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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04
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Change-of-injective-resolution isomorphisms satisfy identity and cocycle laws

Statement

Assume the Axiom of Dependent Choice.

Let I,J,K be supplied injective resolution data on the same domain, and let F:AB be an additive functor between abelian categories. For each ordered pair (S,T) among these data, let θS,T be the change-of-data natural isomorphism whose component at an object A is induced by any comparison extension S(A)T(A) of 1A. Then:

  1. θI,I=1RInF for every n.
  2. θJ,KθI,J=θI,K for every n.

Facts & Assumptions

Given: An object A in the common domain and an integer n.

[L1]

The chosen injective resolutions of the same object are homotopy equivalent under that object (Injective resolutions of the same object are homotopy equivalent under that object).

[L2]

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

[L3]

Reindexing turns cochain homotopies into chain homotopies, and homology then respects both homotopy and composition (Cochain complex in an abelian category, Chain-homotopic maps induce the same map on homology, Homology respects identities and composition).

[L4]

Comparison extensions exist for morphisms between objects in the domain of each supplied injective datum (A morphism has a comparison extension between the supplied injective resolutions).

Proof

technique · direct
1.1

For any ordered pair (S,T), [L1] gives comparison extensions cA:S(A)T(A) and dA:T(A)S(A) of 1A. Their composites extend 1A, so [L2] and [L3] show that the induced cohomology maps are inverse. Any other choice of cA extends the same identity and hence induces the same map. For a morphism u:AB, choose within-data comparison extensions using [L4]. The two composites from S(A) to T(B) both extend u, so [L2] and [L3] give the naturality square. Thus the displayed construction specifies a well-defined natural isomorphism θS,T.

L1L2L3L4givenconstruct
2.1

For the pair (I,I), the identity cochain map on I(A) is a comparison extension of 1A. Any comparison extension used in step 1.1 to define θI,I(A) extends the same identity morphism, so [L2] makes it cochain-homotopic to the identity. By [L3], the induced map on cohomology is therefore the identity on RInF(A).

L2L3step 1.1
2.2

For the triple (I,J,K), the cochain map defining θJ,K(A)θI,J(A) is the composite of two comparison extensions of 1A, while the map defining θI,K(A) is another comparison extension of 1A. By [L2] these are cochain-homotopic, so [L3] gives θJ,K(A)θI,J(A)=θI,K(A).

L2L3step 1.1algebra
3.1

Since A was arbitrary, steps 2.1 and 2.2 prove the identity and cocycle laws for the natural isomorphisms constructed in step 1.1.

step 1.1step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

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Sources