Stacking of geometric braids is a well-defined associative operation on isotopy classes

(a) With $(\gamma \star \beta {)}_j$ given by the two-branch formula of the statement, the two branches agree at $t=\frac{1}{2}$, because the first gives $z_j(1)=q_{\pi (\beta )(j)}$ and the second gives $w_{\pi (\beta )(j)}(0)=q_{\pi (\beta )(j)}$ by [F1]; both branches are composites of continuous maps with the rescalings $t\mapsto 2t$ and $t\mapsto 2t-1$, hence continuous, and the two closed halves of $I$ cover $I$, so $t\mapsto (\gamma \star \beta {)}_j(t)$ is continuous by [L3] and [L5] and takes values in $D^{\circ }$; it is collision-free because on each half the tuple is the collision-free tuple of time-slices of a braid reparametrised by an injective continuous map and the halves meet only at $t=\frac{1}{2}$; the bottom values are $(\gamma \star \beta {)}_j(0)=z_j(0)=q_j$, and the top values are $(\gamma \star \beta {)}_j(1)=w_{\pi (\beta )(j)}(1)=q_{\pi (\gamma )(\pi (\beta )(j))}$, which run through the set $\{q_1,\dots ,q_n\}$; hence $\gamma \star \beta$ is a braid based at $Q$.

Braid isotopy is transitive: given an isotopy $Z$ from $\beta$ to ${\beta }^{\mathrm{\prime }}$ and an isotopy $W$ from ${\beta }^{\mathrm{\prime }}$ to ${\beta }^{\mathrm{\prime }\mathrm{\prime }}$, put $U_j(s,t):=Z_j(2s,t)$ for $s\le \frac{1}{2}$ and $U_j(s,t):=W_j(2s-1,t)$ for $s\ge \frac{1}{2}$; the branches agree at $s=\frac{1}{2}$ because both equal $z_j^{\mathrm{\prime }}(t)$, each is jointly continuous, and the two closed halves of the square cover it, so $U$ is jointly continuous by [L3]; every slice of $U$ is a slice of $Z$ or of $W$, hence a braid based at $Q$, and the boundary slices are $\beta$ and ${\beta }^{\mathrm{\prime }\mathrm{\prime }}$, so $U$ is an isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }\mathrm{\prime }}$.

(e) Let $Z$ be an isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }}$ and fix $j$; if $n=0$ there is no endpoint label and the unique endpoint permutation is fixed; otherwise, for each $k$ put $A_k:=\{s\in I:Z_j(s,1)=q_k\}$. The sets $A_k$ are pairwise disjoint and cover $I$, since every $Z_j(s,1)$ is one of the pairwise distinct points $q_k$ of [F1]; each $A_k$ is closed in $I$, being the preimage under the continuous map $s\mapsto Z_j(s,1)$ of the closed set $\{q_k\}$ by [L4]; each $A_k$ is also open in $I$, because for $s_0\in A_k$ for $n=1$ one already has $A_1=I$; for $n\ge 2$ the distance $\delta :=\min \{\mathrm{\mid }{q}_l-q_k\mathrm{\mid }:l\mathrm{\ne }k\}$ is positive by [F1] and continuity at $s_0$ yields a neighbourhood $V$ of $s_0$ in $I$ with $\mathrm{\mid }{Z}_j(s,1)-q_k\mathrm{\mid }<\delta \mathrm{/}2$ for $s\in V$, which forces $Z_j(s,1)=q_k$ and so $s\in A_k$. Hence the nonempty $A_k$ are pairwise disjoint nonempty clopen subsets of the connected space $I$ by [L4], and if some $A_k$ were nonempty and proper, its open complement ${\bigcup }_{l\mathrm{\ne }k}{A}_l$ would separate $I$ from it; so exactly one $A_k$ is all of $I$. Thus $Z_j(s,1)$, and with it the endpoint permutation of the braid $Z(s,\cdot )$, is independent of $s$ by [F2], and in particular $\pi (\beta )=\pi ({\beta }^{\mathrm{\prime }})$.

(b) The top values computed in step 1.1 satisfy $(\gamma \star \beta {)}_j(1)=q_{(\pi (\gamma )\circ \pi (\beta ))(j)}$ for every $j$, so the bijection $j\mapsto (\pi (\gamma )\circ \pi (\beta ))(j)$ has the defining property of the endpoint permutation of $\gamma \star \beta$ in [F1]; that permutation is unique, whence $\pi (\gamma \star \beta )=\pi (\gamma )\circ \pi (\beta )$.

