The Stacks project

Proof. We will show that an fppf covering is an fpqc covering, and then the rest follows from Lemma 73.7.2. Let $\{ f_ i : U_ i \to U\} _{i \in I}$ be an fppf covering of algebraic spaces over $S$. By definition this means that the $f_ i$ are flat which checks the first condition of Definition 73.9.1. To check the second, let $V \to U$ be a morphism with $V$ affine. We may choose an étale covering $\{ V_{ij} \to V \times _ U U_ i\}$ with $V_{ij}$ affine. Then the compositions $f_{ij} : V_{ij} \to V \times _ U U_ i \to V$ are flat and locally of finite presentation as compositions of such (Morphisms of Spaces, Lemmas 67.28.2, 67.30.3, 67.39.7, and 67.39.8). Hence these morphisms are open (Morphisms of Spaces, Lemma 67.30.6) and we see that $|V| = \bigcup _{i \in I} \bigcup _{j \in J_ i} f_{ij}(|V_{ij}|)$ is an open covering of $|V|$. Since $|V|$ is quasi-compact, this covering has a finite refinement. Say $V_{i_1j_1}, \ldots , V_{i_ Nj_ N}$ do the job. Then $\{ V_{i_ kj_ k} \to V\} _{k = 1, \ldots , N}$ is a standard fpqc covering of $V$ refinining the family $\{ U_ i \times _ U V \to V\}$. This finishes the proof. $\square$

Proof. Part (1) is clear. Consider $g : X' \to X$ and $\{ X_ i \to X\} _{i\in I}$ an fpqc covering as in (3). By Morphisms of Spaces, Lemma 67.30.4 the morphisms $X' \times _ X X_ i \to X'$ are flat. If $h' : Z \to X'$ is a morphism from an affine scheme towards $X'$, then set $h = g \circ h' : Z \to X$. The assumption on $\{ X_ i \to X\} _{i\in I}$ means there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ and morphisms $Z_ j \to X_{i(j)}$ covering $h$ for certain $i(j) \in I$. By the universal property of the fibre product we obtain morphisms $Z_ j \to X' \times _ X X_{i(j)}$ over $h'$ also. Hence $\{ X' \times _ X X_ i \to X'\} _{i\in I}$ is an fpqc covering. This proves (3).

Let $\{ X_ i \to X\} _{i\in I}$ and $\{ X_{ij} \to X_ i\} _{j\in J_ i}$ be as in (2). Let $h : Z \to X$ be a morphism from an affine scheme towards $X$. By assumption there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ and morphisms $h_ j : Z_ j \to X_{i(j)}$ covering $h$ for some indices $i(j) \in I$. By assumption there exist standard fpqc coverings $\{ Z_{j, l} \to Z_ j\} _{l = 1, \ldots , n(j)}$ and morphisms $Z_{j, l} \to X_{i(j)j(l)}$ covering $h_ j$ for some indices $j(l) \in J_{i(j)}$. By Topologies, Lemma 34.9.11 the family $\{ Z_{j, l} \to Z\}$ is a standard fpqc covering. Hence we conclude that $\{ X_{ij} \to X\} _{i \in I, j\in J_ i}$ is an fpqc covering. $\square$

Proof. If $\{ X_ i \to X\} _{i \in I}$ is an fpqc covering, then so is $\{ U \times _ X X_ i \to U\} _{i \in I}$ by Lemma 73.9.3. Assume that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering. Let $h : Z \to X$ be a morphism from an affine scheme towards $X$. Then we see that $U \times _ X Z \to Z$ is a surjective étale morphism of schemes, in particular open. Hence we can find finitely many affine opens $W_1, \ldots , W_ t$ of $U \times _ X Z$ whose images cover $Z$. For each $j$ we may apply the condition that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering to the morphism $W_ j \to U$, and obtain a standard fpqc covering $\{ W_{jl} \to W_ j\}$ which refines $\{ W_ j \times _ X X_ i \to W_ j\} _{i \in I}$. Hence $\{ W_{jl} \to Z\}$ is a standard fpqc covering of $Z$ (see Topologies, Lemma 34.9.11) which refines $\{ Z \times _ X X_ i \to X\}$ and we win. $\square$

Proof. Omitted. Hint: For each $i$ choose a scheme $T_ i$ and a surjective étale morphism $T_ i \to X_ i$. Then check that $\{ T_ i \to X\}$ is an fpqc covering. $\square$