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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Right derived functors relative to supplied data are additive functors

Statement

Assume the Axiom of Dependent Choice.

Let I be a supplied injective resolution datum and F:AB an additive functor between abelian categories. For every nZ, the assignments ARInF(A),uRInF(u) define an additive functor on the domain of I.

Facts & Assumptions

Given: An integer n.

[L1]

Right derived maps are defined from injective comparison extensions (The right derived map relative to supplied resolution data).

[L2]

The cochain comparison-extension construction is available for every morphism, and its induced cohomology map is independent of the chosen extension (A morphism has a comparison extension between the supplied injective resolutions, The induced cohomology map is independent of the chosen injective comparison extension).

[L3]

A cochain complex may be reindexed as a chain complex (Cochain complex in an abelian category).

[L4]

The category of complexes in an additive category is additive, and additive functors apply degreewise to chain maps (The category of complexes in an additive category is additive, An additive functor applies degreewise to complexes and chain maps).

[L5]

Two injective comparison extensions of the same morphism are cochain-homotopic, and after reindexing chain-homotopic maps induce the same map on homology (Injective comparison maps are unique up to cochain homotopy, Chain-homotopic maps induce the same map on homology).

[L6]

An additive functor is a functor that is additive on each hom-group (Additive functor).

Proof

technique · direct
1.1

Identity and composition are proved exactly as on the projective side: choose comparison extensions for the relevant morphisms, compare the extension of a composite or identity with the obvious chain-level candidate, and use [L5] after reindexing by [L3]. Therefore the assignments in [L1] form a functor.

L1L2L3L5givenalgebra
1.2

Let u,v:AB. Choose comparison extensions u~ and v~. By [L4], their degreewise sum is a cochain map and extends u+v, so it is a comparison extension of u+v.

L2L4construct
2.1

Reindexing by [L3], applying F degreewise by [L4], and using homotopy invariance from [L5], the induced cohomology map of u~+v~ equals the sum of the induced cohomology maps of u~ and v~. Hence RInF(u+v)=RInF(u)+RInF(v).

L3L4L5step 1.2algebra
3.1

Steps 1.1 and 2.1 give the functoriality and hom-group additivity required by [L6]. Therefore RInF is an additive functor.

L6step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources