We have the exact sequence:

$\pi_1(\bar{X})^{ab} \to \pi_1(X)^{ab} \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to 0$

and the exact sequence

$\pi_1(\operatorname{Pic}^1 \bar{X})^{ab} \to \pi_1(\operatorname{Pic}^1 X)^{ab} \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to 0$.

The map $X \to Pic^1 X$ induces a map $\pi_1(X)^{ab} \to \pi_1(\operatorname{Pic} X)^{ab}$. Similarly we have a map $\pi_1(\bar{X})^{ab} \to \pi_1(\operatorname{Pic} \bar{X})^{ab}$. These maps form a nice big commutative diagram connecting the two exact sequences.

The map $\pi_1(\bar{X})^{ab} \to \pi_1(\operatorname{Pic} \bar{X})^{ab}$ and $\\operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) $ are both isomorphisms, so the middle map $\pi_1(X)^{ab} \to \pi_1(\operatorname{Pic} X)^{ab}$ is an isomorphism by homological algebra.

We have the exact sequence:

$\pi_1(\bar{X})^{ab} \to \pi_1(X)^{ab} \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to 0$

and the exact sequence

$\pi_1(\operatorname{Pic}^1 \bar{X})^{ab} \to \pi_1(\operatorname{Pic}^1 X)^{ab} \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to 0$.

The map $X \to \operatorname{Pic^1}X$ induces a map $\pi_1(X)^{ab} \to \pi_1(\operatorname{Pic} X)^{ab}$. Similarly we have a map $\pi_1(\bar{X})^{ab} \to \pi_1(\operatorname{Pic} \bar{X})^{ab}$. These maps form a nice big commutative diagram connecting the two exact sequences.

The map $\pi_1(\bar{X})^{ab} \to \pi_1(\operatorname{Pic} \bar{X})^{ab}$ and $\operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) \to \operatorname{Gal}(\bar{\mathbb F}_q/\mathbb F_q) $ are both isomorphisms, so the middle map $\pi_1(X)^{ab} \to \pi_1(\operatorname{Pic} X)^{ab}$ is an isomorphism by homological algebra.