hcju/stacksqa huggingface dataset:
For a discussion of thickenings see More on Morphisms, Section \ref{more-morphisms-section-thickenings}. Let $U' \to X'$ be an \'etale morphism such that $U = U' \times_{X'} X \to X$ is finite \'etale. Then $U' \to X'$ is finite \'etale as well. This follows for example from More on Morphisms, Lemma \ref{more-morphisms-lemma-properties-that-extend-over-thickenings}. Now, if $X \subset X'$ is a finite order thickening then this remark combined with \'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence} proves the lemma. Below we will prove the lemma for general thickenings, but we suggest the reader skip the proof. \medskip\noindent Let $X' = \bigcup X_i'$ be an affine open covering. Set $X_i = X \times_{X'} X_i'$, $X_{ij}' = X'_i \cap X'_j$, $X_{ij} = X \times_{X'} X_{ij}'$, $X_{ijk}' = X'_i \cap X'_j \cap X'_k$, $X_{ijk} = X \times_{X'} X_{ijk}'$. Suppose that we can prove the theorem for each of the thickenings $X_i \subset X'_i$, $X_{ij} \subset X_{ij}'$, and $X_{ijk} \subset X_{ijk}'$. Then the result follows for $X \subset X'$ by relative glueing of schemes, see Constructions, Section \ref{constructions-section-relative-glueing}. Observe that the schemes $X_i'$, $X_{ij}'$, $X_{ijk}'$ are each separated as open subschemes of affine schemes. Repeating the argument one more time we reduce to the case where the schemes $X'_i$, $X_{ij}'$, $X_{ijk}'$ are affine. \medskip\noindent In the affine case we have $X' = \Spec(A')$ and $X = \Spec(A'/I')$ where $I'$ is a locally nilpotent ideal. Then $(A', I')$ is a henselian pair (More on Algebra, Lemma \ref{more-algebra-lemma-locally-nilpotent-henselian}) and the result follows from Lemma \ref{lemma-gabber} (which is much easier in this case).
Let us prove the functor is essentially surjective. Let $U' \to X_{k'}$ be a finite \'etale morphism. Write $k' = \colim A_i$ as a filtered colimit of finite type $k$-algebras. By Limits, Lemma \ref{limits-lemma-descend-finite-presentation} there exists an $i$ and a finitely presented morphism $U_i \to X_{A_i}$ whose base change to $X_{k'}$ is $U'$. After increasing $i$ we may assume that $U_i \to X_{A_i}$ is finite and \'etale (Limits, Lemmas \ref{limits-lemma-descend-finite-finite-presentation} and \ref{limits-lemma-descend-etale}). Since $k$ is algebraically closed we can find a $k$-valued point $t$ in $\Spec(A_i)$. Let $U = (U_i)_t$ be the fibre of $U_i$ over $t$. Let $A_i^h$ be the henselization of $(A_i)_{\mathfrak m}$ where $\mathfrak m$ is the maximal ideal corresponding to the point $t$. By Lemma \ref{lemma-finite-etale-on-proper-over-henselian} we see that $(U_i)_{A_i^h} = U \times \Spec(A_i^h)$ as schemes over $X_{A_i^h}$. Now since $A_i^h$ is algebraic over $A_i$ (see for example discussion in Smoothing Ring Maps, Example \ref{smoothing-example-describe-henselian}) and since $k'$ is algebraically closed we can find a ring map $A_i^h \to k'$ extending the given inclusion $A_i \subset k'$. Hence we conclude that $U'$ is isomorphic to the base change of $U$. The proof of fully faithfulness is exactly the same.
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 $W$ finite \'etale over $Y$ and a morphism $s : X \to W$ over $X$ there is at most one section $t : Y \to W$ such that $s = t \circ f$. Consider two sections $t_1, t_2 : Y \to W$ such that $s = t_1 \circ f = t_2 \circ f$. Since the equalizer of $t_1$ and $t_2$ is closed in $Y$ (Schemes, Lemma \ref{schemes-lemma-where-are-they-equal}) and since $f(X)$ is dense in $Y$ we see that $t_1$ and $t_2$ agree on $Y_{red}$. Then it follows that $t_1$ and $t_2$ have the same image which is an open and closed subscheme of $W$ mapping isomorphically to $Y$ (\'Etale Morphisms, Proposition \ref{etale-proposition-properties-sections}) hence they are equal.
Assume $A$ is strictly henselian. In this case any finite \'etale cover $Y$ of $X$ is isomorphic to a finite disjoint union of copies of $X$. Thus it suffices to prove that any morphism $U \to U \amalg \ldots \amalg U$ over $U$, extends uniquely to a morphism $X \to X \amalg \ldots \amalg X$ over $X$. If $U$ is connected (in particular nonempty), then this is true. \medskip\noindent The general case. 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 $Y$ finite \'etale over $X$ any morphism $s : U \to Y$ over $X$ extends to a morphism $t : X \to Y$ over $X$. Let $A^{sh}$ be the strict henselization of $A$ and denote $X^{sh} = \Spec(A^{sh})$, $U^{sh} = U \times_X X^{sh}$, $Y^{sh} = Y \times_X X^{sh}$. By the first paragraph and our assumption on $A$, we can extend the base change $s^{sh} : U^{sh} \to Y^{sh}$ of $s$ to $t^{sh} : X^{sh} \to Y^{sh}$. Set $A' = A^{sh} \otimes_A A^{sh}$. Then the two pullbacks $t'_1, t'_2$ of $t^{sh}$ to $X' = \Spec(A')$ are extensions of the pullback $s'$ of $s$ to $U' = U \times_X X'$. As $A \to A'$ is flat we see that $U' \subset X'$ is (topologically) dense by going down for $A \to A'$ (Algebra, Lemma \ref{algebra-lemma-flat-going-down}). Thus $t'_1 = t'_2$ by Lemma \ref{lemma-dense-faithful}. Hence $t^{sh}$ descends to a morphism $t : X \to Y$ for example by Descent, Lemma \ref{descent-lemma-fpqc-universal-effective-epimorphisms}.
Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. \medskip\noindent Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ uniquely to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} we find an open $U \subset U'$ containing $x$ and an extension $\varphi' : (Y_1)_{U'} \to (Y_2)_{U'}$ of $\varphi$. Since the underlying topological space of $X$ is Noetherian this finishes the proof by Noetherian induction on the complement of the open over which $\varphi$ is defined.
Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. \medskip\noindent Let $x \in X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. Since every point of $X \setminus U$ is closed $V$ is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ uniquely to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} (this uses that $U$ is retrocompact in $X$) we find an open $U \subset U'_x$ containing $x$ and an extension $\varphi'_x : (Y_1)_{U'_x} \to (Y_2)_{U'_x}$ of $\varphi$. Note that given two points $x, x' \in X \setminus U$ the morphisms $\varphi'_x$ and $\varphi'_{x'}$ agree over $U'_x \cap U'_{x'}$ as $U$ is dense in that open (Lemma \ref{lemma-dense-faithful}). Thus we can extend $\varphi$ to $\bigcup U'_x = X$ as desired.
Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. We will prove existence by showing that we can enlarge $U$ if $U \not = X$ and using Zorn's lemma to finish the proof. \medskip\noindent Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ (uniquely) to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. Choose an affine neighbourhood $W \subset X$ of $x$. Since $U \cap W$ is dense in $W$ it contains the generic points $\eta_1, \ldots, \eta_n$ of $W$. Choose an affine open $W' \subset W \cap U$ containing $\eta_1, \ldots, \eta_n$. Set $V' = W' \times_X \Spec(\mathcal{O}_{X, x})$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} applied to $x \in W \supset W'$ we find an open $W' \subset W'' \subset W$ with $x \in W''$ and a morphism $\varphi'' : (Y_1)_{W''} \to (Y_2)_{W''}$ agreeing with $\varphi$ over $W'$. Since $W'$ is dense in $W'' \cap U$, we see by Lemma \ref{lemma-dense-faithful} that $\varphi$ and $\varphi''$ agree over $U \cap W'$. Thus $\varphi$ and $\varphi''$ glue to a morphism $\varphi'$ over $U' = U \cup W''$ agreeing with $\varphi$ over $U$. Observe that $x \in U'$ so that we've extended $\varphi$ to a strictly larger open. \medskip\noindent Consider the set $\mathcal{S}$ of pairs $(U', \varphi')$ where $U \subset U'$ and $\varphi'$ is an extension of $\varphi$. We endow $\mathcal{S}$ with a partial ordering in the obvious manner. If $(U'_i, \varphi'_i)$ is a totally ordered subset, then it has a maximum $(U', \varphi')$. Just take $U' = \bigcup U'_i$ and let $\varphi' : (Y_1)_{U'} \to (Y_2)_{U'}$ be the morphism agreeing with $\varphi'_i$ over $U'_i$. Thus Zorn's lemma applies and $\mathcal{S}$ has a maximal element. By the argument above we see that this maximal element is an extension of $\varphi$ over all of $X$.
