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I've come across an exercise while self-studying Algebraic Topology:

Prove that if $f: D^n \to D^n$ is a homeomorphism, then $f(S^{n-1}) = S^{n-1}$

I've got some background on Homology theory (I've studied about Mayer-Vietoris sequence and its applications on computing $H_n(S^m)$,the fact that $\mathbb R^n$ is isomorphic to $\mathbb R^m$ if and only if $n=m$ all the way to Jordan's Theorem) but I don't know how to approach it and I don't seem to make any progress.

Any help would be much appreciated!!

Ricky
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Vaggelis Athanasiou
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    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Martin R Sep 05 '23 at 08:03
  • I don't know any algebraic topology, but I don't think any is necessary. In my topology class we proved the following, let X be a topological space, then you can define $ EU_n(X) = { x\in X s.t. x has an open neighborhood homeomorphic to an open subset of \mathbb{R}^n} $, then if $ f : X \longrightarrow Y $ is an homeomorphism then $f(EU_n(X)) = EU_n(Y)$ for all n, and the restriction of $f$ to $EU_n(X)$ is an homeomorphism. Try to prove this, and the result will follow (I think :) ) – Lorenzo1282 Sep 05 '23 at 08:55
  • @LorenzoDeIaco: that works if you can prove that no (or, in fact, not every) point in the boundary of $D^n$ has an open neighborhood homeomorphic to an open neighborhood of $\mathbb{R}^n$… which is not easy, as far as I can tell. – Aphelli Sep 05 '23 at 09:48

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