# Poincare Duality and Integration Isomorphism

Corollary 10.27 in Jeffrey Lee's book "Manifolds and Differential Geometry" states that

If $$M$$ is a connected oriented $$n$$-manifold with finite good cover, then $$H^n_c (M) \simeq \mathbb{R}$$. This isomorphism is given by integration over $$M$$.

It is consequence of Poincare Duality $$PD:H^k(M)\to (H^{n-k}_c(M))^*,$$ but im failing to see how it follows.

For $$k=0$$ we have $$PD:H^0(M)\to (H^{n}_c(M))^*$$. The "integration over M"-map, let's call it $$\Phi$$, is $$[\omega]\mapsto \int_M\omega$$ an element of $$(H^{n}_c(M))^*$$. Hence we can see $$PD(1)=\Phi$$. But why is it an isomorphism?

Assuming you know that $$\Phi \neq 0$$ you can see it formally using the fact that $$H^{k}(M)$$ is a vector space and not just an abelian group, so it has some nice behaviour. In particular since $$H^0(M) \cong \mathbb{R}$$ (recall $$M$$ is connected) Poincare duality gives an isomorphism $$(H^n_c(M))^* \cong \mathbb{R}$$, and it is generated as a vector space by $$\Phi$$. It follows that $$H^n_c(M)$$ is also abstractly isomorphic to $$\mathbb{R}$$ since $$dim(V^*) = dim (V)$$, and then $$\Phi \colon H^n_c(M) \to \mathbb{R}$$ is a non-zero linear transformation between one-dimensional vector spaces so it is an isomorphism.