By Lemma \ref{lemma-thickening} we may replace $X$ by its reduction. Thus we may assume that $X$ is an integral scheme. By Lemma \ref{lemma-functoriality-galois-surjective} the assertion of the lemma translates into the statement that the functors $\textit{F\'Et}_X \to \textit{F\'Et}_U$ and $\textit{F\'Et}_X \to \textit{F\'Et}_\eta$ are fully faithful. \medskip\noindent The result for $\textit{F\'Et}_X \to \textit{F\'Et}_U$ follows from Lemma \ref{lemma-quasi-compact-dense-open-connected-at-infinity} and the fact that for a local ring $A$ which is geometrically unibranch its strict henselization has an irreducible spectrum. See More on Algebra, Lemma \ref{more-algebra-lemma-geometrically-unibranch}. \medskip\noindent Observe that the residue field $\kappa(\eta) = \mathcal{O}_{X, \eta}$ is the filtered colimit of $\mathcal{O}_X(U)$ over $U \subset X$ nonempty open affine. Hence $\textit{F\'Et}_\eta$ is the colimit of the categories $\textit{F\'Et}_U$ over such $U$, see Limits, Lemmas \ref{limits-lemma-descend-finite-presentation}, \ref{limits-lemma-descend-finite-finite-presentation}, and \ref{limits-lemma-descend-etale}. A formal argument then shows that fully faithfulness for $\textit{F\'Et}_X \to \textit{F\'Et}_\eta$ follows from the fully faithfulness of the functors $\textit{F\'Et}_X \to \textit{F\'Et}_U$.
Surjectivity follows from Lemmas \ref{lemma-retrocompact-dense-open-connected-at-infinity-closed} and \ref{lemma-functoriality-galois-surjective}. We can consider the sequence of maps $$ \pi_1(U)  \to \ldots \to \pi_1(X \setminus \{x_1, x_2\}) \to \pi_1(X \setminus \{x_1\}) \to \pi_1(X) $$ A group theory argument then shows it suffices to prove the statement on the kernel in the case $n = 1$ (details omitted). Write $x = x_1$, $U^{sh} = U_1^{sh}$, set $A = \mathcal{O}_{X, x}$, and let $A^{sh}$ be the strict henselization. Consider the diagram $$ \xymatrix{ U \ar[d] & \Spec(A) \setminus \{\mathfrak m\} \ar[l] \ar[d] & U^{sh} \ar[d] \ar[l] \\ X & \Spec(A) \ar[l] & \Spec(A^{sh}) \ar[l] } $$ By Lemma \ref{lemma-functoriality-galois-ses} we have to show finite \'etale morphisms $V \to U$ which pull back to trivial coverings of $U^{sh}$ extend to finite \'etale schemes over $X$. By Lemma \ref{lemma-local-exact-sequence} we know the corresponding statement for finite \'etale schemes over the punctured spectrum of $A$. However, by Limits, Lemma \ref{limits-lemma-glueing-near-closed-point} schemes of finite presentation over $X$ are the same thing as schemes of finite presentation over $U$ and $A$ glued over the punctured spectrum of $A$. This finishes the proof.
Observe that $Y \to X$ is an integral morphism. In each case the morphism $Y \to X$ is locally of finite type by definition. Hence we find that in each case the lemma is finite by Morphisms, Lemma \ref{morphisms-lemma-finite-integral}. In particular we see that (2) is equivalent to (3). An \'etale morphism is unramified, hence (2) implies (1). \medskip\noindent Conversely, assume $Y \to X$ is unramified. Let $x \in X$. We can choose an \'etale neighbourhood $(U, u) \to (X, x)$ such that $$ Y \times_X U = \coprod V_j \longrightarrow U $$ is a disjoint union of closed immersions, see \'Etale Morphisms, Lemma \ref{etale-lemma-finite-unramified-etale-local}. Shrinking we may assume $U$ is quasi-compact. Then $U$ has finitely many irreducible components (Descent, Lemma \ref{descent-lemma-locally-finite-nr-irred-local-fppf}). Since $U$ is normal (Descent, Lemma \ref{descent-lemma-normal-local-smooth}) the irreducible components of $U$ are open and closed (Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}) and we may assume $U$ is irreducible. Then $U$ is an integral scheme whose generic point $\xi$ maps to the generic point of $X$. On the other hand, we know that $Y \times_X U$ is the normalization of $U$ in $\Spec(L) \times_X U$ by More on Morphisms, Lemma \ref{more-morphisms-lemma-normalization-smooth-localization}. Every point of $\Spec(L) \times_X U$ maps to $\xi$. Thus every $V_j$ contains a point mapping to $\xi$ by Morphisms, Lemma \ref{morphisms-lemma-normalization-generic}. Thus $V_j \to U$ is an isomorphism as $U = \overline{\{\xi\}}$. Thus $Y \times_X U \to U$ is \'etale. By Descent, Lemma \ref{descent-lemma-descending-property-etale} we conclude that $Y \to X$ is \'etale over the image of $U \to X$ (an open neighbourhood of $x$).
The scheme $Y$ is normal by Descent, Lemma \ref{descent-lemma-normal-local-smooth}. Since $Y \to X$ is flat every generic point of $Y$ maps to the generic point of $X$ by Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}. Since $Y \to X$ is finite we see that $Y$ has a finite number of irreducible components. Thus $Y$ is the disjoint union of a finite number of integral normal schemes by Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}. Thus if $Y$ is connected, then $Y$ is an integral normal scheme. \medskip\noindent Let $L$ be the function field of $Y$ and let $Y' \to X$ be the normalization of $X$ in $L$. By Morphisms, Lemma \ref{morphisms-lemma-characterize-normalization} we obtain a factorization $Y' \to Y \to X$ and $Y' \to Y$ is the normalization of $Y$ in $L$. Since $Y$ is normal it is clear that $Y' = Y$ (this can also be deduced from Morphisms, Lemma \ref{morphisms-lemma-finite-birational-over-normal}).
Note that $A^{sh}$ is a normal domain, see More on Algebra, Lemma \ref{more-algebra-lemma-henselization-normal}. The map $\pi_1(\Spec(K)) \to \pi_1(X)$ is surjective by Proposition \ref{proposition-normal}. \medskip\noindent Write $X^{sh} = \Spec(A^{sh})$. Let $Y \to X$ be a finite \'etale morphism. Then $Y^{sh} = Y \times_X X^{sh} \to X^{sh}$ is a finite \'etale morphism. Since $A^{sh}$ is strictly henselian we see that $Y^{sh}$ is isomorphic to a disjoint union of copies of $X^{sh}$. Thus the same is true for $Y \times_X \Spec(K^{sh})$. It follows that the composition $\pi_1(\Spec(K^{sh})) \to \pi_1(X)$ is trivial, see Lemma \ref{lemma-composition-trivial}. \medskip\noindent To finish the proof, it suffices according to Lemma \ref{lemma-functoriality-galois-ses} to show the following: Given a finite \'etale morphism $V \to \Spec(K)$ such that $V \times_{\Spec(K)} \Spec(K^{sh})$ is a disjoint union of copies of $\Spec(K^{sh})$, we can find a finite \'etale morphism $Y \to X$ with $V \cong Y \times_X \Spec(K)$ over $\Spec(K)$. Write $V = \Spec(L)$, so $L$ is a finite product of finite separable extensions of $K$. Let $B \subset L$ be the integral closure of $A$ in $L$. If $A \to B$ is \'etale, then we can take $Y = \Spec(B)$ and the proof is complete. By Algebra, Lemma \ref{algebra-lemma-integral-closure-commutes-smooth} (and a limit argument we omit) we see that $B \otimes_A A^{sh}$ is the integral closure of $A^{sh}$ in $L^{sh} = L \otimes_K K^{sh}$. Our assumption is that $L^{sh}$ is a product of copies of $K^{sh}$ and hence $B^{sh}$ is a product of copies of $A^{sh}$. Thus $A^{sh} \to B^{sh}$ is \'etale. As $A \to A^{sh}$ is faithfully flat it follows that $A \to B$ is \'etale (Descent, Lemma \ref{descent-lemma-descending-property-etale}) as desired.
