Lemma 34.10.14. Let \{ f_ i : X_ i \to X\} _{i \in I} be a V covering. Then
is a universally submersive morphism of schemes (Morphisms, Definition 29.24.1).
Lemma 34.10.14. Let \{ f_ i : X_ i \to X\} _{i \in I} be a V covering. Then
is a universally submersive morphism of schemes (Morphisms, Definition 29.24.1).
Proof. We will use without further mention that the base change of a V covering is a V covering (Lemma 34.10.9). In particular it suffices to show that the morphism is submersive. Being submersive is clearly Zariski local on the base. Thus we may assume X is affine. Then \{ X_ i \to X\} can be refined by a standard V covering \{ Y_ j \to X\} . If we can show that \coprod Y_ j \to X is submersive, then since there is a factorization \coprod Y_ j \to \coprod X_ i \to X we conclude that \coprod X_ i \to X is submersive. Set Y = \coprod Y_ j and consider the morphism of affines f : Y \to X. By Lemma 34.10.13 we know that we can lift any specialization x' \leadsto x in X to some specialization y' \leadsto y in Y. Thus if T \subset X is a subset such that f^{-1}(T) is closed in Y, then T \subset X is closed under specialization. Since f^{-1}(T) \subset Y with the reduced induced closed subscheme structure is an affine scheme, we conclude that T \subset X is closed by Algebra, Lemma 10.41.5. Hence f is submersive. \square
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