$f^{-1}(U)$ is regular open set in $X$ for regular open set $U$ in $Y$, whenever $f$ is continuous.
Solution 1:
Not necessarily. Consider the absolute value function $x \mapsto | x |$, and the inverse image of $(0,1)$.
Solution 2:
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.