Lemma 87.19.9. Let $S$ be a scheme. Let $Y$ be an affine formal algebraic space. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable and affine. Then

1. $X$ is an affine formal algebraic space,

2. if $Y$ is countably indexed, then $X$ is countably indexed,

3. if $Y$ is countably indexed and classical, then $X$ is countably indexed and classical,

4. if $Y$ is weakly adic, then $X$ is weakly adic,

5. if $Y$ is adic*, then $X$ is adic*, and

6. if $Y$ is Noetherian and $f$ is (locally) of finite type, then $X$ is Noetherian.

Proof. Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda$ as in Definition 87.9.1. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times _ Y X$ are affine. And $X = \mathop{\mathrm{colim}}\nolimits Y_\lambda \times _ Y X$. Thus $X$ is an affine formal algebraic space.

Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda$ is countable. Then we immediately see that $X$ is countably indexed too.

Proof of (3), (4), and (5). In each of these cases the assumptions imply that $Y$ is a countably indexed affine formal algebraic space (Lemma 87.10.3) and hence $X$ is too by (2). Thus we may write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebras $A$ and $B$, see Lemma 87.10.4. By Lemma 87.9.10 the morphism $f$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 87.19.8 that $\varphi$ is taut. We conclude that (3) follows from Lemma 87.5.9, (4) follows from Lemma 87.7.5, and (5) follows from Lemma 87.6.5.

Proof of (6). Combining (3) with Lemma 87.10.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 87.10.5. First, it tells us the affine schemes $Y_\lambda$ are Noetherian. Then $X_\lambda \to Y_\lambda$ is of finite type, hence $X_\lambda$ is Noetherian too (Morphisms, Lemma 29.15.6). Then the criterion tells us $X$ is Noetherian and the proof is complete. $\square$

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