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Let $M$ be a non-compact differential manifold. It is true that in general $H^q_c(M) \neq H^q(M)$, where $H^q_c$ is the de Rham's cohomolgy with compact support group and $H^q$ is the usual de Rham's cohomology group.

We have just begun the subject, so I don't have much confidence with it. I wanted to ask: is it true that for any non-compact $M$ there exists a $q$ for which $H^q_c (M) \neq H^q(M)$? Or are there any examples of non-compact manifolds for which the two cohomologies are the same for all $q$?

EDIT: ok, from the comments I gathered that if such an example exists it must be non-orientable (a reference to a proof would be nice, even though I think the resul is quite non-elementary). The question still remains open though (that's my main reason to editing: I think this question didn't get enough attention).

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    @DanPetersen: In $f$act, that H^n(M)=0 $f$or $n$o$n$compact is proved in Spivak (Vol1, chapter 8), but he uses a "Problem 20" in Chapter 8 that I think it is not correct (a manifold with two ends and infinite topology should be a counterexample). Do you know an elementary proof (e.g., no Poincare duality) of this without using the Problem 20?2014-10-20

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Let $M$ be a connected non-compact manifold. Then $H_c^0(M)\cong 0$, and $H^0(M)\cong\mathbb{R}$.

Let $f:M\rightarrow \mathbb{R}$ be a function with $df=0$. Then $f$ is constant (this uses the connectedness of $M$). If $f$ is assumed to be compactly supported, this constant must be zero. If $f$ is not assumed to be compactly supported, all constants occur.