CorollaryStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-01

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Two-out-of-three for acyclicity in a short exact sequence of complexes

Statement

In a short exact sequence of complexes $0 \to A_{∙} \to B_{∙} \to C_{∙} \to 0,$ if any two of $A_{∙}$ , $B_{∙}$ , and $C_{∙}$ are acyclic, then so is the third.

Facts & Assumptions

Given: A short exact sequence $0 \to A_{∙} \to B_{∙} \to C_{∙} \to 0$ of complexes.

[L1]

The sequence carries a long exact sequence in homology (The long exact sequence in homology).

[L2]

Acyclic means vanishing homology in every degree (Exactness of a complex at a degree and acyclic complexes).

Proof

technique · direct

1.1L1L2givenalgebra

If $A_{∙}$ and $C_{∙}$ are acyclic, then for every $n$ the exact window $H_{n} (A) \to H_{n} (B) \to H_{n} (C)$ has zero outer terms by [L2]. Exactness from [L1] therefore gives $H_{n} (B) = 0$ for all $n$ .

1.2L1L2givenalgebra

If $A_{∙}$ and $B_{∙}$ are acyclic, then the exact window $H_{n} (B) \to H_{n} (C) \to H_{n - 1} (A)$ has zero outer terms, so $H_{n} (C) = 0$ for all $n$ .

2.1L1L2givenalgebra∎

If $B_{∙}$ and $C_{∙}$ are acyclic, then the exact window $H_{n} (C) \to H_{n - 1} (A) \to H_{n - 1} (B)$ has zero outer terms, so $H_{n - 1} (A) = 0$ for all $n$ . Thus the remaining complex is acyclic in every case.

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Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)
The Stacks Project, Section 12.13: Complexes (standard reference, not scraped)