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This category has commutative monoids and their homomorphisms. Pointwise addition makes each hom-set a commutative monoid, and composition is additive in both variables. Finite cartesian products are also coproducts, using the coordinate injections with zero in the other coordinates and summing images. Therefore it is a semi-additive category, but it is not an additive category. The reflexive pair from the submonoid $\{(m,n):m\le n\le 2m\}$ of ${\mathbb{N}}^2$ to $\mathbb{N}$, using the two projections and section $m\mapsto (m,m)$, has an arrow from $1$ to $2$ and none from $2$ to $1$ in its represented graph. Thus a reflexive pair here need not yield a groupoid.

In an additive category with finite categorical limits and colimits in which every epimorphism is a conormal epimorphism, closure of normal monomorphisms under composition forces every monomorphism to be normal. To prove this, write a monomorphism $m$ as $kg$, where $k=\ker \mathrm{coker}m$, and put $l=\ker \mathrm{coker}g$. If $kl=\ker t$ were normal, $tm=0$ would force $tk=0$, hence $l$ would be an isomorphism. This makes $\mathrm{coker}g=0$, so $g$ is epic as well as monic. An epic monomorphism which is a cokernel is invertible, forcing $m$ to be normal. Contrapositively, any failure of normality provides two normal monomorphisms whose composite is not normal.

Suppose a semi-additive category has countable products and coproducts and the canonical comparison between them is always invertible. For countably many copies of $A$, let $q:A\to {\coprod }_{i}A$ correspond under that comparison to the diagonal into the product, and let $\sigma :{\coprod }_{i}A\to A$ be the fold map. If $s$ shifts the summands by one, coordinatewise comparison proves $q={\nu }_1+sq$ and $\sigma s=\sigma$. Thus $z=\sigma q$ satisfies $z=1_A+z$. In an additive category, cancellation gives $1_A=0$, forcing every hom-set to contain only its zero morphism. The category is equivalent to the one-object, one-arrow category.

An essay arguing against the claim. A good definition can change which mathematical questions are visible. Nevertheless, I would argue that category theory derives its explanatory force from the interaction between definitions and theorems; there is no sound general ordering in which definitions contribute more than theorems.

The strongest case for the contrary view comes from universal properties. A product in a category is characterized by maps into it, rather than by a particular construction of ordered pairs. The same definition applies to sets, groups, and topological spaces. An adjunction goes further: its natural correspondence of arrows puts free constructions, quotient constructions, and several kinds of completion into a common language. Such definitions remove irrelevant choices and expose relationships that would be difficult to recognize from separate constructions.

But a universal-property definition does not supply an object. The definition of a product in a category says what a product would accomplish; an existence theorem says whether the category actually has one. Even uniqueness up to unique isomorphism is a short proof, not an extra clause to be silently supplied by the word “universal”. Given two products, their universal properties produce maps in both directions, and uniqueness forces the two composites to be identities. This elementary theorem makes the definition usable independently of a chosen construction.

The Yoneda lemma illustrates the distinction especially clearly. A representable functor is a definition, but the lemma proves that its natural transformations into any set-valued functor are determined by one element. Its proof evaluates at the identity and reconstructs the whole transformation by functoriality. Consequently the Yoneda embedding is full and faithful: an object's pattern of incoming arrows recovers its morphisms without loss. That is a substantive explanation of why the chosen definition is powerful. Merely calling a functor representable would not establish it.

Similarly, defining a monadic adjunction gives a precise meaning to recovering structured objects as algebras for a monad. It does not decide whether a given forgetful functor is monadic. The precise monadicity theorem translates that difficult classification into a usable test involving adjoint functors, reflected isomorphisms, and specific coequalizers. For compact Hausdorff spaces, the crucial step is proving that a set-split relation is closed. The definition of a monad alone does not contain this topological argument. The theorem connects the algebraic language to an independently meaningful class of spaces.

There is also a danger in praising definitions without hypotheses. A Cartesian closed category has exponential objects, but a category of structured objects need not be cartesian closed merely because underlying sets have function sets. The correct exponential in a functor category generally involves natural transformations out of representable functors, not pointwise function sets. Its existence and universal property require an argument. Theorems also reveal when tempting definitions cannot support the intended construction.

Definitions provide the vocabulary and select the structures worth studying; proofs establish existence, consequences, and limitations. Category theory makes this partnership unusually explicit because its definitions are often expressed through universal properties. Its distinctive achievement is to turn well-chosen definitions into reusable theorems, rather than to make theorems secondary.

An alternative essay arguing for the claim. In category theory, theorems are indispensable, but a carefully chosen definition often performs the greater act of organization. It identifies what many apparently unrelated constructions have in common and makes whole families of proofs possible. In that explanatory sense, definitions can be the subject's most influential contributions.

