If $X$ and $Y$ are Hausdorff, $X$ is compact, and $f:X\to Y$ is continuous and surjective, then $f$ is open.

Question

I was given the following assertion:

If $X$ and $Y$ are Hausdorff, $X$ is compact, and $f:X\to Y$ is continuous and surjective, then $f$ is open.

However, I believe that I have a counterexample:

Define $f:[0,3]\to[0,2]$ by

$$ x\mapsto\begin{cases} x&0\leq x\leq1,\\ 1&1<x<2,\\ x-1&2\leq x\leq3. \end{cases} $$

Then $[0,3]$ and $[0,2]$ are Hausdorff, $[0,3]$ is compact, and $f$ is continuous and surjective. However, $[0,3/2)$ is open in $[0,3]$ but $f([0,3/2))=[0,1]$ is not open in $[0,2]$.

Does this look correct? If so, then is there a similar theorem?

Answer 1

There are some more or less similar theorems:

1. If $X$ is compact and $Y$ is Hausdorff, then $f\colon X\to Y$ is closed.
2. If $f$ is in addtion bijective then $f(U^C)=f(U)^C$ and thus $f$ is open
3. In functional analysis there is the open mapping theorem. For continuous lineare operators the statement is true, there are open if and only if they are surjective.