Lemma 66.41.8. The composition of two morphisms of algebraic spaces which satisfy the (existence part of, resp. uniqueness part of) the valuative criterion satisfies the (existence part of, resp. uniqueness part of) the valuative criterion.

Proof. Let $f : X \to Y$, $g : Y \to Z$ be morphisms of algebraic spaces over the scheme $S$. Consider a solid commutative diagram of the following shape

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

If we have the uniqueness part for $g$, then there exists at most one north-west dotted arrow making the diagram commute. If we also have the uniqueness part for $f$, then we have at most one north-north-west dotted arrow making the diagram commute. The proof in the existence case comes from contemplating the following diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K'') \ar[r] \ar[dd] & \mathop{\mathrm{Spec}}(K') \ar[r] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d]^ f \\ & & & Y \ar[d]^ g \\ \mathop{\mathrm{Spec}}(A'') \ar[r] \ar[rrruu] & \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Z }$

Namely, the existence part for $g$ gives us the extension $K'$, the valuation ring $A'$ and the arrow $\mathop{\mathrm{Spec}}(A') \to Y$, whereupon the existence part for $f$ gives us the extension $K''$, the valuation ring $A''$ and the arrow $\mathop{\mathrm{Spec}}(A'') \to X$. $\square$

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