tree, then $H^1(k,\operatorname{Pic}(\ovl C))=0$, so that the natural map $\operatorname{Br}(k)\to\operatorname{Br}(C)$ is surjective.

Since $k$ has characteristic 0, we have $H^n(k,U)=0$ for $n>0$. Thus (\[a\]) gives an isomorphism $H^1(k,\operatorname{Pic}(\ovl C))=H^1(k,\operatorname{Pic}(\ovl {C'}))$. We have $T=\{1\}$, hence $\operatorname{Pic}(\ovl {C'})= \operatorname{Pic}(\wtl C\times_k\bar k)$. The curve $\wtl C$ is the disjoint union of conics defined over finite extensions of $k$, thus the free abelian group $\operatorname{Pic}(\wtl C\times_k\bar k)$ has a natural $\Gamma_k$-stable ${\mathbb{Z}}$-basis. Hence $H^1(k,\operatorname{Pic}(\wtl C\times_k\bar k))=0$.

Weak approximation {#WA}
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Let $k$ be a number field. Recall that the étale Brauer–Manin set $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is the set of adelic points $(M_v)\in X({\mathbb{A}}_k)$ satisfying the following property: for any torsor $f:Y\to X$ of a finite $k$-group scheme $G$ there exists a $k$-torsor $Z$ of $G$ such that $(M_v)$ is the image of an adelic point in the Brauer–Manin set of $(Y\times_k Z)/G$. Here $G$ acts simultaneously on both factors, and the morphism $(Y\times_k Z)/G\to X$ is induced by $Y\to X$. It is clear that the étale Brauer–Manin obstruction is a functor from the category of varieties over $k$ to the category of sets. Note that $(Y\times_k Z)/G\to X$ is a torsor of an inner form of $G$, called the twist of $Y/X$ by $Z$, see [@Sk 