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In the category of complete join-semilattices, hom-sets have pointwise join as addition and the constant-bottom map as zero, making it a semi-additive category. For any family $A_i$, the cartesian product with coordinatewise joins is also its coproduct. The injections put an element in one coordinate and bottom in all others. A family of maps $f_i:A_i\to B$ extends uniquely by $(a_i)\mapsto {\bigvee }_i{f}_i(a_i)$, since each tuple is the join of its coordinate injections. Thus all set-indexed canonical product-coproduct comparisons are invertible. The two-element chain has distinct identity and zero morphisms, so this category is not trivial and not additive.

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.

A commutative-monoid enrichment gives each hom-set a commutative monoid structure, with zero and addition, so that composition distributes over addition in both variables and annihilates zero. This is called a semi-additive structure in some conventions, including Cambridge Part III paper 119 of 2017. Other conventions reserve semi-additive category for categories with this enrichment and finite biproducts. A one-object category arising from a semiring has the enrichment without generally having those finite products.

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.

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 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 the category of commutative monoids, which is a semi-additive category: hom-sets have pointwise addition, composition is additive in each variable, and finite cartesian products are also coproducts. Explicitly, maps $M_i\to N$ extend from the coordinate injections by $(m_i)\mapsto {\sum }_i{f}_i(m_i)$.

Take the submonoidwith coordinatewise addition, and the two projections $f(m,n)=m$, $g(m,n)=n$. The homomorphism $r(m)=(m,m)$ satisfies $fr=gr=1_B$, so this is a reflexive pair. For $C=\mathbb{N}$, evaluation at $1$ identifies $\mathcal{C}(C,B)$ with $B$ and $\mathcal{C}(C,A)$ with $A$.

The element $(1,2)$ gives an arrow from $1$ to $2$ in the prescribed graph. An inverse would require an element $(2,1)$ of $A$, which does not exist. Therefore no groupoid structure with these source and target maps is possible, regardless of the proposed composition rule:

In the category of complete join-semilattices, define addition of arrows pointwise byand let the zero arrow be the constant-bottom map. These form commutative monoids. Pointwise join of join-preserving maps again preserves all joins, including the empty join, and composition is additive in each variable. Thus the category has the required semi-additive enrichment.

For an arbitrary set-indexed family $(A_i)$, the product poset $P={\prod }_i{A}_i$, with coordinatewise joins, is the product in a category. Define ${\nu }_i:A_i\to P$ by putting an element in coordinate $i$ and bottom in all others. Given join-preserving $f_i:A_i\to B$, defineThis preserves arbitrary joins by interchanging the two joins, and $h{\nu }_i=f_i$. It is unique because every tuple is ${\bigvee }_i{\nu }_i(a_i)$. Therefore $P$ is also the coproduct in a category with these injections, and its canonical coproduct-to-product map is the identity. This proves the biproducts of complete join-semilattices property, even for arbitrary set-indexed families, and in particular makes this a semi-additive category with the hypotheses of part (ii).

The two-element chain has distinct identity and constant-bottom endomorphisms, so the category is not equivalent to $1$. Its addition is idempotent rather than cancellative; for example $1_A+1_A=1_A$ itself supplies an absorbing endomorphism. Hence

In a semi-additive category, hom-sets are commutative monoids and composition is additive in both variables. Write $S={\coprod }_i{A}_i$, $P={\prod }_i{A}_i$, with injections ${\nu }_i$ and projections ${\pi }_i$. The canonical $c:S\to P$ has ${\pi }_{i}c{\nu }_j=1_{A_i}$ when $i=j$ and zero otherwise. DefineBilinearity givesThe universal properties of the product in a category and coproduct in a category imply $cd=1_P$ and $dc=1_S$. For the empty family, $S$ is initial and $P$ terminal. Their endomorphisms are necessarily the identity and the zero morphism; the zero map $P\to S$ therefore also gives inverse composites. Thus the comparison is an isomorphism for every finite family, giving a biproduct:

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.”