5
$\begingroup$

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 $\Bbb R$?

$\endgroup$
  • $\begingroup$ 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 ... $\endgroup$ – DonAntonio Jul 23 '13 at 10:58
  • 2
    $\begingroup$ 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. $\endgroup$ – Etienne Jul 23 '13 at 11:00
  • $\begingroup$ @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. $\endgroup$ – martini Jul 23 '13 at 11:01
  • $\begingroup$ Yes @martini...so? $\endgroup$ – DonAntonio Jul 23 '13 at 11:02
  • $\begingroup$ @martini If the preimages of clsed sets are always open, then the preimages of open sets are always closed. $f^{-1}(A^c)=f^{-1}(A)^c$. $\endgroup$ – Hagen von Eitzen Jul 23 '13 at 11:04
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +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! $\endgroup$ – Amitesh Datta Jul 23 '13 at 11:10
5
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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