A half-circle configuration loop traces an elementary half twist

Compute the round relative path. The addition law [L4] and Euler's identity [L3] give $v(0)=-h$, $v(1)=h$, and $v(t)=-h\exp (i\pi t)$. For $0<t<1$, [L3] and [L5] give $\mathrm{Im}v(t)=-h\sin (\pi t)<0$, since $h>0$ and $0<\pi t<\pi$; the modulus formula gives $\mathrm{\mid }v(t)\mathrm{\mid }=h$. By [L6], $v$ is continuous. The piecewise formula [L2] gives $\mathrm{Im}\rho (t)=-2th<0$ for $0<t\le 1\mathrm{/}2$ and $\mathrm{Im}\rho (t)=2h(t-1)<0$ for $1\mathrm{/}2\le t<1$.

Check the unordered loop. The coordinates $x_j(t)$ are continuous by [L6], [L7], and [L9]. Their moving pair is distinct because $\mathrm{\mid }v(t)\mathrm{\mid }=h>0$ by [L1] and [L3]. Each moving point is at distance $h$ from $m_i$, so lies in $U_i\subset D^{\circ }$; every fixed $q_j$ lies in $D^{\circ }$ and, for $j\mathrm{\notin }\{i,i+1\}$, outside $U_i$ by [L1]. The tuple therefore lies in $F_n(D^{\circ })$ by [L8], and [L10] makes it continuous into that subspace. Its orbit is continuous by [L11]. Since $v(0)=-h$ and $v(1)=h$, the ordered tuple starts at $Q$ and ends with $q_i,q_{i+1}$ exchanged; both orbit endpoints are $[Q]$. Thus ${\alpha }_i$ is an interior based loop by [L12].

Interpolate in the lower half-plane. Set $w_s(t):=(1-s)\rho (t)+sv(t)$ for $s,t\in I$. This is jointly continuous by [L6], [L7], and [L9]. For $0<t<1$, both imaginary parts are strictly negative by step 1.1, so $w_s(t)\mathrm{\ne }0$; at $t=0,1$ the common values are $-h,+h$, also nonzero. The triangle inequality and [L2], [L3] give $\mathrm{\mid }{w}_s(t)\mathrm{\mid }\le (1-s)\mathrm{\mid }\rho (t)\mathrm{\mid }+s\mathrm{\mid }v(t)\mathrm{\mid }\le h<3h\mathrm{/}2$. The tuple with coordinates $x_{s,i}(t)=m_i+w_s(t)$, $x_{s,i+1}(t)=m_i-w_s(t)$, and $x_{s,j}(t)=q_j$ for the other labels stays collision-free and interior: the moving pair is distinct and stays in $U_i$, while all fixed points remain outside $U_i$ by [L1]. Its coordinate maps are continuous, and [L8]–[L10] give a continuous family in the ordered configuration subspace. Each slice is a geometric braid by [L13], and the family is a braid isotopy by [L17], since its bottom tuple is $Q$ and its top set is $Q$ for every $s$. At $s=0$ the braid is ${\sigma }_i$; at $s=1$ it is the round tuple of step 1.2.

Identify the exact traces. The ordered family in step 2.1 is continuous into $F_n(D^{\circ })$ by [L8]–[L10], so its quotient $H(s,t):=[(x_{s,1}(t),\dots ,x_{s,n}(t))]$ is jointly continuous by [L11]. Its endpoints satisfy $H(s,0)=H(s,1)=[Q]$ for every $s$, and every slice is an interior based motion by [L12]. The transpose $(t,u)\mapsto H(u,t)$ is a path homotopy rel endpoints under [L16]. Thus [L15] makes the traces of the two boundary loops braid-isotopic. By [L14], those traces are exactly ${\sigma }_i$ and the round tuple; the latter is the trace of ${\alpha }_i$ by step 1.2. This proves the example.