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

Statement

Assume the Axiom of Dependent Choice.

Let P,Q,R be supplied projective 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 map S(A)T(A) lifting 1A. Then:

  1. θP,P=1LnPF for every n.
  2. θQ,RθP,Q=θP,R for every n.

Facts & Assumptions

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

[L1]

A comparison map between two supplied projective resolutions of A induces the isomorphism θS,T(A), and these objectwise isomorphisms are natural in A (Objectwise comparison of two projective resolution data induces an isomorphism on derived objects, The change-of-projective-resolution isomorphisms are natural).

[L2]

Two projective comparison maps lifting the same morphism are chain-homotopic (Projective comparison maps are unique up to chain homotopy).

[L3]

Chain-homotopic maps induce the same homology map, and homology respects composition (Chain-homotopic maps induce the same map on homology, Homology respects identities and composition).

Proof

technique · direct
1.1

For the pair (P,P), one valid comparison map is the identity chain map on P(A). Any comparison map used to define θP,P(A) also lifts 1A, so [L2] makes it homotopic to the identity chain map. By [L3], the induced homology map is therefore the identity on LnPF(A).

L1L2L3given
1.2

For the triple (P,Q,R), the chain map defining θQ,R(A)θP,Q(A) is the composite of two comparison maps P(A)Q(A)R(A) lifting 1A. The chain map defining θP,R(A) is another comparison map lifting 1A. By [L2] they are homotopic, so [L3] gives equality of the induced homology maps: θQ,R(A)θP,Q(A)=θP,R(A).

L1L2L3algebra
2.1

Since A was arbitrary, steps 1.1 and 1.2 prove the identity and cocycle laws for the natural isomorphisms.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

18 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