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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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Dimension shift for a cohomological delta functor effaced in the middle

Statement

Let T=(Tn,∂) be a cohomological delta functor. For a short exact sequence 0→A→I→C→0 and an integer n>0, the connecting map ∂n−1:Tn−1(C)→Tn(A) has the following properties:

  1. if Tn−1(I)→Tn−1(C) is the zero map, then ∂n−1 is a monomorphism,
  2. if Tn(A)→Tn(I) is the zero map, then ∂n−1 is an epimorphism,
  3. if both conditions hold, then ∂n−1 is an isomorphism.

Facts & Assumptions

Given: A short exact sequence 0→A→I→C→0 and an integer n>0.

[L1]

A cohomological delta functor attaches an exact segment Tn−1(I)→Tn−1(C)→∂n−1Tn(A)→Tn(I) to the given short exact sequence (Cohomological delta functor).

Proof

technique · direct
1.1L1givenalgebra

By [L1], the kernel of ∂n−1 is the image of Tn−1(I)→Tn−1(C). If that map is zero, then ker⁡(∂n−1)=0, so ∂n−1 is monic.

1.2L1givenalgebra

By [L1], the image of ∂n−1 is the kernel of Tn(A)→Tn(I). If the latter map is zero, this kernel is all of Tn(A), so ∂n−1 is epic.

2.1step 1.1step 1.2∎

When both hypotheses hold, steps 1.1 and 1.2 show that ∂n−1 is an isomorphism.

Depends on

Used by

Dependency tree · two levels

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Sources