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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:CnZn(C)Zn1(C) such that in=ϕn1j1, where in:Zn(C)Cn is the cycle inclusion and j1 is the first summand inclusion, and such that the differential is dn=in1π2ϕn, 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
1.1

Define sn1:Cn1Cn by sn1:=ϕn1j2π1ϕn1, where j2 is the second inclusion and π1 is the first projection. Then dnsn1=in1π1ϕn1=ϕn11j1π1ϕn1 by the formula for dn and the compatibility of in1 with ϕn1.

givenalgebra
2.1

Likewise sn2dn1=ϕn11j2π2ϕn1, so dnsn1+sn2dn1=ϕn11(j1π1+j2π2)ϕn1=1Cn1. Thus the identity map is null-homotopic, and [L1] makes C contractible.

L1L2step 1.1algebra

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Dependency tree · two levels

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