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A commutative monoid is cancellative when $a+c=b+c$ implies $a=b$. Cancellation makes the natural map into its Grothendieck group injective.

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.

A nonunit element of a commutative monoid is irreducible if each factorization has a unit factor. In a submonoid of the positive integers under multiplication, the only unit is one. Irreducibility is relative to the allowed factors: a composite integer can be irreducible in a smaller monoid even though it is not a prime number.

A monoid is a semigroup with a two-sided identity element $e$: $e\ast a=a\ast e=a$ for every element $a$. Associativity handles the grouping of a nonempty sequence; the identity element also supplies the result for an empty sequence. This makes the same rules apply to combining zero, one, or many contributions. ^[ref].

The nonnegative integers $\{0,1,2,\dots \}$ with addition form a commutative monoid, with identity element zero. They do not form a group: a positive integer has no additive inverse in that set. All words over an alphabet with concatenation form a free monoid, whose identity element is the empty word. The full transformation monoid consists of all self-maps of a set, with function composition and the identity function.

These examples explain why monoids are useful: accumulation needs an operation and a starting value, but it need not be reversible. Adding counts, joining strings, and composing transformations all share the same grouping and identity laws. Requiring every element to have an inverse element gives a group, a stronger structure suitable for reversible operations.

Yes. A concrete example of nonunique factorization in a congruence submonoid isThe four factors are distinct and each is congruent to one modulo ten. The primes $3,7,13,17$ can be checked by trial division by two and three. Their elementary prime decompositions areFor any one of these products $pq$, where $p,q$ are the indicated distinct primes, its positive divisors are $1,p,q,pq$. This divisor assertion does not require unique factorization: if $d\mid pq$ and $p\nmid d$, then $p\mid pq/d$ by the Euclid lemma proved in (a), so $d\mid q$ and $d$ is either one or $q$. If $p\mid d$, write $d=pe$ and obtain $e\mid q$, giving $d=p$ or $pq$.

None of $3,7,13,17$ is congruent to one modulo ten. Consequently no listed product has a proper nonunit divisor in $S$. Thus each is an irreducible element of a commutative monoid, and the displayed equality proves the claim. Ordinary prime decompositions and irreducibility relative to the smaller commutative monoid $S$ are different notions.

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.

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.”

The substitution $F(a)=e^a$, with $F(-\infty )=0$, is a bijection onto $[0,\infty )$ and obeysThus the log-sum-exp function transfers nonnegative addition to the given set. It is closed, associative and commutative, with identity element $-\infty$. But a finite $a$ has $e^a>0$, and adding a nonnegative number cannot give zero, so no finite $a$ has an inverse element. Therefore it is a commutative monoid but not a group.

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.