# $D^2$ not topological manifold without boundary

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?

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.
• 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. – user408856 Feb 20 at 19:44
• 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. – csprun Feb 20 at 19:49
• 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 – user408856 Feb 20 at 19:52
• How can I easily show that $X\setminus\{1\}$ is star shaped? – user408856 Feb 20 at 19:55