Lemma 66.41.5. Let $S$ be a scheme. Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ satisfies the existence part of the valuative criterion as in Definition 66.41.1,

2. given any commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & Y }$

where $A$ is a valuation ring with field of fractions $K$, there exists a dotted arrow, i.e., $f$ satisfies the existence part of the valuative criterion as in Schemes, Definition 26.20.3.

Proof. We have to show that (1) implies (2). Suppose given a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

as in part (2). By (1) there exists a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

as in Definition 66.41.1 with $K \subset K'$ arbitrary. By Lemma 66.41.4 we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram, i.e., (2) holds. $\square$

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