Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
Can someone please help me on this – mary Apr 26 '12 at 2:26
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
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.