Assume the lemma is not true to get a contradiction. Then there exists a monic irreducible polynomial $P(T) \in \kappa(\mathfrak p)[T]$ of degree $d > 1$. After replacing $P$ by $a^d P(a^{-1}T)$ for suitable $a \in A$ (to clear denominators) we may assume that $P$ is the image of a monic polynomial $Q$ in $A[T]$. Observe that $Q$ is irreducible in $K[T]$. Namely a factorization over $K$ leads to a factorization over $A$ by Algebra, Lemma \ref{algebra-lemma-polynomials-divide} which we could reduce modulo $\mathfrak p$ to get a factorization of $P$. As $K$ is separably closed, $Q$ is not a separable polynomial (Fields, Definition \ref{fields-definition-separable}). Then the characteristic of $K$ is $p > 0$ and $Q$ has vanishing linear term (Fields, Definition \ref{fields-definition-separable}). However, then we can replace $Q$ by $Q + a T$ where $a \in \mathfrak p$ is nonzero to get a contradiction.
Let $(A, \mathfrak m, \kappa)$ be normal local with separably closed fraction field $K$. If $A = K$, then we are done. If not, then the residue field $\kappa$ is algebraically closed by Lemma \ref{lemma-get-algebraic-closure} and it suffices to check that $A$ is henselian. Let $f \in A[T]$ be monic and let $a_0 \in \kappa$ be a root of multiplicity $1$ of the reduction $\overline{f} \in \kappa[T]$. Let $f = \prod f_i$ be the factorization in $K[T]$. By Algebra, Lemma \ref{algebra-lemma-polynomials-divide} we have $f_i \in A[T]$. Thus $a_0$ is a root of $f_i$ for some $i$. After replacing $f$ by $f_i$ we may assume $f$ is irreducible. Then, since the derivative $f'$ cannot be zero in $A[T]$ as $a_0$ is a single root, we conclude that $f$ is linear due to the fact that $K$ is separably algebraically closed. Thus $A$ is henselian, see Algebra, Definition \ref{algebra-definition-henselian}.
We can write $L$ as the filtered colimit of finite Galois extensions of $K$. Hence it suffices to prove this lemma in case $L/K$ is a finite Galois extension, see Algebra, Lemma \ref{algebra-lemma-colimit-colimit-etale}. Since $A = B^G$ as $A$ is integrally closed in $K = L^G$ the result follows from Lemma \ref{lemma-inertia-invariants-etale}.
