Proof of Theorem 2.5.4.
We will focus on the version for the positive cabled crossing, since the other one is analogous. First, we reduce to the case when the object \(Y=Y_1\boxtimes Y_2\) is a generating object of \(\mathrm{BSbim}_m\boxtimes \mathrm{BSbim}_n\). Otherwise, we can decompose into generators: \[Y_1\boxtimes Y_2= (Y_1\boxtimes \mathbf{1}) \circ_1(\mathbf{1}\boxtimes Y_2) = (B_{i_1}\boxtimes \mathbf{1}) \circ_1\cdots \circ_1(B_{i_a}\boxtimes \mathbf{1}) \circ_1(\mathbf{1} \boxtimes B_{j_1})\circ_1\cdots \circ_1(\mathbf{1} \boxtimes B_{j_b})\] and define \[\begin{aligned} \mathrm{slide}_{Y_1,\mathbf{1}}&:= (\mathrm{id}_{\mathrm{swap}_{m,n}(B_{i_1}\circ_1\cdots \circ_1B_{i_{a-1}} \boxtimes \mathbf{1})}\circ_1\mathrm{slide}_{B_{i_{a}},\mathbf{1}}) \circ_2\cdots \circ_2 (\mathrm{slide}_{B_{i_{1}},\mathbf{1}} \circ_1\mathrm{id}_{B_{i_1}\circ_1\cdots \circ_1B_{i_{a-1}} \boxtimes \mathbf{1}}) \\ \mathrm{slide}_{\mathbf{1},Y_2}&:= (\mathrm{id}_{\mathrm{swap}_{m,n}(\mathbf{1}\boxtimes B_{j_1}\circ_1\cdots \circ_1B_{j_{b-1}})}\circ_1\mathrm{slide}_{\mathbf{1}, B_{j_{b}}}) \circ_2\cdots \circ_2 (\mathrm{slide}_{\mathbf{1}, B_{j_{1}}} \circ_1\mathrm{id}_{\mathbf{1}\boxtimes B_{j_1}\circ_1\cdots \circ_1B_{j_{b-1}}}) \\ \mathrm{slide}_{Y_1,Y_2} &:= (\mathrm{id}_{\mathrm{swap}_{m,n}(Y_1\boxtimes \mathbf{1})} \circ_1\mathrm{slide}_{\mathbf{1},Y_2}) \circ_2(\mathrm{slide}_{Y_1,\mathbf{1}} \circ_1\mathrm{id}_{Y_2}) \end{aligned}\]
Now we turn to defining \(\mathrm{slide}_{B,\mathbf{1}_n}\) and \(\mathrm{slide}_{\mathbf{1}_m,B}\), where \(B\) is one of the generating Bott–Samelson bimodules. Here we place subscripts to distinguish the identity bimodules. We first consider the latter situation and reduce it to the case \(m=1\), where the cabled crossing is a Coxeter braid. Indeed, suppose that \(m>1\), then we use the first equality from Lemma 2.2.8 to define \(\mathrm{slide}_{\mathbf{1}_m,B}\) to be the composite: \[\begin{gathered} \nonumber \big((\mathrm{slide}_{\mathbf{1}_1,B}\boxtimes \mathrm{id}_{\mathbf{1}_{m-1}}) \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{m-1-i} \boxtimes X_{1,n}\boxtimes\mathbf{1}_{i}} \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{m-1}\boxtimes X_{1,n}}\big) \circ_2\cdots \\ \circ_2\big(\mathrm{id}_{X_{1,n}\boxtimes \mathbf{1}_{m-1}} \circ_1\cdots \circ_1 (\mathrm{id}_{\mathbf{1}_{m-1-i}} \boxtimes \mathrm{slide}_{\mathbf{1}_1,B}\boxtimes\mathrm{id}_{\mathbf{1}_{i}}) \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{m-1}\boxtimes X_{1,n}}\big) \circ_2\cdots\\ \nonumber \circ_2\big(\mathrm{id}_{X_{1,n}\boxtimes \mathbf{1}_{m-1}} \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{m-1-i} \boxtimes