$f: X \rightarrow Y$ 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 $Y$"
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.