Consider a universal property. Describing a product in a category as an object representing pairs of incoming arrows replaces several different concrete constructions by one invariant pattern. The corresponding uniqueness proof is important but short; the lasting insight is that the same pattern is the right question in every category. Once the definition is available, it becomes meaningful to ask which functors preserve products, whether the products exist in a new category, and how they interact with other universal constructions.

An adjunction is an even more striking example. The existence of free groups, free modules, and a Stone-Čech compactification can be proved separately. The definition of an adjunction recognizes their common relation to forgetful functors and explains why arrows out of a free object are determined by arrows out of its generators. It also packages the relation as a monad. A large collection of isolated existence results becomes an instance of a shared structure, with reusable consequences about composition and categorical limits.

A valuable definition also has to distinguish structures rather than merely rename them. An abelian category is not just an additive category with categorical kernels and categorical cokernels: normality of monomorphisms and epimorphisms matters. The additive example with truncated primary torsion shows why those clauses belong in the definition. The definition therefore preserves exactly the structure on which theorems about images and exact sequences depend.

This view does not make proofs dispensable. The Yoneda lemma justifies the use of representable functors, and the precise monadicity theorem tests when a proposed algebraic description is valid. Rather, it argues that the best categorical definitions have unusually broad explanatory reach: they select the invariant structures to which those proofs apply. If “matter more” means providing the reusable framework in which many theorems become instances of one idea, category-theoretic definitions have a strong claim to that role.

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.

An abelian category is an additive category with kernels in a category and cokernels in a category in which the canonical comparison from coimage to image of every morphism is an isomorphism. This is the standard image-coimage form of the definition, equivalently expressed by normality of every monomorphism and epimorphism. Additivity means abelian-group hom-sets with bilinear composition, a zero object and finite biproducts. A sequence is exact at an object when the image of the incoming morphism is the kernel in a category of the outgoing one.

For $f:X\to Y$, construct $q:X\to \mathrm{coim}f=\mathrm{coker}(\ker f)$ and $j:\mathrm{im}f=\ker (\mathrm{coker}f)\to Y$. The universal properties give $f=j\bar{f}q$: first $f$ kills its kernel in a category and factors through $q$; then the resulting map is killed by $\mathrm{coker}f$ because $q$ is epic, so it factors through $j$. The comparison $\bar{f}$ is invertible by the abelian axiom. Cokernels in a category are epic and kernels in a category monic, directly by uniqueness in their universal properties, so $e=\bar{f}q$ is epic and $j$ monic. Thus $\boxed{f=je\text{ is an epimorphism-monomorphism factorization}}$.

In a short exact sequence $0\to X\stackrel{i}{\to }Y\stackrel{p}{\to }Z\to 0$, we have $i=\ker p$ and $p=\mathrm{coker}i$, up to the canonical isomorphisms. If $r:Y\to X$ retracts $i$, the map $1_Y-ir$ kills $i$ and factors as $sp$ for some $s:Z\to Y$. Composing with $p$ gives $psp=p$, so epicity gives $ps=1_Z$. Conversely if $ps=1_Z$, then $1_Y-sp$ is killed by $p$ and factors as $ir$. Composing with $i$ gives $iri=i$, so monicity gives $ri=1_X$. ConsequentlyIn that case $Y\cong X\oplus Z$.

For the derived-functor requests, use additive right exact functors, as customary in abelian homological algebra. An abelian category with enough projectives permits an augmented projective resolution $\cdots \to P_2\to P_1\to P_0\to X\to 0$, obtained by repeatedly taking an epimorphism from a projective onto the previous kernel in a category. Apply $F$ and defineProjective lifts construct maps of resolutions; the comparison and chain-homotopy results make the construction independent of choices and functorial. For a short exact sequence, the horseshoe construction produces a degreewise split short exact sequence of resolutions, hence of their images under additive $F$. Its homology objects give the long exact sequenceThis supplies the requested sketch; detailed comparison and functoriality proofs are not needed here.

A projective object $X$ has a resolution concentrated in degree zero, so $L_{1}F(X)=0$ for every such $F$. Conversely choose $0\to K\stackrel{i}{\to }P\to X\to 0$ with $P$ projective, and take $F=\mathcal{A}(-,K)$ with target ${\mathbf{Ab}}^{\mathrm{op}}$. Precomposition makes this covariant to the opposite category. The usual left exactness of contravariant Hom, which follows directly from the cokernel in a category universal property, says exactly that this $F$ is right exact in that opposite target. If $L_{1}F(X)=0$, the long exact sequence and $L_{1}F(P)=0$ make $F(K)\to F(P)$ monic in ${\mathbf{Ab}}^{\mathrm{op}}$. Equivalently,is epic, hence surjective, in abelian groups. The identity of $K$ has a preimage, giving a retraction of $i$. The sequence splits, and $X$ is a direct summand of $P$, hence projective: a lifting problem for $X$ extends along the projection $P\to X$, is solved for $P$, and restricts along the summand inclusion. ThereforeThis is projectivity detected by all first left derived functors. The following parts prove the three requested equivalent formulations of a hereditary abelian category with enough projectives.

