Lemma 87.37.1. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Let $T \subset |X_{red}|$ be a closed subset. Then the functor
is an affine formal algebraic space.
Lemma 87.37.1. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Let $T \subset |X_{red}|$ be a closed subset. Then the functor
is an affine formal algebraic space.
Proof. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. Then $X_{\lambda , red} = X_{red}$ and we may and do view $T$ as a closed subset of $|X_\lambda | = |X_{\lambda , red}|$. By Lemma 87.14.1 for each $\lambda $ the completion $(X_\lambda )_{/T}$ is an affine formal algebraic space. The transition morphisms $(X_\lambda )_{/T} \to (X_\mu )_{/T}$ are closed immersions as base changes of the transition morphisms $X_\lambda \to X_\mu $, see Lemma 87.14.4. Also the morphisms $((X_\lambda )_{/T})_{red} \to ((X_\mu )_/T)_{red}$ are isomorphisms by Lemma 87.14.5. Since $X_{/T} = \mathop{\mathrm{colim}}\nolimits (X_\lambda )_{/T}$ we conclude by Lemma 87.36.1. $\square$
Comments (0)