Assume $f:X\dashrightarrow Y$ is dominant rational map (of complex varieties), where $X$ is nonsingular, $Y$ is complete. I want to prove for a general $y$ in $Y$, the closure of the preimage of $y$ is also nonsingular.
In fact, I think I can prove it using Hironaka's theorem: there exists a sequence of smooth blowups $X_n\to X_{n-1} \to \cdots \to X_0=X$ such that $f_n:X_n\to Y$ becomes a morphism. Then, using generic smoothness of morphisms, a general fiber $f_n^{-1}(y)$ is nonsingular, hence in particular, nonsingular at the exceptional divisor, so $\overline{f_{n-1}^{-1}(y)}$ is nonsingular (Edit: I realized this is not correct, by Sasha's example). By induction we will finally prove $\overline{f^{-1}(y)}$ is nonsingular. However, this seems too overkill. Is there easier way to prove it?
This is not true. Let $X = \mathbb{P}^2$ and let $\pi \colon \tilde{X} \to X$ be the blowup of the point $x_0 \in X$. Then $\tilde{X}$ is the Hirzebruch surface $F_1$, in particular there is a $\mathbb{P}^1$-bundle $p \colon \tilde{X} \to \mathbb{P}^1$. Consider the composition $$ f \colon X \stackrel{\pi^{-1}}\dashrightarrow \tilde{X} \stackrel{p}\to \mathbb{P}^1 \stackrel{q}\to \mathbb{P}^1 =: Y, $$ where $q$ is a double covering (branched at two points). Then the closure of the general fiber of $f$ is the union of two lines through $x_0$; in particular it is singular.