Let $X$ be a smooth, projective, and geometrically connected curve over a finite field $\mathbb{F}_q$ and fix a geometric point $\overline{x} : \text{Spec } \overline{\mathbb{F}_q} \to X$. Then there is a fundamental exact sequence $$1 \to \pi_1(\overline{X},\overline{x}) \to \pi_1(X,\overline{x}) \to \text{Gal }(\overline{\mathbb{F}_q}/\mathbb{F}_q) \to 1,$$ where we wrote $\overline{X} = X \times \overline{\mathbb{F}_q}$. Passing to abelianizations gives us an exact sequence $$\pi_1(\overline{X},\overline{x})^{\text{ab}} \to \pi_1(X,\overline{x})^{\text{ab}} \to \text{Gal }(\overline{\mathbb{F}_q}/\mathbb{F}_q) \to 1.$$ I don't think this sequence is exact on the left, since by class field theory the kernel of $\pi_1(X,\overline{x})^{\text{ab}} \to \text{Gal }(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ is finite. So what is the kernel of $\pi_1(\overline{X},\overline{x})^{\text{ab}} \to \pi_1(X,\overline{x})^{\text{ab}}$?

Also, a related question: the Abel-Jacobi map $\overline{X} \to \text{Pic}^1 \ \overline{X}$ induces an isomorphism $\pi_1(\overline{X})^{\text{ab}} \to \pi_1(\text{Pic}^1 \ \overline{X})$. What about $\pi_1(X)^{\text{ab}} \to \pi_1(\text{Pic}^1 \ X)$?

Edit: according to nosr, $\pi_1(\text{Pic}^1 \ X)$ is not necessarily abelian. So the map goes $\pi_1(X)^{\text{ab}} \to \pi_1(\text{Pic}^1 \ X)^{\text{ab}}$, and I am still curious whether it is an isomorphism.

false. More specifically, with $k := \mathbf{F}_q$, although ${\rm{Pic}}^1_{X/k}$ admits a structure of abelian variety (since $k$ is finite and ${\rm{Pic}}^1_{X/k}$ is a torsor for the abelian variety ${\rm{Pic}}^0_{X/k}$), for an abelian variety $A \ne 0$ over a finite field $k$ the group $\pi_1(A)$ isneverabelian. Indeed, for $n$ coprime to char($k$) and $|A(k)|$, $[n]:A \rightarrow A$ is a nontrivial connected finite etale cover withnonontrivial automorphisms! – user28172 Nov 29 '12 at 4:53