Lemma 58.19.3. In Situation 58.19.1. Let $V \to U$ be a finite morphism. Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$, let $X' = \mathop{\mathrm{Spec}}(A^\wedge )$ and let $U'$ and $V'$ be the base changes of $U$ and $V$ to $X'$. If $Y' \to X'$ is a finite morphism such that $V' = Y' \times _{X'} U'$, then there exists a finite morphism $Y \to X$ such that $V = Y \times _ X U$ and $Y' = Y \times _ X X'$.

Proof. This is a straightforward application of More on Algebra, Proposition 15.89.16. Namely, choose generators $f_1, \ldots , f_ t$ of $\mathfrak m$. For each $i$ write $V \times _ U D(f_ i) = \mathop{\mathrm{Spec}}(B_ i)$. For $1 \leq i, j \leq n$ we obtain an isomorphism $\alpha _{ij} : (B_ i)_{f_ j} \to (B_ j)_{f_ i}$ of $A_{f_ if_ j}$-algebras because the spectrum of both represent $V \times _ U D(f_ if_ j)$. Write $Y' = \mathop{\mathrm{Spec}}(B')$. Since $V \times _ U U' = Y \times _{X'} U'$ we get isomorphisms $\alpha _ i : B'_{f_ i} \to B_ i \otimes _ A A^\wedge$. A straightforward argument shows that $(B', B_ i, \alpha _ i, \alpha _{ij})$ is an object of $\text{Glue}(A \to A^\wedge , f_1, \ldots , f_ t)$, see More on Algebra, Remark 15.89.10. Applying the proposition cited above (and using More on Algebra, Remark 15.89.20 to obtain the algebra structure) we find an $A$-algebra $B$ such that $\text{Can}(B)$ is isomorphic to $(B', B_ i, \alpha _ i, \alpha _{ij})$. Setting $Y = \mathop{\mathrm{Spec}}(B)$ we see that $Y \to X$ is a morphism which comes equipped with compatible isomorphisms $V \cong Y \times _ X U$ and $Y' = Y \times _ X X'$ as desired. $\square$

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