Lemma 85.16.1. Let $S$ be a scheme. Let $X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, and flat. Then $X$ is countably indexed if and only if $Y$ is countably indexed.

**Proof.**
Assume $X$ is countably indexed. We write $X = \mathop{\mathrm{colim}}\nolimits X_ n$ as in Lemma 85.6.1. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda $ as in Definition 85.5.1. For every $n$ we can pick a $\lambda _ n$ such that $X_ n \to Y$ factors through $Y_{\lambda _ n}$, see Lemma 85.5.4. On the other hand, for every $\lambda $ the scheme $Y_\lambda \times _ Y X$ is affine (Lemma 85.15.7) and hence $Y_\lambda \times _ Y X \to X$ factors through $X_ n$ for some $n$ (Lemma 85.5.4). Picture

If we can show the dotted arrow exists, then we conclude that $Y = \mathop{\mathrm{colim}}\nolimits Y_{\lambda _ n}$ and $Y$ is countably indexed. To do this we pick a $\mu $ with $\mu \geq \lambda $ and $\mu \geq \lambda _ n$. Thus both $Y_\lambda \to Y$ and $Y_{\lambda _ n} \to Y$ factor through $Y_\mu \to Y$. Say $Y_\mu = \mathop{\mathrm{Spec}}(B_\mu )$, the closed subscheme $Y_\lambda $ corresponds to $J \subset B_\mu $, and the closed subscheme $Y_{\lambda _ n}$ corresponds to $J' \subset B_\mu $. We are trying to show that $J' \subset J$. By the diagram above we know $J'A_\mu \subset JA_\mu $ where $Y_\mu \times _ Y X = \mathop{\mathrm{Spec}}(A_\mu )$. Since $X \to Y$ is surjective and flat the morphism $Y_\lambda \times _ Y X \to Y_\lambda $ is a faithfully flat morphism of affine schemes, hence $B_\mu \to A_\mu $ is faithfully flat. Thus $J' \subset J$ as desired.

Assume $Y$ is countably indexed. Then $X$ is countably indexed by Lemma 85.15.9. $\square$

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