Continuous bijection $f: X \to Y$ from a compact space $X$ to a Hausdorff space $Y$ 
Suppose $X$ is a compact space and $Y$ is Hausdorff such that $f: X \to Y$ is a continuous bijection. Which of the following are true?
(I) $f$ is open.
(II) $f$ is a local homeomorphism.
(III) $f^{-1}$ is continuous.

Some observations and questions:

*

*$Y$ is compact as the continuous image of a compact set is always compact.


*Since $f$ is continuous, the pre-image of every open set in $Y$ is an open set in $X$. But can we be sure that every open set in $X$ is mapped to an open set in $Y$ by $f$? Why or why not?


*Local homeomorphism is a new term for me. Wikipedia says that $f$ is a local homeomorphism if every point of $X$ has a neighborhood (open set containing the point) that is homeomorphic to an open subset of $Y$. I'm not sure if $f$ is locally homeomorphic or not. Any ideas?


*For $f^{-1}$ to be continuous we need that the pre-image of every open set in $X$ is an open set in $Y$ under $f^{-1}$. Is this somehow related to whether or not $f$ is an open map? Well, I think so. If $f$ is open, every open set in $X$ is mapped to an open set in $Y$. And since $f$ is continuous, the pre-image (image under $f^{-1}$) of every open set in $Y$ is an open set in $X$. Thus, if $f$ is open, the open sets in $X$ and $Y$ will be in bijection, and necessarily $f^{-1}$ will be continuous. So I think if (I) is true, (III) immediately follows. Is this correct?

I is true as $f$ is closed, as I showed here, in short: $C \subseteq X$ closed, implies $C$ compact, so $f[C]$ compact and a compact subset of a Hausdorff space is closed, so $f[C]$ is closed.
And a bijection obeys $f[X\setminus O]=Y\setminus f[O]$ so when $O \subseteq X$ is open, $ X\setminus O$ is closed, so its image is closed and so $f[O]= Y\setminus f[X\setminus O]$ is open in $Y$.
So $f$ is an open (and closed) continuous bijection and so a homeomorphism (if $g: Y \to X$ is the inverse map, $g^{-1}[O]=f[O]$ is open in $Y$ for all open $O$ in $X$. So III holds too.
II is then trivial, because we can for each $x \in X$ take $X$ to be a neighbourhood homeomorphic to $Y$ (which is trivially a neighbourhood of $f(x)$). A homeomorphism is trivially a local homeomorphism.
So all follow quite directly from the fact we already have even without a bijection but just continuity: $f$ is a closed map.
