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