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 $F : A \to B$ an additive functor. If $u : A \to B$ is a morphism and ${\tilde{u}}^{∙}, {\hat{u}}^{∙} : I^{∙} (A) \to I^{∙} (B)$ are two injective comparison extensions of $u$ , then for every $n \in Z$ the induced maps on cohomology $H^{n} ⁣ (F ({\tilde{u}}^{∙})), H^{n} ⁣ (F ({\hat{u}}^{∙})) : R_{I}^{n} F (A) \to R_{I}^{n} F (B)$ are equal.

Facts & Assumptions

Given: A morphism $u : A \to B$ and two comparison extensions ${\tilde{u}}^{∙}, {\hat{u}}^{∙}$ 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 $f_{n} - g_{n} = d_{n + 1} s_{n} + s_{n - 1} d_{n}$ (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 $R_{I}^{n} F (A)$ and $R_{I}^{n} F (B)$ are the cohomology objects of the deleted injective resolutions after applying $F$ (Right derived objects relative to supplied injective resolution data).

Proof

technique · direct

1.1

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

L1L2givenconstruct

2.1

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

L3L4step 1.1algebra

3.1

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 $R_{I}^{n} F (A) \to R_{I}^{n} F (B)$ .

L2L5L6step 2.1∎

Depends on

Used by

Dependency tree · two levels

22 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

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 2 `Derived Functors` (standard reference, not scraped)