The $n-1$ homology group of $U\setminus\{x\}\subset \mathbb{R}^n$ where $U$ is open.

Question

Let $U$ be a open subset of $\mathbb{R}^n$ where $n\geq 2$ and let $x\in U$. Show that $H_{n-1}(U\setminus\{x\})$ is not the trivial group.

What I know is that $H_{n-1}(\mathbb{S}^{n-1})=\mathbb{Z}$ and that $U$ is homeomorphic to an open subset of the sphere $\mathbb{S}^{n}$. Can these two facts help?

Answer 1

Let me outline a slightly different solution.

If $A=U$ and $B=\mathbb{R}^n-\{x\}$, then

• $A\cup B=\mathbb{R}^n$
• $B$ deformation retracts to $S^{n-1}$
• $A\cap B=U-\{x\}$

Now what does Mayer-Vietoris tell you about $H_{n-1}$?