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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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The induced cohomology map is independent of the chosen injective comparison extension

Statement

Assume the Axiom of Dependent Choice.

Let I be a supplied injective resolution datum and F:A→B an additive functor. If u:A→B is a morphism and u~∙,u^∙:I∙(A)→I∙(B) are two injective comparison extensions of u, then for every n∈Z the induced maps on cohomology Hn ⁣(F(u~∙)), Hn ⁣(F(u^∙)):RInF(A)→RInF(B) are equal.

Facts & Assumptions

Given: A morphism u:A→B and two comparison extensions u~∙,u^∙ of u.

[L1]

Two injective comparison maps extending the same morphism are cochain-homotopic (Injective comparison maps are unique up to cochain homotopy).

[L2]

A cochain complex is read as a reindexed chain complex by reversing the grading sign (Cochain complex in an abelian category).

[L3]

A chain homotopy is an equation of the form fn−gn=dn+1sn+sn−1dn (A chain homotopy).

[L4]

Additive functors preserve sums and zero morphisms (Additive functor, An additive functor preserves zero morphisms).

[L5]

Chain-homotopic maps induce the same map on homology (Chain-homotopic maps induce the same map on homology).

[L6]

The objects RInF(A) and RInF(B) are the cohomology objects of the deleted injective resolutions after applying F (Right derived objects relative to supplied injective resolution data).

Proof

technique · direct
1.1L1L2givenconstruct

By [L1], the two comparison extensions are cochain-homotopic. Using [L2], read that cochain homotopy as a chain homotopy after reindexing the complexes.

2.1L3L4step 1.1algebra

Applying F to the homotopy equations from [L3] preserves their sum-and- zero form by [L4]. Hence the two reindexed chain maps F(u~∙) and F(u^∙) remain chain-homotopic.

3.1L2L5L6step 2.1∎

By [L5], these two maps induce the same homology map on the reindexed complexes. Translating back through [L2] and [L6], that is exactly equality of the induced maps on cohomology RInF(A)→RInF(B).

Depends on

Used by

Dependency tree · two levels

27 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