hcju/stacksqa huggingface dataset:
The field $\overline{\kappa}$ is the algebraic closure of $\kappa$ by Lemma \ref{lemma-get-algebraic-closure}. Most of the statements immediately follow from the corresponding parts of Lemma \ref{lemma-structure-decomposition}. For example because $\text{Aut}(\overline{\kappa}/\kappa) = \text{Gal}(\kappa^{sep}/\kappa)$ we obtain the first sequence. Then the only other assertion that needs a proof is the fact that with $S$ as in Lemma \ref{lemma-structure-decomposition} the limit $\lim_{n \in S} \mu_n(\overline{\kappa})$ is equal to $\lim_{n\text{ prime to }p} \mu_n(\kappa^{sep})$. To see this it suffices to show that every integer $n$ prime to $p$ divides an element of $S$. Let $\pi \in A$ be a uniformizer and consider the splitting field $L$ of the polynomial $X^n - \pi$. Since the polynomial is separable we see that $L$ is a finite Galois extension of $K$. Choose an embedding $L \to K^{sep}$. Observe that if $B$ is the integral closure of $A$ in $L$, then the ramification index of $A \to B_{\mathfrak m^{sep} \cap B}$ is divisible by $n$ (because $\pi$ has an $n$th root in $B$; in fact the ramification index equals $n$ but we do not need this). Then it follows from the construction of the $S$ in the proof of Lemma \ref{lemma-structure-decomposition} that $n$ divides an element of $S$.
Let $B$ be the strict henselization of $\mathcal{O}_{T, t}$ with respect to $\kappa(t) \subset \kappa(t)^{sep} \subset \kappa(\overline{t})$. Pick $\psi : \overline{t}' \to \Spec(B)$ lifting $\overline{t}' \to T$ as in the construction of the specialization map. Let $s$ and $s'$ denote the images of $t$ and $t'$ in $S$. Let $A$ be the strict henselization of $\mathcal{O}_{S, s}$ with respect to $\kappa(s) \subset \kappa(s)^{sep} \subset \kappa(\overline{s})$. Since $\kappa(\overline{s}) = \kappa(\overline{t})$, by the functoriality of strict henselization (Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial}) we obtain a ring map $A \to B$ fitting into the commutative diagram $$ \xymatrix{ \overline{t}' \ar[r]_-\psi \ar[d] & \Spec(B) \ar[d] \ar[r] & T \ar[d] \\ \overline{s}' \ar[r]^-\varphi & \Spec(A) \ar[r] & S } $$ Here the morphism $\varphi : \overline{s}' \to \Spec(A)$ is simply taken to be the composition $\overline{t}' \to \Spec(B) \to \Spec(A)$. Applying base change we obtain a commutative diagram $$ \xymatrix{ Y_{\overline{t}'} \ar[r] \ar[d] & Y_B \ar[d] \\ X_{\overline{s}'} \ar[r] & X_A } $$ and from the construction of the specialization map the commutativity of this diagram implies the commutativity of the diagram of the lemma.
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa(s) \to \kappa(\overline{s})$. Let $A \to \kappa(\overline{s}')$ be the map used to construct the first specialization map. Let $\mathcal{O}_{S, s'} \to A'$ be the strict henselization constructed using $\kappa(s') \subset \kappa(\overline{s}')$. By functoriality of strict henselization, there is a map $A \to A'$ such that the composition with $A' \to \kappa(\overline{s}')$ is the given map (Algebra, Lemma \ref{algebra-lemma-map-into-henselian-colimit}). Next, let $A' \to \kappa(\overline{s}'')$ be the map used to construct the second specialization map. Then it is clear that the composition of the first and second specialization maps is the specialization map $\pi_1(X_{\overline{s}''}) \to \pi_1(X_{\overline{s}})$ constructed using $A \to A' \to \kappa(\overline{s}'')$.
