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The Stacks project

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 34.9.11 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


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