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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Fundamental Groups of Schemes}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we discuss Grothendieck's fundamental group of a scheme
and applications. A foundational reference is \cite{SGA1}.
A nice introduction is \cite{Lenstra}.
Other references \cite{Murre-lectures} and \cite{Grothendieck-Murre}.
\section{Schemes \'etale over a point}
\label{section-schemes-etale-point}
\noindent
In this section we describe schemes \'etale over the spectrum of a field.
Before we state the result we introduce the category of $G$-sets for a
topological group $G$.
\begin{definition}
\label{definition-G-set-continuous}
Let $G$ be a topological group.
A {\it $G$-set}, sometimes called a {\it discrete $G$-set},
is a set $X$ endowed with a left action $a : G \times X \to X$
such that $a$ is continuous when $X$ is given the discrete topology and
$G \times X$ the product topology.
A {\it morphism of $G$-sets} $f : X \to Y$ is simply any $G$-equivariant
map from $X$ to $Y$.
The category of $G$-sets is denoted {\it $G\textit{-Sets}$}.
\end{definition}
\noindent
The condition that $a : G \times X \to X$ is continuous signifies
simply that the stabilizer of any $x \in X$ is open in $G$.
If $G$ is an abstract group $G$ (i.e., a group but not a topological group)
then this agrees with our preceding definition (see for example
Sites, Example \ref{sites-example-site-on-group})
provided we endow $G$ with the discrete topology.
\medskip\noindent
Recall that if $L/K$ is an infinite Galois extension then the
Galois group $G = \text{Gal}(L/K)$ comes endowed with a canonical
topology, see Fields, Section \ref{fields-section-infinite-galois}.
\begin{lemma}
\label{lemma-sheaves-point}
Let $K$ be a field. Let $K^{sep}$ be a separable closure of $K$.
Consider the profinite group $G = \text{Gal}(K^{sep}/K)$.
The functor
$$
\begin{matrix}
\text{schemes \'etale over }K &
\longrightarrow &
G\textit{-Sets} \\
X/K & \longmapsto &
\Mor_{\Spec(K)}(\Spec(K^{sep}), X)
\end{matrix}
$$
is an equivalence of categories.
\end{lemma}
\begin{proof}
A scheme $X$ over $K$ is \'etale over $K$ if and only if
$X \cong \coprod_{i\in I} \Spec(K_i)$ with
each $K_i$ a finite separable extension of $K$
(Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}).
The functor of the lemma associates to $X$ the $G$-set
$$
\coprod\nolimits_i \Hom_K(K_i, K^{sep})
$$
with its natural left $G$-action. Each element has an open stabilizer
by definition of the topology on $G$. Conversely, any $G$-set $S$
is a disjoint union of its orbits. Say $S = \coprod S_i$. Pick $s_i \in S_i$
and denote $G_i \subset G$ its open stabilizer. By Galois theory
(Fields, Theorem \ref{fields-theorem-inifinite-galois-theory})
the fields $(K^{sep})^{G_i}$ are finite separable field extensions of $K$, and
hence the scheme
$$
\coprod\nolimits_i \Spec((K^{sep})^{G_i})
$$
is \'etale over $K$. This gives an inverse to the functor of the lemma.
Some details omitted.
\end{proof}
\begin{remark}
\label{remark-covering-surjective}
Under the correspondence of Lemma \ref{lemma-sheaves-point},
the coverings in the small \'etale site
$\Spec(K)_\etale$ of $K$ correspond to surjective families of
maps in $G\textit{-Sets}$.
\end{remark}
\section{Galois categories}
\label{section-galois}
\noindent
In this section we discuss some of the material the reader can
find in \cite[Expos\'e V, Sections 4, 5, and 6]{SGA1}.
\medskip\noindent
Let $F : \mathcal{C} \to \textit{Sets}$ be a functor.
Recall that by our conventions categories have a set of objects and
for any pair of objects a set of morphisms. There is a canonical
injective map
\begin{equation}
\label{equation-embedding-product}
\text{Aut}(F)
\longrightarrow
\prod\nolimits_{X \in \Ob(\mathcal{C})} \text{Aut}(F(X))
\end{equation}
For a set $E$ we endow $\text{Aut}(E)$ with the compact open topology, see
Topology, Example \ref{topology-example-automorphisms-of-a-set}.
Of course this is the discrete topology when $E$ is finite, which
is the case of interest in this section\footnote{When we discuss the
pro-\'etale fundamental group the general case will be of interest.}.
We endow $\text{Aut}(F)$ with the topology induced from the
product topology on the right hand side of (\ref{equation-embedding-product}).
In particular, the action maps
$$
\text{Aut}(F) \times F(X) \longrightarrow F(X)
$$
are continuous when $F(X)$ is given the discrete topology because this
is true for the action maps $\text{Aut}(E) \times E \to E$ for any set $E$.
The universal property of our topology on $\text{Aut}(F)$ is the following:
suppose that $G$ is a topological group and $G \to \text{Aut}(F)$
is a group homomorphism such that the induced actions $G \times F(X) \to F(X)$
are continuous for all $X \in \Ob(\mathcal{C})$ where $F(X)$ has
the discrete topology. Then $G \to \text{Aut}(F)$ is continuous.
\medskip\noindent
The following lemma tells us that the group of automorphisms of a functor
to the category of finite sets is automatically a profinite group.
\begin{lemma}
\label{lemma-aut-inverse-limit}
Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$
be a functor. The map (\ref{equation-embedding-product}) identifies
$\text{Aut}(F)$ with a closed subgroup of
$\prod_{X \in \Ob(\mathcal{C})} \text{Aut}(F(X))$.
