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I'm starting to study Algebraic Topology. After doing some problems and studying the theory I've arrived at:

Let $D^2$ be the unit disk in $R^2$, $\partial D^2$ the topological boundary of $D^2$ (i.e. $S^1$, the unit circle) and $f:D^2\to D^2$ an homeomorphism. Then $f(\partial D^2)=\partial D^2$.

This should be true because $D^2-\partial D^2\approx f(D^2-\partial D^2)$ and since $D^2-\partial D^2$ is path connected, $f(D^2-\partial D^2)$ is path connected as well but if some point in the boundary is mapped to the interior of the disk then it seems that $f(D^2-\partial D^2)$ has two path connected components instead of just one.

This might not be true, and if that's the case provide an example. Also try to keep the answers as elementary as possible, assuming only knowledge of point-set topology. I'm also interested in the same problem but in $D^n$.

Zero
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  • Why don't you take the identity map from $D^2$ to $D^2$ –  Sep 02 '15 at 16:20
  • What do you mean? The identity obviously satisfies what I'm trying to prove. – Zero Sep 02 '15 at 16:21
  • You're, essentially, asking about the invariance of domain theorem. As you note this particular case can be solved with the Jordan curve theorem, but one needs to work much harder for other manifolds with boundary, especially in higher dimensions. –  Sep 02 '15 at 16:22
  • Okay sorry i thought you meant "an homeomorphism" –  Sep 02 '15 at 16:22

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I'm not sure whether this counts as an elementary proof, but you can do it in this way:

Assume there is a point $x$ on the boundary that is mapped into the interior of $D^2$. Then the restriction of $f$ induces a homeomorphism $D^2 \setminus \{x\} \to D^2 \setminus \{f(x)\}$. But $D^2\setminus \{x\}$ is simply connected, while $D^2\setminus \{f(x)\}$ is not.

The more general case can be found in this thread.

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Dominik
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  • Can you clarify what does $f(x)$ map to in the above case? Does it mapping to the interior of $D^2$ mean that the boundary $\partial f(D^2)$ now has a hole, because (at least) one $f(x)$ was mapped to the interior, rather than the boundary. – mavavilj Feb 16 '17 at 14:02
  • $f(x)$ is just some point in the interior of $D^2$. Since $f(D^2) = D^2$, the boundary of $f(D^2)$ is the same as the boundary of $D^2$. But this has nothing to do with the proof - we look at $D^2\setminus{x}$ resp. $D^2\setminus{f(x)}$. – Dominik Feb 16 '17 at 14:14