Closed surjective map How to prove that every closed surjective map is open? (Exercise from book Borisovich "General topology")
Thank you very much!
 A: 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 $[0,1]$.
A: 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 $f(X\setminus U)=Y\setminus f(U)$.
