## 56.3 Galois categories

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

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

56.3.0.1
\begin{equation} \label{pione-equation-embedding-product} \text{Aut}(F) \longrightarrow \prod \nolimits _{X \in \mathop{\mathrm{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 (56.3.0.1). 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ 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 56.3.1. Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor. The map (56.3.0.1) identifies $\text{Aut}(F)$ with a closed subgroup of $\prod _{X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})} \text{Aut}(F(X))$. 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 56.3.2. Let $G$ be a topological group. An important example will be the forgetful functor

56.3.2.1
\begin{equation} \label{pione-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 56.2.1.

Let $G$ be a topological group. The profinite completion of $G$ will be the profinite group

$G^\wedge = \mathop{\mathrm{lim}}\nolimits _{U \subset G\text{ open, normal, finite index}} G/U$

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 56.3.3. Let $G$ be a topological group. The automorphism group of the functor (56.3.2.1) endowed with its profinite topology from Lemma 56.3.1 is the profinite completion of $G$.

Proof. Denote $F_ G$ the functor (56.3.2.1). 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 56.3.1 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. 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 56.3.4. 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 (56.3.2.1).

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

$\mathop{\mathrm{lim}}\nolimits _{U \subset G\text{ open, normal, finite idex}} F(G/U)$

is nonempty (Categories, Lemma 4.21.7). Pick $\gamma = (\gamma _ U)$ an element in this limit. Denote $F_ G$ the functor (56.3.2.1). We can identify $F_ G$ with the functor

$X \longmapsto \mathop{\mathrm{colim}}\nolimits _ U \mathop{Mor}\nolimits (G/U, X)$

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 56.3.5. 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. By Lemma 56.3.1 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

56.3.5.1
\begin{equation} \label{pione-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 $\text{Aut}(F)$.

The purpose of defining Galois categories is to single out those pairs $(\mathcal{C}, F)$ for which the functor (56.3.5.1) is an equivalence. Our definition of a Galois category is as follows.

Definition 56.3.6. 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and

4. $F$ reflects isomorphisms and is exact.

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

Lemma 56.3.7. 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 56.3.8. 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 56.3.7 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 56.3.9. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 56.3.5. For any connected $X$ in $\mathcal{C}$ the action of $G$ on $F(X)$ is transitive.

Proof. We will use the results of Lemma 56.3.7 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 56.3.8 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

$F'(X) = \mathop{\mathrm{colim}}\nolimits _ I \mathop{Mor}\nolimits _\mathcal {C}(X_ i, X)$

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 56.3.8. 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

$A = \mathop{\mathrm{lim}}\nolimits _ I A_ i$

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 56.3.10. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 56.3.5. The functor $F : \mathcal{C} \to \textit{Finite-}G\textit{-Sets}$ (56.3.5.1) an equivalence.

Proof. We will use the results of Lemma 56.3.7 without further mention. In particular we know the functor is faithful. By Lemma 56.3.9 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

$\mathop{\mathrm{Ker}}(G \longrightarrow \prod \nolimits _ i \text{Aut}(F(X_ i)))$

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 56.3.8. 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 56.3.11. 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 56.3.10 and Lemma 56.3.3 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 56.3.4. 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. $\square$

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

Comment #2346 by Xuezha on

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

Comment #2415 by on

Comment #4255 by awllower on

I do not understand what is the object $X^n$ in the proof of lemma 0BN2. It seems to come out of nowhere, endowed with a mysterious action of $S_n$. Could someone explain what is it? Thanks.

Comment #4256 by awllower on

I do not understand what is the object $X^n$ in the proof of lemma 0BN2. It seems to come out of nowhere, endowed with a mysterious action of $S_n$. Could someone explain what is it? Thanks.

Comment #4257 by awllower on

I am so sorry. I didn't think through enough and didn't realize that $X^n$ is just the product of $n$ copies of $X$, with the obvious action of $S_n$. I am sorry for any inconvenience thus caused. Also feel free to delete the comments to make the site cleaner if that is desired. Thanks in any case for this great site and effort.

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