Lemma 86.9.5. Let $S$ be a scheme. Let $X$ be a sheaf on $(\mathit{Sch}/S)_{fppf}$. Then $X$ is an affine formal algebraic space if and only if the following hold

1. any morphism $U \to X$ where $U$ is an affine scheme over $S$ factors through a morphism $T \to X$ which is representable and a thickening with $T$ an affine scheme over $S$, and

2. a set theoretic condition as in Remark 86.11.5.

Proof. It follows from Lemmas 86.9.3 and 86.9.4 that an affine formal algebraic space satisfies (1) and (2). In order to prove the converse we may assume $X$ is not empty. Let $\Lambda$ be the category of representable morphisms $T \to X$ which are thickenings where $T$ is an affine scheme over $S$. This category is directed. Since $X$ is not empty, $\Lambda$ contains at least one object. If $T \to X$ and $T' \to X$ are in $\Lambda$, then we can factor $T \amalg T' \to X$ through $T'' \to X$ in $\Lambda$. Between any two objects of $\Lambda$ there is a unique arrow or none. Thus $\Lambda$ is a directed set and by assumption $X = \mathop{\mathrm{colim}}\nolimits _{T \to X\text{ in }\Lambda } T$. To finish the proof we need to show that any arrow $T \to T'$ in $\Lambda$ is a thickening. This is true because $T' \to X$ is a monomorphism of sheaves, so that $T = T \times _{T'} T' = T \times _ X T'$ and hence the morphism $T \to T'$ equals the projection $T \times _ X T' \to T'$ which is a thickening because $T \to X$ is a thickening. $\square$

Comment #1942 by Brian Conrad on

At the end of the proof it would be appropriate to insert a sentence or two to justify that the transition maps in $\Lambda$ are thickenings. (This is easy, but ought to be addressed.)

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