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

\[ \coprod \nolimits _{i \in I} f_ i : \coprod \nolimits _{i \in I} X_ i \longrightarrow X \]

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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