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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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    @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.2011-06-19

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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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    @AlexanderThumm How do you know that a disc without a boundary is convex?2018-09-09
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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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    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.2011-06-19