# Top Cohomology group of a “punctured” manifold is zero?

The following question is from my algebraic topology exam which I was unable to solve.

Let $X$ be a orientable connected closed $n$-manifold. Let $p \in X$. Show that $H^n (X-{p},R)=0$, where $R$ is some ring.

I observed that since $X$ is orientable so $X-p$ is also orientable hence $H_n(X-p,R) \cong R$. I think that Poincare duality may help us to prove this result but I am unable to prove it. Any ideas?

#### Solutions Collecting From Web of "Top Cohomology group of a “punctured” manifold is zero?"

There is the long exact sequence in homology of the pair $(X, X \setminus p)$ (everything is with $\mathbb{Z}$-coefficients unless indicated otherwise):
$$0 \to H_n(X \setminus p) \to H_n(X) \to H_n(X, X \setminus p) \to H_{n-1}(X \setminus p) \to H_{n-1}(X) \to 0$$
(recall that the local homology groups satisfy, by excision, $H_n(X, X \setminus p) = \mathbb{Z}$ and $H_k(X, X \setminus p) = 0$ for $k \neq n$).

The fundamental class $[X] \in H_n(X)$ is sent to a generator of the local homology group $H_n(X, X \setminus p) = \mathbb{Z}$ (either by definition, or cf. Hatcher, Theorem 3.26 for example). Hence $H_n(X) \to H_n(X, X \setminus p)$ is an isomorphism and thus $H_n(X \setminus p) = 0$ and $H_{n-1}(X \setminus p) \cong H_n(X)$.

The universal coefficient theorem reads
$$H^n(X \setminus p; R) \cong \operatorname{Hom}_{\mathbb{Z}}(H_n(X \setminus p), R) \oplus \operatorname{Ext}^1_{\mathbb{Z}}(H_{n-1}(X \setminus p), R).$$
We already know the first summand vanishes. By Poincaré duality and universal coefficients, $H_{n-1}(X) \cong H^1(X) \cong \hom(H_1(X), \mathbb{Z})$. Thus $H_{n-1}(X)$ is torsion-free, and it is also finitely generated because $X$ is compact, hence it is free as an abelian group (as $\mathbb{Z}$ is a PID). So the second summand vanishes too ($H_{n-1}(X \setminus p) = H_{n-1}(X)$ is free) and finally $H^n(X \setminus p) = 0$.