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.

  • $\begingroup$ Are you sure that $f$ is a closed map? If $A\subseteq X$ is closed then it is compact hence is sent by the continuous $f$ to a compact $f(A)\subseteq Y$. But there is no guarantee that $f(A)$ is closed. So I see no reason to believe that $f$ is a closed map. $\endgroup$ – drhab Apr 5 '18 at 12:23
  • $\begingroup$ @drhab. Yah..I am sorry. $f$ may not be a closed map. $\endgroup$ – abcdmath Apr 5 '18 at 12:31

Take the identity function on set $X$ where the domain is equipped with discrete topology and the codomain with indiscrete topology.

That is a continuous bijection.

If $X$ has exactly $2$ elements then $X$ is compact but the singletons (which are open and closed) will be sent to sets that are not open and are not closed.

So the function is not open and is not closed.


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