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I want to prove that the two dimensional disk $D^2$ is not a topological manifold without boundary, i.e. there is a point $x\in D^2$ such that $x$ has no neighbourhood $U$ with $U$ homeomorphic to $\mathbb{R}^2$.

I choose $x=1\in\partial D^2$. Without loss of generality, let $U$ be a contractible open neighbourhood and suppose $U$ is homeomorphic to $\mathbb{R}^2$ with $f:U\to\mathbb{R}^2$ a homeomorphism. Is it true that $U\setminus\{1\}$ is contractible? And what is the reason it is so?

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Why not pick $U$ to be a small open disk intersected with $D^2$? E.g., $U = B_\epsilon(x) \cap D^2$. Then $U\setminus \{x\}$ is contractible because it is star-shaped.

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  • $\begingroup$ But I have to show that there does not exist an open neighbourhood of 1 homeomorphic to $\mathbb{R}^2$, so I can not pick it. $\endgroup$ – user408856 Feb 20 at 19:44
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    $\begingroup$ Yes, but any open neighborhood of $1$ contains one of this form. And no open subset of $\mathbb{R}^2$ (the image under the proposed homeomorphism of this small open disk) has the property that removing a point gives you a contractible set. $\endgroup$ – csprun Feb 20 at 19:49
  • $\begingroup$ So, if I read correctly, this is what you are saying: Let $U$ be an open neighbourhood of 1 homeomorphic to $\mathbb{R}^2$, with $f$ the homeomorphism. Then, there exists $\epsilon>0$ such that $X:=B_\epsilon(1)\cap D^2\subset U$. By restricting $f$, we get a homeomorphism between $X$ and some open $V$ in $\mathbb{R}^2$. Then, $X\setminus\{1\}$ is homeomorphic to $V\setminus\{f(1)\}$, but $X\setminus\{1\}$ is contractible since it is star shaped and $V\setminus\{f(1)\}$ is not contractible $\endgroup$ – user408856 Feb 20 at 19:52
  • $\begingroup$ @James That's right. $\endgroup$ – csprun Feb 20 at 19:54
  • $\begingroup$ How can I easily show that $X\setminus\{1\}$ is star shaped? $\endgroup$ – user408856 Feb 20 at 19:55

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