$f^{-1}(U)$ is regular open set in $X$ for regular open set $U$ in $Y$, whenever $f$ is continuous. Let $f$ be a continuous function from space $X$ to space $Y$. If $U$ is regular open set in $Y$, it it true that $f^{-1}(U)$ is a regular open set in $X$?

As arjafi explains, the answer is 'no.' Further to that, adding smoothness conditions won't help either. For example, consider $$f: \mathbb{R} \rightarrow \mathbb{R}$$
$$f(x) = x^2$$
Then:


*

*$f^{-1}(0,1) = (-1,0) \cup (0,1),$ so preimages of regular open sets needn't be regular open.

*$f^{-1}[-1,0] = \{0\},$ so preimages of regular closed sets needn't be regular closed.

