PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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A natural transformation induces natural transformations of right derived functors

Statement

Assume the Axiom of Dependent Choice.

Let $I$ be a supplied injective resolution datum, let $F, G : A \to B$ be additive functors, and let $α : F \Rightarrow G$ be a natural transformation. Then for every $n \in Z$ the maps $R_{I}^{n} (α)_{A} : = H^{n} ⁣ (α_{I (A)_{d e l}}) : R_{I}^{n} F (A) \to R_{I}^{n} G (A)$ define a natural transformation $R_{I}^{n} (α) : R_{I}^{n} F \Rightarrow R_{I}^{n} G .$

Facts & Assumptions

Given: An integer $n$ .

[L1]

A natural transformation is objectwise and satisfies the naturality equation on every morphism (Natural transformation and its components).

[L2]

A cochain complex is read as a reindexed chain complex (Cochain complex in an abelian category).

[L3]

Additive functors apply degreewise to chain maps, and every chain map induces a homology map (An additive functor applies degreewise to complexes and chain maps, A chain map induces a well-defined map on homology).

[L4]

Right derived maps are induced by comparison extensions on the chosen injective resolutions (The right derived map relative to supplied resolution data).

[L5]

The source and target assignments are already functors (Right derived functors relative to supplied data are additive functors).

Proof

technique · direct

1.1

For each object $A$ , the components $α_{I^{k} (A)}$ commute with the cochain differentials by [L1], so they form a cochain map $F (I (A)_{d e l}) \to G (I (A)_{d e l})$ . Reindexing by [L2] turns this into a chain map.

L1L2givenalgebra

2.1

By [L3], step 1.1 induces a map on homology of the reindexed complexes, hence on cohomology: $R_{I}^{n} (α)_{A} : R_{I}^{n} F (A) \to R_{I}^{n} G (A)$ .

L2L3step 1.1construct

3.1

Let $u : A \to B$ , and choose a comparison extension ${\tilde{u}}^{∙} : I^{∙} (A) \to I^{∙} (B)$ . Naturality in [L1] gives degreewise commutative squares with the maps $α_{I^{k} (A)}$ and $α_{I^{k} (B)}$ . Passing to cohomology and translating through [L4] gives $R_{I}^{n} G (u) \circ R_{I}^{n} (α)_{A} = R_{I}^{n} (α)_{B} \circ R_{I}^{n} F (u) .$ Thus the components from step 2.1 are natural.

L1L4L5step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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)