Punctured open set is not contractible

Question

Let $U\subseteq\mathbb{R}^2$ be an open subset, and let $x\in U$. Then $U\setminus\{x\}$ is not contractible. A space $X$ is called contractible if the identity map on $X$ is homotopic to a constant map $X\to X$. So, I have to show that $Id_{U\setminus\{x\}}$ can not be homotopic to a constant map. What are tools to show that maps are not homotopic?

Answer 1

You can use the fundamental groups to solve this, here's a way :

Find a small open ball $B$ centered at $x$ included in $U$ ($U$ is open, so that exists).

Prove that if $y\in B\setminus\{x\}$, then the map induced by the inclusion $\pi_1(B\setminus\{x\}, y)\to \pi_1(\mathbb{R}^2\setminus\{x\}, y)$ is non zero. This will use some knowledge of the fundamental group of $\mathbb{R}^2 \setminus\{x\}$ and thus of $S^1$.

Prove that this map factors through $\pi_1(U\setminus\{x\}, y)$, and conclude.

Note that when you learn about higher homotopy groups, you will be able to adapt this proof to $\mathbb{R}^n$