In particular, if $F(X)$ is finite for all $X$, then
$\text{Aut}(F)$ is a profinite group.
\end{lemma}
\begin{proof}
Let $\xi = (\gamma_X) \in \prod \text{Aut}(F(X))$ be an element not in
$\text{Aut}(F)$. Then there exists a morphism $f : X \to X'$ of $\mathcal{C}$
and an element $x \in F(X)$ such that
$F(f)(\gamma_X(x)) \not = \gamma_{X'}(F(f)(x))$.
Consider the open neighbourhood
$U = \{\gamma \in \text{Aut}(F(X)) \mid \gamma(x) = \gamma_X(x)\}$
of $\gamma_X$ and the open neighbourhood
$U' = \{\gamma' \in \text{Aut}(F(X')) \mid \gamma'(F(f)(x)) =
\gamma_{X'}(F(f)(x))\}$.
Then
$U \times U' \times \prod_{X'' \not = X, X'} \text{Aut}(F(X''))$
is an open neighbourhood of $\xi$ not meeting $\text{Aut}(F)$.
The final statement follows from the fact that
$\prod \text{Aut}(F(X))$ is a profinite space if each $F(X)$ is finite.
\end{proof}
\begin{example}
\label{example-galois-category-G-sets}
Let $G$ be a topological group. An important example will be the
forgetful functor
\begin{equation}
\label{equation-forgetful}
\textit{Finite-}G\textit{-Sets} \longrightarrow \textit{Sets}
\end{equation}
where $\textit{Finite-}G\textit{-Sets}$ is the full subcategory of
$G\textit{-Sets}$ whose objects are the finite $G$-sets.
The category $G\textit{-Sets}$ of $G$-sets is defined in
Definition \ref{definition-G-set-continuous}.
\end{example}
\noindent
Let $G$ be a topological group. The {\it profinite completion} of $G$
will be the profinite group
$$
G^\wedge =
\lim_{U \subset G\text{ open, normal, finite index}} G/U
$$
with its profinite topology. Observe that the limit is cofiltered
as a finite intersection of open, normal subgroups of finite index
is another. The universal property of the profinite completion is
that any continuous map $G \to H$ to a profinite group $H$ factors
canonically as $G \to G^\wedge \to H$.
\begin{lemma}
\label{lemma-single-out-profinite}
Let $G$ be a topological group. The automorphism group of the functor
(\ref{equation-forgetful}) endowed with its profinite topology from
Lemma \ref{lemma-aut-inverse-limit} is the profinite completion of $G$.
\end{lemma}
\begin{proof}
Denote $F_G$ the functor (\ref{equation-forgetful}). Any morphism
$X \to Y$ in $\textit{Finite-}G\textit{-Sets}$ commutes with the action
of $G$. Thus any $g \in G$ defines an automorphism of $F_G$ and
we obtain a canonical homomorphism $G \to \text{Aut}(F_G)$ of groups.
Observe that any finite $G$-set $X$ is a finite disjoint union of
$G$-sets of the form $G/H_i$ with canonical $G$-action where
$H_i \subset G$ is an open subgroup of finite index. Then
$U_i = \bigcap gH_ig^{-1}$ is open, normal, and has finite index.
Moreover $U_i$ acts trivially on $G/H_i$ hence
$U = \bigcap U_i$ acts trivially on $F_G(X)$. Hence the action
$G \times F_G(X) \to F_G(X)$ is continuous. By the universal
property of the topology on $\text{Aut}(F_G)$ the map
$G \to \text{Aut}(F_G)$ is continuous.
By Lemma \ref{lemma-aut-inverse-limit} and the universal property
of profinite completion there is an induced
continuous group homomorphism
$$
G^\wedge \longrightarrow \text{Aut}(F_G)
$$
Moreover, since $G/U$ acts faithfully on $G/U$ this map is
injective. If the image is dense, then the map is surjective and hence a
homeomorphism by Topology, Lemma \ref{topology-lemma-bijective-map}.
\medskip\noindent
Let $\gamma \in \text{Aut}(F_G)$ and let $X \in \Ob(\mathcal{C})$.
We will show there is a $g \in G$ such that $\gamma$ and $g$
induce the same action on $F_G(X)$. This will finish the proof.
As before we see that $X$ is a finite disjoint union of $G/H_i$.
With $U_i$ and $U$ as above, the finite $G$-set $Y = G/U$
surjects onto $G/H_i$ for all $i$ and hence it suffices to
find $g \in G$ such that $\gamma$ and $g$ induce the same action
on $F_G(G/U) = G/U$. Let $e \in G$ be the neutral element and
say that $\gamma(eU) = g_0U$ for some $g_0 \in G$. For any
$g_1 \in G$ the morphism
$$
R_{g_1} : G/U \longrightarrow G/U,\quad gU \longmapsto gg_1U
$$
of $\textit{Finite-}G\textit{-Sets}$ commutes with the action of
$\gamma$. Hence
$$
\gamma(g_1U) = \gamma(R_{g_1}(eU)) = R_{g_1}(\gamma(eU)) =
R_{g_1}(g_0U) = g_0g_1U
$$
Thus we see that $g = g_0$ works.
\end{proof}
\noindent
Recall that an exact functor is one which commutes with all
finite limits and finite colimits. In particular such a functor
commutes with equalizers, coequalizers, fibred products,
pushouts, etc.
\begin{lemma}
\label{lemma-second-fundamental-functor}
Let $G$ be a topological group. Let
$F : \textit{Finite-}G\textit{-Sets} \to \textit{Sets}$
be an exact functor with $F(X)$ finite for all $X$.
Then $F$ is isomorphic to the functor (\ref{equation-forgetful}).
\end{lemma}
\begin{proof}
Let $X$ be a nonempty object of $\textit{Finite-}G\textit{-Sets}$.
The diagram
$$
\xymatrix{
X \ar[r] \ar[d] & \{*\} \ar[d] \\
\{*\} \ar[r] & \{*\}
}
$$
is cocartesian. Hence we conclude that $F(X)$ is nonempty.
Let $U \subset G$ be an open, normal subgroup with finite index.
Observe that
$$
G/U \times G/U = \coprod\nolimits_{gU \in G/U} G/U
$$
where the summand corresponding to $gU$ corresponds to the orbit of
$(eU, gU)$ on the left hand side. Then we see that
$$
F(G/U) \times F(G/U) = F(G/U \times G/U) = \coprod\nolimits_{gU \in G/U} F(G/U)
$$
Hence $|F(G/U)| = |G/U|$ as $F(G/U)$ is nonempty. Thus we see that
$$
\lim_{U \subset G\text{ open, normal, finite idex}} F(G/U)
$$
is nonempty (Categories, Lemma \ref{categories-lemma-nonempty-limit}).
Pick $\gamma = (\gamma_U)$ an element in this limit.
Denote $F_G$ the functor (\ref{equation-forgetful}). We can identify
$F_G$ with the functor
$$
X \longmapsto \colim_U \Mor(G/U, X)
$$
where $f : G/U \to X$ corresponds to $f(eU) \in X = F_G(X)$
(details omitted). Hence the element $\gamma$ determines
a well defined map
$$
t : F_G \longrightarrow F
$$
Namely, given $x \in X$ choose $U$ and $f : G/U \to X$ sending
$eU$ to $x$ and then set $t_X(x) = F(f)(\gamma_U)$.
We will show that $t$ induces a bijective map
$t_{G/U} : F_G(G/U) \to F(G/U)$ for any $U$.
This implies in a straightforward manner that $t$
is an isomorphism (details omitted).
Since $|F_G(G/U)| = |F(G/U)|$ it suffices to show
that $t_{G/U}$ is surjective. The image contains at least
one element, namely
$t_{G/U}(eU) = F(\text{id}_{G/U})(\gamma_U) = \gamma_U$.
For $g \in G$ denote $R_g : G/U \to G/U$ right multiplication.
Then set of fixed points of $F(R_g) : F(G/U) \to F(G/U)$
is equal to $F(\emptyset) = \emptyset$ if $g \not \in U$ because $F$
commutes with equalizers. It follows that if
$g_1, \ldots, g_{|G/U|}$ is a system of representatives
for $G/U$, then the elements $F(R_{g_i})(\gamma_U)$ are pairwise distinct
and hence fill out $F(G/U)$. Then
$$
t_{G/U}(g_iU) = F(R_{g_i})(\gamma_U)
$$
and the proof is complete.
\end{proof}
\begin{example}
\label{example-from-C-F-to-G-sets}
Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$
be a functor such that $F(X)$ is finite for all $X \in \Ob(\mathcal{C})$.
By Lemma \ref{lemma-aut-inverse-limit} we see that $G = \text{Aut}(F)$
comes endowed with the structure of a profinite topological group in a
canonical manner. We obtain a functor
\begin{equation}
\label{equation-remember}
\mathcal{C} \longrightarrow \textit{Finite-}G\textit{-Sets},\quad
X \longmapsto F(X)
\end{equation}
where $F(X)$ is endowed with the induced action of $G$. This action
is continuous by our construction of the topology on $\text{Aut}(F)$.
\end{example}
\noindent
The purpose of defining Galois categories is to single out those
pairs $(\mathcal{C}, F)$ for which the functor (\ref{equation-remember})
is an equivalence. Our definition of a Galois category is as follows.
\begin{definition}
\label{definition-galois-category}
\begin{reference}
Different from the definition in \cite[Expos\'e V, Definition 5.1]{SGA1}.
Compare with \cite[Definition 7.2.1]{BS}.
\end{reference}
Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$
be a functor. The pair $(\mathcal{C}, F)$ is a {\it Galois category} if
\begin{enumerate}
\item $\mathcal{C}$ has finite limits and finite colimits,
\item
\label{item-connected-components}
every object of $\mathcal{C}$ is a finite (possibly empty)
coproduct of connected objects,
\item $F(X)$ is finite for all $X \in \Ob(\mathcal{C})$, and
\item $F$ reflects isomorphisms\footnote{Namely, given a morphism
$f$ of $\mathcal{C}$ if $F(f)$ is an isomorphism, then
$f$ is an isomorphism.} and is exact\footnote{This means that
$F$ commutes with finite limits and colimits, see
Categories, Section \ref{categories-section-exact-functor}.}.
\end{enumerate}
Here we say $X \in \Ob(\mathcal{C})$ is connected if
it is not initial and for any monomorphism $Y \to X$
either $Y$ is initial or $Y \to X$ is an isomorphism.
\end{definition}
\noindent
{\bf Warning:} This definition is not the same (although eventually we'll
see it is equivalent) as the definition given in most references.
