# Generic smoothness in the analytic setting.

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?

• Just to confirm, you mean complex-analytic spaces, right? Let me also point out that the first result you mention is false in positive characteristic. Commented Mar 5, 2021 at 8:14
• @KReiser Thanks, that was a bit sloppy. I corrected it. Commented Mar 5, 2021 at 8:33
• You should assume that $f$ is proper, otherwise, the claim is already false for holomorphic functions of one variable. Once you assume properness, the claim is a consequence of Remmert's theorem on proper analytic maps. But, perhaps, you want to assume less than properness? Commented Mar 6, 2021 at 7:37

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).

My answer was removed twice. The first time was for some unknown reasons by @Elliot Yu and @Leucippus. The second time was after my complaint to the Stacksexchange support. The answer was deleted “per my request”.

Theorem Assume that $$X$$ is smooth and $$Y$$ is irreducible. Let $$f:X\rightarrow Y$$ be a proper holomorphic map. Then the complement of the set $$V:=\{y\in Y: X_y \text{ is a manifold}\}$$ is a proper analytic subset of $$Y$$.
The complement of $$V$$ is just a negligible set without any extra assumptions. Takumi’s answer gives a stronger claim when $$f$$ is flat. I'm not sure if it holds or not.