Lemma 37.68.7. Let $f : X \to Y$ be a separated, locally quasi-finite, and universally open morphism of schemes. Let $n_{X/Y}$ be as in Lemma 37.68.5. If $n_{X/Y}$ attains a maximum $d < \infty$, then the set

$Y_ d = \{ y \in Y \mid n_{X/Y}(y) = d\}$

is open in $Y$ and the morphism $f^{-1}(Y_ d) \to Y_ d$ is finite.

Proof. The openness of $Y_ d$ is immediate from Lemma 37.68.6. To prove finiteness over $Y_ d$ we redo the argument of the proof of that lemma. Namely, let $y \in Y_ d$. Then there are at most $d$ points of $X$ lying over $y$. Say $x_1, \ldots , x_ n$ are the points of $X$ lying over $y$. Apply Lemma 37.37.5 to get an étale neighbourhood $(U, u) \to (Y, y)$ and a decomposition

$U \times _ Y X = W \amalg \ \coprod \nolimits _{i = 1, \ldots , n} \ \coprod \nolimits _{j = 1, \ldots , m_ i} V_{i, j}$

as in locus citatus. Observe that $d = n_{X/Y}(y) = \sum _ i m_ i$ in this situation; some details omitted. Since $f$ is universally open, we see that $V_{i, j} \to U$ is open for all $i, j$. Hence after shrinking $U$ we may assume $V_{i, j} \to U$ is surjective for all $i, j$ and we may assume $U$ maps into $W$. This proves that $n_{U \times _ Y X/U} \geq \sum _ i m_ i = d$. Since the construction of $n_{X/Y}$ is compatible with base change we know that $n_{U \times _ Y X/U} = d$. This means that $W$ has to be empty and we conclude that $U \times _ Y X \to U$ is finite. By Descent, Lemma 35.20.23 this implies that $X \to Y$ is finite over the image of the open morphism $U \to Y$. In other words, we see that $f$ is finite over an open neighbourhood of $y$ as desired. $\square$

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