Far commutativity of elementary geometric half twists

(a), ${\mathrm{\Sigma }}_{ij}$ is a braid. Each motion of ${\mathrm{\Sigma }}_{ij}$ is either the constant $q_k$ or one of $m_i\pm \rho$, $m_j\pm \rho$, hence continuous with values in $U_i$ respectively $U_j$, and $U_i\cup U_j\subseteq D^{\circ }$ by [F2]; the only nonconstant pairs are $\{i,i+1\}$, which are antipodal about $m_i$ and therefore distinct because $\rho (t)\mathrm{\ne }0$ for every $t$ by [F2], and $\{j,j+1\}$, likewise antipodal about $m_j$; the two supports are disjoint and the remaining motions are the constant $q_k$ with $q_k\mathrm{\notin }{U}_i\cup U_j$, so all $n$ motions are pairwise distinct at every height; the bottom values are $m_i+\rho (0)=m_i+(-h,0)=q_i$, $m_i-\rho (0)=q_{i+1}$ and likewise for $j$, the remaining values being the base points themselves; the top values are $m_i+\rho (1)=q_{i+1}$, $m_i-\rho (1)=q_i$ and likewise for $j$, so the top values run through $\{q_1,\dots ,q_n\}$ and the endpoint permutation of ${\mathrm{\Sigma }}_{ij}$ is the product of the transpositions of $i,i+1$ and of $j,j+1$.

The interpolation of the two time windows. For $s\in I$ put ${\alpha }_s(t):=0$ for $t\le \frac{1-s}{2}$ and ${\alpha }_s(t):=\frac{t-(1-s)\mathrm{/}2}{(1+s)\mathrm{/}2}$ for $t\ge \frac{1-s}{2}$, and ${\beta }_s(t):=\frac{t}{(1+s)\mathrm{/}2}$ for $t\le \frac{1+s}{2}$, ${\beta }_s(t):=1$ for $t\ge \frac{1+s}{2}$; the two branches of each definition agree at the switch point, so ${\alpha }_s,{\beta }_s$ are continuous, nondecreasing and map $I$ onto $I$ with ${\alpha }_s(0)={\beta }_s(0)=0$ and ${\alpha }_s(1)={\beta }_s(1)=1$, the map $(s,t)\mapsto ({\alpha }_s(t),{\beta }_s(t))$ is jointly continuous because the switch points depend continuously on $s$ and the two branches agree there, and ${\alpha }_0(t)=\max (0,2t-1)$, ${\beta }_0(t)=\min (2t,1)$, ${\alpha }_1={\beta }_1={\mathrm{id}}_I$.

Pasting two isotopies. If $Z$ is a braid isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }}$ and $W$ one from ${\beta }^{\mathrm{\prime }}$ to ${\beta }^{\mathrm{\prime }\mathrm{\prime }}$, then $U_j(s,t):=Z_j(2s,t)$ for $s\le \frac{1}{2}$ and $U_j(s,t):=W_j(2s-1,t)$ for $s\ge \frac{1}{2}$ is jointly continuous by [L6] because the branches agree at $s=\frac{1}{2}$ and the two closed halves of the square cover it, every slice of $U$ is a slice of $Z$ or of $W$ and hence a braid based at $Q$ by [F4], and its boundary slices are $\beta$ and ${\beta }^{\mathrm{\prime }\mathrm{\prime }}$; so braid isotopy is transitive.

