0
$\begingroup$

We know a continuous bijection from a compact space to a Hausdorff space is always a homeomorphism.

But I am wondering what happens if we switch the domain and codomain. Is a continuous bijection function from a Hausdorff space to a compact space a homeomorphism?

What I think this is not true. Consider the example $f:\mathbb{R}\to [-\frac{\pi}{2},\frac{\pi}{2}]$ defined by $f(x)=\tan(x)$, then $f$ is a continuous bijection but $f^{-1}$ is not continuous.

Am I right? Thank you for any comments.

$\endgroup$
  • $\begingroup$ Wait, what is your $f$? $\endgroup$ – IAmNoOne Dec 8 '14 at 4:47
  • $\begingroup$ sorry, my $f$ is the $tan(x)$, I forget to put it in. $\endgroup$ – user138017 Dec 8 '14 at 4:48
  • 1
    $\begingroup$ I guess you mean $\tan^{-1}$? But still it is not onto. $\endgroup$ – user99914 Dec 8 '14 at 4:49
  • $\begingroup$ oh, yes, I meant $tan^{-1}$ Thank you for point it out. I miss the endpoint, yes, that is not onto. $\endgroup$ – user138017 Dec 8 '14 at 4:53
  • 1
    $\begingroup$ Note that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, since closed subsets of a compact space are compact, are thus sent to compact subsets of the Hausdorff codomain, which are thus closed. Then, since the function is continuous, bijective, and closed, it's a homeomorphism. $\endgroup$ – Nishant Dec 8 '14 at 5:24
4
$\begingroup$

My attempt:

Consider the identity map $i:(\mathbb R,\tau_d)\rightarrow (\mathbb R,\tau_f)$

$(\mathbb R,\tau_d)$ is the real line with discrete topology and $(\mathbb R,\tau_f)$ is real line with cofinite topology.One is compact and other is not

$\endgroup$
3
$\begingroup$

Consider $f: [0,2\pi) \to \mathbb S^1$, $t\mapsto (\cos t ,\sin t)$. (Is there a continuous bijective map from $\mathbb R$ to $[-\pi/2, \pi/2]$?)

$\endgroup$
  • $\begingroup$ is this a counterexample to the question? is my example OK? Thanks a lot. $\endgroup$ – user138017 Dec 8 '14 at 4:47
  • $\begingroup$ @user138017: Yes, that's a counterexample. I don't think your $f$ can be found. $\endgroup$ – user99914 Dec 8 '14 at 4:48
  • $\begingroup$ sorry, I just realized I have not put my $f$ in the statement. I meant $f(x)=tan(x)$ for my example. $\endgroup$ – user138017 Dec 8 '14 at 4:50
1
$\begingroup$

No, take any open interval $(a,b)$ on the Reals and map it to any $[c,d]$. Then $(a,b)$ is disconnected by removing one point, but you can remove the endpoints of $[a,b]$ without disconnecting it. k-connectedness is a topological property.

More simply, given $f: (a,b) \rightarrow [c,d]$, if f is a homeomorphism, then $f^{-1}$ is continuous, but maps the compact interval $[c,d]$ into the non-compact interval $(a,b)$.

$\endgroup$
0
$\begingroup$

The canonical example: on any set $X$ the discrete topology (every set is open) is Hausdorff and the codiscrete topology (no set is open besides $\emptyset$ and $X$ itself) is compact. If $X$ has more than one point the identity map from $X$ with the discrete topology to $X$ with the codiscrete topology is a continuous bijection, but its inverse is not continuous.

$\endgroup$

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.