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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The preconnecting arrow annihilates boundaries

Statement

For a short exact sequence of complexes, the composite $B_{n} (C) \to Z_{n} (C) \overset{{\tilde{\partial}}_{n}}{\to} H_{n - 1} (A)$ is zero.

Facts & Assumptions

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

[L1]

The preconnecting arrow is ${\tilde{\partial}}_{n} = δ_{n}^{s n a k e} q_{n}$ , where $q_{n} : Z_{n} (C) \to H_{n} (C)$ is the homology quotient (The preconnecting arrow on cycles).

[L2]

The homology quotient $q_{n}$ is the cokernel of the canonical boundary-to-cycle map $β_{n} : B_{n} (C) \to Z_{n} (C)$ (Homology object of a chain complex).

Proof

technique · direct

1.1L2given

Because $q_{n}$ is the cokernel of $β_{n}$ , one has $q_{n} β_{n} = 0$ .

2.1L1step 1.1algebra∎

Using [L1], ${\tilde{\partial}}_{n} β_{n} = δ_{n}^{s n a k e} q_{n} β_{n} = 0.$ Thus the composite from the boundary subobject to $H_{n - 1} (A)$ is the zero morphism.

The preconnecting arrow annihilates boundaries

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources