PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Zero cocycles in the Hom complex are chain maps

Statement

Let $u = (u_{n})_{n}$ be a degree- $0$ element of $\underset{}{Hom} (C, D)_{0}$ . Then $\partial u = 0$ if and only if $u : C_{∙} \to D_{∙}$ is a chain map.

Facts & Assumptions

Given: A degree- $0$ graded morphism $u = (u_{n})_{n} : C_{∙} \to D_{∙}$ .

[L1]

In degree $0$ , the Hom-complex differential is $(\partial u)_{n} = d_{n}^{D} u_{n} - u_{n - 1} d_{n}^{C}$ (The Hom complex of chain complexes).

[L2]

A chain map is a degreewise family satisfying $d_{n}^{D} u_{n} = u_{n - 1} d_{n}^{C}$ for every $n$ (Chain map).

Proof

technique · direct

1.1L1givenalgebra

By [L1], the equality $\partial u = 0$ means exactly that $d_{n}^{D} u_{n} - u_{n - 1} d_{n}^{C} = 0$ for every $n \in Z$ .

2.1L2step 1.1algebra∎

Rewriting the equality in step 1.1 gives $d_{n}^{D} u_{n} = u_{n - 1} d_{n}^{C}$ for every $n$ , and [L2] is precisely this condition. Thus $\partial u = 0$ if and only if $u$ is a chain map.

Zero cocycles in the Hom complex are chain maps

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources