Suppose I have a continuous map $f:X\rightarrow Y$. Then one can wonder, whether for every open set $U\subset X$ the set 
$U':=\{x\in X|f^{-1}(f(x))\subset U\}$ is open again. This is not true in general, as the example 


$U:=\{(x,y)|\quad |y|<\frac{1}{x^2+1}\}\subset \mathbb{R}^2$ and $f:\mathbb{R}^2\rightarrow \mathbb{R} \qquad (x,y)\mapsto y$


shows. Here $U'$ is just $\mathbb{R}\times \{0\}$. 
So the question is: Which continuous maps $f:X\rightarrow Y$ have the property, that for every open set $U\subset X$, the set $U'$ is also open?



Finding saturated open sets
The definition of $U'$ doesn't depend on the Topology on $Y$. 
Given any map $f:X\rightarrow Y$, we can consider the equivalence relation $x'\sim x$, iff $f(x)=f(x')$ on $X$ and factor the map $f$ as $X\rightarrow X/\sim \quad \rightarrow Y$. Call the first $f'$. Note that,the second map is injective. Hence we get the same set $U'$ for any set $U$, if we replace $f$ by $f'$.


So we might assume without restriction, that $f:X\rightarrow X/\sim $ is a quotient map. Then the upper statement is equivalent to saying, that $f$ is closed.


Suppose $f$ is closed. Then one gets for the complements $U'^c=f^{-1}(f(U^c))$. And hence $U'^c$ must be closed.
There is the [closed map lemma](http://en.wikipedia.org/wiki/Open_and_closed_maps), which shows, that under the nice conditions mentioned in the comments above properness is also sufficient.


Now suppose $f:X\rightarrow X/\sim$ is not closed and a quotient map. Then there is a closed set $A\subset X$, such that its image is not closed. By definition of the quotient topology, a subset of $X/\sim$ is closed, if and only if its preimage is closed. So $f^{-1}(f(A))$ is not closed and hence $U=A^c$ gives the desired counterexample.


So we get: A map $f:X\rightarrow Y$ has the property, if and only if the induced map $X\rightarrow X/\sim$ is closed.


