Lemma 58.3.8. 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 58.3.7 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. $\square$

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).