By construction the group $D$ acts on $(A^{sep})_{\mathfrak m^{sep}}$ over $A$. By the uniqueness of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ given the map on residue fields (Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial}) we see that the image of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ is contained in $((A^{sep})_{\mathfrak m^{sep}})^I$. On the other hand, Lemma \ref{lemma-inertial-invariants-unramified} shows that $((A^{sep})_{\mathfrak m^{sep}})^I$ is a filtered colimit of \'etale extensions of $A$. Since $A^{sh}$ is the maximal such extension, we conclude that $A^{sh} = ((A^{sep})_{\mathfrak m^{sep}})^I$. Hence $K^{sh} = (K^{sep})^I$. \medskip\noindent Recall that $I$ is the kernel of a surjective map $D \to \text{Aut}(\kappa(\mathfrak m^{sep})/\kappa)$, see More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric-galois}. We have $\text{Aut}(\kappa(\mathfrak m^{sep})/\kappa) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)$ as we have seen above that these fields are the algebraic and separable algebraic closures of $\kappa$. On the other hand, any automorphism of $A^{sh}$ over $A$ is an automorphism of $A^{sh}$ over $A^h$ by the uniqueness in Algebra, Lemma \ref{algebra-lemma-henselian-functorial}. Furthermore, $A^{sh}$ is the colimit of finite \'etale extensions $A^h \subset A'$ which correspond $1$-to-$1$ with finite separable extension $\kappa'/\kappa$, see Algebra, Remark \ref{algebra-remark-construct-sh-from-h}. Thus $$ \text{Aut}(A^{sh}/A) = \text{Aut}(A^{sh}/A^h) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa) $$ Let $\kappa \subset \kappa'$ be a finite Galois extension with Galois group $G$. Let $A^h \subset A'$ be the finite \'etale extension corresponding to $\kappa \subset \kappa'$ by Algebra, Lemma \ref{algebra-lemma-henselian-cat-finite-etale}. Then it follows that $(A')^G = A^h$ by looking at fraction fields and degrees (small detail omitted). Taking the colimit we conclude that $(A^{sh})^{\text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)} = A^h$. Combining all of the above, we find $A^h = ((A^{sep})_{\mathfrak m^{sep}})^D$. Hence $K^h = (K^{sep})^D$. \medskip\noindent Since $U$, $U^h$, $U^{sh}$ are the spectra of the fields $K$, $K^h$, $K^{sh}$ we see that the top lines of the diagrams correspond via Lemma \ref{lemma-fundamental-group-Galois-group}. By Lemma \ref{lemma-gabber} we have $\pi_1(X^h) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)$. The exactness of the sequence $1 \to I \to D \to \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa) \to 1$ was pointed out above. By Proposition \ref{proposition-normal} we see that $\pi_1(X) = \text{Gal}(M/K)$. Finally, the statement on the kernel of $\text{Gal}(K^{sep}/K) \to \text{Gal}(M/K) = \pi_1(X)$ follows from Lemma \ref{lemma-local-exact-sequence-normal}. This finishes the proof.
The equivalence of (2) and (3) follows from Lemma \ref{lemma-identify-inertia} which tells us that $I_y$ is conjugate to $\text{Gal}(K^{sep}/K_x^{sh})$ if $y$ lies over $x$. By Lemma \ref{lemma-local-exact-sequence-normal} we see that $\text{Gal}(K^{sep}/K_x^{sh})$ maps trivially to $\pi_1(\Spec(\mathcal{O}_{X, x}))$ and therefore the subgroup $N \subset G = \text{Gal}(K^{sep}/K)$ of (2) and (3) is contained in the kernel of $G \longrightarrow \pi_1(X)$. \medskip\noindent To prove the other inclusion, since $N$ is normal, it suffices to prove: given $N \subset U \subset G$ with $U$ open normal, the quotient map $G \to G/U$ factors through $\pi_1(X)$. In other words, if $L/K$ is the Galois extension corresponding to $U$, then we have to show that $X$ is unramified in $L$ (Section \ref{section-normal}, especially Proposition \ref{proposition-normal}). It suffices to do this when $X$ is affine (we do this so we can refer to algebra results in the rest of the proof). Let $Y \to X$ be the normalization of $X$ in $L$. The inclusion $L \subset K^{sep}$ induces a morphism $\pi : X^{sep} \to Y$. For $y \in X^{sep}$ the inertia group of $\pi(y)$ in $\text{Gal}(L/K)$ is the image of $I_y$ in $\text{Gal}(L/K)$; this follows from More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric-galois-compare}. Since $N \subset U$ all these inertia groups are trivial. We conclude that $Y \to X$ is \'etale by applying Lemma \ref{lemma-inertia-invariants-etale}. (Alternative: you can use Lemma \ref{lemma-local-exact-sequence-normal} to see that the pullback of $Y$ to $\Spec(\mathcal{O}_{X, x})$ is \'etale for all $x \in X$ and then conclude from there with a bit more work.)