of $\bar x$ in $C'$. Each geometric point $\bar y$ of $\wtl C$ belongs to exactly one geometric connected component of $\wtl C$, and we denote by $\O_{\bar y}$ the strict henselisation of the local ring of $\bar y$ in its geometric connected component. By the construction of $C'$ we have an exact sequence $$0\lrar \O_{\bar x}\lrar k(\bar x)\times
\prod_{\nu'(\bar y)=\bar x}\O_{\bar y}\lrar 
\prod_{\nu'(\bar y)=\bar x} k(\bar y)\lrar 0,$$ where $\O_{\bar y}\to k(\bar y)$ is the reduction modulo the maximal ideal of $\O_{\bar y}$, and $k(\bar x)\to k(\bar y)$ is the multiplication by $-1$. We obtain an exact sequence of abelian groups $$1\lrar \O_{\bar x}^*\lrar k(\bar x)^*\times
\prod_{\nu'(\bar y)=\bar x}\O_{\bar y}^*\lrar 
\prod_{\nu'(\bar y)=\bar x} k(\bar y)^*\lrar 1.$$ Using [@EC Cor. II.3.5 (a), (c)] one sees that this is the sequence of stalks of (\[ee1\]) at $\bar x$, so that $(i)$ is proved.

To prove $(ii)$ consider the exact sequence $$0\to{\mathbb{G}}_{m,C}\to \nu''_*{\mathbb{G}}_{m,C'}\to \nu''_*{\mathbb{G}}_{m,C'}/{\mathbb{G}}_{m,C}
\to 0.$$ Since $\nu''$ is an isomorphism away from $i(\Pi)$, the restriction of the sheaf $\nu''_*{\mathbb{G}}_{m,C'}/{\mathbb{G}}_{m,C}$ to $C\setminus i(\Pi)$ is zero, hence $\nu''_*{\mathbb{G}}_{m,C'}/{\mathbb{G}}_{m,C}=i_*{\mcal{U}}$ for some sheaf ${\mcal{U}}$ on $\Pi$. To see that ${\mcal{U}}$ is a unipotent group scheme it is enough to check the stalks at geometric points. 