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:

Let $r:B\to A$ split the reflexive pair $f,g$. Fix $C$, and take objects $u:C\to B$ and arrows $a:C\to A$ with source $fa$ and target $ga$. Since $\mathcal{C}$ is an additive category, its hom-sets are abelian groups, and composition is additive.

The identity at $u$ is $ru$. For $a:u\to v$ and $b:v\to w$, defineIts source is $fa-v+fb=u-v+v=u$ and its target is $ga-v+gb=v-v+w=w$. The two identity laws follow from $a-rv+rv=a$ and $ru-ru+a=a$. For $c:w\to x$, both parenthesizations of the triple composite are $a-rv+b-rw+c$, so composition is associative.

Define the inverse byIts source is $u+v-u=v$ and its target is $u+v-v=u$. Moreover $a^{-1}\star a=ru$ and $a\star a^{-1}=rv$. Thus these data form a groupoid, proving the represented form of reflexive pair in an additive category is an internal groupoid:

Use countably many copies of $A$, and let $c:S={\coprod }_{i\ge 1}A\to P={\prod }_{i\ge 1}A$ be the assumed isomorphism. Write $p_i={\pi }_{i}c:S\to A$, let $\delta :A\to P$ be the diagonal with every component $1_A$, and put $q=c^{-1}\delta$. Thus $p_{i}q=1_A$ for every $i$. Let $s:S\to S$ shift the injections, $s{\nu }_i={\nu }_{i+1}$, and let $\sigma :S\to A$ be the fold map, $\sigma {\nu }_i=1_A$.

The coordinate relations give $p_{1}s=0$ and $p_{i+1}s=p_i$, while $p_1{\nu }_1=1_A$ and $p_{i+1}{\nu }_1=0$. Therefore $q$ and ${\nu }_1+sq$ have identical $p_i$-coordinates. Since $c$ is invertible, these coordinates jointly distinguish arrows into $S$, soThe coproduct property gives $\sigma s=\sigma$, because both have value $1_A$ on each injection. Consequently $z=\sigma q$ satisfiesThis proves the absorbing-endomorphism assertion without assuming an infinite addition operation on hom-sets.

If the category is an additive category, each hom-set is an abelian group, so cancellation gives $1_A=0$. Any arrow $f:A\to B$ is then $f{1}_A=f0=0$. Every hom-set has exactly one arrow, and every object is isomorphic to the zero object. This is countable biproducts in an additive category force triviality:

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 Lawvere theory is a small category with finite products generated by one object $[1]$, so every object is a finite power $[n]$. A model in a finite-product category $\mathcal{C}$ is a finite-product-preserving functor from the theory to $\mathcal{C}$. For a finitary monad $T$ on sets, take the opposite of the full subcategory of its Eilenberg-Moore category on the finitely generated free algebras; the resulting finite-power category is its Lawvere theory, and its set-valued models are $T$-algebras.

If ${\mathbf{Set}}^T$ is additive, its full subcategory of finite free algebras is additive, and passing to the opposite preserves biproducts and abelian-group enrichment. Thus the associated theory is an additive category. A product-preserving model sends the abelian-group object $[1]$ and its addition, zero, and inverse maps to an internal abelian group in $\mathcal{C}$.

In an additive theory the product $[n]=[1{]}^n$ is also a biproduct. If ${\iota }_i:[1]\to [n]$ and ${\pi }_i:[n]\to [1]$ are its injections and projections, every $n$-ary operation decomposes uniquely aswhich in a model reads $\alpha (x_1,\dots ,x_n)={\sum }_i{\alpha }_i(x_i)$. Put $R=\mathrm{End}([1])$, with addition from the enrichment and multiplication from composition. The unary operations give every model an $R$-module structure, and the displayed decomposition says that all operations are exactly $R$-linear combinations. Conversely every $R$-module interprets them this way. Hence ${\mathbf{Set}}^T\simeq R\text{-}\mathbf{Mod}$, up to the conventional choice of left versus right modules.

A category is preadditive if every hom-set is an abelian group and composition is additive in each variable. Zero morphisms and differences of parallel morphisms are therefore defined. A zero object or finite biproducts need not exist; these are extra requirements in an additive category.

If $f,g:A\rightrightarrows B$ have common splitting $r$ in an additive category, define the composite of $x,y:C\to A$ with $gx=fy$ byThe identity at $b:C\to B$ is $rb$, and the inverse of $x$ is $rfx+rgx-x$. These formulas make the reflexive pair an internal groupoid.