Lemma 87.14.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Then the functor
is a formal algebraic space.
Lemma 87.14.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Then the functor
is a formal algebraic space.
Proof. Denote $F$ the functor. Let $\{ U_ i \to U\} $ be an fppf covering. Then $\coprod |U_ i| \to |U|$ is surjective. Since $X$ is an fppf sheaf, it follows that $F$ is an fppf sheaf.
Let $\{ g_ i : X_ i \to X\} $ be an étale covering such that $X_ i$ is affine for all $i$, see Properties of Spaces, Lemma 66.6.1. The morphisms $F \times _ X X_ i \to F$ are étale (see Spaces, Lemma 65.5.5) and the map $\coprod F \times _ X X_ i \to F$ is a surjection of sheaves. Thus it suffices to prove that $F \times _ X X_ i$ is an affine formal algebraic space. A $U$-valued point of $F \times _ X X_ i$ is a morphism $U \to X_ i$ whose image is contained in the closed subset $g_ i^{-1}(T) \subset |X_ i|$. Thus this follows from Lemma 87.14.1. $\square$
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