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

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De rham cohomology is smooth homotopy invariant

Statement

A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

Facts & Assumptions

Given: Smooth maps $f : M \to N$ , $g : N \to M$ with smooth homotopies $g f ≃ {id}_{M}$ and $f g ≃ {id}_{N}$ .

[F1]

Smoothly homotopic maps induce the same de rham map: Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.

[F2]

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F : M \to N$ and $G : N \to P$ , $(G \circ F)^{*} = F^{*} \circ G^{*}$ , and ${id}_{M}^{*} = {id}_{H^{k} (M)}$ .

[F3]

Pullback is a homomorphism of de rham cohomology algebras: Smooth pullback induces a unital graded real algebra homomorphism $H_{d R}^{∙} (N) \to H_{d R}^{∙} (M)$ .

Proof

technique · direct

1.1F1F2given

By smooth-homotopy invariance and functoriality, $f^{*} g^{*} = (g f)^{*} = ({id}_{M})^{*} = {id}_{H^{∙} (M)}$ .

2.1F1F2F3step 1.1given∎

The other homotopy gives $g^{*} f^{*} = (f g)^{*} = {id}_{H^{∙} (N)}$ . Both maps preserve multiplication and units, so these two identities exhibit inverse graded algebra homomorphisms. The conclusion includes empty manifolds, since a homotopy equivalence to an empty manifold forces both to be empty.

Source locator

Lee, Theorem 17.11, pp.445–446; this item assumes smooth homotopies explicitly, so does not use a continuous smoothing theorem.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition