in $(\ref{ee3})$. Let $\Lambda=\coprod_\lambda \operatorname{Spec}(k(\lambda))$ be the decomposition into the disjoint union of connected components, so that $\wtl C=\coprod_\lambda \wtl C_\lambda$, where $\wtl C_\lambda$ is a geometrically integral, smooth, projective curve over the field $k(\lambda)$. Then there is an exact sequence $$0\lrar \operatorname{Br}(C)\lrar \operatorname{Br}(\Pi)\oplus\bigoplus_{\lambda\in\Lambda}
\operatorname{Br}(\wtl C_\lambda) \lrar \operatorname{Br}(\Psi),
\label{ee4}$$ where the maps are the composition of canonical maps $$\operatorname{Br}(\wtl C_\lambda)\to \operatorname{Br}(\wtl C_\lambda\cap \Psi)
\to \operatorname{Br}(\Psi),$$ and the opposite of the restriction map $\operatorname{Br}(\Pi)\to \operatorname{Br}(\Psi)$.

Let $\pi$ range over the irreducible components of $\Pi$, so that ${\mcal{U}}=\prod_{\pi} i_{\pi*}(U_\pi)$, where $U_\pi$ is a commutative unipotent group over the field $k(\pi)$. Since $i_*$ is an exact functor [@EC Cor. II.3.6], we have $H^n(C,i_*{\mcal{U}})=H^n(\Pi,{\mcal{U}})=\prod_\pi H^n(k(\pi),U_\pi)$. The field $k$ has characteristic 0, and it is well known that this implies that any commutative unipotent group has zero cohomology in degree $n>0$. (Such a group has a composition series with factors ${\mathbb{G}}_a$, and $H^n(k,{\mathbb{G}}_a)=0$ for any $n>0$, see [@CL X, Prop. 1].) Thus the long exact sequence of cohomology groups associated to (\[ee2\]) gives rise to an isomorphism $\operatorname{Br}(C)=H^2(C,{\mathbb{G}}_{m,C}) \tilde\lrar
H^2(C,\nu''_*{\mathbb{G}}_{m,C'})$. Since $\nu''$ is finite, the functor $\nu''_*$ is exact [@EC Cor. II.3.6], so we obtain an isomorphism $\operatorname{Br}(C)\tilde\lrar \operatorname{Br}(C')$. We now apply similar arguments to 