0PBJ Lemma 7.4.25. Let θ∈Hom𝒮M∙(Z)(I,J)\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I,J) and θ′∈Hom𝒮M∙(Z)(I′,I)\theta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I^{\prime},I) such that θ∘θ′\theta\circ\theta^{\prime} is a braid. Let I0I_{0} be a finite subset of M∖(I∪I′∪J)M\setminus(I\cup I^{\prime}\cup J). If |θ|+|θ′|=|θ∘θ′||\theta|+|\theta^{\prime}|=|\theta\circ\theta^{\prime}|, then (θ⊠idI0)⋅(θ′⊠idI0)=(θ⋅θ′)⊠idI0(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})\cdot(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=(\theta\cdot\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}}.
0PBK Proof. Let s′∈I′s^{\prime}\in I^{\prime}. Since |θθ′(s′)|+|θs′′|=|θθ′(s′)∘θs′′||\theta_{\theta^{\prime}(s^{\prime})}|+|\theta^{\prime}_{s^{\prime}}|=|\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}}|, it follows that i(θθ′(s′),idi)+i(θs′′,idi)=i(θθ′(s′)∘θs′′,idi)i(\theta_{\theta^{\prime}(s^{\prime})},\operatorname{id}\nolimits_{i})+i(\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})=i(\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i}) for all i∈I0i\in I_{0}. As a consequence, i((θ∘θ′)⊠idI0)−i(θ⊠idI0)−i(θ′⊠idI0)=i(θ∘θ′)−i(θ)−i(θ′).i((\theta\circ\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime}). The lemma follows now from Lemma 7.4.9. ∎