Namely, in \cite[Expos\'e V, Definition 5.1]{SGA1} a Galois category is
defined to be a category equivalent to $\textit{Finite-}G\textit{-Sets}$
for some profinite group $G$. Then Grothendieck characterizes
Galois categories by a list of axioms (G1) -- (G6) which are weaker
than our axioms above. The motivation for our choice is to stress the
existence of finite limits and finite colimits and exactness of the
functor $F$. The price we'll pay for this later is that we'll have
to work a bit harder to apply the results of this section.
\begin{lemma}
\label{lemma-epi-mono}
Let $(\mathcal{C}, F)$ be a Galois category. Let
$X \to Y \in \text{Arrows}(\mathcal{C})$. Then
\begin{enumerate}
\item $F$ is faithful,
\item $X \to Y$ is a monomorphism
$\Leftrightarrow F(X) \to F(Y)$ is injective,
\item $X \to Y$ is an epimorphism
$\Leftrightarrow F(X) \to F(Y)$ is surjective,
\item an object $A$ of $\mathcal{C}$ is initial if and only if
$F(A) = \emptyset$,
\item an object $Z$ of $\mathcal{C}$ is final if and only if
$F(Z)$ is a singleton,
\item if $X$ and $Y$ are connected, then $X \to Y$ is an epimorphism,
\item
\label{item-one-element}
if $X$ is connected and $a, b : X \to Y$ are two morphisms
then $a = b$ as soon as $F(a)$ and $F(b)$ agree on one element of $F(X)$,
\item if $X = \coprod_{i = 1, \ldots, n} X_i$ and
$Y = \coprod_{j = 1, \ldots, m} Y_j$ where $X_i$, $Y_j$ are connected,
then there is map $\alpha : \{1, \ldots, n\} \to \{1, \ldots, m\}$
such that $X \to Y$ comes from a collection of morphisms
$X_i \to Y_{\alpha(i)}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). Suppose $a, b : X \to Y$ with $F(a) = F(b)$.
Let $E$ be the equalizer of $a$ and $b$. Then $F(E) = F(X)$
and we see that $E = X$ because $F$ reflects isomorphisms.
\medskip\noindent
Proof of (2). This is true because $F$ turns the morphism $X \to X \times_Y X$
into the map $F(X) \to F(X) \times_{F(Y)} F(X)$ and $F$ reflects isomorphisms.
\medskip\noindent
Proof of (3). This is true because $F$ turns the morphism $Y \amalg_X Y \to Y$
into the map $F(Y) \amalg_{F(X)} F(Y) \to F(Y)$ and $F$ reflects isomorphisms.
\medskip\noindent
Proof of (4). There exists an initial object $A$ and certainly
$F(A) = \emptyset$. On the other hand, if $X$ is an object with
$F(X) = \emptyset$, then the unique map $A \to X$ induces a bijection
$F(A) \to F(X)$ and hence $A \to X$ is an isomorphism.
\medskip\noindent
Proof of (5). There exists a final object $Z$ and certainly
$F(Z)$ is a singleton. On the other hand, if $X$ is an object with
$F(X)$ a singleton, then the unique map $X \to Z$ induces a bijection
$F(X) \to F(Z)$ and hence $X \to Z$ is an isomorphism.
\medskip\noindent
Proof of (6). The equalizer $E$ of the two maps $Y \to Y \amalg_X Y$ is not
an initial object of $\mathcal{C}$ because $X \to Y$ factors through $E$
and $F(X) \not = \emptyset$. Hence $E = Y$ and we conclude.
\medskip\noindent
Proof of (\ref{item-one-element}).
The equalizer $E$ of $a$ and $b$ comes with a monomorphism
$E \to X$ and $F(E) \subset F(X)$ is the set of elements where
$F(a)$ and $F(b)$ agree. To finish use that either $E$ is initial
or $E = X$.
\medskip\noindent
Proof of (8). For each $i, j$ we see that $E_{ij} = X_i \times_Y Y_j$
is either initial or equal to $X_i$. Picking $s \in F(X_i)$
we see that $E_{ij} = X_i$ if and only if $s$ maps to an element
of $F(Y_j) \subset F(Y)$, hence this happens for a unique $j = \alpha(i)$.
\end{proof}
\noindent
By the lemma above we see that, given a connected object $X$ of a
Galois category $(\mathcal{C}, F)$, the automorphism group
$\text{Aut}(X)$ has order at most $|F(X)|$. Namely, given $s \in F(X)$
and $g \in \text{Aut}(X)$ we see that $g(s) = s$ if and only
if $g = \text{id}_X$ by (\ref{item-one-element}).
We say $X$ is {\it Galois} if equality holds.
Equivalently, $X$ is Galois if it is connected and
$\text{Aut}(X)$ acts transitively on $F(X)$.
\begin{lemma}
\label{lemma-galois}
Let $(\mathcal{C}, F)$ be a Galois category. For any connected object $X$
of $\mathcal{C}$ there exists a Galois object $Y$ and a morphism $Y \to X$.
\end{lemma}
\begin{proof}
We will use the results of Lemma \ref{lemma-epi-mono} without further mention.
Let $n = |F(X)|$. Consider $X^n$ endowed with its natural action of
$S_n$. Let
$$
X^n = \coprod\nolimits_{t \in T} Z_t
$$
be the decomposition into connected objects. Pick a $t$ such that
$F(Z_t)$ contains $(s_1, \ldots, s_n)$ with $s_i$ pairwise distinct.
If $(s'_1, \ldots, s'_n) \in F(Z_t)$ is another element, then we
claim $s'_i$ are pairwise distinct as well. Namely, if not, say
$s'_i = s'_j$, then $Z_t$ is the image of an connected component of
$X^{n - 1}$ under the diagonal morphism
$$
\Delta_{ij} : X^{n - 1} \longrightarrow X^n
$$
Since morphisms of connected objects are epimorphisms and induce
surjections after applying $F$ it would follow that $s_i = s_j$
which is not the case.
\medskip\noindent
Let $G \subset S_n$ be the subgroup of elements with $g(Z_t) = Z_t$.
