The Stacks project

Proof. The implication (2) $\Rightarrow $ (1) follows from the more general Topologies, Lemma 34.10.14 and our definition of h covers. Assume $\coprod X_ i \to X$ is universally submersive. We will show that $\{ X_ i \to X\} $ can be refined by a ph covering; this will suffice by Topologies, Lemma 34.8.7 and our definition of h coverings. The argument will be the same as the one used in the proof of Lemma 38.34.1.

Choose a generic flatness stratification

as in More on Morphisms, Lemma 37.54.2 for the finitely presented morphism

We are going to use all the properties of the stratification without further mention. By construction the base change of each $f_ i$ to $U_ k = S_ k \setminus S_{k + 1}$ is flat. Denote $Y_ k$ the scheme theoretic closure of $U_ k$ in $S_ k$. Since $U_ k \to S_ k$ is a quasi-compact open immersion (all schemes in this paragraph are Noetherian), we see that $U_ k \subset Y_ k$ is a quasi-compact dense (and scheme theoretically dense) open immersion, see Morphisms, Lemma 29.6.3. The morphism $\coprod _{k = 0, \ldots , t - 1} Y_ k \to X$ is finite surjective, hence $\{ Y_ k \to X\} $ is a ph covering. By the transitivity property of ph coverings (Topologies, Lemma 34.8.8) it suffices to show that the pullback of the covering $\{ X_ i \to X\} $ to each $Y_ k$ can be refined by a ph covering. This reduces us to the case described in the next paragraph.

Assume $\coprod X_ i \to X$ is universally submersive and there is a dense open $U \subset X$ such that $X_ i \times _ X U \to U$ is flat for all $i$. By Theorem 38.30.7 there is a $U$-admissible blowup $X' \to X$ such that the strict transform $f'_ i : X'_ i \to X'$ of $f_ i$ is flat for all $i$. Observe that the projective (hence closed) morphism $X' \to X$ is surjective as $U \subset X$ is dense and as $U$ is identified with an open of $X'$. After replacing $X'$ by a further $U$-admissible blowup if necessary, we may also assume $U \subset X'$ is dense (see Remark 38.30.1). Hence for every point $x \in X'$ there is a discrete valuation ring $A$ and a morphism $g : \mathop{\mathrm{Spec}}(A) \to X'$ such that the generic point of $\mathop{\mathrm{Spec}}(A)$ maps into $U$ and the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $x$, see Limits, Lemma 32.15.1. Set

Since $\coprod X_ i \to X$ is universally submersive, there is a specialization $w' \leadsto w$ in $W$ such that $w'$ maps to the generic point of $\mathop{\mathrm{Spec}}(A)$ and $w$ maps to the closed point of $\mathop{\mathrm{Spec}}(A)$. (If not, then the closed fibre of $W \to \mathop{\mathrm{Spec}}(A)$ is stable under generalizations, hence open, which contradicts the fact that $W \to \mathop{\mathrm{Spec}}(A)$ is submersive.) Say $w' \in \mathop{\mathrm{Spec}}(A) \times _ X X_ i$ so of course $w \in \mathop{\mathrm{Spec}}(A) \times _ X X_ i$ as well. Let $x'_ i \leadsto x_ i$ be the image of $w' \leadsto w$ in $X' \times _ X X_ i$. Since $x'_ i \in X'_ i$ and since $X'_ i \subset X' \times _ X X_ i$ is a closed subscheme we see that $x_ i \in X'_ i$. Since $x_ i$ maps to $x \in X'$ we conclude that $\coprod X'_ i \to X'$ is surjective! In particular $\{ X'_ i \to X'\} $ is an fppf covering. But an fppf covering is a ph covering (More on Morphisms, Lemma 37.48.7). Since $X' \to X$ is proper surjective, we conclude that $\{ X'_ i \to X\} $ is a ph covering and the proof is complete. $\square$