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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.

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Dear Justin, You have an exact sequence of groups with the right hand term being abelian, in fact procyclic, and you are trying to understand the associated right exact sequence of abelianizations. This is just an elementary exercise in group theory; if you do it, you will find that the map $\pi_1(\overline{X})^{ab} \to \pi_1(X)^{ab}$ has image equal to the Galois coinvariants of the source. Incidentally, you might find it profitable to rewrite your abelianized sequence in terms of etale (co)homology; you will then find that it is part of the Hochschild--Serre spectral sequence that ... –  Emerton Nov 29 '12 at 4:00
    
... compares etale cohomology of $\overline{X}$ to that of $X$. Regards, –  Emerton Nov 29 '12 at 4:19
    
Justin, for your final question you are tacitly assuming that $\pi_1({\rm{Pic}}^1_{X/k})$ is abelian, but is 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 $\pi_1(A)$ is never abelian. Indeed, for $n$ coprime to char($k$) and $|A(k)|$, $[n]:A \rightarrow A$ is a nontrivial connected finite etale cover with no nontrivial automorphisms! –  user28172 Nov 29 '12 at 4:53
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1 Answer

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.

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The diagram chase proves surjectivity of the map of interest at the end, but to prove injectivity in this way it seems you need to know the 2nd right exact sequence displayed is actually short exact (i.e., injective on the left side). This is a question concerning abelian varieties, or more generally smooth proper $\mathbf{F}_q$-schemes with a rational point (such as ${\rm{Pic}}^1_{X/\mathbf{F}_q}$. –  user28172 Nov 29 '12 at 11:07
    
To fix the argument, you just need to check that both right exact sequences have the same kernel term at the left (under the isomorphism between left sides), and this follows from the concrete description indicated in Emerton's comment. –  user28172 Nov 29 '12 at 14:36
    
Yes, that makes sense. –  Will Sawin Nov 29 '12 at 16:36
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