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I'm working through Massey's "Basic Course in AT." One of the problems is prove that a homeomorphism of the closed disk maps the boundary to the boundary and the interior to the interior. How would one prove this? I can't seem to get this one problem.

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Hint: if there's a homeomorphism from X to X that maps a point x to y, then this induces a homeomorphism from X\{x} to X\{y}. – Chris Eagle Jun 19 '11 at 19:57
Start with the fact that homeomorphisms map open sets to open sets. Take a point p in the interior--take it, please!--and consider its image h(p). The image of the interior of the disk D (i.e., an open disk) is an open set, so h(D) is also open, and h(p) is in h(D), which is open. You can show something similar for points in the exterior of CL(D), i.e., they are mapped to the exterior of Cl(D). Use the def. of a bdry. pt as having 'hoods (neighborhoods) that intersect both the interior and exterior, to show the image has the same properties. – gary Jun 19 '11 at 19:59
@gary Looks like your proof only shows that any homeomorphism of the ambient Euclidian space maps interior of the disk to the interior. – Grigory M Jun 19 '11 at 20:12
Chris, I see where you're going. I know that my ultimate contradiction will come from the fact that I would have an isomorphism $0=\pi(D)=\pi(S^{1})=\mathbb{Z}$. I believe this comes from the homeomorphism inducing an homeomorphism between the disk and the punctured disk which can be deformation retracted to the circle. However I'm not sure how I get to remove the point. If I map an interior point to the boundary what prevents that "hole" from being filled by a boundary point? – SingularDegenerate Jun 19 '11 at 20:24
@Grigory:I am trying to show that if every 'hood of a point x intersects both Int(D) and Ext(D), the same will be the case for every 'hood of h(x). So if points in Int(D) are sent to Int(D), and points in Ext(D) are sent to Ext(D), then, for any 'hood Up of a bdry point p, points of Up in Int(D) will be sent to points in Int(h(D)), and the points of Up in Ext(D) will be sent to Ext(h(D)). So if x is a bdry point of D, so that every 'hood Ux of x intersects both Int(D) and Ext(D), the same will be the case for every 'hood h(Ux) of h(x), i.e., h(Ux) will intersect both Int(h(D)), and Ext(h(D)). – gary Jun 19 '11 at 20:50

This answer extends on Chris Eagles comment:

Let $D^n \subset \mathbb R^n$ denote the $n$-dimensional closed unit disk, that is $D^n = \{ x \in \mathbb R^n \;|\; |x|\leq 1 \}$, with boundary $\partial D^n = S^{n-1} = \{ x \in \mathbb R^n \;|\; |x| = 1 \}$ the $(n-1)$-dimensional sphere.

Let $f: D^n \to D^n$ be a homeomorphism that maps $x \in \partial D^n$ to $f(x) \in D^n \setminus \partial D^n$. Obviously $f$ induces a homeomorphism $\tilde{f}: D^n \setminus \{ x\} \to D^n \setminus \{ f(x) \}$.

Since $x \in \partial D^n$, we have that $D^n \setminus \{ x\}$ is convex and therefore homotopy equivalent to a point. On the other hand we can construct a homotopy equivalence $D^n \setminus \{ f(x) \} \simeq \partial D^n = S^{n-1}$ since $D^n$ is compact and radially convex wrt. a neighborhood of $f(x)$. Thus we get $\{pt\} \simeq D^n \setminus \{ x\} \cong D^n \setminus \{ f(x)\} \simeq S^{n-1}$, which is a contradiction by your technique of choice. For example $\pi_{n-1}(\{pt\}) \not \cong \pi_{n-1}(S^{n-1})$.

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I like this proof but I don't understand the step where you say that $D^n \backslash \{ f(x) \} \simeq S^{n-1}$. I mean, I "know" that $R^n \backslash \{ * \} \simeq S^{n-1}$ or $D^n \backslash \{ * \} \simeq S^{n-1}$. But where do you use that $D^n$ is compact and radially convex? – Rudy the Reindeer Jun 20 '11 at 11:37
Well it is more a technical idea. If you want to construct a homotopy equivalence of $D^n \setminus \{ y\} \simeq S^{n-1}$ where $y$ is an interior point of $D^n$, it is probably easiest to use the following lemma: Let $U \subset \mathbb R^n$ be compact, radially convex wrt. an open subset $V \subset U$ and $y\in V$, then there is a homeomorphism $f: \mathbb R^n \to \mathbb R^n$ satisfying $f(U) = D^n$ and $f(y) = 0$. – Alexander Thumm Jun 20 '11 at 16:46
You can of course construct that homotopy equivalence directly. This idea was the first thing that came to my mind as I was writing the answer. Thinking about it for some more time, it is in fact a bit of an overkill for this problem :D – Alexander Thumm Jun 20 '11 at 16:57
If you know that lemma then what you wrote is less work than constructing the deformation retract ; ) – Rudy the Reindeer Jun 20 '11 at 20:36

Still can't comment, so I'll add this to Chris's helpful hint: you should be able to calculate the fundamental group of the closed unit disk minus a boundary point.