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa(s) \to \kappa(\overline{s})$. Let $A \to \kappa(\overline{s}')$ be the map used to construct $sp$. Let $R \subset \kappa(\overline{s}')$ be a valuation ring with fraction field $\kappa(\overline{s}')$ dominating the image of $A$. See Algebra, Lemma \ref{algebra-lemma-dominate}. Observe that $R$ is strictly henselian for example by Lemma \ref{lemma-normal-local-domain-separablly-closed-fraction-field} and Algebra, Lemma \ref{algebra-lemma-valuation-ring-normal}. Then the lemma is clear.
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa(s) \to \kappa(\overline{s})$. Let $A \to \kappa(\overline{s}')$ be the map used to construct $sp$. Let $R \subset \kappa(\overline{s}')$ be a discrete valuation ring dominating the image of $A$, see Algebra, Lemma \ref{algebra-lemma-exists-dvr}. Choose a diagram of fields $$ \xymatrix{ \kappa(\overline{s}) \ar[r] & k \\ A/\mathfrak m_A \ar[r] \ar[u] & R/\mathfrak m_R \ar[u] } $$ with $k$ algebraically closed. Let $R^{sh}$ be the strict henselization of $R$ constructed using $R \to k$. Then $R^{sh}$ is a discrete valuation ring by More on Algebra, Lemma \ref{more-algebra-lemma-henselization-dvr}. Denote $\eta, o$ the generic and closed point of $\Spec(R^{sh})$. Since the diagram of schemes $$ \xymatrix{ \overline{\eta} \ar[d] \ar[r] & \Spec(R^{sh}) \ar[d] & \Spec(k) \ar[d] \ar[l] \\ \overline{s}' \ar[r] & \Spec(A) & \overline{s} \ar[l] } $$ commutes, we obtain a commutative diagram $$ \xymatrix{ \pi_1(X_{\overline{\eta}}) \ar[d] \ar[r]_{sp_{R^{sh}}} & \pi_1(X_o) \ar[d] \\ \pi_1(X_{\overline{s}'}) \ar[r]^{sp} & X_{\overline{s}} } $$ of specialization maps by the construction of these maps. Since the vertical arrows are isomorphisms (Lemma \ref{lemma-finite-etale-invariant-over-proper}), this proves the lemma.
Since the category of finite \'etale coverings has an internal hom (Lemma \ref{lemma-internal-hom-finite-etale}) it suffices to prove the following: Given $U$ finite \'etale over $X$ and a morphism $t : Y \to U$ over $X$ there exists a unique section $s : X \to U$ such that $t = s|_Y$. Picture $$ \xymatrix{ & U \ar[d]^f \\ Y \ar[r] \ar[ru] & X \ar@{..>}@/^1em/[u] } $$ Finding the dotted arrow $s$ is the same thing as finding an $\mathcal{O}_X$-algebra map $$ s^\sharp : f_*\mathcal{O}_U \longrightarrow \mathcal{O}_X $$ which reduces modulo the ideal sheaf of $Y$ to the given algebra map $t^\sharp : f_*\mathcal{O}_U \to \mathcal{O}_Y$. By Lemma \ref{lemma-thickening} we can lift $t$ uniquely to a compatible system of maps $t_n : Y_n \to U$ and hence a map $$ \lim t_n^\sharp : f_*\mathcal{O}_U \longrightarrow \lim \mathcal{O}_{Y_n} $$ of sheaves of algebras on $X$. Since $f_*\mathcal{O}_U$ is a finite locally free $\mathcal{O}_X$-module, we conclude that we get a unique $\mathcal{O}_X$-module map $\sigma : f_*\mathcal{O}_U \to \mathcal{O}_X$ whose completion is $\lim t_n^\sharp$. To see that $\sigma$ is an algebra homomorphism, we need to check that the diagram $$ \xymatrix{ f_*\mathcal{O}_U \otimes_{\mathcal{O}_X} f_*\mathcal{O}_U \ar[r] \ar[d]_{\sigma \otimes \sigma} & f_*\mathcal{O}_U \ar[d]^\sigma \\ \mathcal{O}_X \otimes_{\mathcal{O}_X} \mathcal{O}_X \ar[r] & \mathcal{O}_X } $$ commutes. For every $n$ we know this diagram commutes after restricting to $Y_n$, i.e., the diagram commutes after applying the completion functor. Hence by faithfulness of the completion functor we conclude.
