Lemma 88.23.10. In the situation above. If $X$ is locally Noetherian, $f$ is proper, and $U' \to U$ is surjective, then $f_{/T}$ is rig-surjective.
Proof. (The statement makes sense by Lemma 88.23.1 and Formal Spaces, Lemma 87.20.8.) Let $R$ be a complete discrete valuation ring with fraction field $K$. Let $p : \text{Spf}(R) \to X_{/T}$ be an adic morphism of formal algebraic spaces. By Formal Spaces, Lemma 87.33.4 the composition $\text{Spf}(R) \to X_{/T} \to X$ corresponds to a morphism $q : \mathop{\mathrm{Spec}}(R) \to X$ which maps $\mathop{\mathrm{Spec}}(K)$ into $U$. Since $U' \to U$ is proper and surjective we see that $\mathop{\mathrm{Spec}}(K) \times _ U U'$ is nonempty and proper over $K$. Hence we can choose a field extension $K'/K$ and a commutative diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U' \ar[r] \ar[d] & X' \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r] & U \ar[r] & X } \]
Let $R' \subset K'$ be a discrete valuation ring dominating $R$ with fraction field $K'$, see Algebra, Lemma 10.119.13. Since $\mathop{\mathrm{Spec}}(K) \to X$ extends to $\mathop{\mathrm{Spec}}(R) \to X$ we see by the valuative criterion of properness (Morphisms of Spaces, Lemma 67.44.1) that we can extend our $K'$-point of $U'$ to a morphism $\mathop{\mathrm{Spec}}(R') \to X'$ over $\mathop{\mathrm{Spec}}(R) \to X$. It follows that the inverse image of $T'$ in $\mathop{\mathrm{Spec}}(R')$ is the closed point and we find an adic morphism $\text{Spf}((R')^\wedge ) \to X'_{/T'}$ lifting $p$ as desired (note that $(R')^\wedge $ is a complete discrete valuation ring by More on Algebra, Lemma 15.43.5). $\square$
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