the Galois group $\Gamma_{k_v}$, and hence it intersects $C_0$ in a $k_v$-point. This contradiction proves (1).

By (1) the natural map $\operatorname{Br}(k)\to\operatorname{Br}(C)$ has a retraction, and so is injective, but it is also surjective by Corollary \[C1\]. This proves (2).

Let us prove (3). Let $G$ be a finite $k$-group scheme, and let $\T \to C$ be a torsor of $G$. Fix a $k$-point $P$ in $C$. The fibre $\T_P$ is then a $k$-torsor. The twist of $\T$ by $\T_P$ is the quotient of $\T\times_k \T_P$ by the diagonal action of $G$. This is a $C$-torsor of an inner form of $G$ such that the fibre at $P$ has a $k$-point, namely the quotient by $G$ of the diagonal in $\T_P\times_k \T_P$. Thus, twisting $\T$ by a $k$-torsor of $G$, and replacing $G$ by an inner form we can assume that $\T_P$ contains a $k$-point $Q$. Since all irreducible components of $\ovl C$ are homeomorphic to ${\mathbb{P}}^1_{\bar k}$, the torsor $\ovl\T\to \ovl C$ trivialises over each component of $\ovl C$. But $X(C)$ is a tree, and this implies that the torsor $\ovl\T\to \ovl C$ is trivial, that is, $\ovl \T\simeq (C\times_k G)\times_k\bar k$. The connected component of $\ovl\T$ that contains 