LemmaStatement: AI-generatedProof: AI-generated

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.

Homotopic maps act equally on cohomology

PDF p. 4: Definition 2.2

Statement

If cochain maps $f, g : C \to D$ are cochain-homotopic, then $H^{i} (f) = H^{i} (g)$ for every integer $i$ . Here a cochain homotopy is a family $h^{i} : C^{i} \to D^{i - 1}$ with $f^{i} - g^{i} = d_{D}^{i - 1} h^{i} + h^{i + 1} d_{C}^{i}$ .

Facts & Assumptions

[F1]

Cochain maps induce maps on cohomology by applying the map to cocycle representatives (Cohomology maps induced by cochain maps).

Proof

Given: Cochain maps $f, g$ and a cochain homotopy $h$ between them.

Goal: Prove equality of their induced cohomology maps.

1.1given

For a cocycle $x \in C^{i}$ , the homotopy identity gives $f^{i} (x) - g^{i} (x) = d_{D}^{i - 1} h^{i} (x)$ because $d_{C}^{i} (x) = 0$ . This difference is a coboundary.

2.1F1step 1.1∎

Every cohomology class has a cocycle representative, and by [F1] the induced maps take it to the classes of $f^{i} (x)$ and $g^{i} (x)$ . These classes agree by step 1.1.

Depends on

Cohomology maps induced by cochain maps

Used by · 1

Cohomology descends to the derived category

Dependency tree · 3 levels

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