Theorem
de Rham theorem
Integration identifies de Rham cohomology naturally with singular cohomology with real coefficients.
Statement
For every smooth manifold , integration over smooth singular simplices defines a cochain map
The de Rham theorem states that the induced map
is an isomorphism for every . Thus the cohomology of the de Rham complex of smooth forms agrees naturally with singular cohomology with real coefficients. No orientation or compactness assumption on is required.
Naturality and products
For a smooth map , pullback of forms and pullback of singular cochains commute with the de Rham isomorphism. On cohomology, the isomorphism also identifies wedge products of forms with cup products, so it is an isomorphism of graded real algebras, not merely of graded vector spaces Bott and Tu, Chapter I.
Proof architecture
One proof uses the Poincaré lemma to establish local exactness and partitions of unity to pass from local to global information. Another proceeds by showing that both theories satisfy compatible Mayer–Vietoris sequences, checking the result on contractible coordinate neighborhoods, and then gluing. Smooth singular cochains compute the same cohomology as ordinary singular cochains.
Coefficients and compact supports
The target is real singular cohomology. The theorem does not identify de Rham cohomology with integral cohomology: torsion information disappears over . There is a separate compact-support version relating compactly supported de Rham cohomology to singular cohomology with compact supports; its support conditions should not be silently inserted into the ordinary theorem.
References
- Raoul Bott and Loring W. Tu, Differential Forms in Algebraic Topology, Springer, 1982. DOI record. Relevant: Chapter I, the de Rham theorem, products, and Mayer–Vietoris method.
- Loring W. Tu, An Introduction to Manifolds, 2nd ed., Springer, 2011. DOI record. Relevant: the chapter on de Rham theory.