with respect to the unit circle ; they are thus mapped to a line if the original line passes through 0, and otherwise to a circle passing through the origin. The induced action of $G(e,e,r)$ is given by $s_1 : z \mapsto \frac{1}{z}$, $g_1 : z \mapsto \zeta^{-1} z$, $g_2 : z \mapsto \zeta z$. The images of $t_1$ and $t_2$ are depicted in figure \[figt1t2\]. The images of $t_2$ and of the positive detour from $(b_1,b_2)$ to $(\zeta^{-2} b_2,\zeta^2 b_1)$ are then clearly homotopic (see figure \[figt2detour\]), and the first coordinate of both paths is easily checked to remain in a simply connected region of $\C \setminus \{ 0 \}$. With the same argument, using the relation $t_{i+1} = t_i^{-1} t_0 t_1$ and possibly using $(z_1,z_2) \mapsto z_1/z_2$ instead of $(z_1,z_2)
\mapsto z_2/z_1$, we get that each $t_i$ is (homotopic to) the positive detour from $(b_1,b_2)$ to $(\zeta^{-i}b_2,\zeta^i b_1)$. We thus got the following

Let $\underline{b} = (b_1,\dots,b_r) \in \mathcal{M}(e,r)$ with $0 < b_1 \ll b_2 \ll \dots \ll b_r$. Then ${{\mathrm{B}}}(e,e,r) = \pi_1( \mathcal{M}(e,r)/G(e,e,r),\underline{b})$ is generated by braided reflections $t_0,\dots,t_{e-1},s_3,\dots,s_r$ which are positive detours from $\underline{b}$ to their images under the corresponding reflection. Under ${{\mathrm{B}}}(e,e,r) \onto
G(e,e,r)$, $t_i$ 