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In David Mumford's book Algebraic Geometry I, Complex Projective Varieties the proof of (3.25) Specialization principle on page 53 contains an argument I not understand.

General assumptions: all our varieties are over $\mathbb{C}$. The statement is:

(3.25) Specialization principle. Let $Z \subset \mathbb{P}^n \times \mathbb{P}^m $ be $r$-dimensionally subvariety and $X =p_1(Z) \subset \mathbb{P}^n$ for $p_1: \mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^n $. Suppose $\operatorname{dim} X=r $ and let

$$\phi= \text{ res } p_1: Z \to X $$

is almost everywhere (= on an open set) finite to one. [... ]

Then the map

$$ F: X_1 \to \mathbb{N}, x \mapsto \# \phi^{-1}(x)$$

where $X_1 := \{x \in X \ \vert \ X \text{ smooth at } x \text{ and } \phi^{-1}(x) \text{ finite } \}$ is lower semi-continuous in the Zariski topology on $X_1$.

Lower semi-continuous means that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is closed.

The first part of the proof shows that $ F: X_1 \to \mathbb{N}$ is lower semi-continuous in the classical topology, recall a smooth complex variety can be canonically endowed with classical analytical topology considering it as a complex manifold.

Mumford remarks that we can observe that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructible, ie a union of finite number of locally closed sets in Zariski topology.

Then Mumford claims that this implies that $F$ is lower semi-continuous in the Zariski topology.

That's the point I not understand. Does anybody see why the implication $F$ lower semi-continuous with resp classical topology and that $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructible imply $F$ lower semi-continuous with resp Zariski topology?

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First, for lower semi-continuity, we should consider sublevel set, i.e. $$\{x\in X:\ \#\phi^{-1}(x)\leq n\}$$ Now, since $\#\phi^{-1}(x)$ is a constructible function, the above subset, denoted as A, satisfies that \begin{equation} A=\bigcup_{i=1}^N (Z_{i,1}-Z_{i,2}) \end{equation} where $Z_{i,j}$'s are all Zariski closed subsets. Also, from the lower semi-continuity of $\#\phi^{-1}(x)$ in the classical topology, A is closed in the classical topology. Taking the classical closure of the above equation on $A$, we get $$A=\bigcup_{i=1}^N Z_{i,1}$$ which is Zariski closed. This shows that $\#\phi^{-1}(x)$ is also lower semi-continuous in the Zariski topology.

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  • $\begingroup$ do you maybe mean in your penultimate sentence that we take "Zariski closure" of $A$ and not "classical closure"? Indeed, since $A$ is assumed to be already closed in classical topology, the latter action would not change $A$, or do I missing your point? $\endgroup$
    – user267839
    Commented Mar 11, 2023 at 12:43
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    $\begingroup$ @user267839 No, since A is classical closed when we take the classical closure of A, it is unchanged. But the right-hand side is changed when we take the classical closure, which is also Zariski closed. This leads to the Zariski closeness of A. $\endgroup$
    – Drizzle
    Commented Mar 12, 2023 at 4:27

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