# Is a continuous bijection function from a hausdorff space to a compact space a homeomorphism?

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.

• Wait, what is your $f$? – IAmNoOne Dec 8 '14 at 4:47
• sorry, my $f$ is the $tan(x)$, I forget to put it in. – user138017 Dec 8 '14 at 4:48
• I guess you mean $\tan^{-1}$? But still it is not onto. – user99914 Dec 8 '14 at 4:49
• oh, yes, I meant $tan^{-1}$ Thank you for point it out. I miss the endpoint, yes, that is not onto. – user138017 Dec 8 '14 at 4:53
• 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. – Nishant Dec 8 '14 at 5:24

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

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 $[-\pi/2, \pi/2]$?)

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

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 $(a,b)$.

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.