# Map $\mathrm{Aut}_X(X_j)\to\mathrm{Aut}_X(X_i)$ in defining fundamental group of a scheme

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?

• 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 $X_i = X_j/ Aut_{X_i}(X_j)$ Jul 26, 2019 at 1:29

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 $$\tau$$.