LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The Hom cochain differential squares to zero

Statement

If $C$ is a chain complex and $G$ a module, then the differential $δ^{n} (f) = f \circ d_{n + 1}$ on ${Hom}_{R} (C_{n}, G)$ satisfies $δ^{n + 1} δ^{n} = 0$ .

Proof

Given: $f : C_{n} \to G$ and the chain differential $d$ .

1.1

By definition, $δ^{n + 1} (δ^{n} f) = (f \circ d_{n + 1}) \circ d_{n + 2} = f \circ (d_{n + 1} d_{n + 2})$ .

given

2.1

The chain-complex identity $d_{n + 1} d_{n + 2} = 0$ makes this composite zero, so the Hom groups form a cochain complex.

step 1.1algebra∎

Depends on

The Hom cochain complex of a chain complex

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Weibel, An Introduction to Homological Algebra (standard reference, not scraped)