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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.