0PCP Proof. The counit of the adjunction is ε=κ1∘mult\varepsilon=\kappa_{1}\circ\mathrm{mult}. Let γ∈Rξ−∙(T,S)\gamma\in R_{\xi^{-}}^{\bullet}(T,S). Let η\eta be the map defined in the lemma. We have η(idT)=∑x∈ξ~−1(T)(ξ~([−1→x])⊠idT∖{ξ~(x)})⊗(ξ~([x→1])⊠idT∖{ξ~(x)}),\eta(\operatorname{id}\nolimits_{T})=\sum_{x\in\tilde{\xi}^{-1}(T)}(\tilde{\xi}([-1\to x])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\otimes(\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}}), hence (id⊗ε)∘(η⊗id)(γ)=∑x∈ξ~−1(T)(ξ~([−1→x])⊠idT∖{ξ~(x)})⋅κ1((ξ~([x→1])⊠idT∖{ξ~(x)})⋅γ)(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=\sum_{x\in\tilde{\xi}^{-1}(T)}(\tilde{\xi}([-1\to x])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\kappa_{1}\bigl((\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\gamma\bigr) Let xx be the unique element of ξ~−1(χ(γ)(ξ−(−1)))\tilde{\xi}^{-1}(\chi(\gamma)(\xi^{-}(-1))). We have γξ−(−1)=ξ~([−1→x])\gamma_{\xi^{-}(-1)}=\tilde{\xi}([-1\to x]) and κ1((ξ~([x→1])⊠idT∖{ξ~(x)})⋅γ)=γ|S\kappa_{1}\bigl((\tilde{\xi}([x\to 1])\boxtimes\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}(x)\}})\cdot\gamma\bigr)=\gamma_{|S}, hence (id⊗ε)∘(η⊗id)(γ)=(γξ−(−1)⊠idT∖{χ(γ)(ξ−(−1))})⊗γ|S(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=(\gamma_{\xi^{-}(-1)}\boxtimes\operatorname{id}\nolimits_{T\setminus\{\chi(\gamma)(\xi^{-}(-1))\}})\otimes\gamma_{|S} We deduce that mult∘(id⊗ε)∘(η⊗id)(γ)=γ\mathrm{mult}\circ(\operatorname{id}\nolimits\otimes\varepsilon)\circ(\eta\otimes\operatorname{id}\nolimits)(\gamma)=\gamma and the lemma follows. ∎