The restriction functor is fully faithful by Lemma \ref{lemma-restriction-fully-faithful}. \medskip\noindent Let $U_1 \to Y$ be a finite \'etale morphism. To finish the proof we will show that $U_1$ is in the essential image of the restriction functor. \medskip\noindent For $n \geq 1$ let $Y_n$ be the $n$th infinitesimal neighbourhood of $Y$. By Lemma \ref{lemma-thickening} there is a unique finite \'etale morphism $\pi_n : U_n \to Y_n$ whose base change to $Y = Y_1$ recovers $U_1 \to Y_1$. Consider the sheaves $\mathcal{F}_n = \pi_{n, *}\mathcal{O}_{U_n}$. We may and do view $\mathcal{F}_n$ as an $\mathcal{O}_X$-module on $X$ wich is locally isomorphic to $(\mathcal{O}_X/f^{n + 1}\mathcal{O}_X)^{\oplus r}$. This $(\mathcal{F}_n)$ is a finite locally free object of $\textit{Coh}(X, \mathcal{I})$. By assumption there exists a finite locally free $\mathcal{O}_X$-module $\mathcal{F}$ and a compatible system of isomorphisms $$ \mathcal{F}/\mathcal{I}^n\mathcal{F} \to \mathcal{F}_n $$ of $\mathcal{O}_X$-modules. \medskip\noindent To construct an algebra structure on $\mathcal{F}$ consider the multiplication maps $\mathcal{F}_n \otimes_{\mathcal{O}_X} \mathcal{F}_n \to \mathcal{F}_n$ coming from the fact that $\mathcal{F}_n = \pi_{n, *}\mathcal{O}_{U_n}$ are sheaves of algebras. These define a map $$ (\mathcal{F}\otimes_{\mathcal{O}_X} \mathcal{F})^\wedge \longrightarrow \mathcal{F}^\wedge $$ in the category $\textit{Coh}(X, \mathcal{I})$. Hence by assumption we may assume there is a map $\mu : \mathcal{F}\otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{F}$ whose restriction to $Y_n$ gives the multiplication maps above. After possibly shrinking further we may assume $\mu$ defines a commutative $\mathcal{O}_X$-algebra structure on $\mathcal{F}$ compatible with the given algebra structures on $\mathcal{F}_n$. Setting $$ U = \underline{\Spec}_X((\mathcal{F}, \mu)) $$ we obtain a finite locally free scheme $\pi : U \to X$ whose restriction to $Y$ is isomorphic to $U_1$. The the discriminant of $\pi$ is the zero set of the section $$ \det(Q_\pi) : \mathcal{O}_X \longrightarrow \wedge^{top}(\pi_*\mathcal{O}_U)^{\otimes -2} $$ constructed in Discriminants, Section \ref{discriminant-section-discriminant}. Since the restriction of this to $Y_n$ is an isomorphism for all $n$ by Discriminants, Lemma \ref{discriminant-lemma-discriminant} we conclude that it is an isomorphism. Thus $\pi$ is \'etale by Discriminants, Lemma \ref{discriminant-lemma-discriminant}.
