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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 , with smooth homotopies and .
Smoothly homotopic maps induce the same de rham map: Smoothly homotopic smooth maps induce equal maps on de Rham cohomology in every degree.
De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth and , , and .
Pullback is a homomorphism of de rham cohomology algebras: Smooth pullback induces a unital graded real algebra homomorphism .
Proof
By smooth-homotopy invariance and functoriality, .
The other homotopy gives . 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
Used by
Dependency tree · two levels
12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)