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Show that every homeomorphism $f: D^2\to D^2$ restricts to a homeomorphism $f_{|\partial D^2}: \partial D^2\to\partial D^2$

I want to proof the following statement. I have to show, that $f_{|\partial D^2}$ is a bijection and continuous and $f^{-1}_{|\partial D^2}$ is also continuous.

That $f_{|\partial D^2}$ is injective is clear, since $f$ is injective and we just observe $f$ on a subset of its preimage, but we would need that $f_{|\partial D^2}$ is surjective first.

How can I proof, that this function stays surjective? Let $y\in\partial D^2$. I have to find $x\in \partial D^2$ such that $f(x)=y$.

I tried to suppose, that $x\notin\partial D^2$ and then take on open set $U\ni y$ and tried to observe $f^{-1}(U)\ni x$ (open).

That $f_{\partial D^2}$ is continuous is implied by $f$ beeing continuous. So when I have that my function is bijective I can deduce it is a homeomorphism, since $\partial D^2$ is compact and Hausdorff.

What do you think?

Thanks in advance.

Cornman
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    You need to know that the boundary necessarily maps to the boundary. Good tips here: https://math.stackexchange.com/questions/1079235/boundary-points-of-a-manifold – Randall Oct 14 '18 at 01:43
  • @Randall Thank you for the link, but the answers did not help me. I am not familiar with manifolds so far. Do you know a source for a proof too? – Cornman Oct 14 '18 at 02:36
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    @Cornman: What tools do you know? Do you know fundamental group? – Cheerful Parsnip Oct 14 '18 at 03:03
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    @CheerfulParsnip Yes, I do know the fundamental group. – Cornman Oct 14 '18 at 03:36
  • Oh, then this just got a lot easier – Randall Oct 14 '18 at 03:43
  • @Cornman: If you delete a point from the interior of the disk, then the fundamental group is $\mathbb Z$. If you delete a point from the boundary, the fundamental group is $0$. So any homeomorphism of the disk will have to send boundary points to boundary points. – Cheerful Parsnip Oct 14 '18 at 03:49
  • @CheerfulParsnip Thank you. How could I show that? If $x\in D^{2\circ}$ (the interior of $D^2$) then $D^{2}-x\simeq S^1$ and I know, that $\pi_1(S^1)=\mathbb{Z}$. If $x\in\partial D^2$, then $D^2-x\simeq D^2$? – Cornman Oct 14 '18 at 04:09
  • @Cornman: If $x$ is on the boundary, then for $D^2\setminus {x}$, you can show directly that the fundamental group is trivial since any loop in the space can be homotoped to the center. – Cheerful Parsnip Oct 14 '18 at 04:14
  • @CheerfulParsnip What do you mean by "homotoped to the center"? And how do I conclude, that $S^1\stackrel{\simeq} S^1$ from this? Excuse me, I am not that much into the fundamental group, but I try to learn and get a hang of it at the moment. – Cornman Oct 14 '18 at 04:24
  • @CheerfulParsnip I know, that the image would have the same fundamental group under a homeomorphism, so it has to be $S^1$ too, right? – Cornman Oct 14 '18 at 04:25
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    @Cornman: The fundamental group is homotopy classes of loops. If you remove a boundary point, any loop is homotopic within the space to the constant loop at the center. So by definition, the fundamental group is trivial. – Cheerful Parsnip Oct 14 '18 at 05:04
  • @CheerfulParsnip You should give an official answer. – Paul Frost Oct 14 '18 at 22:43

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