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Dylan, I understand your hint. I get the obvious contradiction when a point of the interior goes to the boundary or a point of the boundary goes to the interior. What I don't get is why you can't have a point P on the boundary interchanged with a point Q in the interior thereby leaving no hole and having neither the interior go to the interior nor the boundary to the boundary. Thoughts? Anyways, I appreciate the help y'alls are giving me. This would really bother me if I didn't finally get the problem! :) – SingularDegenerate Jun 19 '11 at 21:01
I'm not sure how any filling could occur! In the situation you mention, we would still obtain a homeomorphism from $D \setminus \{P\}$ to $D \setminus \{Q\}$. Where does your proof break down? – Dylan Moreland Jun 19 '11 at 21:17
If P is on the boundary and goes to the interior then we have a hole on the boundary. If Q is in the interior and goes to the boundary then we have a hole in the interior. Now if we have P on the boundary and Q on the interior and we place P where Q is and Q where P was then the hypotheses are satisfied: the boundary doesn't go to the boundary and the interior doesn't go to the interior. Plus we have no hole since the hole left by Q is filled by P and the hole left by P is filled by Q. Then our homeomorphism is between the disk and the disk. – SingularDegenerate Jun 19 '11 at 21:24
@MathKid: If what you say is that you swap $P$ and $Q$ then that's not a homeomorphism. Everybody seems to point you to the same conclusion and you don't seem to get it. If the boundary is not mapped to boundary and interior not mapped to interior, then you could construct two non homeomorphic sets which should be homeomorphic by our assumptions, and therefore we reach a contradiction. – Beni Bogosel Jun 19 '11 at 23:26

A homeomorphism $\phi:D \to D$ maps open sets to open sets. Pick a point $P$ in the interior of the disk and a disk $D(P,\delta)$ which lies inside the given disk $D$. If $P$ would be mapped on the boundary of $D$ then $\phi(P)$ would not be in the interior of $\phi(D(P,\delta))$; contradiction with the fact that $\phi(P) \in \phi(D(P,\delta))$ and $\phi(D(P,\delta))$ is open.

This means that interior points are mapped to the interior of the disk.

Pick now $Q$ on the boundary of $D$. If $\phi(Q)$ is on the interior of $D$, then by a similar argument with the one above, since $\phi^{-1}$ is also a homeomorphism it follows that $\phi^{-1}(\phi(Q))=Q$ is in the interior of $D$. Contradiction.

Therefore $\phi$ maps the boundary of the disk onto the boundary of the disk.

[edit] As mentioned in the comments, I used the standard euclidean topology, not the induced one, and that might be a problem. I think that the best solution still remains the one mentioned in the first comment:

That if the homeomorphism $\phi : D\to D$ maps $P$ to $Q$ then $D\setminus\{P\}$ and $D\setminus \{Q\}$ are homeomorphic. If $P$ is on the boundary and $Q$ is in the interior (or the other way around) we get two domains which are not homeomorphic: one is simply connected, one isn't (it has a "hole")

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"then \phi(P)$ would bot be in the interior of..." -- why exactly? it's not true in the induced topology on the disk, I believe. Looks like your proof only shows that any homeomorphism of the ambient Euclidian space preserving the disk maps interior of the disk to the interior (and boundary to boundary). – Grigory M Jun 19 '11 at 20:14
Beni, thank you for your quick response, but I think there might be a flaw. Remember that open sets can be on the boundary as well. It's simply that the basis for these open sets on the boundary are the homeomorphic images of half spaces. Therefore, following your logic we would still have phi mapping open sets to open sets. Wouldn't we? – SingularDegenerate Jun 19 '11 at 20:22
How do we deal with the case where our homeomorphism just exchanges P and Q thereby leaving no "hole"? – SingularDegenerate Jun 19 '11 at 20:51
I think the case I presented covers all others. You want to prove that the homeomorphism maps boundary points to boundary points and interior points to interior points. Assume it doesn't. Then it either maps a boundary point to an interior point or the other way around. This means that the disk minus a boundary point (no holes) is homeomorphic with the disk minus one interior point (one hole): contradiction. – Beni Bogosel Jun 19 '11 at 20:56

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