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:

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

2. 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?

3. 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?

4. 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?

• Although your question isn’t really a duplicate of this one, the answers there essentially answer yours. In particular, the answer to your boldface question is yes. – Brian M. Scott Aug 19 at 18:42

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.