# Is a surjective continuous map with compact domain is open?

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.

• 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. – drhab Apr 5 '18 at 12:23
• @drhab. Yah..I am sorry. $f$ may not be a closed map. – 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.

It is well-known that if $$f\colon X\to Y$$ is a continuous surjection, $$X$$ is compact and $$Y$$ is Hausdorff, then the function $$f$$ is closed.1 (In particular, if we add the requirement that $$f$$ is a bijection, we get that $$f$$ is also a homeomorphism.)

It might be tempting to ask what happens if we change closed map to open map in this result. I.e., we get the question: Is it true that a continuous surjection from a compact space to Hausdorff space is open?

Let me add also a counterexample for this. (I.e., an example where the target space is Hausdorff. This was not required in the original question.) Naturally, to get such an example we need a function which is not bijective. (Otherwise we would get a homeomorphism - which is an open map.)

Let us consider the unit interval $$I=[0,1]$$ and the unit circle $$S=\{(\cos x,\sin x); x\in[0,2\pi]\}$$. Then we have a very natural map $$f\colon I\to S$$ $$f\colon x\mapsto (\cos x,\sin x).$$ This map is surjective and continuous. (In fact, it is a quotient map.) However, it is not open. For example, the set $$[0,1/2)$$ is open $$I$$, but the image $$f[U]$$ is not open in $$S$$. (No open neighborhood the point $$f(0)=(1,0)$$ is contained in $$f[U]$$.)