If $X, Y$ are non-singular algebraic varieties over an algebraically closed field of characteristic 0, and $f: X \to Y$ is a morphism, then there exists a Zariski-open, non-empty subset $U \subset Y$, such that $f^{-1}(U) \to U$ is a smooth morphism.
I wonder: Is the same true if $X, Y$ are non-singular complex analytic spaces?
I know that Sard's theorem shows that "most" fibers are smooth, in the sense that $f(\operatorname{Sing}(f)) \subset Y$ has Lebesgue measure $0$. But is $f(\operatorname{Sing}(f))$ also contained in a Zariski-closed subset?
If this is true, what is a good reference?
Here is a reference. I have translated the statement and changed the wording slightly.
Theorem [Bănică, Théorème, (ii)]. Let $f\colon X \to Y$ be a morphism of complex analytic spaces. If $f$ is flat, then the set $$U = \bigl\{ x \in X \bigm\vert X_{f(x)}\ \text{is regular at}\ x\bigr\}$$ is open in $X$, and the complement of $U$ is analytic in $X$.
For proper flat morphisms, we have:
Corollary [Bănică, Corollaire, (ii)]. Let $f\colon X \to Y$ be a proper morphism of complex analytic spaces. If $f$ is flat, then the set $$V = \bigl\{ y \in Y \bigm\vert X_{y}\ \text{is a manifold}\bigr\}$$ is open in $Y$, and the complement of $V$ is analytic in $Y$.
As Moishe Kohan suggests, the Corollary follows from the Theorem using Remmert's theorem.
To obtain a statement that does not mention flatness, you can use Frisch's theorem [Frisch, Théorème (IV, 9)] (see [Kiehl, Satz 4] for an alternative proof), which says that the locus in $X$ at which $f$ is not flat is closed and analytic. For a textbook account of Frisch's theorem, see [Bănică–Stănăşilă, Theorem 4.5] (their proof follows [Kiehl]).
Finally, you can see [Bingener and Flenner, Corollary 2.1] for an English reference proving variants of these results (they also prove a version for real analytic spaces).