Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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Ab-enriched categories are exactly preadditive categories

Statement

A category enriched in Ab with tensor product ⊗Z is exactly a preadditive category.

Facts & Assumptions

Given: A category or enriched category with object set Ob⁡.

[L1]

An Ab-enriched category has abelian-group hom-objects, composition morphisms A(B,C)⊗ZA(A,B)→A(A,C), and identity morphisms Z→A(A,A) (Enriched category over a monoidal base).

[L2]

A preadditive category is a category whose hom-sets are abelian groups and whose composition is bilinear (Preadditive category).

[L3]

Ab is monoidal under ⊗Z, and morphisms out of X⊗ZY are exactly bilinear maps out of X×Y (Abelian groups are monoidal under the tensor product).

[L4]

In a preadditive category, the hom-bifunctor already takes values in abelian groups (The hom-bifunctor of a preadditive category takes values in abelian groups).

Proof

technique · direct
1.1L1L2L3given

Assume A is Ab-enriched. By [L1], each hom-object A(A,B) is an abelian group. The composition morphism is a morphism in Ab out of the tensor product, so by [L3] it is exactly a bilinear map A(B,C)×A(A,B)→A(A,C). The identity map Z→A(A,A) selects the identity element in the endomorphism group. Thus the underlying ordinary category has abelian-group homs and bilinear composition, so [L2] makes it preadditive.

1.2L2L3L4algebra

Conversely, let C be preadditive. By [L4], each hom-set is an abelian group and the two-variable hom-assignment is additive in each variable. So for every triple A,B,C, the bilinear composition map C(B,C)×C(A,B)→C(A,C) transposes uniquely, by [L3], to a group homomorphism C(B,C)⊗ZC(A,B)→C(A,C). Sending 1∈Z to the identity morphism of A gives the required unit map Z→C(A,A). The ordinary associativity and unit laws are exactly the enriched ones after this transposition.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 show that the two notions encode the same data.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources