0PD1 Lemma 8.2.5. We have an isomorphism νn:Rξ2−∙(−,−,en)Lξ1+∙(−,−,en)→∼Dn\nu_{n}:R_{\xi_{2}^{-}}^{\bullet}(-,-,e^{n})L^{\bullet}_{\xi_{1}^{+}}(-,-,e^{n})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D_{n}. In particular, we have an isomorphism ν1:Rξ2−∙Lξ1+∙→∼D1=A1=C1\nu_{1}:R_{\xi_{2}^{-}}^{\bullet}L^{\bullet}_{\xi_{1}^{+}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D_{1}=A_{1}=C_{1} and Cn=C1∗n=A1∗nC_{n}=C_{1}^{\ast n}=A_{1}^{\ast n}.
0PD2 Proof. Let β∧α∈Hom(−⊔(−n,−1),T)∧Hom(S,−⊔(1,n))\beta\wedge\alpha\in\operatorname{Hom}\nolimits(-\sqcup(-n,-1),T)\wedge\operatorname{Hom}\nolimits(S,-\sqcup(1,n)). We have β∧α=β′∧α′\beta\wedge\alpha=\beta^{\prime}\wedge\alpha^{\prime} where β′=β|(−n,−1)⊠id\beta^{\prime}=\beta_{|(-n,-1)}\boxtimes\operatorname{id}\nolimits and α′=(β|−⊠id(1,n))⋅α\alpha^{\prime}=(\beta_{|-}\boxtimes\operatorname{id}\nolimits_{(1,n)})\cdot\alpha. If νn(β′∧α′)=0\nu_{n}(\beta^{\prime}\wedge\alpha^{\prime})=0, then β′=α′=0\beta^{\prime}=\alpha^{\prime}=0 (cf the beginning of §7.4.10). Now νn\nu_{n} has an inverse given by σ↦(id⊠σ|(−n,−1))∧σ|S\sigma\mapsto(\operatorname{id}\nolimits\boxtimes\sigma_{|(-n,-1)})\wedge\sigma_{|S}. ∎