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

Question

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

Answer 1

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.