is mapped to $(z_1,z_2,\dots,z_r) \mapsto
(\zeta^{-i} z_2,\zeta^i z_1,\dots,z_r)$, and $t_0,s_3,\dots,
s_r$ are mapped to the successive transpositions of $\mathfrak{S}_r$ in that order. These generators provide a presentation of ${{\mathrm{B}}}(e,e,r)$ with the relations (2)-(5) of page , and with (1) replaced by $t_i t_{i+1} = t_j t_{j+1}$.

We notice that the slight change in the presentation is meaningless in monoid-theoretic terms, as both monoids are isomorphic under $t_i \mapsto t_{- i}$, but it is not in topological terms, as $t_1 t_0 t_1^{-1}$ is *not* homotopic to a detour from $\underline{b}$ to its image (see figure \[figt2plus\]).

Let $S_0 = \{ t_i,s_3,\dots,s_r \}$. The subgroup of ${{\mathrm{B}}}(e,e,r) = \pi_1(\mathcal{M}(e,r)/G(e,e,r)$ generated by $S_0$ is a parabolic subgroup in the sense of [@BMR], and can be naturally identified with the braid group on $r$ strands as the fundamental group of $\{ (z_1,\dots,z_r) \ | \ z_i \neq z_j, z_1+\dots+z_r = 0 \}/\mathfrak{S}_r$, with base point $(-\zeta^{-i}(b_1+b_2+\dots + b_r),b_2,\dots,b_r)$, in such a way that the elements of $S_0$ are identified with positive detours.

The parabolic subgroup of $G(e,e,r)$ defined as the fixer of $(\zeta^i, 1,1 , \dots, 1)$ is obviously conjugated to the one fixing $(1,1,\dots,1)$, the latter being the natural $\mathfrak{S}_r  \subset G(e,e,r)$. We 