Show that if $N$ is connected non-compact, orientable n-dimensional manifold, then $H_{n}(N; \mathbb{Z}) = 0.$

Question

Show that if $N$ is a connected, non-compact, orientable $n$-dimensional manifold, then $H_{n}(N; \mathbb{Z}) = 0$.

I got the following hint (which I was also asked to prove):

$H_{n}(N; \mathbb{Z}) \cong H_{c}^{0}(N; \mathbb{Z})$ and $H_{c}^{0}(N; \mathbb{Z})$ is the kernel of $\delta\colon C_{c}^{0}(N; \mathbb{Z}) \rightarrow C_{c}^{1}(N; \mathbb{Z}),$ which is generated by constant functions.

Then I was asked which constant functions have compact support.

Could anyone help me in proving the hint please, and tell me which constant functions have compact support?

Any other hints for the main problem will be greatly appreciated.

Answer 1

I will give some further hints and outlines while trying to still leave some work for you. (All coefficients are in $\mathbb{Z}$.)

First, $H_n(N) \cong H_c^0(N)$ by Poincare Duality (here we use orientability).

Since there are no coboundaries in degree $0$, $H^0_c(N) = ker(\delta)$.

So what is $ker(\delta)$? Describing it depends on the specific version of cohomology you're working with, but here's how it looks for singular cohomology. Recall that $C_0(N)$ is the free abelian group generated by all functions $\sigma_0\colon\Delta^0 \to N$, and that $C^0(N)= Hom(C_0(N),\mathbb{Z})$ and so in particular $\varphi\in C^0(N)$ is determined by a (not necessarily continuous) function $N\to \mathbb{Z}$. The coboundary $\delta(\varphi) \in C^1(N)$ is defined by assigning to any $1$-simplex $\sigma_1\colon \Delta^1 \to N$ the value $\varphi(\partial_1\sigma_1) - \varphi(\partial_0\sigma_1)$. What does this tell you if $\delta(\varphi) = 0$? (Hint: since we've assumed $N$ is connected and every manifold is locally-path connected it follows that $N$ is path connected. Also note that this argument doesn't depend on the support of $\varphi$, indeed both $H^0(N)$ and $H^0_c(N)$ are generated by constant functions.)

Now how does this solve the problem? Suppose $N$ is non-compact and let $\varphi\in H^0_c(N)$ be a cocycle, so in particular it is constant. By definition there is a compact $K\subset N$ such that $\varphi(\sigma_0) = 0$ for any $\sigma_0\colon \Delta^0\to N$ whose image is not in $K$, so if $N$ is not compact then the set of points where $\varphi$ vanishes is not empty. Since $\varphi$ is constant, of course that means it is the $0$ function.