7
$\begingroup$

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.

$\endgroup$
1
  • 6
    $\begingroup$ 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 $S^n$) $\endgroup$ Nov 1 '12 at 16:46
3
$\begingroup$

Hint: look at the topic Invariance of Domain

$\endgroup$
5
  • $\begingroup$ I'm sorry, could you elaborate a bit more on your argument? I can't seem to work it out on my own. $\endgroup$ Nov 1 '12 at 16:59
  • 2
    $\begingroup$ @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. $\endgroup$
    – JSchlather
    Nov 1 '12 at 18:53
  • $\begingroup$ @JacobSchlather, thank you for your comment! $\endgroup$ Nov 1 '12 at 19:11
  • $\begingroup$ @JacobSchlather Thank you $\endgroup$
    – Marso
    Nov 2 '12 at 5:12
  • $\begingroup$ @MattN. Hope you have got it $\endgroup$
    – Marso
    Nov 2 '12 at 5:12
3
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ This result uses the homology groups of spheres, so wouldn't it be easier to make the same conclusion using homology? $\endgroup$ Nov 1 '12 at 17:03
  • $\begingroup$ @espen180 Yes, one could of course also argue using homology groups. I added it to my answer, thanks a lot for mentioning it! $\endgroup$ Nov 1 '12 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.