Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\Bbb R$?

share|cite|improve this question
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 – 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
up vote 6 down vote accepted

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.

share|cite|improve this answer
+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.


The collection of sets $f^{-1}(\{y\})$ for $y\in Y$ constitutes a non-trivial separation of $X$ if $f$ is not constant.


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.


See Haigen von Eitzen's excellent answer and try to generalize it to a proof!


I hope this helps!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.