Lemma 80.11.3. Let S be a scheme. Let \{ X_ i \to X\} _{i \in I} be an fppf covering of algebraic spaces over S.
Fppf descent data for algebraic spaces are effective.
Proof. Proof of (1). By Descent on Spaces, Lemma 74.23.1 this translates into the statement that an fppf sheaf F endowed with a map F \to X is an algebraic space provided that each F \times _ X X_ i is an algebraic space. The restriction on the cardinality of I implies that coproducts of algebraic spaces indexed by I are algebraic spaces, see Spaces, Lemma 65.8.4 and Sets, Lemma 3.9.9. The morphism
is representable by algebraic spaces (as the base change of \coprod X_ i \to X, see Lemma 80.3.3), and surjective, flat, and locally of finite presentation (as the base change of \coprod X_ i \to X, see Lemma 80.4.2). Hence part (1) follows from Theorem 80.10.1.
Proof of (2). First we apply Descent on Spaces, Lemma 74.23.1 to obtain an fppf sheaf F endowed with a map F \to X such that F \times _ X X_ i = Y_ i for all i \in I. Our goal is to show that F is an algebraic space. Choose a scheme U and a surjective étale morphism U \to X. Then F' = U \times _ X F \to F is representable, surjective, and étale as the base change of U \to X. By Theorem 80.10.1 it suffices to show that F' = U \times _ X F is an algebraic space. We may choose an fppf covering \{ U_ j \to U\} _{j \in J} where U_ j is a scheme refining the fppf covering \{ X_ i \times _ X U \to U\} _{i \in I}, see Topologies on Spaces, Lemma 73.7.4. Thus we get a map a : J \to I and for each j a morphism U_ j \to X_{a(j)} over X. Then we see that U_ j \times _ U F' = U_ j \times _{X_{a(j)}} Y_{a(j)} is of finite type over U_ j. Hence F' is an algebraic space by Lemma 80.11.2. \square
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