The Stacks project

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. $\square$

Proof. Denote $F_ G$ the functor (58.3.2.1). 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 58.3.1 and the universal property of profinite completion there is an induced continuous group homomorphism

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

of $\textit{Finite-}G\textit{-Sets}$ commutes with the action of $\gamma $. Hence

Thus we see that $g = g_0$ works. $\square$

Proof. Let $X$ be a nonempty object of $\textit{Finite-}G\textit{-Sets}$. The diagram

is cocartesian. Hence we conclude that $F(X)$ is nonempty. Let $U \subset G$ be an open, normal subgroup with finite index. Observe that

where the summand corresponding to $gU$ corresponds to the orbit of $(eU, gU)$ on the left hand side. Then we see that

Hence $|F(G/U)| = |G/U|$ as $F(G/U)$ is nonempty. Thus we see that

is nonempty (Categories, Lemma 4.21.7). Pick $\gamma = (\gamma _ U)$ an element in this limit. Denote $F_ G$ the functor (58.3.2.1). We can identify $F_ G$ with the functor

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

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

and the proof is complete. $\square$

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

Proof. We will use the results of Lemma 58.3.7 without further mention. Let $n = |F(X)|$. Consider $X^ n$ endowed with its natural action of $S_ n$. Let

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

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

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. $\square$

Proof. We will use the results of Lemma 58.3.7 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 58.3.8 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

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 58.3.8. 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

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

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. $\square$

Proof. We will use the results of Lemma 58.3.7 without further mention. In particular we know the functor is faithful. By Lemma 58.3.9 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

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 58.3.8. 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

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. $\square$

Proof. By Proposition 58.3.10 and Lemma 58.3.3 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 58.3.4. 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. $\square$