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Let $X$ and $Y$ be two general topological spaces. Is the following statement true?

For any open $U\subset X$, there exists an open $V\subset Y$ and a continuous map $f:X\rightarrow Y$, such that $f^{-1}V=U$.

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  • $\begingroup$ Take $Y$ to be finite, and $X$ to have an infinite number of open sets. $\endgroup$ – copper.hat Jan 16 '19 at 14:57
  • $\begingroup$ But there are potentially many $f$'s? $\endgroup$ – Uchiha Jan 16 '19 at 14:59
  • $\begingroup$ You are correct, sloppy thinking on my part. $\endgroup$ – copper.hat Jan 16 '19 at 15:00
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If $Y$ only has one element then $f^{-1}(V)\in\{\varnothing,X\}$ for any open set $V$.

So if $X$ has a non-trivial open set then it does not work.

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  • $\begingroup$ Why is $f^{-1}(V)=\{\emptyset,X\}$? $\endgroup$ – SmileyCraft Jan 16 '19 at 15:03
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    $\begingroup$ @SmileyCraft (typo, repaired) Because $V=\varnothing$ or $V=Y=\{y\}$ and $f:X\to Y$ can only be the constant function. $\endgroup$ – drhab Jan 16 '19 at 15:05
  • $\begingroup$ Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element? $\endgroup$ – Uchiha Jan 16 '19 at 15:08
  • $\begingroup$ Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)\in\{\varnothing,X\}$ for every open $V$, and $Y$ can have as much elements as you want. $\endgroup$ – drhab Jan 16 '19 at 15:14
  • $\begingroup$ Right, it seems the question is only interesting if more regularity is assumed on the topology of Y. $\endgroup$ – Uchiha Jan 16 '19 at 15:25

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