The isotopy. Define $Z_k(s,t):=m_i+\rho ({\alpha }_s(t))$ for $k=i$, $Z_k(s,t):=m_i-\rho ({\alpha }_s(t))$ for $k=i+1$, $Z_k(s,t):=m_j+\rho ({\beta }_s(t))$ for $k=j$, $Z_k(s,t):=m_j-\rho ({\beta }_s(t))$ for $k=j+1$, and $Z_k(s,t):=q_k$ otherwise. Every $Z_k$ is a composite of jointly continuous maps by step 1.2 and [F2], hence jointly continuous; for fixed $s$ the tuple $Z(s,\cdot )$ is collision-free and takes the base values at $t=0$ and the setwise base values at $t=1$ by the same computations as in step 1.1, with ${\alpha }_s,{\beta }_s$ in place of the identity: at $t=0$ one has ${\alpha }_s(0)={\beta }_s(0)=0$ so the four moving labels sit at $q_i,q_{i+1},q_j,q_{j+1}$, while at $t=1$ one has ${\alpha }_s(1)={\beta }_s(1)=1$ so they sit at the same four points with the two neighbouring labels interchanged.

The two ends of the isotopy. At $s=0$ the formulas of step 1.2 give $Z_k(0,t)=m_i+\rho (\max (0,2t-1))$ for $k=i$ and $Z_k(0,t)=m_i-\rho (\max (0,2t-1))$ for $k=i+1$, which is the motion of ${\sigma }_i$ run during the second half of the height interval and held at $q_i$ respectively $q_{i+1}$ during the first half, while $Z_k(0,t)=m_j\pm \rho (\min (2t,1))$ for the labels $k=j,j+1$ runs ${\sigma }_j$ during the first half and holds it at the swapped base points during the second half, and all other labels are constant; comparing with the stacking formula $(\gamma \star \beta {)}_k(t)=z_k(2t)$ for $t\le \frac{1}{2}$, $=w_{\pi (\beta )(k)}(2t-1)$ for $t\ge \frac{1}{2}$ of [F3], with $\beta ={\sigma }_j$, $\gamma ={\sigma }_i$ and $\pi ({\sigma }_j)$ the transposition of $j,j+1$, shows $Z(0,\cdot )={\sigma }_i\star {\sigma }_j$. At $s=1$ one has ${\alpha }_1={\beta }_1={\mathrm{id}}_I$ by step 1.2 and step 2.1, so $Z(1,\cdot )={\mathrm{\Sigma }}_{ij}$ by the definitions of step 1.1.

(a), first stacking. Step 2.1 exhibits $Z$ as a tuple of jointly continuous maps whose every slice is a braid based at $Q$, and step 3.1 identifies its boundary slices as ${\sigma }_i\star {\sigma }_j$ and ${\mathrm{\Sigma }}_{ij}$; hence $Z$ is a braid isotopy from ${\sigma }_i\star {\sigma }_j$ to ${\mathrm{\Sigma }}_{ij}$ in the sense of [F4], that is ${\sigma }_i\star {\sigma }_j\sim {\mathrm{\Sigma }}_{ij}$.

(a), second stacking. Interchanging the roles of the two pairs, that is replacing $({\alpha }_s,i,U_i)$ by $({\beta }_s,j,U_j)$ and conversely throughout steps 1.2, 2.1 and 3.1, yields in the same way a braid isotopy whose first boundary slice is ${\sigma }_j\star {\sigma }_i$, the pair $\{i,i+1\}$ now executing its half twist during the second half of the height interval and the pair $\{j,j+1\}$ during the first, and whose second boundary slice is again ${\mathrm{\Sigma }}_{ij}$; hence ${\sigma }_j\star {\sigma }_i\sim {\mathrm{\Sigma }}_{ij}$.

Steps 4.1 and 4.2 give ${\sigma }_i\star {\sigma }_j\sim {\mathrm{\Sigma }}_{ij}\sim {\sigma }_j\star {\sigma }_i$, so transitivity of braid isotopy in step 1.3 yields ${\sigma }_i\star {\sigma }_j\sim {\sigma }_j\star {\sigma }_i$; passing to isotopy classes with [F5] gives $[{\sigma }_i][{\sigma }_j]=[{\sigma }_i\star {\sigma }_j]=[{\sigma }_j\star {\sigma }_i]=[{\sigma }_j][{\sigma }_i]$, which is (b), while (a) is steps 1.1, 4.1 and 4.2. ∎