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?

  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Commented Mar 5, 2021 at 8:14
  • $\begingroup$ @KReiser Thanks, that was a bit sloppy. I corrected it. $\endgroup$ Commented Mar 5, 2021 at 8:33
  • $\begingroup$ 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? $\endgroup$ Commented Mar 6, 2021 at 7:37

2 Answers 2


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”.

Let me answer your question for the last time.

The answer to your question is yes. In fact, we have a stronger result.

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 $Y$.

This result can be found on Page 108 of Several Complex Variables VII by Grauert, Peternell, and Remmert (link 1, link 2).

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.

  • $\begingroup$ I did find the paper of Bănică in our library and scanned it. So if you're interested I can make it accessible to you. $\endgroup$ Commented Oct 24, 2022 at 12:01
  • $\begingroup$ @red_trumpet Great! Could you please send it to me by email? $\endgroup$ Commented Oct 24, 2022 at 12:46

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