Compact cohomology group of connected n-dimensional connected oriented manifold

Question

I know how to show $H_c^n(M)\simeq\mathbb{R}$, where M is a oriented connected n-dimensional manifold, by showing the integration map is isomorphism. However, I found in the book that this is a consequence of Poincaré duality. I don't know how to deduce this result from that theorem. Can any one help?

Answer 1

I do not know what version of Poincare duality you've learnt, but the following version of Poincare duality is quit standard, I think:

If $M$ is as you said, then the cap product with the fundamental class $\cap [M]$ is an isomorphism between $H^k_c(M)$ and $H_{n-k}(M)$.

Then the isomorphism follows trivially by $H_0(M)=\mathbb{R}^{\mathrm{number\;of\;connected\;components}}\cong \mathbb{R}$.

Answer 2

It's a special case.

In the cohomology form of Poincare duality defines a homeomorphism: $$H_c^n(M) \simeq H^0(M)$$ In fact the cup product is just the integration map you mentioned $$H_c^n(M) \times H^0(M) \rightarrow \mathbb{R}:\ (\omega, f) \rightarrow \int_Mf\cdot\omega$$

Now $H^0(M) = \mathbb{R}$ as any closed 0-form (functions with derivative 0) are just constants on any connected $M.$