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49.3. Galois categories

In this section we discuss some of the material the reader can find in [SGA1, Exposé V, Sections 4, 5, and 6].

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} \tag{49.3.0.1} \text{Aut}(F) \longrightarrow \prod\nolimits_{X \in \mathop{\rm Ob}\nolimits(\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 5.30.2. Of course this is the discrete topology when $E$ is finite, which is the case of interest in this section1. We endow $\text{Aut}(F)$ with the topology induced from the product topology on the right hand side of (49.3.0.1). 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 $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$ where $F(X)$ has the discrete topology. Then $G \to \text{Aut}(F)$ is continuous.

The following lemma tells us that the group of automorphisms of a functor to the category of finite sets is automatically a profinite group.

Lemma 49.3.1. Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor. The map (49.3.0.1) identifies $\text{Aut}(F)$ with a closed subgroup of $\prod_{X \in \mathop{\rm Ob}\nolimits(\mathcal{C})} \text{Aut}(F(X))$. In particular, if $F(X)$ is finite for all $X$, then $\text{Aut}(F)$ is a profinite group.

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 is follows from the fact that $\prod \text{Aut}(F(X))$ is a profinite space if each $F(X)$ is finite. $\square$

Example 49.3.2. Let $G$ be a topological group. An important example will be the forgetful functor \begin{equation} \tag{49.3.2.1} \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 49.2.1.

Let $G$ be a topological group. The profinite completion of $G$ will be the profinite group $$ G^\wedge = \mathop{\rm lim}\nolimits_{U \subset G\text{ open, normal, finite index}} G/U $$ 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 $G \to G^\wedge \to H$.

Lemma 49.3.3. Let $G$ be a topological group. The automorphism group of the functor (49.3.2.1) endowed with its profinite topology from Lemma 49.3.1 is the profinite completion of $G$.

Proof. Denote $F_G$ the functor (49.3.2.1). 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 49.3.1 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 5.17.8.

Let $\gamma \in \text{Aut}(F_G)$ and let $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$. 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. $\square$

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.

Lemma 49.3.4. 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 (49.3.2.1).

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 $$ \mathop{\rm lim}\nolimits_{U \subset G\text{ open, normal, finite idex}} F(G/U) $$ is nonempty (Categories, Lemma 4.21.7). Pick $\gamma = (\gamma_U)$ an element in this limit. Denote $F_G$ the functor (49.3.2.1). We can identify $F_G$ with the functor $$ X \longmapsto \mathop{\rm colim}\nolimits_U \mathop{\rm Mor}\nolimits(G/U, X) $$ 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. $\square$

Example 49.3.5. 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 $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$. By Lemma 49.3.1 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} \tag{49.3.5.1} \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 $\text{Aut}(F)$.

The purpose of defining Galois categories is to single out those pairs $(\mathcal{C}, F)$ for which the functor (49.3.5.1) is an equivalence. Our definition of a Galois category is as follows.

Definition 49.3.6. Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor. The pair $(\mathcal{C}, F)$ is a Galois category if

  1. $\mathcal{C}$ has finite limits and finite colimits,
  2. every object of $\mathcal{C}$ is a finite (possibly empty) coproduct of connected objects,
  3. $F(X)$ is finite for all $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$, and
  4. $F$ reflects isomorphisms and is exact.

Here we say $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$ 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.

Warning: This definition is not the same (although eventually we'll see it is equivalent) as the definition given in most references. Namely, in [SGA1, Exposé V, Definition 5.1] 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.

Lemma 49.3.7. Let $(\mathcal{C}, F)$ be a Galois category. Let $X \to Y \in \text{Arrows}(\mathcal{C})$. Then

  1. $F$ is faithful,
  2. $X \to Y$ is a monomorphism $\Leftrightarrow F(X) \to F(Y)$ is injective,
  3. $X \to Y$ is an epimorphism $\Leftrightarrow F(X) \to F(Y)$ is surjective,
  4. an object $A$ of $\mathcal{C}$ is initial if and only if $F(A) = \emptyset$,
  5. an object $Z$ of $\mathcal{C}$ is final if and only if $F(Z)$ is a singleton,
  6. if $X$ and $Y$ are connected, then $X \to Y$ is an epimorphism,
  7. 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 $F(X)$,
  8. 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 $X_i \to Y_{\alpha(i)}$.

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.

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.

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.

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.

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.

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.

Proof of (7). 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 $E = X$.

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 $j = \alpha(i)$. $\square$

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 (7). We say $X$ is Galois if equality holds. Equivalently, $X$ is Galois if it is connected and $\text{Aut}(X)$ acts transitively on $F(X)$.

Lemma 49.3.8. 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 $Y \to X$.

Proof. We will use the results of Lemma 49.3.7 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.

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. $\square$

Here is a key lemma.

Lemma 49.3.9. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 49.3.5. For any connected $X$ in $\mathcal{C}$ the action of $G$ on $F(X)$ is transitive.

Proof. We will use the results of Lemma 49.3.7 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 4.21.1) 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 49.3.8 applied to a connected component of $X_{i_1} \times X_{i_2}$. Then $Y \cong X_i$ for some $i \in I$ and $i \geq i_1$, $i \geq I_2$.

We claim that the functor $F$ is isomorphic to the functor $F'$ which sends $X$ to $$ F'(X) = \mathop{\rm colim}\nolimits_I \mathop{\rm Mor}\nolimits_\mathcal{C}(X_i, X) $$ 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 (7) 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 49.3.8. 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.

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 $$ A = \mathop{\rm lim}\nolimits_I A_i $$ 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 4.21.7.

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. $\square$

Proposition 49.3.10. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 49.3.5. The functor $F : \mathcal{C} \to \textit{Finite-}G\textit{-Sets}$ (49.3.5.1) an equivalence.

Proof. We will use the results of Lemma 49.3.7 without further mention. In particular we know the functor is faithful. By Lemma 49.3.9 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.

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 $$ \mathop{\rm Ker}(G \longrightarrow \prod\nolimits_i \text{Aut}(F(X_i))) $$ 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 49.3.8. 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. $\square$

Lemma 49.3.11. 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.

Proof. By Proposition 49.3.10 and Lemma 49.3.3 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 49.3.4. 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. $\square$

  1. When we discuss the pro-étale fundamental group the general case will be of interest.

The code snippet corresponding to this tag is a part of the file pione.tex and is located in lines 124–703 (see updates for more information).

\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 is 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 and is exact.
\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}

Comments (2)

Comment #2346 by Xuezha on January 10, 2017 a 10:45 pm UTC

In the definition of profinite completion of G, there's a small typo: it should be finite index. (n was missing)

Comment #2415 by Johan (site) on February 17, 2017 a 1:47 pm UTC

Thanks. Fixed here. If you want your name listed, please give both parts of your name.

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