Looking at the action of $S_n$ on
$$
F(X)^n = F(X^n) = \coprod\nolimits_{t' \in T} F(Z_{t'})
$$
we see that $G = \{g \in S_n \mid g(s_1, \ldots, s_n) \in F(Z_t)\}$.
Now pick a second element $(s'_1, \ldots, s'_n) \in F(Z_t)$.
Above we have seen that $s'_i$ are pairwise distinct. Thus we can
find a $g \in S_n$ with $g(s_1, \ldots, s_n) = (s'_1, \ldots, s'_n)$.
In other words, the action of $G$ on $F(Z_t)$ is transitive and
the proof is complete.
\end{proof}
\noindent
Here is a key lemma.
\begin{lemma}
\label{lemma-tame}
\begin{reference}
Compare with \cite[Definition 7.2.4]{BS}.
\end{reference}
Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$
be as in Example \ref{example-from-C-F-to-G-sets}. For any connected
$X$ in $\mathcal{C}$ the action of $G$ on $F(X)$ is transitive.
\end{lemma}
\begin{proof}
We will use the results of Lemma \ref{lemma-epi-mono} without further mention.
Let $I$ be the set of isomorphism classes of Galois objects in $\mathcal{C}$.
For each $i \in I$ let $X_i$ be a representative of the isomorphism class.
Choose $\gamma_i \in F(X_i)$ for each $i \in I$.
We define a partial ordering on $I$ by setting $i \geq i'$ if
and only if there is a morphism $f_{ii'} : X_i \to X_{i'}$.
Given such a morphism we can post-compose by an automorphism
$X_{i'} \to X_{i'}$ to assure that $F(f_{ii'})(\gamma_i) = \gamma_{i'}$.
With this normalization the morphism $f_{ii'}$ is unique.
Observe that $I$ is a directed partially ordered set:
(Categories, Definition \ref{categories-definition-directed-set})
if $i_1, i_2 \in I$ there exists a Galois object $Y$ and a morphism
$Y \to X_{i_1} \times X_{i_2}$ by Lemma \ref{lemma-galois} applied
to a connected component of $X_{i_1} \times X_{i_2}$.
Then $Y \cong X_i$ for some $i \in I$ and $i \geq i_1$, $i \geq I_2$.
\medskip\noindent
We claim that the functor $F$ is isomorphic to the functor $F'$
which sends $X$ to
$$
F'(X) = \colim_I \Mor_\mathcal{C}(X_i, X)
$$
via the transformation of functors $t : F' \to F$ defined as follows:
given $f : X_i \to X$ we set $t_X(f) = F(f)(\gamma_i)$.
Using (\ref{item-one-element}) we find that $t_X$ is injective.
To show surjectivity, let $\gamma \in F(X)$. Then we can immediately
reduce to the case where $X$ is connected by the definition of
a Galois category. Then we may assume $X$ is Galois by
Lemma \ref{lemma-galois}. In this case $X$ is isomorphic to $X_i$
for some $i$ and we can choose the isomorphism $X_i \to X$ such
that $\gamma_i$ maps to $\gamma$ (by definition of Galois objects).
We conclude that $t$ is an isomorphism.
\medskip\noindent
Set $A_i = \text{Aut}(X_i)$.
We claim that for $i \geq i'$ there is a canonical map
$h_{ii'} : A_i \to A_{i'}$ such that for all $a \in A_i$
the diagram
$$
\xymatrix{
X_i \ar[d]_a \ar[r]_{f_{ii'}} & X_{i'} \ar[d]^{h_{ii'}(a)} \\
X_i \ar[r]^{f_{ii'}} & X_{i'}
}
$$
commutes. Namely, just let $h_{ii'}(a) = a' : X_{i'} \to X_{i'}$
be the unique automorphism such that
$F(a')(\gamma_{i'}) = F(f_{ii'} \circ a)(\gamma_i)$.
As before this makes the diagram commute and moreover the choice
is unique.
It follows that
$h_{i'i''} \circ h_{ii'} = h_{ii''}$
if $i \geq i' \geq i''$.
Since $F(X_i) \to F(X_{i'})$ is surjective we see that
$A_i \to A_{i'}$ is surjective.
Taking the inverse limit we obtain a group
$$
A = \lim_I A_i
$$
This is a profinite group since the automorphism groups are finite.
The map $A \to A_i$ is surjective for all $i$ by
Categories, Lemma \ref{categories-lemma-nonempty-limit}.
\medskip\noindent
Since elements of $A$ act on the inverse system $X_i$ we get an action of
$A$ (on the right) on $F'$ by pre-composing. In other words, we get
a homomorphism $A^{opp} \to G$. Since $A \to A_i$ is surjective we conclude
that $G$ acts transitively on $F(X_i)$ for all $i$. Since every connected
object is dominated by one of the $X_i$ we conclude the lemma is true.
\end{proof}
\begin{proposition}
\label{proposition-galois}
\begin{reference}
This is a weak version of \cite[Expos\'e V]{SGA1}.
The proof is borrowed from \cite[Theorem 7.2.5]{BS}.
\end{reference}
Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$
be as in Example \ref{example-from-C-F-to-G-sets}. The functor
$F : \mathcal{C} \to \textit{Finite-}G\textit{-Sets}$
(\ref{equation-remember}) an equivalence.
\end{proposition}
\begin{proof}
We will use the results of Lemma \ref{lemma-epi-mono} without further mention.
In particular we know the functor is faithful.
