Lemma 87.29.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces. Assume $Y$ countably indexed. The following are equivalent

1. $f$ is locally of finite type,

2. $f$ is of finite type,

3. $f$ corresponds to a morphism $B \to A$ of $\textit{WAdm}^{count}$ (Section 87.21) satisfying the equivalent conditions of Lemma 87.29.6.

Proof. Since $X$ and $Y$ are affine it is clear that conditions (1) and (2) are equivalent. In cases (1) and (2) the morphism $f$ is representable by algebraic spaces by definition, hence affine by Lemma 87.19.7. Thus if (1) or (2) holds we see that $X$ is countably indexed by Lemma 87.19.9. Write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for topological $S$-algebras $A$ and $B$ in $\textit{WAdm}^{count}$, see Lemma 87.10.4. By Lemma 87.9.10 we see that $f$ corresponds to a continuous map $B \to A$. Hence now the result follows from Lemma 87.29.2. $\square$

Comment #1974 by Brian Conrad on

The applicability of the Lemma at the end of the 2nd sentence of the proof requires knowing that $f$ is representable and affine. So it would help the reader if it is said that under (1) and (2) the map $f$ is representable in algebraic spaces by the definitions and thus it is also affine by Tag 0AKN.

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