Lemma 37.76.6. 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 37.76.4. 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
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