TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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De rham homotopy formula for a smooth homotopy

Statement

If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

Facts & Assumptions

Given: A smooth homotopy $F$ and a smooth form $ω$ on $N$ .

[F1]

De rham homotopy formula on a product: For endpoint inclusions $i_{t} : M \to M \times [0, 1]$ , $i_{1}^{*} - i_{0}^{*} = d K + K d$ on smooth forms of every degree.

[F2]

The exterior derivative commutes with pullback: For every smooth map $F : M \to N$ and every form $ω$ on $N$ , $d (F^{*} ω) = F^{*} (d ω) .$

Proof

technique · direct

1.1F1given

Apply the product identity to the smooth form $F^{*} ω$ : $i_{1}^{*} F^{*} ω - i_{0}^{*} F^{*} ω = d K F^{*} ω + K d F^{*} ω$ . Evaluation on tangent tuples shows $i_{t}^{*} F^{*} ω = F_{t}^{*} ω$ , including functions.

2.1F2step 1.1∎

Naturality gives $d F^{*} ω = F^{*} d ω$ . Substitution yields $F_{1}^{*} ω - F_{0}^{*} ω = d (K F^{*} ω) + K F^{*} d ω$ , the required identity for every degree; zero terms require no separate extension.

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

De rham homotopy formula for a smooth homotopy

Statement

Facts & Assumptions

Proof

Source locator

Depends on

Dependency tree · 2 levels

Sources