define the composite of $a$ followed by $b$ byThese expressions use addition and subtraction in the hom-set of the preadditive category. The reflexive pair identities $fr=gr=1_B$ giveso the composite has the required endpoints. Also $f(a^{-1})=ga$ and $g(a^{-1})=fa$.

The unit identities follow directly: $a\circ rfa=rfa+a-rfa=a$ and $rga\circ a=a+rga-rga=a$. If $ga=fb$ and $gb=fc$, either parenthesization of three arrows equals $a+b+c-rga-rgb$, so composition is associative. The inverse identities areTherefore these operations define a groupoid on the specified object and arrow sets. Precomposing by $k:C^{\prime }\to C$ respects each expression, by additivity of composition, so the groupoids are natural in $C$.

Every reflexive pair therefore has the required groupoid structure on generalized points. This is the reflexive-pair groupoid formula in a preadditive category. If the composable-arrow pullback in a category $A{\times }_{g,B,f}A$ exists, its projections $p,q$ give the actual internal composition map $p+q-rgp$. The printed generalized-point definition does not assume that this pullback exists, and preadditivity alone would not ensure it.

A preadditive category has an abelian group structure on every hom-set, with composition additive in each variable. Neither a zero object nor biproducts are part of this definition.

Fix $C$ and use the reflexive pair $f,g:A\rightrightarrows B$, $r:B\to A$, with $fr=gr=1_B$. Take objects $x:C\to B$ and arrows $a:C\to A$, with source $fa$, target $ga$, and identity at $x$ equal to $rx$. For composable $a,b$, so $ga=fb$, defineThe preadditive category axioms giveThus the formula has the required endpoints. The identities satisfy $(rga)\circ a=a$ and $a\circ (rfa)=a$, using $gr=fr=1_B$.

For $ga=fb$ and $gb=fc$, both ways of composing three arrows equalIndeed $g(b\circ a)=gb$, while expanding $c\circ b$ and then composing with $a$ gives the same expression. Hence composition is associative.

Every arrow has inverseIts source is $ga$ and its target is $fa$. Substituting in the composition formula gives $a^{-1}\circ a=rfa$ and $a\circ a^{-1}=rga$. Therefore this is a groupoid.

For $u:C^{\prime }\to C$, precomposition by $u$ preserves sources, targets, identities, composition and inverses by bilinearity. Thus the construction is natural in $C$, giving the requested internal groupoid structure in its hom-set formulation. If the composable-arrow pullback in a category exists, the same formula defines its internal composition morphism. The reflexive-pair groupoid formula in a preadditive category requires no extra additive-category hypotheses.

A morphism $e:A\to B$ in a pointed category is pseudo-epic when, for every $t:B\to Z$, the equation $te=0$ forces $t=0$. If a categorical cokernel exists, this is equivalent to that cokernel having a zero object as codomain. Indeed all zero composites then factor uniquely through zero; conversely the cokernel itself must be zero, and being both zero and epic forces its codomain to have identity zero, hence be a zero object. Every epimorphism is pseudo-epic. In a preadditive category, subtraction shows the converse: $ue=ve$ implies $(u-v)e=0$, hence $u=v$. In a general pointed category no subtraction is available.

For a reflexive pair $f,g:A\rightrightarrows B$ with section $r$ in a preadditive category, each hom-set diagram defines a groupoid. Arrows $a,b:C\to A$ compose when $ga=fb$, by $b\circ a=a+b-rga$. Identity at $x:C\to B$ is $rx$, and the inverse of $a$ is $rfa+rga-a$. Bilinearity proves the endpoint, identity and associativity laws and compatibility with precomposition in $C$. In the hom-set formulation this does not assume that the composable-arrow pullback exists.

A semi-additive category has a commutative-monoid enrichment and finite products in a category and coproducts in a category. Composition preserves addition and zero in each variable. Every finite product in a category is canonically a coproduct in a category and conversely, producing finite biproducts. Some authors use "semi-additive structure" for the commutative-monoid enrichment alone; that weaker convention also permits one-object categories without finite products in a category.

A preadditive category instead has abelian-group-valued hom sets. Semi-additivity alone does not require additive inverses.