# Top deRham cohomology group of a compact orientable manifold is 1-dimensional

Let $M$ be a compact smooth orientable manifold of dimension $n$. I am looking for a simple proof that $H_{dR}^n(M) \cong \mathbb R$. Equivalently, an $n$-form which integrates to 0 is exact. I can show this via a rather indirect argument as follows: we know $H_{dR}^n(M) \cong H^n(M, \mathbb R)$, where $H^n$ denotes the singular cohomology. By the universal coefficient theorem (and the fact that $\mathbb R$ is a field) this is isomorphic to $Hom(H_n(M, \mathbb Z) , \mathbb R)$. From the (rather lengthy) proof in Section 3.3 of Hatcher's Algebraic Topology, we find that $H_n(M, \mathbb Z)$ is isomorphic to $\mathbb Z$, and so $Hom(H_n(M, \mathbb Z) , \mathbb R) \cong \mathbb R$. However, it seems like there should be a simpler way to prove this. Does anyone know of one?

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You probably want $M$ to be connected. –  Zhen Lin Nov 3 '12 at 23:30
Integrate!! It is fairly straightforward to check that $\omega \mapsto \int_M \omega$ is the desired isomorphism. –  Matt Nov 3 '12 at 23:30
I know that the isomorphism takes this form; the part I am having trouble with is seeing why it is injective. –  user15464 Nov 3 '12 at 23:41
I don't think there is a simple argument. (Standard proof of Poincare duality uses Mayer–Vietoris for induction by covering, I believe.) –  Grigory M Nov 5 '12 at 15:02