Observe that $\mathcal{V}$ is a directed set, so the colimits are as in Categories, Section \ref{categories-section-directed-colimits}. The rest of the argument is almost exactly the same as the argument in the proof of Lemma \ref{lemma-restriction-fully-faithful}; we urge the reader to skip it. \medskip\noindent Since the category of finite \'etale coverings has an internal hom (Lemma \ref{lemma-internal-hom-finite-etale}) it suffices to prove the following: Given $U$ finite \'etale over $V \in \mathcal{V}$ and a morphism $t : Y \to U$ over $V$ there exists a $V' \geq V$ and a morphism $s : V' \to U$ over $V$ such that $t = s|_Y$. Picture $$ \xymatrix{ & & U \ar[d]^f \\ Y \ar[r] \ar[rru] & V' \ar@{..>}[ru] \ar[r] & V } $$ Finding the dotted arrow $s$ is the same thing as finding an $\mathcal{O}_{V'}$-algebra map $$ s^\sharp : f_*\mathcal{O}_U|_{V'} \longrightarrow \mathcal{O}_{V'} $$ which reduces modulo the ideal sheaf of $Y$ to the given algebra map $t^\sharp : f_*\mathcal{O}_U \to \mathcal{O}_Y$. By Lemma \ref{lemma-thickening} we can lift $t$ uniquely to a compatible system of maps $t_n : Y_n \to U$ and hence a map $$ \lim t_n^\sharp : f_*\mathcal{O}_U \longrightarrow \lim \mathcal{O}_{Y_n} $$ of sheaves of algebras on $V$. Observe that $f_*\mathcal{O}_U$ is a finite locally free $\mathcal{O}_V$-module. Hence we get a $V' \geq V$ a map $\sigma : f_*\mathcal{O}_U|_{V'} \to \mathcal{O}_{V'}$ whose completion is $\lim t_n^\sharp$. To see that $\sigma$ is an algebra homomorphism, we need to check that the diagram $$ \xymatrix{ (f_*\mathcal{O}_U \otimes_{\mathcal{O}_V} f_*\mathcal{O}_U)|_{V'} \ar[r] \ar[d]_{\sigma \otimes \sigma} & f_*\mathcal{O}_U|_{V'} \ar[d]^\sigma \\ \mathcal{O}_{V'} \otimes_{\mathcal{O}_{V'}} \mathcal{O}_{V'} \ar[r] & \mathcal{O}_{V'} } $$ commutes. For every $n$ we know this diagram commutes after restricting to $Y_n$, i.e., the diagram commutes after applying the completion functor. Hence by faithfulness of the completion functor we deduce that there exists a $V'' \geq V'$ such that $\sigma|_{V''}$ is an algebra homomorphism as desired.
Observe that $\mathcal{V}$ is a directed set, so the colimits are as in Categories, Section \ref{categories-section-directed-colimits}. The rest of the argument is almost exactly the same as the argument in the proof of Lemma \ref{lemma-restriction-equivalence}; we urge the reader to skip it. \medskip\noindent The restriction functor is fully faithful by Lemma \ref{lemma-restriction-fully-faithful-general}. \medskip\noindent Let $U_1 \to Y$ be a finite \'etale morphism. To finish the proof we will show that $U_1$ is in the essential image of the restriction functor. \medskip\noindent For $n \geq 1$ let $Y_n$ be the $n$th infinitesimal neighbourhood of $Y$. By Lemma \ref{lemma-thickening} there is a unique finite \'etale morphism $\pi_n : U_n \to Y_n$ whose base change to $Y = Y_1$ recovers $U_1 \to Y_1$. Consider the sheaves $\mathcal{F}_n = \pi_{n, *}\mathcal{O}_{U_n}$. We may and do view $\mathcal{F}_n$ as an $\mathcal{O}_X$-module on $X$ wich is locally isomorphic to $(\mathcal{O}_X/f^{n + 1}\mathcal{O}_X)^{\oplus r}$. This $(\mathcal{F}_n)$ is a finite locally free object of $\textit{Coh}(X, \mathcal{I})$. By assumption there exists a $V \in \mathcal{V}$ and a finite locally free $\mathcal{O}_V$-module $\mathcal{F}$ and a compatible system of isomorphisms $$ \mathcal{F}/\mathcal{I}^n\mathcal{F} \to \mathcal{F}_n $$ of $\mathcal{O}_V$-modules. \medskip\noindent To construct an algebra structure on $\mathcal{F}$ consider the multiplication maps $\mathcal{F}_n \otimes_{\mathcal{O}_V} \mathcal{F}_n \to \mathcal{F}_n$ coming from the fact that $\mathcal{F}_n = \pi_{n, *}\mathcal{O}_{U_n}$ are sheaves of algebras. These define a map $$ (\mathcal{F}\otimes_{\mathcal{O}_V} \mathcal{F})^\wedge \longrightarrow \mathcal{F}^\wedge $$ in the category $\textit{Coh}(X, \mathcal{I})$. Hence by assumption after shrinking $V$ we may assume there is a map $\mu : \mathcal{F}\otimes_{\mathcal{O}_V} \mathcal{F} \to \mathcal{F}$ whose restriction to $Y_n$ gives the multiplication maps above. After possibly shrinking further we may assume $\mu$ defines a commutative $\mathcal{O}_V$-algebra structure on $\mathcal{F}$ compatible with the given algebra structures on $\mathcal{F}_n$. Setting $$ U = \underline{\Spec}_V((\mathcal{F}, \mu)) $$ we obtain a finite locally free scheme over $V$ whose restriction to $Y$ is isomorphic to $U_1$. It follows that $U \to V$ is \'etale at all points lying over $Y$, see More on Morphisms, Lemma \ref{more-morphisms-lemma-check-smoothness-on-infinitesimal-nbhds}. Thus after shrinking $V$ once more we may assume $U \to V$ is finite \'etale. This finishes the proof.
Let $a, b : U \to U'$ be two morphisms of schemes finite \'etale over $X$ whose restriction to $Y$ are the same. The image of a connected component of $U$ is an connected component of $X$; this follows from Topology, Lemma \ref{topology-lemma-finite-fibre-connected-components} applied to the restriction of $U \to X$ to a connected component of $X$. Hence the image of every connected component of $U$ meets $Y$ by assumption. We conclude that $a = b$ after restriction to each connected component of $U$ by \'Etale Morphisms, Proposition \ref{etale-proposition-equality}. Since the equalizer of $a$ and $b$ is an open subscheme of $U$ (as the diagonal of $U'$ over $X$ is open) we conclude.
The pushout exists by More on Morphisms, Proposition \ref{more-morphisms-proposition-pushout-along-closed-immersion-and-integral}. The functor is given by sending a scheme $U$ finite \'etale over the pushout to the base changes $Y' = U \times_{Y \amalg_Z X} Y$ and $X' = U \times_{Y \amalg_Z X} X$ and the natural isomorphism $Y' \times_Y Z \to X' \times_X Z$ over $Z$. To prove this functor is an equivalence we use More on Morphisms, Lemma \ref{more-morphisms-lemma-pushout-functor-equivalence-flat} to construct a quasi-inverse functor. The only thing left to prove is to show that given a morphism $U \to Y \amalg_Z X$ which is separated, quasi-finite and \'etale such that $X' \to X$ and $Y' \to Y$ are finite, then $U \to Y \amalg_Z X$ is finite. This can either be deduced from the corresponding algebra fact (More on Algebra, Lemma \ref{more-algebra-lemma-finite-module-over-fibre-product}) or it can be seen because $$ X' \amalg Y' \to U $$ is surjective and $X'$ and $Y'$ are proper over $Y \amalg_Z X$ (this uses the description of the pushout in More on Morphisms, Proposition \ref{more-morphisms-proposition-pushout-along-closed-immersion-and-integral}) and then we can apply Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-image-is-proper} to conclude that $U$ is proper over $Y \amalg_Z X$. Since a quasi-finite and proper morphism is finite (More on Morphisms, Lemma \ref{more-morphisms-lemma-characterize-finite}) we win.
This is a straightforward application of More on Algebra, Proposition \ref{more-algebra-proposition-equivalence}. Namely, choose generators $f_1, \ldots, f_t$ of $\mathfrak m$. For each $i$ write $V \times_U D(f_i) = \Spec(B_i)$. 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 $Y' = \Spec(B')$. 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 \ref{more-algebra-remark-glueing-data}. Applying the proposition cited above (and using More on Algebra, Remark \ref{more-algebra-remark-formal-glueing-algebras} 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 $Y = \Spec(B)$ 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.
Assume $\textit{F\'Et}_{U'} \longrightarrow \textit{F\'Et}_{U'_0}$ is a fully faithful. Since $X' \to X$ is faithfully flat, it is immediate that the functor $V \to V_0 = V \times_U U_0$ is faithful. Since the category of finite \'etale coverings has an internal hom (Lemma \ref{lemma-internal-hom-finite-etale}) it suffices to prove the following: Given $V$ finite \'etale over $U$ we have $$ \Mor_U(U, V) = \Mor_{U_0}(U_0, V_0) $$ The we assume we have a morphism $s_0 : U_0 \to V_0$ over $U_0$ and we will produce a morphism $s : U \to V$ over $U$. \medskip\noindent By our assumption there does exist a morphism $s' : U' \to V'$ whose restriction to $V'_0$ is the base change $s'_0$ of $s_0$. Since $V' \to U'$ is finite \'etale this means that $V' = s'(U') \amalg W'$ for some $W' \to U'$ finite and \'etale. Choose a finite morphism $Z' \to X'$ such that $W' = Z' \times_{X'} U'$. This is possible by Zariski's main theorem in the form stated in More on Morphisms, Lemma \ref{more-morphisms-lemma-quasi-finite-separated-pass-through-finite} (small detail omitted). Then $$ V' = s'(U') \amalg W' \longrightarrow X' \amalg Z' = Y' $$ is an open immersion such that $V' = Y' \times_{X'} U'$. By Lemma \ref{lemma-fill-in-missing} we can find $Y \to X$ finite such that $V = Y \times_X U$ and $Y' = Y \times_X X'$. Write $Y = \Spec(B)$ so that $Y' = \Spec(B \otimes_A A^\wedge)$. Then $B \otimes_A A^\wedge$ has an idempotent $e'$ corresponding to the open and closed subscheme $X'$ of $Y' = X' \amalg Z'$. \medskip\noindent The case $A$ is henselian (slightly easier). The image $\overline{e}$ of $e'$ in $B \otimes_A \kappa(\mathfrak m) = B/\mathfrak mB$ lifts to an idempotent $e$ of $B$ as $A$ is henselian (because $B$ is a product of local rings by Algebra, Lemma \ref{algebra-lemma-characterize-henselian}). Then we see that $e$ maps to $e'$ by uniqueness of lifts of idempotents (using that $B \otimes_A A^\wedge$ is a product of local rings). Let $Y_1 \subset Y$ be the open and closed subscheme corresponding to $e$. Then $Y_1 \times_X X' = s'(X')$ which implies that $Y_1 \to X$ is an isomorphism (by faithfully flat descent) and gives the desired section. \medskip\noindent The case where $(A, (f))$ is a henselian pair. Here we use that $s'$ is a lift of $s'_0$. Namely, let $Y_{0, 1} \subset Y_0 = Y \times_X X_0$ be the closure of $s_0(U_0) \subset V_0 = Y_0 \times_{X_0} U_0$. As $X' \to X$ is flat, the base change $Y'_{0, 1} \subset Y'_0$ is the closure of $s'_0(U'_0)$ which is equal to $X'_0 \subset Y'_0$ (see Morphisms, Lemma \ref{morphisms-lemma-flat-base-change-scheme-theoretic-image}). Since $Y'_0 \to Y_0$ is submersive (Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}) we conclude that $Y_{0, 1}$ is open and closed in $Y_0$. Let $e_0 \in B/fB$ be the corresponding idempotent. By More on Algebra, Lemma \ref{more-algebra-lemma-characterize-henselian-pair} we can lift $e_0$ to an idempotent $e \in B$. Then we conclude as before.
Let $A'$ be the $\mathfrak m$-adic completion of $A$. We will show that the hypotheses remain true for $A'$. This is clear for conditions (a) and (b). Condition (c)(ii) is preserved by Local Cohomology, Lemma \ref{local-cohomology-lemma-change-completion}. Next, assume (c)(i) holds. Since $A$ is universally catenary (Dualizing Complexes, Lemma \ref{dualizing-lemma-universally-catenary}) we see that every irreducible component of $\Spec(A')$ not contained in $V(f)$ has dimension $\geq 3$, see More on Algebra, Proposition \ref{more-algebra-proposition-ratliff}. Since $A \to A'$ is flat with Gorenstein fibres, the condition that $A_f$ is $(S_2)$ implies that $A'_f$ is $(S_2)$. References used: Dualizing Complexes, Section \ref{dualizing-section-formal-fibres}, More on Algebra, Section \ref{more-algebra-section-properties-formal-fibres}, and Algebra, Lemma \ref{algebra-lemma-Sk-goes-up}. Thus by Lemma \ref{lemma-fully-faithful-henselian-completion} we may assume that $A$ is a Noetherian complete local ring. \medskip\noindent Assume $A$ is a complete local ring in addition to the other assumptions. By Lemma \ref{lemma-restriction-fully-faithful} the result follows from Algebraic and Formal Geometry, Lemmas \ref{algebraization-lemma-fully-faithful-simple-one} and \ref{algebraization-lemma-fully-faithful-simple-two}.
By Lemma \ref{lemma-fully-faithful-henselian-completion} we may assume that $A$ is a Noetherian complete local ring. (The assumptions carry over; use Dualizing Complexes, Lemma \ref{dualizing-lemma-torsion-change-rings}.) By Lemma \ref{lemma-restriction-fully-faithful} the result follows from Algebraic and Formal Geometry, Lemma \ref{algebraization-lemma-fully-faithful-alternative}.
Set $V = \pi^{-1}(U)$ and denote $\pi' : V \to U$ the restriction of $\pi$. Consider the $\mathcal{O}_X$-module map $$ \pi_*\mathcal{O}_Y \longrightarrow j_*\pi'_*\mathcal{O}_V $$ where $j : U \to X$ is the inclusion morphism. We claim Divisors, Lemma \ref{divisors-lemma-depth-2-hartog} applies to this map. If so, then $B = \Gamma(Y, \mathcal{O}_Y)$ and we see that the lemma holds. Let $x \in X$ be the closed point. It suffices to show that $\text{depth}((\pi_*\mathcal{O}_Y)_x) \geq 2$. Let $y_1, \ldots, y_n \in Y$ be the points mapping to $x$. By Algebra, Lemma \ref{algebra-lemma-depth-goes-down-finite} it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_i}) \geq 2$ for $i = 1, \ldots, n$. Since this is the assumption of the lemma the proof is complete.
Denote $\pi : V \to U$ the given finite \'etale morphism. Assume $Y$ as in (1) exists. Let $x \in X$ be the point corresponding to $\mathfrak m$. Let $y \in Y$ be a point mapping to $x$. We claim that $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$. This is true because $Y \to X$ is \'etale and hence $A = \mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ have the same depth (Algebra, Lemma \ref{algebra-lemma-apply-grothendieck}). Hence Lemma \ref{lemma-sections-over-punctured-spec} applies and $Y = \Spec(B)$. \medskip\noindent The implication (2) $\Rightarrow$ (1) is easier and the details are omitted.