8
$\begingroup$

Looking at the definition of an homeomorphism, this question came to my mind.

$\endgroup$
  • 1
    $\begingroup$ Did you mean homeomorphism, rather than a homomorphism? $\endgroup$ – Asaf Karagila Sep 8 '13 at 8:33
  • 1
    $\begingroup$ "What about the function of my edit?" The function of your edit is not a function on $\mathbb R$, it is a function on a disconnected subset of $\mathbb R$. $\endgroup$ – TonyK Sep 8 '13 at 8:49
  • 1
    $\begingroup$ No, it's not obvious. It looks like you are exhibiting this function as a counter-example. I suggest that you undo your edit and post a new question. $\endgroup$ – TonyK Sep 8 '13 at 9:01
  • 1
    $\begingroup$ @lolo: Now it is totally unclear what your question is. If you assume that the domain is open then the positive answer follows from invariance of domain theorem. If you do not require domain to be open then the answer is negative. $\endgroup$ – Moishe Kohan Sep 8 '13 at 9:49
  • 3
    $\begingroup$ Dear @closevoters, could you please tell me where is an answer to this question in the above link? I couldn't find it. $\endgroup$ – njguliyev Sep 8 '13 at 19:58
8
$\begingroup$

No. Hint: A bijective continuous function on $\mathbb{R}$ must be monotonic.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

The answer is no, as can be seen from the Invariance of domain. The answer remains no even when $\mathbb R$ is replaced with $\mathbb R^n$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @lolo what about it? Its inverse is still continuous, so it is a homeomorphism as well. $\endgroup$ – user1337 Sep 8 '13 at 9:35