# $f: X \rightarrow Y$ is a continuous surjective function, $Y$ hausdorff and $X$ compact. proof that $f$ is open..

$f: X \rightarrow Y$ is a continuous surjective function, $Y$ hausdorff and $X$ compact.
proof that $f$ is an open map..

"A function $f:X \rightarrow Y$ is an open map if whenever $U$ is an open subset of $X$, then $f(U)$ is an open subset of $Y$"

My Attempt
i see that $f(X-A)$ is closed in $Y$ when $A$ is an open set in $X$, but i can't conclude that $f(A)$ is an open subset of $Y$ using the fact that $f$ is only surjective ..

Any hint will be appreciated.

• Former $U$ is the latter $A$? – ajotatxe Jun 22 '15 at 21:29
• It's not generally true. Was it "closed map" perhaps? – Daniel Fischer Jun 22 '15 at 21:29
• @DanielFischer Ty for the comment .. in the paper says "open" ime.unicamp.br/~posgrad/NOVO/Matem%C3%A1tica/Mestrado/2006.pdf .. its in the page 4 .. exercice 2 .. portugues to english .. is aa open map – Francisco Jun 22 '15 at 21:39
• @DanielFischer, do you have a simple counter-example ? – Sylvain L. Jun 22 '15 at 21:39
• @SylvainL. $X = Y = [0,1]$, and $$f(x) = \begin{cases} 2x &, x \leqslant \frac{1}{2} \\ 1 &, x > \frac{1}{2}\end{cases}$$ – Daniel Fischer Jun 22 '15 at 21:40

Let $X=[0,3]\cup[4,7]$ and $Y=[0,3]$, both with the relative topology from the usual topology on $\mathbb{R}$. Define $$f(x)=\begin{cases} x & \text{if x\in[0,3]}\\ 2 & \text{if x\in[4,7]} \end{cases}$$ The set $A=(1,2)\cup(5,6)$ is open in $X$, but $f(A)=(1,2]$ is not open in $Y$.

• #@!$@!#$@! these tests from unicamp !! second time they put a miss print and a lost my time ! !!!! ty very much !! – Francisco Jun 22 '15 at 21:55

These fact can be combined to be a proof for a closed map.

1. $f$ is a closed map $\iff$ any closed set is sent to a closed set.

2. Closed subset of compact space is compact.

3. Continuous map sends a compact set to a compact set.

4. Compact set in the Hausdorff space is closed.

• Ty for the answer.. but how do you pass to "open" when you have f closed ? – Francisco Jun 22 '15 at 21:40
• The only possible way is to make $f(U)=f(X\setminus U^c)$, $U$: any open in $X$, which we stack, since generally $f(X\setminus U^c) \supset f(X) \setminus f(U^c)$ – Arch Jun 22 '15 at 21:51