# Does there always exist a continuous map saturating a given open set?

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 $$f^{-1}V=U$$.

• Take $Y$ to be finite, and $X$ to have an infinite number of open sets. – copper.hat Jan 16 '19 at 14:57
• But there are potentially many $f$'s? – Uchiha Jan 16 '19 at 14:59
• You are correct, sloppy thinking on my part. – copper.hat Jan 16 '19 at 15:00

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.
• Why is $f^{-1}(V)=\{\emptyset,X\}$? – SmileyCraft Jan 16 '19 at 15:03
• @SmileyCraft (typo, repaired) Because $V=\varnothing$ or $V=Y=\{y\}$ and $f:X\to Y$ can only be the constant function. – drhab Jan 16 '19 at 15:05
• Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element? – Uchiha Jan 16 '19 at 15:08
• 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. – drhab Jan 16 '19 at 15:14