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Morphisms between finite biproducts correspond to matrices

Statement

Let C be additive, let A=i=1mAi, and let B=j=1nBj. Then

C(A,B)i=1mj=1nC(Ai,Bj)

by the map sending f:AB to the matrix of entries fji:=pjfii. This is an isomorphism of abelian groups.

Facts & Assumptions

Given: An additive category with finite biproducts A=iAi and B=jBj.

[L1]

An additive category is preadditive, so hom-sets are abelian groups with finite sums (Additive category).

[L2]

On a biproduct, the identity is the sum of injection-projection terms (On a biproduct, the injections and projections satisfy the identity-sum relation).

[L3]

Finite biproducts are canonically associative and commutative, so the bracketing of the finite sums does not matter (Biproducts are associative, commutative, and unital up to canonical isomorphism).

Proof

technique · direct
1.1

Define Φ:C(A,B)i,jC(Ai,Bj) by Φ(f)=(pjfii)j,i. This is a group homomorphism because each pj()ii is additive by bilinearity in [L1].

L1
1.2

For a matrix (uji) of morphisms uji:AiBj, define Ψ((uji)):=i,jijujipi:AB. This finite sum is legitimate by [L1] and [L3].

L1L3construct
2.1

Using the zero equations and [L2], one gets pkΨ((uji))i=i,jpkijujipii=uk. Hence ΦΨ is the identity on the matrix product.

L1L2step 1.2algebra
2.2

For f:AB, the identity-sum relation on both source and target gives f=1Bf1A=(jijpj)f(iiipi)=i,jij(pjfii)pi=Ψ(Φ(f)). So ΨΦ=1C(A,B).

L1L2step 1.2
3.1

Thus Φ and Ψ are inverse group homomorphisms, giving the asserted matrix description of C(A,B).

step 1.1step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

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Sources