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This is what I have so far:

Assume $S^n$ is homeomorphic to $S^m$. Also, assume $m≠n$. So, let $m>n$.

From here I am not sure what is implied. Of course in this problem $S^k$ is defined as:

$S^k=\lbrace (x_0,x_1,⋯,x_{k+1}):x_0^2+x_1^2+⋯+x_{k+1}^2=1 \rbrace$ with subspace topology.

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Can you prove that $\mathbb R^m$ and $\mathbb R^n$ are not homeomorphic when $m\neq n$? Because it follows directly from there when you note that $S^n\setminus \{x\} \cong \mathbb R^n$ (and $x$ is any point of $S^n$) –  Thomas Andrews Nov 1 '12 at 16:46
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2 Answers

up vote 3 down vote accepted

Hint: look at the topic Invariance of Domain

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I'm sorry, could you elaborate a bit more on your argument? I can't seem to work it out on my own. –  Matt N. Nov 1 '12 at 16:59
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@MattN. It's the same argument as given in the comments. $S^n \setminus \{x\} \cong \mathbb R^n$ so $S^n \cong S^m$ implies $\mathbb R^n \cong \mathbb R^m$, which gives $n=m$ by invariance of domain. –  JSchlather Nov 1 '12 at 18:53
    
@JacobSchlather, thank you for your comment! –  Matt N. Nov 1 '12 at 19:11
    
@JacobSchlather Thank you –  Bunuelian Trick Nov 2 '12 at 5:12
    
@MattN. Hope you have got it –  Bunuelian Trick Nov 2 '12 at 5:12
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A non-elementary argument would be that spaces with non-identical homotopy groups are not homeomorphic.

Assume $n < m$. Then $\pi_n (S^n) = \mathbb Z$ (see this article) but $\pi_n(S^m) = 0$. Hence $S^n$ and $S^m$ are not homeomorphic if $n \neq m$.

Alternatively, as pointed out in the comments, you could use homology groups: $H_n(S^n) = \mathbb Z$ but $H_n(S^m) = 0$. But homeomorphic spaces have isomorphic homology groups.

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This result uses the homology groups of spheres, so wouldn't it be easier to make the same conclusion using homology? –  espen180 Nov 1 '12 at 17:03
    
@espen180 Yes, one could of course also argue using homology groups. I added it to my answer, thanks a lot for mentioning it! –  Matt N. Nov 1 '12 at 17:11
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