Galois stable elements of the Picard group of a curve and the rational divisors Let $C$ be a (smooth,proper) curve over a field $k$. Let $\operatorname{Div}_C(k)$ be the free abelian group generated by the closed points of $C/k$ and $k(C)^\times$ be the group of rational functions. We have an injective map:
$$i \colon \operatorname{Div}_C(k)/k(C)^\times \to \left(\operatorname{Div}_C(\overline{k})/\overline{k}(C)^\times\right)^{\operatorname{Gal}(\overline{k}/k)}$$
When is this map surjective? I think this is definitely true if the Brauer group $H^2(k,\mathbb G_m) = 0$, as it follows from the four-term exact sequence
$$0 \to \overline{k}^\times \to \overline{k}(C)^\times \to \operatorname{Div}_C(\overline{k}) \to \operatorname{Pic}_C(\overline{k})\to 0.$$
Q1. Are there other situations where this is true? Also, what are some examples where the map is not injective?
Q2. In particular, I saw a comment somewhere that if $k$ is a number field and $C$ has a point locally for every completion $k_v$, then the map is indeed surjective. Is this true and if so, how does one see it?
Q3. Is the map surjective if $C$ has a rational point?
 A: It is always injective, and the cokernel is the group of Brauer classes split by $k(C)$, see On the period-index probem in light of the section conjecture by J. Stix. Equivalently, those are the classes of Brauer-Severi varieties $P$ with a morphism $C\to P$.
If $k$ is a number field and $C$ has local points everywhere, it cannot map to a non-trivial Brauer-Severi variety (a Brauer-Severi variety with local points everywhere is trivial). Also, if C has a rational point it cannot map to a non-trivial Brauer-Severi variety.
A: A partial (I have a few questions at the end) proof to Q.2,3 communicated to me by Ananth Shankar:
Suppose $\mathscr L$ is a line bundle corresponding to a Galois stable element in $Pic_C(\overline{k})$ and assume also that it is very ample.
Let $f: C \to \mathbb P(H^0(C))$ be the induced map on global sections defined over $\overline{k}$. This map is in fact Galois stable by assumption and therefore we can descent to:
$$f: C \to B$$
where $B$ is, a priori, only a twist of projective space over $k$. However, if $C$ has a k-rational point, we can push it forward to get a rational point on $B$ which forces the twist to be trivial and then the pull back of a divisor gives a rational divisor on $C$ in the same class as $\mathscr L$.
Similarly, if $C$ has a $k_v$ rational point for every completion $v$ of k, then $B$ is the trivial twist for every $k_v$ and by the Hasse local-global theorem for twists of projective space, this forces $B$ to be projective space over $k$ and we conclude as before.
Q: What if $\mathscr L$ is not very ample? Is the cokernel of the map $i$ torsion-free? If so, we can deal with $\mathscr L$ not of degree $0$. Finally, what if $\mathscr L$ is degree 0
