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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?

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  • $\begingroup$ Do you know about fundamental groups or homology groups ? $\endgroup$ – Max Feb 16 at 11:24
  • $\begingroup$ We have only treated fundamental groups $\endgroup$ – user408856 Feb 16 at 11:27
  • $\begingroup$ @James Excellent. Then take any base point in the punctured space and any closed path around the point $\;x\;$ . That path cannot be "deformed" to a point and thus the Fund. Group isn't trivial. $\endgroup$ – DonAntonio Feb 16 at 11:41
  • $\begingroup$ contractible implies simply connected $\endgroup$ – 0x539 Feb 16 at 11:53
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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 $\mathbb{R}^n$

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  • $\begingroup$ What do you mean with, "The induced inclusion is non zero"? If $[\gamma]\neq 0$, then $i_*([\gamma])=[i\circ\gamma]\neq 0$? Could you add some details? $\endgroup$ – user408856 Feb 16 at 11:54
  • $\begingroup$ The map induced by the inclusion. And no I didn't say injective (well actually it will be), it just means it's not rhe zero map. You could prove that it's actually an isomorphism if that's easier for you (it is true) $\endgroup$ – Max Feb 16 at 12:02
  • $\begingroup$ I don't really get what you mean $\endgroup$ – user408856 Feb 16 at 12:19
  • $\begingroup$ The inclusion $(B\setminus\{x\},y)\to (\mathbb{R}^2,y)$ induces a morphism on $\pi_1$. The point is to prove that it's not the $0$ morphism; or if you can, prove that it's an isomorphism $\endgroup$ – Max Feb 16 at 12:35
  • $\begingroup$ I have shown that $\pi_1(B\setminus\{x\})\cong\pi_1(\mathbb{R}^2\setminus\{x\},y)$, how can I conclude that this maps factors through $\pi_1(U\setminus\{x\},y)?$ $\endgroup$ – user408856 Feb 18 at 13:36

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