Theorem
Homotopy invariance of de Rham cohomology
Smoothly homotopic maps induce the same pullback homomorphism on de Rham cohomology.
Statement
Let be smooth maps between smooth manifolds. If and are joined by a smooth homotopy, then their pullbacks induce the same homomorphism on every de Rham cohomology group:
Indeed, the associated de Rham homotopy operator satisfies
Hence the difference of the endpoint pullbacks is cochain-homotopic to zero and vanishes on cohomology. The conclusion holds in every degree and is natural with respect to composition by smooth maps.
Consequences
A smooth homotopy equivalence induces an isomorphism on de Rham cohomology. In particular, a smooth deformation retract has the same de Rham cohomology as the retract. This makes a contravariant functor on the smooth homotopy category, not merely on the category of smooth manifolds Bott–Tu, Chapter I, §4.
Canonical example
The radial homotopy contracts to the origin. Homotopy invariance therefore gives
At the level of forms, the same homotopy operator supplies primitives for closed forms of positive degree, which is the usual star-shaped-domain proof of the Poincaré lemma.
Scope
The theorem concerns smooth homotopies and ordinary de Rham cohomology. A continuous homotopy between smooth maps can be replaced by a smooth one under the standard smoothing theorem, but that extra step is not part of the chain-homotopy formula. Compact-support, relative, and boundary-condition variants require the homotopy operator to preserve the relevant support or restriction conditions Tu, Chapter 17.
References
- Raoul Bott and Loring W. Tu, Differential Forms in Algebraic Topology, Springer, 1982. Springer DOI record. Relevant: Chapter I, §4, homotopy operators and homotopy invariance.
- Loring W. Tu, An Introduction to Manifolds, 2nd ed., Springer, 2011. Springer DOI record. Relevant: Chapter 17, homotopy operator and induced maps on de Rham cohomology.