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Maybe someone has a quick answer. Thanks.

For noncompact manifolds, is the De Rham cohomology isomorphic to the singular cohomology? Is the De Rham cohomology defined with the cochain of compactly supported differential forms isomorphic to the singular cohomology with compact support?

The references I checked on De Rham isomorphism state the theorem only for compact manifolds.

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    $\begingroup$ Bott-Tu handles the general case assuming only the existence of a "good cover", i.e. a cover by contractible open sets, such that all multiple intersections are contractible. $\endgroup$
    – ThiKu
    Commented Oct 29, 2013 at 17:23
  • $\begingroup$ Actually Bott-Tu only proves the isomorphism between deRham and Cech cohomology, but literally the same argument (replacing differential forms by singular cochains everywhere) proves the isomorphism between singular and Cech cohomology. $\endgroup$
    – ThiKu
    Commented Oct 29, 2013 at 17:25

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The answer is Yes. You can find a proof (via sheaves theory) for example in Demailly's book Chapter IV (6.7) for ordinaire cohomology and (7.10) for compactly supported cohomology http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf

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  • $\begingroup$ Thanks. Demailly's book uses Weil's proof. It's abstract but elegant. $\endgroup$
    – John
    Commented Oct 30, 2013 at 6:44
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In Lee smooth manifolds (Theorem 16.12) the de Rham isomorphism is proved for any smooth manifold. There are some comments on compactly supported cohomology and Poincare Duality in the exercises.

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