The Stacks project

Lemma 86.15.2. Let $S$ be a scheme. Let $X, Y$ be formal algebraic spaces over $S$ and let $Z$ be a sheaf whose diagonal is representable by algebraic spaces. Let $X \to Z$ and $Y \to Z$ be maps of sheaves. Then $X \times _ Z Y$ is a formal algebraic space.

Proof. Choose $\{ X_ i \to X\} $ and $\{ Y_ j \to Y\} $ as in Definition 86.11.1. Then $\{ X_ i \times _ Z Y_ j \to X \times _ Z Y\} $ is a family of maps which are representable by algebraic spaces and ├ętale. Thus Lemma 86.15.1 tells us it suffices to show that $X \times _ Z Y$ is a formal algebraic space when $X$ and $Y$ are affine formal algebraic spaces.

Assume $X$ and $Y$ are affine formal algebraic spaces. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ and $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ as in Definition 86.9.1. Then $X \times _ Z Y = \mathop{\mathrm{colim}}\nolimits X_\lambda \times _ Z Y_\mu $. Each $X_\lambda \times _ Z Y_\mu $ is an algebraic space. For $\lambda \leq \lambda '$ and $\mu \leq \mu '$ the morphism

\[ X_\lambda \times _ Z Y_\mu \to X_\lambda \times _ Z Y_{\mu '} \to X_{\lambda '} \times _ Z Y_{\mu '} \]

is a thickening as a composition of base changes of thickenings. Thus we conclude by applying Lemma 86.13.1. $\square$

Comments (2)

Comment #1558 by Matthew Emerton on

I think the lemma is missing the hypothesis that there are maps .

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