I am looking at Milne's etale cohomology where the fundamental group is defined; fixing some geometric point $\overline x\to X$ we have that the functor $\mathbf{FEt}/X\to\mathbf{Set}$ given by $\hom_X(\overline x,-)$ is strictly pro-representable by some system $\tilde X=(X_i,\phi_{ij})$, where we may assume $X_i$ are all Galois over $X$ (in case this matters). The following is then stated:

Given $j\ge i$ we can define a map $\psi_{ij}:\mathrm{Aut}_X(X_j)\to\mathrm{Aut}_X(X_i)$ by requiring that $\psi_{ij}(\sigma)f_i=\phi_{ij}\sigma f_j$. Then we define $\pi_1(X,\overline x)=\varprojlim\mathrm{Aut}_X(X_i)$.

I see why there is at most one map $\psi_{ij}$ satisfying the given condition, but I don't see why such a map exists in the first place; can anybody explain?

  • 1
    $\begingroup$ Did you look at his example $X =X_1= \Bbb{C}^*, \overline{x} = 1$, each $X_n$ is another copy of $ \Bbb{C}^*$ such that the map $X_{nm} \to X_m$ is $x \mapsto x^n$ then $Aut_X(X_n)= \langle e^{2i \pi /n}\rangle\cong \Bbb{Z/nZ}, \pi_1(X,1)\cong \varprojlim \Bbb{Z/nZ}= \hat{\Bbb{Z}}$. So in your $j \ge i$, $\ge$ would be a partial order saying $X_i = X_j/ Aut_{X_i}(X_j)$ $\endgroup$
    – reuns
    Jul 26, 2019 at 1:29

1 Answer 1


Consider Proposition 3.2.8 of Lei Fu's Étale Cohomology Theory.

Proposition 3.2.8. Let $(S, \gamma)$ be a pointed connected Noetherian scheme, $X$ a connected étale covering space of $S$, $X(\gamma)$ the set of geometric points in $X$ lying above $\gamma$, and $G = \text{Aut}(X/S)^\circ$. The following conditions are equivalent:

(i) $X/G \cong S$, that is, $X$ is a Galois covering of $S$.

(ii) $G$ acts transitively on $X(\gamma)$.

(iii) $G$ and $X(\gamma)$ have the same number of element.

By (ii), if $(Y, \bar y)/(X, \bar x)$ is Galois, then for any two geometric points $y_1$ and $y_2$ over $\bar x$, there is one and only one automorphism $\sigma$ of $Y/X$ such that $\sigma(y_1)=\sigma(y_2)$. Let $(Y', \bar y')/(X, \bar x)$ be another Galois covering such that we have an $X$-morphism $f: Y' \to Y$ mapping $\bar y'$ to $\bar y$. For any $\sigma$ in $\text{Aut}(Y'/X)$, both $\bar y$ and $f(\sigma(\bar y'))$ are points in $Y$ over $\bar x$. So there exists a unique automorphism $\tau$ of $Y/X$ such that $\tau(\bar y)=f(\sigma(\bar y'))$. We thus get the homomorphism $\text{Aut}(Y'/X) \to \text{Aut}(Y/X)$ by mapping $\sigma$ to $\tau$.


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