By Lemma \ref{lemma-tame} we know that for any connected $X$ the
action of $G$ on $F(X)$ is transitive. Hence $F$ preserves
the decomposition into connected components (existence of which is
an axiom of a Galois category). Let $X$ and $Y$ be objects and let
$s : F(X) \to F(Y)$ be a map. Then the graph
$\Gamma_s \subset F(X) \times F(Y)$ of $s$
is a union of connected components. Hence there exists a
union of connected components $Z$ of $X \times Y$,
which comes equipped with a monomorphism $Z \to X \times Y$,
with $F(Z) = \Gamma_s$. Since $F(Z) \to F(X)$ is bijective
we see that $Z \to X$ is an isomorphism and we conclude
that $s = F(f)$ where $f : X \cong Z \to Y$ is the composition.
Hence $F$ is fully faithful.
\medskip\noindent
To finish the proof we show that $F$ is essentially surjective.
It suffices to show that $G/H$ is in the essential image for
any open subgroup $H \subset G$ of finite index.
By definition of the topology on $G$ there exists a finite
collection of objects $X_i$ such that
$$
\Ker(G \longrightarrow \prod\nolimits_i \text{Aut}(F(X_i)))
$$
is contained in $H$. We may assume $X_i$ is connected
for all $i$. We can choose a Galois object $Y$ mapping
to a connected component of $\prod X_i$ using
Lemma \ref{lemma-galois}. Choose an isomorphism $F(Y) = G/U$
in $G\textit{-sets}$ for some open subgroup $U \subset G$.
As $Y$ is Galois, the group
$\text{Aut}(Y) = \text{Aut}_{G\textit{-Sets}}(G/U)$ acts transitively
on $F(Y) = G/U$. This implies that $U$ is normal. Since
$F(Y)$ surjects onto $F(X_i)$ for each $i$ we see that
$U \subset H$. Let $M \subset \text{Aut}(Y)$ be the finite subgroup
corresponding to
$$
(H/U)^{opp} \subset (G/U)^{opp} = \text{Aut}_{G\textit{-Sets}}(G/U)
= \text{Aut}(Y).
$$
Set $X = Y/M$, i.e., $X$ is the coequalizer
of the arrows $m : Y \to Y$, $m \in M$.
Since $F$ is exact we see that $F(X) = G/H$ and the
proof is complete.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-galois}
Let $(\mathcal{C}, F)$ and $(\mathcal{C}', F')$ be Galois categories.
Let $H : \mathcal{C} \to \mathcal{C}'$ be an exact functor.
There exists an isomorphism $t : F' \circ H \to F$.
The choice of $t$ determines a continuous homomorphism
$h : G' = \text{Aut}(F') \to \text{Aut}(F) = G$ and
a $2$-commutative diagram
$$
\xymatrix{
\mathcal{C} \ar[r]_H \ar[d] & \mathcal{C}' \ar[d] \\
\textit{Finite-}G\textit{-Sets} \ar[r]^h &
\textit{Finite-}G'\textit{-Sets}
}
$$
The map $h$ is independent of $t$ up
to an inner automorphism of $G$.
Conversely, given a continuous homomorphism $h : G' \to G$ there
is an exact functor $H : \mathcal{C} \to \mathcal{C}'$ and an
isomorphism $t$ recovering $h$ as above.
\end{lemma}
\begin{proof}
By Proposition \ref{proposition-galois} and
Lemma \ref{lemma-single-out-profinite} we may assume
$\mathcal{C} = \textit{Finite-}G\textit{-Sets}$ and $F$ is the
forgetful functor and similarly for $\mathcal{C}'$. Thus the existence of
$t$ follows from Lemma \ref{lemma-second-fundamental-functor}. The map $h$
comes from transport of structure via $t$. The commutativity of the
diagram is obvious. Uniqueness of $h$ up to inner conjugation by
an element of $G$ comes from the fact that the choice of $t$ is
unique up to an element of $G$. The final statement is straightforward.
\end{proof}
\section{Functors and homomorphisms}
\label{section-translation}
\noindent
Let $(\mathcal{C}, F)$, $(\mathcal{C}', F')$, $(\mathcal{C}'', F'')$
be Galois categories. Set $G = \text{Aut}(F)$, $G' = \text{Aut}(F')$, and
$G'' = \text{Aut}(F'')$. Let $H : \mathcal{C} \to \mathcal{C}'$
and $H' : \mathcal{C}' \to \mathcal{C}''$ be exact functors.
Let $h : G' \to G$ and $h' : G'' \to G'$ be the corresponding
continuous homomorphism as in Lemma \ref{lemma-functoriality-galois}.
In this section we consider the corresponding $2$-commutative diagram
\begin{equation}
\label{equation-translation}
\vcenter{
\xymatrix{
\mathcal{C} \ar[r]_H \ar[d] &
\mathcal{C}' \ar[r]_{H'} \ar[d] &
\mathcal{C}'' \ar[d] \\
\textit{Finite-}G\textit{-Sets} \ar[r]^h &
\textit{Finite-}G'\textit{-Sets} \ar[r]^{h'} &
\textit{Finite-}G''\textit{-Sets}
}
}
\end{equation}
and we relate exactness properties of the sequence
$1 \to G'' \to G' \to G \to 1$ to properties of the functors $H$ and $H'$.
\begin{lemma}
\label{lemma-functoriality-galois-surjective}
In diagram (\ref{equation-translation}) the following are equivalent
\begin{enumerate}
\item $h : G' \to G$ is surjective,
\item $H : \mathcal{C} \to \mathcal{C}'$ is fully faithful,
\item if $X \in \Ob(\mathcal{C})$ is connected, then $H(X)$ is connected,
\item if $X \in \Ob(\mathcal{C})$ is connected and there is
a morphism $*' \to H(X)$ in $\mathcal{C}'$, then
there is a morphism $* \to X$, and
\item for any object $X$ of $\mathcal{C}$ the map
$\Mor_\mathcal{C}(*, X) \to \Mor_{\mathcal{C}'}(*', H(X))$
is bijective.
