Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31
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A degreewise split exact complex with compatible splittings is contractible

Statement

Let C∙ be an acyclic chain complex. Suppose that for every n there is an isomorphism ϕn:Cn→∼Zn(C)⊕Zn−1(C) such that in=ϕn−1j1, where in:Zn(C)↪Cn is the cycle inclusion and j1 is the first summand inclusion, and such that the differential is dn=in−1π2ϕn, where π2 is the second projection. Then C∙ is contractible.

Facts & Assumptions

Given: An acyclic chain complex C∙ and isomorphisms ϕn as in the statement.

[L1]

A contractible complex is one whose identity map is null-homotopic (A contractible complex).

[L2]

Acyclic means exact at every degree (Exactness of a complex at a degree and acyclic complexes).

Proof

technique · direct
1.1givenalgebra

Define sn−1:Cn−1→Cn by sn−1:=ϕn−1j2π1ϕn−1, where j2 is the second inclusion and π1 is the first projection. Then dnsn−1=in−1π1ϕn−1=ϕn−1−1j1π1ϕn−1 by the formula for dn and the compatibility of in−1 with ϕn−1.

2.1L1L2step 1.1algebra∎

Likewise sn−2dn−1=ϕn−1−1j2π2ϕn−1, so dnsn−1+sn−2dn−1=ϕn−1−1(j1π1+j2π2)ϕn−1=1Cn−1. Thus the identity map is null-homotopic, and [L1] makes C∙ contractible.

Depends on

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Sources