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

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Smoothly homotopic maps induce the same de rham map

Statement

Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.

Facts & Assumptions

Given: Smooth maps $f_{0}, f_{1} : M \to N$ joined by a smooth homotopy $F$ .

[F1]

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

[F2]

Pullback induces a well defined map on de rham cohomology: For a smooth $F : M \to N$ , the formula $F^{*} [ω] = [F^{*} ω]$ defines a linear map $H_{d R}^{k} (N) \to H_{d R}^{k} (M)$ for every integer $k$ .

[F3]

Chain-homotopic maps induce the same map on homology: If $f, g : C_{∙} \to D_{∙}$ are chain-homotopic chain maps, then for every $n \in Z$ , $H_{n} (f) = H_{n} (g) : H_{n} (C) \to H_{n} (D) .$

Proof

technique · direct

1.1

For every closed $k$ -form $ω$ , the homotopy formula gives $f_{1}^{*} ω - f_{0}^{*} ω = d (K F^{*} ω)$ , since $d ω = 0$ . For $k = 0$ the primitive term is zero, so the functions themselves agree.

F1given

2.1

The two pullbacks therefore give the same quotient class. Equivalently, under $C_{n} = Ω^{- n}$ the operator $K F^{*}$ has degree $+ 1$ and the formula in [F1] is the chain-homotopy identity; the chain-homotopy theorem gives equality on $H_{- k} = H^{k}$ . The well-defined maps on those classes are exactly the induced de Rham maps.

F1F2F3step 1.1∎

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.

Depends on

Used by

Dependency tree · two levels

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Sources

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)