\end{enumerate}
Here $*$ and $*'$ are final objects of $\mathcal{C}$ and $\mathcal{C}'$.
\end{lemma}
\begin{proof}
The implications (5) $\Rightarrow$ (4) and (2) $\Rightarrow$ (5) are clear.
\medskip\noindent
Assume (3). Let $X$ be a connected object of $\mathcal{C}$ and let
$*' \to H(X)$ be a morphism. Since $H(X)$ is connected by (3)
we see that $*' \to H(X)$ is an isomorphism. Hence the $G'$-set
corresponding to $H(X)$ has exactly one element, which means the
$G$-set corresponding to $X$ has one element which means $X$ is
isomorphic to the final object of $\mathcal{C}$, in particular
there is a map $* \to X$. In this way we see that (3) $\Rightarrow$ (4).
\medskip\noindent
If (1) is true, then the functor
$\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}$
is fully faithful: in this case a map of $G$-sets commutes with the
action of $G$ if and only if it commutes with the action of $G'$.
Thus (1) $\Rightarrow$ (2).
\medskip\noindent
If (1) is true, then for a $G$-set $X$ the $G$-orbits and $G'$-orbits
agree. Thus (1) $\Rightarrow$ (3).
\medskip\noindent
To finish the proof it suffices to show that (4) implies (1).
If (1) is false, i.e., if $h$ is not surjective, then there is
an open subgroup $U \subset G$ containing $h(G')$ which is not
equal to $G$. Then the finite $G$-set $M = G/U$ has a transitive
action but $G'$ has a fixed point. The object $X$ of $\mathcal{C}$
corresponding to $M$ would contradict (3). In this way we see that
(3) $\Rightarrow$ (1) and the proof is complete.
\end{proof}
\begin{lemma}
\label{lemma-composition-trivial}
In diagram (\ref{equation-translation}) the following are equivalent
\begin{enumerate}
\item $h \circ h'$ is trivial, and
\item the image of $H' \circ H$ consists of objects isomorphic to finite
coproducts of final objects.
\end{enumerate}
\end{lemma}
\begin{proof}
We may replace $H$ and $H'$ by the canonical functors
$\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}
\to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$.
Then we are saying that the action of $G''$ on every $G$-set is trivial
if and only if the homomorphism $G'' \to G$ is trivial. This is clear.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-galois-ses}
In diagram (\ref{equation-translation}) the following are equivalent
\begin{enumerate}
\item the sequence $G'' \xrightarrow{h'} G' \xrightarrow{h} G \to 1$
is exact in the following sense: $h$ is surjective, $h \circ h'$ is trivial,
and $\Ker(h)$ is the smallest closed normal subgroup containing $\Im(h')$,
\item $H$ is fully faithful and an object $X'$ of $\mathcal{C}'$ is in
the essential image of $H$ if and only if $H'(X')$ is isomorphic to a
finite coproduct of final objects, and
\item $H$ is fully faithful, $H \circ H'$ sends every object to a finite
coproduct of final objects, and for an object $X'$ of $\mathcal{C}'$
such that $H'(X')$ is a finite coproduct of final objects there exists
an object $X$ of $\mathcal{C}$ and an epimorphism $H(X) \to X'$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemmas \ref{lemma-functoriality-galois-surjective} and
\ref{lemma-composition-trivial} we may assume that
$H$ is fully faithful, $h$ is surjective, $H' \circ H$ maps
objects to disjoint unions of the final object, and $h \circ h'$
is trivial. Let $N \subset G'$ be the smallest closed normal
subgroup containing the image of $h'$. It is clear that
$N \subset \Ker(h)$.
We may assume the functors $H$ and $H'$ are the canonical functors
$\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}
\to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$.
\medskip\noindent
Suppose that (2) holds. This means that for a finite $G'$-set $X'$
such that $G''$ acts trivially, the action of $G'$ factors through $G$.
Apply this to $X' = G'/U'N$ where $U'$ is a small open subgroup of $G'$.
Then we see that $\Ker(h) \subset U'N$ for all $U'$. Since $N$ is closed
this implies $\Ker(h) \subset N$, i.e., (1) holds.
\medskip\noindent
Suppose that (1) holds. This means that $N = \Ker(h)$. Let $X'$ be a
finite $G'$-set such that $G''$ acts trivially. This means that
$\Ker(G' \to \text{Aut}(X'))$ is a closed normal subgroup containing
$\Im(h')$. Hence $N = \Ker(h)$ is contained in it and the $G'$-action
on $X'$ factors through $G$, i.e., (2) holds.
\medskip\noindent
Suppose that (3) holds. This means that for a finite $G'$-set $X'$
such that $G''$ acts trivially, there is a surjection of $G'$-sets
$X \to X'$ where $X$ is a $G$-set. Clearly this means the action of
$G'$ on $X'$ factors through $G$, i.e., (2) holds.
\medskip\noindent
The implication (2) $\Rightarrow$ (3) is immediate. This finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-galois-injective}
In diagram (\ref{equation-translation}) the following are equivalent
\begin{enumerate}
\item $h'$ is injective, and
\item for every connected object $X''$ of $\mathcal{C}''$
there exists an object $X'$ of $\mathcal{C}'$ and a diagram
$$
X'' \leftarrow Y'' \rightarrow H(X')
$$
in $\mathcal{C}''$ where $Y'' \to X''$ is an epimorphism and
$Y'' \to H(X')$ is a monomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
We may replace $H'$ by the corresponding functor between the categories
of finite $G'$-sets and finite $G''$-sets.
