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

$\xymatrix{ Y_\lambda \times _ Y X \ar[r] \ar[d] & X_ n \ar[r] \ar[d] & X \ar[d] \\ Y_\lambda \ar@{..>}[r] \ar@/_1pc/[rr] & Y_{\lambda _ n} \ar[r] & Y }$

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. $\square$

Comment #1955 by Brian Conrad on

There are a lot of arrows in the diagram pointing the wrong way: both left horizontal arrows and the lower curved arrow, while the upper curves arrow plays no role and so can be removed. Maybe reorganize the diagram?

Likewise, the contained being proved is that $J' \subset J$, not that $J \subset J'$, and the diagram gives that $J' A_{\mu} \subset J A_{\mu}$ (rather than the other way around).

Comment #1958 by jojo on

In the statement $S$ doesn't seem to be used explicitly.

Comment #2009 by on

@#1955 Thanks very much. Indeed it was very confusing. I've simplified the diagram and now it is clearer. See changes.

@#1958 You are in good company with your complaint. But please read the post on the stacks project blog with title "Base scheme for spaces" for the explanation.

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