Lemma 73.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.

1. If $X' \to X$ is an isomorphism then $\{ X' \to X\}$ is an fpqc covering of $X$.

2. If $\{ X_ i \to X\} _{i\in I}$ is an fpqc covering and for each $i$ we have an fpqc covering $\{ X_{ij} \to X_ i\} _{j\in J_ i}$, then $\{ X_{ij} \to X\} _{i \in I, j\in J_ i}$ is an fpqc covering.

3. If $\{ X_ i \to X\} _{i\in I}$ is an fpqc covering and $X' \to X$ is a morphism of algebraic spaces then $\{ X' \times _ X X_ i \to X'\} _{i\in I}$ is an fpqc covering.

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 34.9.10 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. $\square$

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