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In a pointed category with finite product in a category and coproducts in a category and invertible canonical maps $c:A+B\to A\times B$, the unique commutative-monoid enrichment is $f+g=[1_B,1_B]c_{B,B}^{-1}⟨f,g⟩$. Associativity and commutativity follow by comparing fold maps on triple and swapped coproduct injections. Composition distributes by the universal properties. For uniqueness, bilinearity forces $i_1{p}_1+i_2{p}_2=1$ on each biproduct, and composing with the fold and the paired morphism forces the addition formula.

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.

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 semiring has a commutative monoid operation $(S,+,0)$ and a monoid operation $(S,\cdot ,1)$, with multiplication distributing over addition on both sides and $0$ absorbing multiplication. Additive inverses are not required. The natural numbers with their usual addition and multiplication are a basic example. Regarding multiplication as composition yields a one-object category with commutative-monoid enrichment.

For a multiplicative monoid automorphism $\varphi :\mathbb{N}\to \mathbb{N}$ fixing $0$, define $a\oplus b={\varphi }^{-1}(\varphi (a)+\varphi (b))$. This transports a semiring structure to the fixed underlying multiplication, hence a commutative-monoid enrichment on its one-object category. Swapping prime $2$ with an odd prime $p$ gives $1\oplus 1=p$. Infinitely many such primes therefore give infinitely many distinct additions, although the transported enriched categories are isomorphic.

In a pointed category, the zero morphism $0_{A,B}:A\to B$ is the composite through its zero object. Composition on either side by any morphism remains zero. In a commutative-monoid enrichment on a pointed category, the additive zero must coincide with this zero morphism: morphisms to and from the zero object belong to singleton hom-sets.