The Stacks project

Proof. Since the category of finite étale coverings has an internal hom (Lemma 58.5.4) it suffices to prove the following: Given $W$ finite étale 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 26.21.5) 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 $Y$ (Étale Morphisms, Proposition 41.6.1) hence they are equal. $\square$

Proof. Assume $A$ is strictly henselian. In this case any finite étale 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.

The general case. Since the category of finite étale coverings has an internal hom (Lemma 58.5.4) it suffices to prove the following: Given $Y$ finite étale 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 $X^{sh} = \mathop{\mathrm{Spec}}(A^{sh})$, $U^{sh} = U \times _ X X^{sh}$, $Y^{sh} = Y \times _ X X^{sh}$. 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 $X' = \mathop{\mathrm{Spec}}(A')$ 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 10.39.19). Thus $t'_1 = t'_2$ by Lemma 58.10.1. Hence $t^{sh}$ descends to a morphism $t : X \to Y$ for example by Descent, Lemma 35.13.7. $\square$

Proof. Let $Y_1, Y_2$ be finite étale 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 58.10.1.

Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ uniquely to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. By Limits, Lemma 32.20.3 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. $\square$

Proof. Let $Y_1, Y_2$ be finite étale 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 58.10.1.

Let $x \in X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. Since every point of $X \setminus U$ is closed $V$ is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ uniquely to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. By Limits, Lemma 32.20.3 (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 58.10.1). Thus we can extend $\varphi $ to $\bigcup U'_ x = X$ as desired. $\square$

Proof. Let $Y_1, Y_2$ be finite étale 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 58.10.1. We will prove existence by showing that we can enlarge $U$ if $U \not= X$ and using Zorn's lemma to finish the proof.

Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ (uniquely) to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. 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 $V' = W' \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Limits, Lemma 32.20.3 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 58.10.1 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.

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 $X$. $\square$

Proof. The map $\pi _1(U) \to \pi _1(X)$ is surjective by Lemmas 58.10.2 and 58.4.1.

Write $X^{sh} = \mathop{\mathrm{Spec}}(A^{sh})$. Let $Y \to X$ be a finite étale morphism. Then $Y^{sh} = Y \times _ X X^{sh} \to X^{sh}$ is a finite étale 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 $Y \times _ X U^{sh}$. It follows that the composition $\pi _1(U^{sh}) \to \pi _1(U) \to \pi _1(X)$ is trivial, see Lemma 58.4.2.

To finish the proof, it suffices according to Lemma 58.4.3 to show the following: Given a finite étale morphism $V \to U$ such that $V \times _ U U^{sh}$ is a disjoint union of copies of $U^{sh}$, we can find a finite étale morphism $Y \to X$ with $V \cong Y \times _ X U$ over $U$. The assumption implies that there exists a finite étale morphism $Y^{sh} \to X^{sh}$ and an isomorphism $V \times _ U U^{sh} \cong Y^{sh} \times _{X^{sh}} U^{sh}$. Consider the following diagram

Since $U \subset X$ is quasi-compact by assumption, all the downward arrows are quasi-compact open immersions. Let $\xi \in X^{sh} \times _ X X^{sh}$ be a point not in $U^{sh} \times _ U U^{sh}$. Then $\xi $ lies over the closed point $x^{sh}$ of $X^{sh}$. Consider the local ring homomorphism

determined by the first projection $X^{sh} \times _ X X^{sh}$. This is a filtered colimit of local homomorphisms which are localizations étale ring maps. Since $A^{sh}$ is strictly henselian, we conclude that it is an isomorphism. Since this holds for every $\xi $ in the complement it follows there are no specializations among these points and hence every such $\xi $ is a closed point (you can also prove this directly). As the local ring at $\xi $ is isomorphic to $A^{sh}$, it is strictly henselian and has connected punctured spectrum. Similarly for points $\xi $ of $X^{sh} \times _ X X^{sh} \times _ X X^{sh}$ not in $U^{sh} \times _ U U^{sh} \times _ U U^{sh}$. It follows from Lemma 58.10.4 that pullback along the vertical arrows induce fully faithful functors on the categories of finite étale schemes. Thus the canonical descent datum on $V \times _ U U^{sh}$ relative to the fpqc covering $\{ U^{sh} \to U\} $ translates into a descent datum for $Y^{sh}$ relative to the fpqc covering $\{ X^{sh} \to X\} $. Since $Y^{sh} \to X^{sh}$ is finite hence affine, this descent datum is effective (Descent, Lemma 35.37.1). Thus we get an affine morphism $Y \to X$ and an isomorphism $Y \times _ X X^{sh} \to Y^{sh}$ compatible with descent data. By fully faithfulness of descent data (as in Descent, Lemma 35.35.11) we get an isomorphism $V \to U \times _ X Y$. Finally, $Y \to X$ is finite étale as $Y^{sh} \to X^{sh}$ is, see Descent, Lemmas 35.23.29 and 35.23.23. $\square$

Proof. By Lemma 58.8.3 we may replace $X$ by its reduction. Thus we may assume that $X$ is an integral scheme. By Lemma 58.4.1 the assertion of the lemma translates into the statement that the functors $\textit{FÉt}_ X \to \textit{FÉt}_ U$ and $\textit{FÉt}_ X \to \textit{FÉt}_\eta $ are fully faithful.

The result for $\textit{FÉt}_ X \to \textit{FÉt}_ U$ follows from Lemma 58.10.5 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 15.106.5.

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{FÉt}_\eta $ is the colimit of the categories $\textit{FÉt}_ U$ over such $U$, see Limits, Lemmas 32.10.1, 32.8.3, and 32.8.10. A formal argument then shows that fully faithfulness for $\textit{FÉt}_ X \to \textit{FÉt}_\eta $ follows from the fully faithfulness of the functors $\textit{FÉt}_ X \to \textit{FÉt}_ U$. $\square$

Proof. Surjectivity follows from Lemmas 58.10.4 and 58.4.1. We can consider the sequence of maps

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

By Lemma 58.4.3 we have to show finite étale morphisms $V \to U$ which pull back to trivial coverings of $U^{sh}$ extend to finite étale schemes over $X$. By Lemma 58.10.6 we know the corresponding statement for finite étale schemes over the punctured spectrum of $A$. However, by Limits, Lemma 32.20.1 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. $\square$