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Fppf descent data for algebraic spaces are effective.

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 $S$.

  1. If $I$ is countable1, then any descent datum for algebraic spaces relative to $\{ X_ i \to X\} $ is effective.

  2. Any descent datum $(Y_ i, \varphi _{ij})$ relative to $\{ X_ i \to X\} _{i \in I}$ (Descent on Spaces, Definition 74.22.3) with $Y_ i \to X_ i$ of finite type is 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

\[ \coprod F \times _ X X_ i \longrightarrow F \]

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 ├ętale morphism $U \to X$. Then $F' = U \times _ X F \to F$ is representable, surjective, and ├ętale 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. $\square$

[1] The restriction on countablility can be ignored by those who do not care about set theoretical issues. We can allow larger index sets here if we can bound the size of the algebraic spaces which we are descending. See for example Lemma 80.11.2.

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