The Stacks project

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. $\square$


Comments (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BN2. Beware of the difference between the letter 'O' and the digit '0'.