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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.”