Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-01
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The connecting morphism vanishes for a chain-split short exact sequence

Statement

Let 0→A∙→iB∙→pC∙→0 be a short exact sequence of complexes. If it admits either a chain section s:C∙→B∙ with ps=1C or a chain retraction r:B∙→A∙ with ri=1A, then ∂n=0:Hn(C)→Hn−1(A) for every n.

Facts & Assumptions

Given: A short exact sequence 0→A∙→iB∙→pC∙→0 of complexes.

[L1]

A chain map is a degreewise morphism commuting with the differentials (Chain map).

[L2]

The connecting morphism is induced from the preconnecting arrow on cycles (The connecting morphism in homology, The preconnecting arrow on cycles).

[L3]

The associated homology sequence is exact (The long exact sequence in homology).

Proof

technique · direct
1.1L1L3givenalgebra

Suppose first that there is a chain section s with ps=1C. Functoriality of homology gives Hn(p)Hn(s)=1Hn(C), so Hn(p) is epic. Exactness in [L3] gives ker⁡(∂n)=im⁡(Hn(p))=Hn(C), hence ∂n=0.

2.1L1L2L3givenalgebra∎

Suppose instead that there is a chain retraction r with ri=1A. Then Hn−1(r)Hn−1(i)=1Hn−1(A), so Hn−1(i) is monic. Exactness in [L3] gives im⁡(∂n)=ker⁡(Hn−1(i))=0, and a morphism with zero image in an abelian category is zero. Thus the connecting morphism vanishes in either chain-split situation.

Depends on

Used by

Dependency tree · two levels

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Sources