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I already asked this on math.SE, but didn't receive any response.

The following question arose when studying Hwang and Oguiso's Characteristic foliation on the discriminant hypersurface of a holomorphic Langrangian fibration.

Let $f: M \to B$ be a proper flat map of complex manifolds with connected fibers, and assume that the set of critical values $D \subset B$ is (set-theoretically¹) a hypersurface. Set $Y = f^{-1}(D)_{\text{red}}$, let $b \in D$ be a general point, and set $Y_b = f^{-1}(b)_{\text{red}} = f^{-1}(b) \cap Y$ as the reduced fiber over $b$.

How do I argue formally for the following statement?

After shrinking $B$ around $b$ (analytically), we may assume that the map $$Z \mapsto Z \cap Y_b$$ induces a bijection between the irreducible components $Z$ of $Y$ and the irreducible components of $Y_b$.

Intuitively, I think that because $b$ is general, the neighboring singular fibers $Y_{b'}$ for $b' \in D$ near $b$ "look alike", so they have the same number of irreducible components, which correspond to the irreducible components of $Y_b$, and together they form the irreducible components of $Y$. But I don't know how to make this more rigorous.

¹ I always think of $D$ as reduced. I don't know if there is any more natural scheme structure to endow it with.

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The next example shows that this is not true without shrinking to an analytic neighborhood of $b$ (see the comment of Jason Starr below).

Consider the universal conic --- the incidence hypersurface $$ M \subset \mathbb{P}^5 \times \mathbb{P}^2 $$ (where the first factor is considered as the space of conics on the second factor), and the first projection $$ f \colon M \to \mathbb{P}^5 =: B. $$ Then for general point $b$ in the discriminant hypersurface $D \subset \mathbb{P}^5$ the conic $Y_b$ is the union of two lines; in particular it has two irreducible components. On the other hand, the second projection $$ Y = f^{-1}(D) \to \mathbb{P}^2 $$ is a locally trivial fibration (because it is equivariant for the natural action of $\mathrm{PGL}_3$, whose action on $\mathbb{P}^2$ is transitive) with irreducible fibers, hence $Y$ is also irreducible.

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    $\begingroup$ I read the question differently: shrink to an open analytic neighborhood of $b$, and then let $Z$ be one of the "irreducible components" (i.e., closure of connected component of complement of singular locus) in the intersection of $Y$ with the inverse image of that open analytic neighborhood. For this question, the answer is positive, roughly for the reason the OP writes: the map to $D$ from the open complement of the singular locus of $Y$ is a smooth fibration over a Zariski dense open in $D$. $\endgroup$
    – Jason Starr
    Commented Nov 2, 2022 at 10:10
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    $\begingroup$ @JasonStarr: You are right, of course (I missed somehow the suggestion to shrink to an analytic neigborhood). Indeed, the only problem is the monodromy permuting irreducible components of the fibers, and since $D$ is smooth at $b$, the local fundamental group of $D$ is trivial, and the local monodromy is trivial as well. I edited the answer accordingly. $\endgroup$
    – Sasha
    Commented Nov 2, 2022 at 10:32
  • $\begingroup$ Thanks for the example, but @JasonStarr is correct. Can you elaborate on "the only problem is the monodromy permuting components"? I see that this would be a problem. But I don't see how to argue in the absence of monodromy. $\endgroup$
    – red_trumpet
    Commented Nov 2, 2022 at 13:11
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    $\begingroup$ @red_trumpet: As Jason explained, if you remove the singular locus of $Y$, locally around $b$ it will be a smooth fibration $Y_0 \to D$, and the former irreducible components of fibers are now connected components. Therefore, it factors as $Y_0 \to D' \to D$, where $Y_0 \to D'$ has connected fibers and $D' \to D$ is an unramified covering. If the fundamental group is trivial, the covering is trivial as weill, $D' = \sqcup D$, hence $Y_0$ is also a disjoint union of as many components as you need. $\endgroup$
    – Sasha
    Commented Nov 3, 2022 at 4:36

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