# Pre-images of closed sets are open

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 $\Bbb R$?

• I can't see why you think $\,g\circ f\,$ is cont.: we know that for any closed $\,A\subset Z\;$, then $\;g^{-1}(A)\subset Y\,$ is open, but we don't know whether $\,f^{-1}\left(g^{-1}(A)\right)=(g\circ f)^{-1}(A)\subset X\,$ is closed ... – DonAntonio Jul 23 '13 at 10:58
• Such a map does not exist. For any $x\in \mathbb R$, the set $O_x=f^{-1}(\{ x\})$ should be open end nonempty; but these sets are pairwise disjoint. – Etienne Jul 23 '13 at 11:00
• @DonAntonio $g^{-1}(A)$ is open in $Y$, so $Y - g^{-1}(A)$ is closed, hence $$X - (g\circ f)^{-1}(A) = f^{-1}(Y-g^{-1}(A))$$ is open. – martini Jul 23 '13 at 11:01
• Yes @martini...so? – DonAntonio Jul 23 '13 at 11:02
• @martini If the preimages of clsed sets are always open, then the preimages of open sets are always closed. $f^{-1}(A^c)=f^{-1}(A)^c$. – Hagen von Eitzen Jul 23 '13 at 11:04

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.

• +1 I think this is a great answer. Clearly, you're a prolific contributor to math stackexchange as well. I'd just like to take this opportunity to thank you for your excellent contributions here! – Amitesh Datta Jul 23 '13 at 11:10

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!