Proof.
Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 86.9.1. 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 86.10.3) 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 86.10.4. By Lemma 86.9.10 the morphism $f$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 86.19.8 that $\varphi $ is taut. We conclude that (3) follows from Lemma 86.5.9, (4) follows from Lemma 86.7.5, and (5) follows from Lemma 86.6.5.
Proof of (6). Combining (3) with Lemma 86.10.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 86.10.5. 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.
$\square$
Comments (0)