PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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.

Identities and composites of chain maps are chain maps

Statement

For every chain complex $C_{∙}$ , the identity family $1_{C_{n}}$ is a chain map $1_{C} : C \to C$ . If $f : C \to D$ and $g : D \to E$ are chain maps, then the componentwise composite $(g_{n} f_{n})_{n}$ is a chain map $g \circ f : C \to E$ .

Facts & Assumptions

Given: Chain complexes $C_{∙}, D_{∙}, E_{∙}$ and chain maps $f : C \to D$ , $g : D \to E$ .

[L1]

A chain map is a degree-zero family commuting with the differentials (Chain map).

Proof

technique · direct

1.1L1givenalgebra

For the identity family, $d_{n}^{C} \circ 1_{C_{n}} = 1_{C_{n - 1}} \circ d_{n}^{C}$ holds trivially in every degree, so [L1] makes $1_{C}$ a chain map.

2.1L1givenalgebra∎

Since $f$ and $g$ are chain maps, [L1] gives $d_{n}^{E} g_{n} = g_{n - 1} d_{n}^{D}$ and $d_{n}^{D} f_{n} = f_{n - 1} d_{n}^{C}$ . Therefore $d_{n}^{E} (g_{n} f_{n}) = g_{n - 1} (d_{n}^{D} f_{n}) = g_{n - 1} f_{n - 1} d_{n}^{C},$ so the composite family again satisfies [L1].

Identities and composites of chain maps are chain maps

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources