Lemma 37.74.6. Let $f : X \to Y$ be a separated, locally quasi-finite morphism. Let $w : X \to \mathbf{Z}_{> 0}$ be a positive weighting of $f$. Assume $\int _ w f$ attains its maximum $d$ and let $Y_ d \subset Y$ be the open set of points $y$ with $(\int _ f w)(y) = d$. Then the morphism $f^{-1}(Y_ d) \to Y_ d$ is finite.

Proof. Observe that $Y_ d$ is open by Lemma 37.74.4. Let $y \in Y_ d$. Say $x_1, \ldots , x_ n$ are the points of $X$ lying over $y$. Apply Lemma 37.41.5 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 $d = \sum w(v_{i, j})$ where $w(v_{i, j}) = w(x_ i)$. Since $\int _{V_{i, j} \to U} w|_{V_{i, j}}$ is locally constant by definition, we may after shrinking $U$ assume these functions are constant with value $w(v_{i, j})$. We conclude that

$\textstyle {\int }_{U \times _ Y X \to U} w|_{U \times _ Y X} = \textstyle {\int }_{W \to U} w|_ W + \sum \textstyle {\int }_{V_{i, j} \to U} w|_{V_{i, j}} = \textstyle {\int }_{W \to U} w|_ W + (\int _ f w)(y)$

This is $\geq (\int _ f w)(y) = d$ and we conclude that $W$ must be the emptyset. Thus $U \times _ Y X \to U$ is finite. By Descent, Lemma 35.23.23 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).