Alphabeta Math
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:MN, g:NM with smooth homotopies gfidM and fgidN.

[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:MN and G:NP, (GF)=FG, and idM=idHk(M).

[F3]

Pullback is a homomorphism of de rham cohomology algebras: Smooth pullback induces a unital graded real algebra homomorphism HdR(N)HdR(M).

Proof

technique · direct
1.1

By smooth-homotopy invariance and functoriality, fg=(gf)=(idM)=idH(M).

F1F2given
2.1

The other homotopy gives gf=(fg)=idH(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.

F1F2F3step 1.1given

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

Used by

Dependency tree · two levels

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Sources