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The category $\mathbf{Part}$ has sets as objects and partial functions as morphisms. Composition is defined where both successive functions are defined. The nowhere-defined map is a zero morphism, and the empty set is its sole actual zero object. Adjoining a tagged basepoint turns a partial function into a total basepoint-preserving function, giving an equivalence of categories with the category of pointed sets.

The category of pointed sets and basepoint-preserving functions is a pointed category. It is equivalent to the category of partial functions by adjoining or deleting the basepoint. It is not isomorphic to the category of all actual sets and partial functions: there are many distinct singleton zero objects, whereas the empty set is the sole zero object in that category.

An isomorphism of categories is a functor with a strictly inverse functor. It is equivalently bijective on objects and on each hom-set. An equivalence of categories only requires inverse composites up to invertible natural transformations. For example, the category of partial functions and the category of pointed sets are equivalent but their actual object collections prevent an isomorphism: the former has one zero object, the empty set, while the latter has distinct singleton pointed set objects that are all zero objects.

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.

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.

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.

In a pointed category, a zero object defines zero morphisms between all objects. A categorical cokernel of $f:A\to B$ is a map $g:B\to C$ with $gf=0$ such that every $u:B\to X$ with $uf=0$ factors uniquely as $u=\bar{u}g$. Equivalently it is the coequalizer of $f$ and the zero map, so $g$ is an epimorphism.

Write $a:A\to A^{\prime }$, $b:B\to B^{\prime }$, and $c:C\to C^{\prime }$ for the vertical arrows, with $g^{\prime }=\mathrm{coker}{f}^{\prime }$ and $cg=g^{\prime }b$. The left pushout in a category applied to the compatible pair $g:B\to C$ and $0:A^{\prime }\to C$ gives $h:B^{\prime }\to C$ satisfying $hb=g$ and $h{f}^{\prime }=0$. The categorical cokernel property of $g^{\prime }$ then gives $d:C^{\prime }\to C$ with $d{g}^{\prime }=h$.

Now $dcg=d{g}^{\prime }b=hb=g$. Cancel the epimorphism $g$ to obtain $dc=1_C$. To prove the other identity, the two arrows $cd{g}^{\prime },g^{\prime }:B^{\prime }\to C^{\prime }$ agree after $b$, because $cd{g}^{\prime }b=cg=g^{\prime }b$, and after $f^{\prime }$, because both composites are zero. The pushout in a category uniqueness clause gives $cd{g}^{\prime }=g^{\prime }$. Cancelling the epimorphism $g^{\prime }$ gives $cd=1_{C^{\prime }}$.

Thus $c$ is an isomorphism: cokernel invariance under pushout holds already in pointed categories with the indicated cokernels.

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.

An equivalence of categories consists of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ together with invertible natural transformations $GF\cong 1_{\mathcal{C}}$ and $FG\cong 1_{\mathcal{D}}$. A strict isomorphism of categories instead requires a functor with an inverse whose composites are literally identity functors.

First suppose $F$ belongs to an equivalence of categories. The isomorphisms $FGD\cong D$ give essential surjectivity. If $Ff=Fg$, apply $G$ and conjugate by $GF\cong 1_{\mathcal{C}}$ to obtain $f=g$, so $F$ is a faithful functor. Similarly $G$ is a faithful functor. Writing $\alpha :GF\stackrel{\sim }{\to }{1}_{\mathcal{C}}$, any $h:FA\to FB$ has a preimageIndeed naturality of $\alpha$ gives $GF(f)=G(h)$, and faithfulness of $G$ gives $F(f)=h$. Thus $F$ is a full and faithful functor.

Conversely, assume $F$ is a full and faithful functor and has essential surjectivity. Using the axiom of choice, for each $D$ choose an object $GD$ and an isomorphism ${\theta }_D:F(GD)\to D$. Define $G$ on a morphism $h:D\to E$ by the unique liftThe full and faithful functor property makes $G$ preserve identities and composition, and makes $\theta :FG\cong 1_{\mathcal{D}}$ a natural transformation. For $A\in \mathcal{C}$, lift ${\theta }_{FA}:FGFA\to FA$ uniquely to ${\alpha }_A:GFA\to A$. Lifting its inverse shows that ${\alpha }_A$ is invertible. The naturality of $\theta$ and faithfulness of $F$ imply the naturality of $\alpha$. This provesFor large categories, this choice argument is understood in a fixed universe, or with the corresponding class-choice convention; ordinary set-sized axiom of choice suffices for small categories.

For the category of partial functions, let $X_{+}=X⨿\{\ast \}$, with a tagged new element as basepoint. Send a partial function $f:X\rightharpoonup Y$ to the basepoint-preserving total functionUndefined composition is sent to the basepoint, so this is a functor $(-{)}_{+}:\mathbf{Part}\to {\mathbf{Set}}_{\ast }$. Restriction away from the basepoint recovers each partial function uniquely; hence it is a full and faithful functor. Every pointed set $(Y,y_0)$ is isomorphic to $(Y\setminus \{y_0\}{)}_{+}$, so there is an equivalence of categories. This particular equivalence can also be constructed explicitly, without choice, by deleting and adjoining the basepoint.

These actual categories are equivalent but not isomorphic. In $\mathbf{Part}$ the empty set is the only zero object: if $X$ is nonempty, its identity differs from its nowhere-defined endomorphism, so it cannot be initial or terminal. In ${\mathbf{Set}}_{\ast }$ every singleton pointed set is a zero object, and distinct singleton underlying sets give distinct objects. An isomorphism of categories is a bijection on objects preserving zero objects; it cannot take one such object onto several. This uses the categories of all actual sets, as in the paper, rather than chosen skeletal categories of representatives.

A skeletal category has no distinct isomorphic objects. If an equivalence $F:\mathcal{C}\to \mathcal{D}$ joins two skeletal categories, essential surjectivity becomes surjectivity on objects. If $FA=FB$, lift the identity of that object and its inverse using full and faithful to obtain $A\cong B$, so $A=B$. Thus $F$ is bijective on objects and on every hom-set. Its inverse on objects and morphisms is a strictly inverse functor, proving that it is an isomorphism of categories.

Under the axiom of choice, choose one object from each isomorphism class of a small category. The full subcategory on those objects is a skeleton of a category, and its inclusion is a full and faithful functor with essential surjectivity, hence an equivalence of categories.

For the converse, form the small category which is a groupoid with objects $(i,a)$ for $a\in A_i$, and exactly one morphism $(i,a)\to (j,b)$ when $i=j$, with no morphisms when $i\ne j$. Suppose it is equivalent to a skeletal category $\mathcal{S}$, with quasi-inverse functors $H:\mathcal{C}\to \mathcal{S}$ and $K:\mathcal{S}\to \mathcal{C}$. For each $i$, the objects $H(i,a)$ for $a\in A_i$ are isomorphic, hence all equal to a uniquely determined $s_i$. The isomorphism $KH(i,a)\cong (i,a)$ ensures that $K(s_i)=(i,a_i)$ for some $a_i\in A_i$. The rule $i\mapsto a_i$ is a choice function. In particular, no representative in $A_i$ had to be chosen to define $s_i$, since it is unique. Consequently

A pointed category has a zero object, both initial and terminal. Factoring through that object gives a zero morphism $0_{A,B}$ between any two objects. In this paper a semi-additive structure means commutative-monoid enrichment: every hom-set is a commutative monoid with additive zero, and composition distributes over addition in both variables. No existence of finite products in a category is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite biproducts, that example would not be a semi-additive category.

Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a coproduct in a category, with injectionsWrite ${\nabla }_B:B\times B\to B$ for the morphism whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the biproduct-induced addition of morphisms byThe zero is the existing zero morphism. We verify the laws from the universal properties, without assuming addition in advance.

All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the zero object. Thus both $(B\times B)\times B$ and $B\times (B\times B)$ are ternary coproducts in a category. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $⟨f,g,h⟩$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with ${\nabla }_B$ is ${\nabla }_B$, giving commutativity. Finally $⟨f,0⟩=i_{1}f$, so $f+0=f$, and similarly $0+f=f$.

Precomposition is additive because $⟨f,g⟩u=⟨fu,gu⟩$. For $v:B\to C$, the morphisms $v{\nabla }_B$ and ${\nabla }_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a zero morphism is zero. This proves the commutative-monoid enrichment.

It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the zero object belongs to a singleton hom-set. The projections of $i_1{p}_1+i_2{p}_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, soFor $A=B$, composing on the left by ${\nabla }_B$ and on the right by $⟨f,g⟩$ forces the displayed formula for $f+g$. Thus the biproducts determine exactly one such enrichment.

For the last part, the underlying one-object category has endomorphism monoid $(\mathbb{N},\cdot )$, including $0$, and identity $1$. Its usual addition gives a commutative-monoid enrichment. Any permutation $\sigma$ of the prime numbers extends, by unique prime factorization, to a multiplicative monoid automorphism ${\varphi }_{\sigma }:\mathbb{N}\to \mathbb{N}$, fixing $0,1$. Transport addition byThis is a commutative monoid operation with zero $0$, and multiplication distributes over it: applying ${\varphi }_{\sigma }$ reduces each distributive law to the ordinary one in $\mathbb{N}$. Thus each operation supplies a semi-additive structure on the same fixed composition law.

For each odd prime $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. ThenDifferent choices of $p$ give different additions, and there are infinitely many prime numbers. Hence there are infinitely many distinct semi-additive structures, even though these transported structures are isomorphic as enriched categories. The underlying one-object category has no terminal object, because its endomorphism set is not a singleton, so it indeed has no finite products in a category.

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: