The Stacks project

Lemma 87.22.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume

  1. $X$ and $Y$ are locally Noetherian,

  2. $f$ locally of finite type,

  3. $\Delta _ f : X \to X \times _ Y X$ is rig-├ętale and rig-surjective.

Then $f$ is rig surjective if and only if every adic morphism $\text{Spf}(R) \to Y$ where $R$ is a complete discrete valuation ring lifts to a morphism $\text{Spf}(R) \to X$.

Proof. One direction is trivial. For the other, suppose that $\text{Spf}(R) \to Y$ is an adic morphism such that there exists an extension of complete discrete valuation rings $R \subset R'$ with $\text{Spf}(R') \to \text{Spf}(R) \to X$ factoring through $Y$. Consider the fibre product diagram

\[ \xymatrix{ \text{Spf}(R') \ar[r] \ar[rd] & \text{Spf}(R) \times _ Y X \ar[r] \ar[d]^ p & X \ar[d]^ f \\ & \text{Spf}(R) \ar[r] & Y } \]

The morphism $p$ is locally of finite type as a base change of $f$, see Formal Spaces, Lemma 86.24.4. The diagonal morphism $\Delta _ p$ is the base change of $\Delta _ f$ and hence is rig-├ętale and rig-surjective. By Lemma 87.22.2 the flat locus of $\text{Spf}(R) \times _ Y X$ over $R$ is either $\emptyset $ or equal to $\text{Spf}(R)$. However, since $\text{Spf}(R')$ factors through it we conclude it is not empty and hence we get a morphism $\text{Spf}(R) \to \text{Spf}(R) \times _ Y X \to X$ as desired. $\square$

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