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I've heard that Brouwer's invariance of domain problem is very difficult to prove. This confuses me, though, since it seems like a proof along the following lines should work:

Consider the sphere $S^m$ embedded in $\mathbb{R}^n$ via a mapping $\phi$ (which is a homeomorphism onto the image), where $m>n-1$. Show that $\mathbb{R}^n\setminus \phi(S^m)$ is still connected (maybe this is harder than I'm giving it credit for). Then it is easy to see that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic for $n>m$ from the simple fact that $S^{m-1}$ disconnects $\mathbb{R}^m$ but any homeomorphic embedding of it does not disconnect $\mathbb{R}^n$.

Where does this go wrong?

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    $\begingroup$ Brouwers invariance of domain theorem states more than just that $\Bbb{R}^n$ is not homeomorphic to $\Bbb{R}^m$. Also I think that indeed you are not giving enough credit to the difficulty of the connectedness claim. $\endgroup$
    – PhoemueX
    Oct 11 '21 at 13:08
  • $\begingroup$ The rest of the statement of Brouwer’s invariance of domain theorem follows from some relatively elementary simple arguments though, doesn’t it? $\endgroup$
    – exfret
    Oct 11 '21 at 13:33
  • $\begingroup$ Every way that I know of to prove that connectedness claim is significantly harder than the easiest proofs that $\mathbb{R}^m$ and $\mathbb{R}^n$ are not homeomorphic. $\endgroup$ Oct 11 '21 at 15:52
  • $\begingroup$ Ah so it is indeed the part I’m not giving enough credit to. $\endgroup$
    – exfret
    Oct 11 '21 at 16:35
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The lowest dimensional case $n=1,m=2$ already fails: $\Bbb R^m\setminus f[S^1]$ is disconnected (although this is hard to show, look up Jordan's curve theorem), not connected. The sphere disconnection fact is in fact very hard, and you underestimate that hardness IMO.

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  • $\begingroup$ We don’t need the Jordan curve theorem since it is enough to show that the regular embedding of $S^1$ into $\mathbb{R}^2$ disconnects it. It then remains to show that any embedding of $S^1$ into higher dimensional space does not disconnect it. $\endgroup$
    – exfret
    Oct 11 '21 at 13:25
  • $\begingroup$ @exfret that’s exactly what the Jordan curve theorem says. What is a regular embedding anyway? $\endgroup$ Oct 11 '21 at 13:27
  • $\begingroup$ I mean, just the set $\{(x,y)\vert x^2+y^2=1\}$ $\endgroup$
    – exfret
    Oct 11 '21 at 13:28
  • $\begingroup$ Also, I don’t see how your example is a counterexample. 2 is not less than or equal to 1. (I mixed up $n$’s and $m$’s so maybe this is the source of confusion). $\endgroup$
    – exfret
    Oct 11 '21 at 13:30
  • $\begingroup$ You want the version where we have ahomeomorphic copy of $S^1$, otherwise it won't work. $\endgroup$ Oct 11 '21 at 13:30

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