# Closed surjective map

How to prove that every closed surjective map is open? (Exercise from book Borisovich "General topology")

Thank you very much!

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I will show a counterexample showing that the assertion is false. Let us consider the following function $h:[0,1]\rightarrow [0,1],$

h(x) = \begin{array}{ccc} \frac{3}{2}x & \text{if} & x\in \lbrack 0,\frac{1}{3}] \\ \frac{1}{2} & \text{if} & x\in \lbrack \frac{1}{3},\frac{2}{3}] \\ \frac{3}{2}x-\frac{1}{2} & \text{if} & x\in \lbrack \frac{2}{3},1]. \end{array}

Here we consider $[0,1]$ to be a topological space with the topology generated by the metric $d\left( x,y\right) =|x-y|$. Now the function is clearly continuous and surjective. It is also closed which follows from compactness. On the other hand the interval $\left( \frac{1}{3},\frac{2}{3}% \right)$ is an open set in $[0,1]$, and $h[\left( \frac{1}{3},\frac{2}{3}% \right) ]=\left\{ \frac{1}{2}\right\}$ which is not open in $[0,1]$.

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Let $f:X\to Y$ be closed and surjective, and assume we have given a $U\subseteq X$ open subset. Use complement, and prove that $f(U)$ is open.
Update: It is indeed not that trivial, moreover, not even true (see comments below). The problem is that, though $f(U)\cup f(X\setminus U)=Y$ by surjectivity, these 2 sets may intersect, so we cannot simply conclude $f(X\setminus U)=Y\setminus f(U)$.
This is exactly what I did, but I don't see how it can be trivially deduced from this. I see that the image of $U$ in union with image of it's complement must give the whole $Y$. But they can intersect, because we are not given that map is injective. Could you, please, give a more detailed proof. –  Sergey Finsky Nov 17 '12 at 14:46