0PDX Proof. Assume first α|U∖{χ(α)−1(ξ1+(1))}=id\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}=\operatorname{id}\nolimits and β|S=id\beta_{|S}=\operatorname{id}\nolimits. We have ρ(α⊗β)\displaystyle\rho(\alpha\otimes\beta) =ε1Rξ1−Lξ1+∘Lξ1+τLξ1+(α⊗β⊗η1(idS))\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\circ L_{\xi_{1}^{+}}\tau L_{\xi_{1}^{+}}(\alpha\otimes\beta\otimes\eta_{1}(\operatorname{id}\nolimits_{S})) =ε1Rξ1−Lξ1+(∑x∈ξ~1−1(S)α⊗τ((βξ1−(−1)⊠id)⊗(id⊠ξ~1([−1→x])))⊗(ξ~1([x→1])⊠id))\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in\tilde{\xi}_{1}^{-1}(S)}\alpha\otimes\tau\Bigl((\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\Bigr)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr) =ε1Rξ1−Lξ1+(∑x∈Iα⊗(id⊠ξ~1([−1→x]))⊗(βξ1−(−1)⊠id)⊗(ξ~1([x→1])⊠id))\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in I}\alpha\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\otimes(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr) =δ1(βξ1−(−1)⊠id)⊗(αχ(α)−1(ξ1+(1))⊠id)\displaystyle=\delta_{1}(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes(\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}\boxtimes\operatorname{id}\nolimits) where I={x∈ξ~1−1(S)|(ξ~1([−1→x])⊠(βξ1−(−1)∘[ξ1−(−2)→ξ1−(−1)])⋅τ≠0}I=\{x\in\tilde{\xi}_{1}^{-1}(S)\ |\ (\tilde{\xi}_{1}([-1\to x])\boxtimes(\beta_{\xi_{1}^{-}(-1)}\circ[\xi_{1}^{-}(-2)\to\xi_{1}^{-}(-1)])\cdot\tau\neq 0\}. Since ρ\rho is a morphism of (𝒮M(Z),𝒮M(Z))({\mathcal{S}}_{M}(Z),{\mathcal{S}}_{M}(Z))-bimodules, the general result follows using the decompositions α=(α|U∖{χ(α)−1(ξ1+(1))}⊠idξ1+(1))⋅(id⊠αχ(α)−1(ξ1+(1)))\alpha=(\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{+}(1)})\cdot(\operatorname{id}\nolimits\boxtimes\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}) and β=(id⊠βξ1−(−1))⋅(β|S⊠idξ1−(−1))\beta=(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\cdot(\beta_{|S}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)}). ∎