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An additive category is a category which is
(sometimes called a pre-additive category–this means that each hom-set carries the structure of an abelian group and composition is bilinear)
which admits finite coproducts
(and hence, by prop. below, finite products which coincide with the coproducts, hence finite biproducts).
The natural morphisms between additive categories are additive functors.
A pre-abelian category is an additive category which also has kernels and cokernels. Equivalently, it is an Ab-enriched category with all finite limits and finite colimits. An especially important sort of additive category is an abelian category, which is a pre-abelian one satisfying the extra exactness property that all monomorphisms are kernels and all epimorphisms are cokernels. See at additive and abelian categories for more.
The Ab-enrichment of an additive category does not have to be given a priori. Every semiadditive category (a category with finite biproducts) is automatically enriched over commutative monoids (as described at biproduct), so an additive category may be defined as a category with finite biproducts whose hom-monoids happen to be groups. (The requirement that the hom-monoids be groups can even be stated in elementary terms without discussing enrichment at all, but to do so is not very enlightening.) Note that the entire $Ab$-enriched structure follows automatically for abelian categories.
Some authors use additive category to simply mean an Ab-enriched category, with no further assumptions. It can also be used to mean a $CMon$-enriched (commutative monoid enriched) category, with or without assumptions of products.
In an Ab-enriched category (or even just a $CMon$-enriched category), a finite product is also a coproduct, and dually.
This statement includes the zero-ary case: any terminal object is also an initial object, hence a zero object (and dually), hence every additive category has a zero object.
More precisely, for $\{X_i\}_{i \in I}$ a finite set of objects in an Ab-enriched category, the unique morphism
whose components are identities for $i = j$ and are zero otherwise is an isomorphism.
Consider first the zero-ary case. Given an initial object $\emptyset$ and a terminal object $\ast$, observe that since the hom-sets $Hom(\emptyset,\emptyset)$ and $Hom(\ast,\ast)$ by definition contain a single element, this element has to be the zero element in the abelian group structure. But it also has to be the identity morphism, and hence $id_\emptyset = 0$ and $id_{\ast} = 0$. It follows that the 0-element in $Hom(\ast, \emptyset)$ is a left and right inverse to the unique element in $Hom(\emptyset,\ast)$, and so this is an isomorphism
Consider now the case of binary (co-)products. Using the existence of the zero object, hence of zero morphisms, then in addition to its canonical projection maps $p_i \colon X_1 \times X_2 \to X_i$, any binary product also receives “injection” maps $X_i \to X_1 \times X_2$, and dually for the coproduct:
Observe some basic compatibility of the $Ab$-enrichment with the product:
First, for $(\alpha_1,\beta_1), (\alpha_2, \beta_2)\colon R \to X_1 \times X_2$ then
(using that the projections $p_1$ and $p_2$ are linear and by the universal property of the product).
Second, $(id,0) \circ p_1$ and $(0,id) \circ p_2$ are two projections on $X_1\times X_2$ whose sum is the identity:
(We may check this, via the Yoneda lemma on generalized elements: for $(\alpha, \beta) \colon R \to X_1\times X_2$ any morphism, then $(id,0)\circ p_1 \circ (\alpha,\beta) = (\alpha,0)$ and $(0,id)\circ p_2\circ (\alpha,\beta) = (0,\beta)$, so the statement follows with equation $(\star)$.)
Now observe that for $f_i \;\colon\; X_i \to Q$ any two morphisms, the sum
gives a morphism of cocones
Moreover, this is unique: suppose $\phi'$ is another morphism filling this diagram, then, by using equation $(\star \star)$, we get
and hence $\phi = \phi'$. This means that $X_1\times X_2$ satisfies the universal property of a coproduct.
By a dual argument, the binary coproduct $X_1 \sqcup X_2$ is seen to also satisfy the universal property of the binary product. By induction, this implies the statement for all finite (co-)products.
Such products which are also coproducts as in prop. are sometimes called biproducts or direct sums; they are absolute limits for Ab-enrichment.
The coincidence of products with biproducts in prop. does not extend to infinite products and coproducts.) In fact, an Ab-enriched category is Cauchy complete just when it is additive and moreover its idempotents split.
Conversely:
A semiadditive category is a category that has all finite products which, moreover, are biproducts in that they coincide with finite coproducts as in def. .
In a semiadditive category, def. , the hom-sets acquire the structure of commutative monoids by defining the sum of two morphisms $f,g \;\colon\; X \longrightarrow Y$ to be
With respect to this operation, composition is bilinear.
The associativity and commutativity of $+$ follows directly from the corresponding properties of $\oplus$. Bilinearity of composition follows from naturality of the diagonal $\Delta_X$ and codiagonal $\nabla_X$:
Given an additive category according to def. , then the enrichement in commutative monoids which is induced on it via prop. and prop. from its underlying semiadditive category structure coincides with the original enrichment.
By the proof of prop. , the codiagonal on any object in an additive category is the sum of the two projections:
Therefore (checking on generalized elements, as in the proof of prop. ) for all morphisms $f,g \colon X \to Y$ we have commuting squares of the form
Prop. says that being an additive category is an extra property on a category, not extra structure. We may ask whether a given category is additive or not, without specifying with respect to which abelian group structure on the hom-sets.
Discussion of model category structures on additive categories is around def. 4.3 of
Last revised on April 20, 2017 at 17:03:05. See the history of this page for a list of all contributions to it.