I am reading Milne's lectures on etale cohomology. In the note he states:
Proposition 6.16 Assume $X$ is connected, and let $\bar{x}$ be a geometric point of $X$. The map $\mathcal{F} \mapsto \mathcal{F}_{\bar{x}}$ defines an equivalence between the category of locally constant sheaves of sets (resp. abelian groups) with finite stalks on $X$ and the category of finite $\pi_1(X, \bar{x})$-sets (resp. modules).
PROOF. In fact, one shows that, if $Z \rightarrow X$ is a finite étale map, then the sheaf $\mathcal{F}_Z$ is locally constant with finite stalks, and every such sheaf is of this form for some $Z$; more precisely, $Z \mapsto \mathcal{F}_Z$ defines an equivalence from FEt/ $X$ to the category of locally constant sheaves on $X_{\text {et }}$ with finite stalks. Now we can apply Theorem 3.1 to prove the proposition.
Where the sheaf is : An $X$-scheme $Z$ defines a contravariant functor: $$ \mathcal{F}: \text { Et } / X \rightarrow \text { Sets, } \quad \mathcal{F}(U)=\operatorname{Hom}_X(U, Z) . $$
Does he mean that the etale stalks of the sheaf is $\operatorname{Hom}_X(\bar{x}, Z)$? Because the theorem states
Define $F:$ FEt $/ X \rightarrow$ Sets to be the functor sending $(Y, \pi)$ to the set of $\bar{x}$-valued points of $Y$ lying over $x$, so $F(Y)=\operatorname{Hom}_X(\bar{x}, Y)$.
THEOREM 3.1 The functor $Y \mapsto F(Y)$ defines an equivalence from the category of finite étale coverings of $X$ to the category of finite discrete $\pi_1(X, \bar{x})$-sets.
I don't see why this should be ture.
There are two things to prove.
Suppose we have two $X$-maps $f_i: U_i \rightarrow Z$ with $X$-maps $x_i: \overline{x} \rightarrow U_i$ such that $f_1 \circ x_1=f_2 \circ x_2$.
Then we have a map $(f_1,f_2): U_1 \times_X U_2 \rightarrow Z \times_X Z$, and the inverse image $W$ of the diagonal is an open subscheme of $U_1 \times_X U_2$ (hence is étale over $X$) and “contains” $(x_1,x_2)$. In particular, it’s non-empty, so that $f_1$ and $f_2$ are the same on the stalk of $\overline{x}$.
It follows that $\operatorname{Hom}_X(-,Z)_{\overline{x}} \rightarrow \operatorname{Hom}_X(\overline{x},Z)$ is injective.
To show that it is surjective, one needs to show that any map $\overline{x} \rightarrow Z$ extends to some $g: U \rightarrow Z$, for some $h: \overline{x} \rightarrow U$.
Take $U=Z$ and $g=\mathrm{id}$, $h$ being the given map $\overline{x} \rightarrow Z$.