TITLE: Pre-images of closed sets are open
QUESTION [5 upvotes]: Let $X$ and $Y$ be two topological spaces and let $f$ be such a map that $f^{-1}(A)$ is open in $X$ for any closed $A$. Note that if $X\stackrel{f}{\longrightarrow}Y\stackrel{g}\longrightarrow  Z$ are two such maps, then $g\circ f$ is continuous. Perhaps, it is a trivial task - but is there an example of such surjective map from $\Bbb R$ to $\Bbb R$?

REPLY [7 votes]: Let $f\colon \mathbb R\to\mathbb R$ have the given property.
Since points are closed, we get pairwise disjoint open sets $f^{-1}(x)$. If $x$ is in the image of $f$, then $f^{-1}(x)$ contains some open interval and hence a rational number. We conclude that $f^{-1}(x)\ne\emptyset$ only for countably many $x$, i.e. $f$ is not surjective.

REPLY [5 votes]: Theorem 1
Let $f:X\to Y$ be such a map of topological spaces with $X$ connected and $Y$ a $T_1$-space. Then $f$ is constant.
Proof
The collection of sets $f^{-1}(\{y\})$ for $y\in Y$ constitutes a non-trivial separation of $X$ if $f$ is not constant.
Q.E.D.
Theorem 2 (based on Hagen von Eitzen's answer)
Let $f:X\to Y$ be such a map of topological spaces with $X$ separable and $Y$ an uncountable $T_1$-space. Then $f$ is not surjective.
Proof
See Haigen von Eitzen's excellent answer and try to generalize it to a proof!
Q.E.D.
I hope this helps!