Let $f:X→Y$ be a continuous surjective map and $X$ is compact. Is $f$ is an open map?
$"$A function $f : X → Y$ is open if for any open set $U$ in $X$, the image $f(U)$ is open in $Y$.$"$
Since $f$ is continuous and $X$ is compact then image of $X$ under $f$ will be compact. But how can I prove that $f$ is an open map? If not please help me with a counterexample.