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
Comments (0)
There are also: