Lemma 58.19.4. In Situation 58.19.1 assume $A$ is henselian or more generally that $(A, (f))$ is a henselian pair. Let $A^\wedge $ be the $\mathfrak m$-adic completion of $A$, let $X' = \mathop{\mathrm{Spec}}(A^\wedge )$ and let $U'$ and $U'_0$ be the base changes of $U$ and $U_0$ to $X'$. If $\textit{FÉt}_{U'} \to \textit{FÉt}_{U'_0}$ is fully faithful, then $\textit{FÉt}_ U \to \textit{FÉt}_{U_0}$ is fully faithful.
Proof. Assume $\textit{FÉt}_{U'} \longrightarrow \textit{FÉt}_{U'_0}$ 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 étale coverings has an internal hom (Lemma 58.5.4) it suffices to prove the following: Given $V$ finite étale over $U$ we have
\[ \mathop{\mathrm{Mor}}\nolimits _ U(U, V) = \mathop{\mathrm{Mor}}\nolimits _{U_0}(U_0, V_0) \]
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 $U$.
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 étale this means that $V' = s'(U') \amalg W'$ for some $W' \to U'$ finite and étale. 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 37.43.3 (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 58.19.3 we can find $Y \to X$ finite such that $V = Y \times _ X U$ and $Y' = Y \times _ X X'$. Write $Y = \mathop{\mathrm{Spec}}(B)$ so that $Y' = \mathop{\mathrm{Spec}}(B \otimes _ A A^\wedge )$. Then $B \otimes _ A A^\wedge $ has an idempotent $e'$ corresponding to the open and closed subscheme $X'$ of $Y' = X' \amalg Z'$.
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 10.153.3). 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.
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 29.25.16). Since $Y'_0 \to Y_0$ is submersive (Morphisms, Lemma 29.25.12) 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 15.11.6 we can lift $e_0$ to an idempotent $e \in B$. Then we conclude as before. $\square$
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