X_{1,n}\boxtimes\mathbf{1}_{i}} \circ_1\cdots \circ_1 (\mathrm{id}_{\mathbf{1}_{m-1}}\boxtimes \mathrm{slide}_{\mathbf{1}_1,B}) \big) \end{gathered}\] For the other case, we first choose chain maps \(\varphi\) and \(\varphi^{-1}\) realising the first homotopy equivalence in Lemma 2.2.8. Then we define \(\mathrm{slide}_{B,\mathbf{1}_n}\) as the composition: \[\begin{gathered} \nonumber \varphi^{-1} \circ_2\big((\mathbf{1}_{n-1}\boxtimes \mathrm{slide}_{B,\mathbf{1}_1}) \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{i} \boxtimes X_{m,1}\boxtimes\mathbf{1}_{n-1-i}} \circ_1\cdots \circ_1 \mathrm{id}_{X_{m,1}\boxtimes \mathbf{1}_{n-1}}\big) \circ_2\cdots \\ \circ_2\big(\mathrm{id}_{\mathbf{1}_{n-1}\boxtimes X_{m,1}} \circ_1\cdots \circ_1 (\mathrm{id}_{\mathbf{1}_{i}} \boxtimes \mathrm{slide}_{B,\mathbf{1}_1}\boxtimes\mathrm{id}_{\mathbf{1}_{n-1-i}}) \circ_1\cdots \circ_1 \mathrm{id}_{X_{m,1}\boxtimes \mathbf{1}_{n-1}}\big) \circ_2\cdots \\ \nonumber \circ_2\big(\mathrm{id}_{\mathbf{1}_{n-1}\boxtimes X_{m,1}} \circ_1\cdots \circ_1 \mathrm{id}_{\mathbf{1}_{i} \boxtimes X_{m,1}\boxtimes\mathbf{1}_{n-1-i}} \circ_1\cdots \circ_1 (\mathrm{slide}_{B,\mathbf{1}_1}\boxtimes \mathbf{1}_{n-1}) \big)\circ_2\varphi \end{gathered}\] It remains to construct \(\mathrm{slide}_{\mathbf{1}_1,B_i}\) and \(\mathrm{slide}_{B_j,\mathbf{1}_1}\) where \(B_i\) is a generating object of \(\mathrm{BSbim}_n\) and \(B_j\) is a generating object of \(\mathrm{BSbim}_m\). Now we reduce this problem to the cases when \(n=2\) and \(m=2\) respectively. We define \(\mathrm{slide}_{\mathbf{1}_1,B_i}\) as the composite: \[\begin{aligned} F(\sigma_{n}\cdots\sigma_{1}) \circ_1B_i =& F(\sigma_{n}\cdots\sigma_{i+1})\circ_1F(\sigma_{i}\sigma_{i-1})\circ_1F(\sigma_{i-2}\cdots\sigma_{1})\circ_1B_i \\ \rightarrow &F(\sigma_{n}\cdots\sigma_{i+1})\circ_1F(\sigma_{i}\sigma_{i-1}) \circ_1B_i \circ_1F(\sigma_{i-2}\cdots\sigma_{1}) \\ \xrightarrow{\mathrm{slide}} &F(\sigma_{n}\cdots\sigma_{i+1})\circ_1B_{i-1} \circ_1F(\sigma_{i}\sigma_{i-1}) \circ_1F(\sigma_{i-2}\cdots\sigma_{1}) \\ \rightarrow& B_{i-1} \circ_1F(\sigma_{n}\cdots\sigma_{i+1})\circ_1F(\sigma_{i}\sigma_{i-1}) \circ_1F(\sigma_{i-2}\cdots\sigma_{1})\\ =& B_{i-1} \circ_1F(\sigma_{n}\cdots\sigma_{1}) \end{aligned}\] where the unlabelled maps are far-commutativity isomorphisms and the labelled arrow is given by \(\mathrm{id}\circ_1\mathrm{slide}_{\mathbf{1}_1,B_1} \circ_1\mathrm{id}\), which is determined by the \(n=2\) case. The reduction of \(\mathrm{slide}_{B_j,\mathbf{1}_1}\) to the case \(m=2\) is completely analogous.
Thus we reduced the problem to the statement from Lemma 2.5.5. By construction, all slide maps constructed in this proof are homotopy equivalences. ◻