Lemma 87.9.2. Let S be a scheme. If X is an affine formal algebraic space over S, then the diagonal morphism \Delta : X \to X \times _ S X is representable and a closed immersion.
Proof. Suppose given U \to X and V \to X where U, V are schemes over S. Let us show that U \times _ X V is representable. Write X = \mathop{\mathrm{colim}}\nolimits X_\lambda as in Definition 87.9.1. The discussion above shows that Zariski locally on U and V the morphisms factors through some X_\lambda . In this case U \times _ X V = U \times _{X_\lambda } V which is a scheme. Thus the diagonal is representable, see Spaces, Lemma 65.5.10. Given (a, b) : W \to X \times _ S X where W is a scheme over S consider the map X \times _{\Delta , X \times _ S X, (a, b)} W \to W. As before locally on W the morphisms a and b map into the affine scheme X_\lambda for some \lambda and then we get the morphism X_\lambda \times _{\Delta _\lambda , X_\lambda \times _ S X_\lambda , (a, b)} W \to W. This is the base change of \Delta _\lambda : X_\lambda \to X_\lambda \times _ S X_\lambda which is a closed immersion as X_\lambda \to S is separated (because X_\lambda is affine). Thus X \to X \times _ S X is a closed immersion. \square
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