Triple concatenations. Let $\sigma :=\pi (\beta )$ and $\tau :=\pi (\gamma )$, and for $0<a<b<1$ let ${\mathrm{\Theta }}^{a,b}$ be the tuple that equals $z_j(t\mathrm{/}a)$ for $0\le t\le a$, equals $w_{\sigma (j)}(\frac{t-a}{b-a})$ for $a\le t\le b$, and equals $v_{\tau \sigma (j)}(\frac{t-b}{1-b})$ for $b\le t\le 1$, where $v_j$ are the motions of $\delta$; the three branches agree at the break points by [F1], each is continuous, and the three closed pieces of $I$ cover $I$, so each ${\mathrm{\Theta }}_j^{a,b}$ is continuous by [L3] and [L5]; collision-freeness and the endpoint conditions are checked exactly as in step 1.1, so ${\mathrm{\Theta }}^{a,b}$ is a braid based at $Q$, and reading off the two bracketings gives $(\delta \star \gamma )\star \beta ={\mathrm{\Theta }}^{1\mathrm{/}2,3\mathrm{/}4}$ and $\delta \star (\gamma \star \beta )={\mathrm{\Theta }}^{1\mathrm{/}4,1\mathrm{/}2}$, since in both the three motions occur in the order $\beta ,\gamma ,\delta$ with couplings $\sigma$ then $\tau$ and only the two height breaks differ.

(c) Let $Z$ be an isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }}$ and $W$ an isotopy from $\gamma$ to ${\gamma }^{\mathrm{\prime }}$, and put $U_j(s,t):=Z_j(s,2t)$ for $t\le \frac{1}{2}$ and $U_j(s,t):=W_{\pi (\beta )(j)}(s,2t-1)$ for $t\ge \frac{1}{2}$; the two branches agree at $t=\frac{1}{2}$ because $Z_j(s,1)=q_{\pi (\beta )(j)}=W_{\pi (\beta )(j)}(s,0)$ for every $s$ by [F1] and [F2], and step 1.3 shows that the coupling $\pi (\beta )$ is the endpoint permutation of every braid $Z(s,\cdot )$, so no $s$-dependent relabelling is needed; joint continuity follows from [L3] and [L5], each slice $U(s,\cdot )$ is the stacking of the braids $Z(s,\cdot )$ and $W(s,\cdot )$, hence a braid by step 1.1, and the boundary slices are $\gamma \star \beta$ and ${\gamma }^{\mathrm{\prime }}\star {\beta }^{\mathrm{\prime }}$; hence $\gamma \star \beta \sim {\gamma }^{\mathrm{\prime }}\star {\beta }^{\mathrm{\prime }}$.

Moving the breaks is an isotopy. Let $s\mapsto (a_s,b_s)$ be continuous with $0<a_s<b_s<1$, $a_0=\frac{1}{2}$, $b_0=\frac{3}{4}$, $a_1=\frac{1}{4}$, $b_1=\frac{1}{2}$, for instance the linear interpolation of the two pairs; then $(s,t)\mapsto {\mathrm{\Theta }}_j^{a_s,b_s}(t)$ is jointly continuous, because the three regions $\{t\le a_s\}$, $\{a_s\le t\le b_s\}$, $\{t\ge b_s\}$ are closed and cover the square and on each the formula is a composite of continuous maps with the positive denominators $a_s$, $b_s-a_s$, $1-b_s$ bounded below on the compact parameter interval; each slice is a braid by step 2.2 and the boundary slices are ${\mathrm{\Theta }}^{1\mathrm{/}2,3\mathrm{/}4}$ and ${\mathrm{\Theta }}^{1\mathrm{/}4,1\mathrm{/}2}$, so those two braids are braid-isotopic.

(d) Steps 2.2 and 3.1 exhibit the two bracketings $(\delta \star \gamma )\star \beta$ and $\delta \star (\gamma \star \beta )$ of the triple as braid-isotopic representatives, and step 2.3 shows that the isotopy class of a stacking depends only on the isotopy classes of its two factors; hence $(\delta \star \gamma )\star \beta \sim \delta \star (\gamma \star \beta )$ for the given braids, and the induced operation on classes is associative.

Assertions (a), (b), (c), (d) and (e) are steps 1.1, 2.1, 2.3, 4.1 and 1.3, so the stacking of classes $[\gamma ][\beta ]=[\gamma \star \beta ]$ is a well-defined associative binary operation on the set of braid isotopy classes based at $Q$, and $\pi (\delta \star \gamma \star \beta )=\pi (\delta )\pi (\gamma )\pi (\beta )$ follows by applying step 2.1 twice. ∎