Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The horseshoe construction stays short exact after applying a right exact functor

Statement

Assume the Axiom of Dependent Choice.

Let F:A→B be a right exact functor between abelian categories, and let 0→A′→A→A′′→0 be a short exact sequence in A. If 0→P∙′→H∙→P∙′′→0 is a horseshoe short exact sequence of projective resolutions of A′,A,A′′, then 0→F(P∙,del′)→F(H∙,del)→F(P∙,del′′)→0 is a short exact sequence of complexes in B.

Facts & Assumptions

Given: A horseshoe short exact sequence of projective resolutions over 0→A′→A→A′′→0.

[L2]

The horseshoe lemma produces a degreewise split short exact sequence of projective resolutions (The horseshoe lemma for projective resolutions).

[L3]

Additive functors apply degreewise to complexes and chain maps (An additive functor applies degreewise to complexes and chain maps).

[L4]

Exactness of a sequence of complexes is equivalent to exactness in each degree, and that is the definition of a short exact sequence of complexes (A sequence of chain maps is exact exactly when it is exact degreewise, Short exact sequence of complexes).

Proof

technique · direct
1.1L1L2givenalgebra

By [L2], each degree of the horseshoe row is a split short exact sequence 0→Pn′→Hn→Pn′′→0. Because F is additive by [L1], it preserves the biproduct decomposition carried by that split sequence, so each degree remains exact after applying F.

2.1L3L4step 1.1construct∎

By [L3], the degreewise images from step 1.1 assemble into a sequence of chain maps 0→F(P∙,del′)→F(H∙,del)→F(P∙,del′′)→0. Since it is exact in every degree, [L4] identifies it as a short exact sequence of complexes.

Depends on

Used by

Dependency tree · two levels

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Sources