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

general-topology

$f: X \rightarrow Y$ 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.

Best Answer

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 $Y$.

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