# DeRham Cohomology

Let $p$ and $q$ be two points of $\mathbb{R}^n$ where let $n\geq 1$. Then

$$\dim H^k(\mathbb R^n - p - q) = \begin{cases}0, &\text{ if }k\text{ is not equal to }n-1,\\ 2,&\text{ if }k = n-1.\end{cases}$$

I'm trying to prove this, but I've thought of letting $S = \{p,q\}$ and using the fact that the open set of $\mathbb R^{n+1} - S\times \mathbb{R}^1$ of $\mathbb R^{n+1}$ is homologically trivial in all dimensions.

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Is this a homework question? –  Ehsan M. Kermani Apr 26 '12 at 2:15
No, but just some extra practice problem –  mary Apr 26 '12 at 2:16
What do you know about DeRham Cohomology? –  Ryan Budney Apr 26 '12 at 2:32
Pretty much, it boiled down to understanding the Mayer Vietoris theorem, and homotopy equivalence theorem –  mary Apr 26 '12 at 2:37

Use Mayer-Vietoris with $U = \mathbb R^n - \{p\}, V = \mathbb R^n - \{q\}$. Then $U \cap V = \mathbb R^n -\{p,q\}$ and $U \cup V = \mathbb R^n$. Then Mayer-Vietoris gives $$H^k(\mathbb R^n) \to H^k(U) \oplus H^k(V) \to H^k(\mathbb R^n - \{p,q\})\to H^{k+1}(\mathbb R^n).$$ But $H^*(\mathbb R^n)$ is trivial so you get $H^*(\mathbb R^n - \{p,q\}) \simeq H^k(U) \oplus H^k(V)$. But $U$ and $V$ are homotopic to $S^{n-1}$, giving you the desired result.

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