Class field theory for function fields and a curious statement Let $X_0$ be a smooth curve over a finite field $\mathbb{F}_q$, and let $X$ be the base-change to the algebraic closure. I read that, according to class field theory in function fields, 
"the image of $\pi_1(X, a)$ in the abelianized locally compact Weil group of $X_0$ is canonically isomorphic to the divisor class group $Pic^0$ of divisors of degree zero, which are rational over $\mathbb{F}_q$"
(Source: Kiehl and Weissauer's book on the Weil conjectures. Deligne makes this assertion in Weil II.)
What is the justification for this? I thought "class field theory for function fields" would give an isomorphism of the abelianized absolute Galois group of the function field of $X_0$ and the group of idele classes of $X_0$. It's not obvious to me how this relates to the statement claimed.
 A: In brief, a (local or global) Weil group intends to accommodate abelian extensions of all extensions of a (global or local) field, not "only" the abelian extensions of a fixed base. Thus, smashing it down to make an assertion about extensions of a given (global or local) field requires the intervention of suitable abelianization. 
Talking about $Pic^0$ is just the algebraic geometry way of talking about a class group, with some rationality questions intervening. The $\pi_1(X,a)$ is the geometric 'absolute Galois group', so includes some things that wouldn't be rational over the fixed base.
In short, as in the question, indeed, the classfield theory over a fixed (e.g., global) base can be formulated in terms of the idele class group of that base. The fancier assertion involving Weil group and $\pi_1$ and such is saying the same thing, but is obliged to mash things down, to drop/forget some things to make the same assertion.
The fact that Weil groups (or, Weil-Deligne, etc.) exist and behave reasonably is a fairly subtle fact about the simultaneous classfield theory of all extensions of a given base, and has content. However, that content adds little to the assertions about extensions of a fixed base.
"Class formations" and such axiomatize the cohomological formulation of classfield theory (wherein the reciprocity laws are cup products...), and give us "Weil groups", eventually.
A: There is a standard exact sequence $1 \to \pi_1(X,a) \to \pi_1(X_0,a) \to 
Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q) \to 1,$  given by covariant functoriality of $\pi_1$.
How can one think about $\pi_1(X_0,a)$?  Well, it is a quotient of the absolute Galois group of the function field $K(X_0)$ of $X_0$, and can be thought of as the
Galois group of the maximal Galois extension of $K(X_0)$ which is everywhere
unramified.  Thus $\pi_1(X_0,a)^{ab}$ is the Galois group of the maximal abelian extension of $K(X_0)$ which is everywhere unramified.
Class field theory gives a description of this everywhere unramified abelian extension in terms of class groups/ideles.  When you sort it out in this context, you will find that $\pi_1(X_0,a)^{ab}$ gets identified with the profinite completion of the $\mathbb{F}_q$-rational points of $Pic(X_0)$.    Concretely,
$Pic^0(X_0)(\mathbb F_q)$ is finite, and the degree map induces an exact sequence
$$0 \to Pic^0(X_0)(\mathbb F_q) \to Pic(X_0)(\mathbb F_q) \to \mathbb Z \to 0.$$
Taking profinite completions just replaces $\mathbb Z$ by $\hat{\mathbb Z}$.
Now going back to the exact sequence at the beginning of the answer, since
$Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$ is abelian, there is an induced
sequence
$$\pi_1(X,a) \to \pi_1(X_0,a)^{ab} \to Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)
\to 1.$$ 
Since $Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$ is a copy of $\hat{\mathbb{Z}}$, generated by $Frob_q$, we may rewrite this as 
$$\pi_1(X,a) \to \pi_1(X_0,a)^{ab} \to \hat{\mathbb Z} \to 0.$$
Now one checks that under the identification of $\pi_1(X_0,a)^{ab}$ 
and the profinite completion of $Pic(X_0)(\mathbb F_q)$, the map to
$\hat{\mathbb Z}$ of this last exact sequence is the same as the degree map.
Comparing exact sequences, we see that the image of $\pi_1(X,a)$ in
$\pi_1(X_0,a)^{ab}$ is equal to $Pic^0(X_0)(\mathbb{F}_q)$, which is what
is being asserted.
[Note: The reason that Weil groups are invoked are that, in this context,
passing from Galois groups to Weil groups simply replaces $\hat{\mathbb Z}$
by $\mathbb Z$.  So $\pi_1(X_0,a)^{ab}$ is a Galois group, while 
$Pic(X_0)(\mathbb{F}_q)$ is the corresponding Weil group.  The latter has
a surjection to $\mathbb Z$, while the former has a surjection to 
$\hat{\mathbb Z}$.  If you compute the quotient of the idele class group pertaining to the maximal abelian everywhere unramified extension, you will
naturally get $Pic(X_0)(\mathbb F_q)$, i.e. the Weil group.  In the function
field case the Artin reciprocity map from quotients of the idele class group
to abelian Galois groups is injective, with image being the Weil group.
By constrast, in the number field case the reciprocity map is surjective,
but has a kernel, coming from the contribution at the archimedean primes.
A careful discussion of Weil groups, including the distinction between
the number field and function field cases, can be found in Tate's Numer
theoretic background article in the Corvalis proceedings.]
