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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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Extending a partial comparison homotopy by one degree

Statement

Let f∙,g∙:P∙→Q∙ be augmentation-preserving maps of projective resolutions lifting the same object morphism. Suppose h0,…,hn−1 have already been chosen so that fk−gk=dk+1Qhk+hk−1dkP holds for every k<n (with h−1=0). Then there exists hn:Pn→Qn+1 extending the homotopy identity to degree n.

Facts & Assumptions

Given: Projective resolutions P∙→A, Q∙→B, two comparison maps f∙,g∙ lifting the same object morphism, and a partial chain homotopy through degree n−1.

[L1]

A chain homotopy is given by the equation fn−gn=dn+1Qhn+hn−1dnP (A chain homotopy).

[L2]

Cycle objects are kernels of the differentials (Cycle and boundary subobjects of a complex).

[L3]

Projective objects lift across epimorphisms (Projective object).

Proof

technique · direct
1.1L1L2givenalgebra

Put cn:=fn−gn−hn−1dnP, with h−1=0 when n=0. Using the chain-map identities for f∙ and g∙ and the already verified lower-degree homotopy equations, one gets dnQcn=0. Thus cn lands in Zn(Q) by [L2]; when n=0, the common augmentation condition on f0 and g0 says exactly that c0 lands in ker⁡(εQ)=Z0(Q).

2.1L3step 1.1construct∎

Exactness of Q∙ makes Qn+1↠Zn(Q) epic, and Pn is projective. By [L3], lift cn to a map hn:Pn→Qn+1. Then dn+1Qhn=cn, which is precisely the degree-n homotopy equation from [L1].

Depends on

Used by

Dependency tree · two levels

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Sources