Compare with [Definition 7.2.4, BS].

Lemma 58.3.9. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 58.3.5. For any connected $X$ in $\mathcal{C}$ the action of $G$ on $F(X)$ is transitive.

Proof. We will use the results of Lemma 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

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

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 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

$\xymatrix{ X_ i \ar[d]_ a \ar[r]_{f_{ii'}} & X_{i'} \ar[d]^{h_{ii'}(a)} \\ X_ i \ar[r]^{f_{ii'}} & X_{i'} }$

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. $\square$

There are also:

• 7 comment(s) on Section 58.3: Galois categories

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).