\medskip\noindent
Assume $h' : G'' \to G'$ is injective. Let $H'' \subset G''$
be an open subgroup. Since the topology on $G''$ is the induced
topology from $G'$ there exists an open subgroup $H' \subset G'$
such that $(h')^{-1}(H') \subset H''$.
Then the desired diagram is
$$
G''/H'' \leftarrow G''/(h')^{-1}(H') \rightarrow G'/H'
$$
Conversely, assume (2) holds for the functor
$\textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$.
Let $g'' \in \Ker(h')$. Pick any open subgroup $H'' \subset G''$.
By assumption there exists a finite $G'$-set $X'$ and a diagram
$$
G''/H'' \leftarrow Y'' \rightarrow X'
$$
of $G''$-sets with the left arrow surjective and the right arrow injective.
Since $g''$ is in the kernel of $h'$ we see that $g''$ acts trivially on $X'$.
Hence $g''$ acts trivially on $Y''$ and hence trivially on $G''/H''$.
Thus $g'' \in H''$. As this holds for all open subgroups we conclude
that $g''$ is the identity element as desired.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-galois-normal}
In diagram (\ref{equation-translation}) the following are equivalent
\begin{enumerate}
\item the image of $h'$ is normal, and
\item for every connected object $X'$ of $\mathcal{C}'$ such that
there is a morphism from the final object of $\mathcal{C}''$
to $H'(X')$ we have that $H'(X')$ is isomorphic to a finite coproduct
of final objects.
\end{enumerate}
\end{lemma}
\begin{proof}
This translates into the following statement for the continuous
group homomorphism $h' : G'' \to G'$: the image of $h'$ is normal
if and only if every open subgroup $U' \subset G'$ which
contains $h'(G'')$ also contains every conjugate of $h'(G'')$.
The result follows easily from this; some details omitted.
\end{proof}
\section{Finite \'etale morphisms}
\label{section-finite-etale}
\noindent
In this section we prove enough basic results on finite \'etale
morphisms to be able to construct the \'etale fundamental group.
\medskip\noindent
Let $X$ be a scheme. We will use the notation $\textit{F\'Et}_X$
to denote the category of schemes finite and \'etale over $X$.
Thus
\begin{enumerate}
\item an object of $\textit{F\'Et}_X$ is a finite \'etale morphism
$Y \to X$ with target $X$, and
\item a morphism in $\textit{F\'Et}_X$
from $Y \to X$ to $Y' \to X$ is a morphism $Y \to Y'$ making
the diagram
$$
\xymatrix{
Y \ar[rr] \ar[rd] & & Y' \ar[ld] \\
& X
}
$$
commute.
\end{enumerate}
We will often call an object of $\textit{F\'Et}_X$ a
{\it finite \'etale cover} of $X$ (even if $Y$ is empty).
It turns out that there is a stack $p : \textit{F\'Et} \to \Sch$
over the category of schemes whose fibre over $X$ is the category
$\textit{F\'Et}_X$ just defined. See Examples of Stacks, Section
\ref{examples-stacks-section-finite-etale}.
\begin{example}
\label{example-finite-etale-geometric-point}
Let $k$ be an algebraically closed field and $X = \Spec(k)$. In this case
$\textit{F\'Et}_X$ is equivalent to the category of finite sets. This works
more generally when $k$ is separably algebraically closed. The reason is
that a scheme \'etale over $k$ is the disjoint union of spectra of
fields finite separable over $k$, see
Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}.
\end{example}
\begin{lemma}
\label{lemma-finite-etale-covers-limits-colimits}
Let $X$ be a scheme. The category $\textit{F\'Et}_X$ has finite limits and
finite colimits and for any morphism $X' \to X$ the base change functor
$\textit{F\'Et}_X \to \textit{F\'Et}_{X'}$ is exact.
\end{lemma}
\begin{proof}
Finite limits and left exactness. By
Categories, Lemma \ref{categories-lemma-finite-limits-exist}
it suffices to show that $\textit{F\'Et}_X$ has a final object
and fibred products. This is clear because the category of
all schemes over $X$ has a final object (namely $X$) and fibred products.
Also, fibred products of schemes finite \'etale over $X$ are
finite \'etale over $X$. Moreover, it is clear that base
change commutes with these operations and hence base change
is left exact (Categories, Lemma
\ref{categories-lemma-characterize-left-exact}).
\medskip\noindent
Finite colimits and right exactness. By
Categories, Lemma \ref{categories-lemma-colimits-exist}
it suffices to show that $\textit{F\'Et}_X$ has finite
coproducts and coequalizers. Finite coproducts are given
by disjoint unions (the empty coproduct is the empty scheme).
Let $a, b : Z \to Y$ be two morphisms of $\textit{F\'Et}_X$.
Since $Z \to X$ and $Y \to X$ are finite \'etale we can write
$Z = \underline{\Spec}(\mathcal{C})$ and $Y = \underline{\Spec}(\mathcal{B})$
for some finite locally free $\mathcal{O}_X$-algebras $\mathcal{C}$
and $\mathcal{B}$. The morphisms $a, b$ induce two maps
$a^\sharp, b^\sharp : \mathcal{B} \to \mathcal{C}$.
Let $\mathcal{A} = \text{Eq}(a^\sharp, b^\sharp)$ be their
equalizer. If
$$
\underline{\Spec}(\mathcal{A}) \longrightarrow X
$$