0PDZ Proof. It is immediate that d(f1)=0d(f_{1})=0. For the second equality, consider α∈Rξ2−∙(T,U)\alpha\in R^{\bullet}_{\xi_{2}^{-}}(T,U) and β∈Hom𝒮∙(Zξ)(S,U)\beta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,U). Since α⊗β=(id⊠αξ2−(−1))⊗(α|U⋅β)\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|U}\cdot\beta), we can assume that α|U=id\alpha_{|U}=\operatorname{id}\nolimits. We have d(f2)(α⊗β)=α⋅(d(β⊠[ξ1−(−1)→ξ2−(−1)])+d(β)⊠[ξ1−(−1)→ξ2−(−1)]).d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\bigl(d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]\bigr). We have d(β⊠[ξ1−(−1)→ξ2−(−1)])\displaystyle d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]) =d((id⊠[ξ1−(−1)→ξ2−(−1)])⋅(β⊠idξ1−(−1)))\displaystyle=d\bigl((\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})\bigr) =(id⊠[ξ1+(1)→ξ2−(−1)])⋅d(idU⊠[ξ1−(−1)→ξ1+(1)])⋅(β⊠idξ1−(−1))\displaystyle=(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\cdot d(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{-}(-1)\to\xi_{1}^{+}(1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)}) +(id⊠[ξ1−(−1)→ξ2−(−1)])⋅(d(β)⊠idξ1−(−1))\displaystyle\ \ \ \ +(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(d(\beta)\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)}) =(idU⊠[ξ1+(1)→ξ2−(−1)])∑x∈ξ~1−1(U)(id⊠ξ~1([x→1])⊠ξ~1([−1→x]))\displaystyle=(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([x\to 1])\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr) ⋅(β⊠idξ1−(−1))+d(β)⊠[ξ1−(−1)→ξ2−(−1)]\displaystyle\ \ \ \ \cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)] hence d(f2)(α⊗β)=α⋅∑x∈ξ~1−1(U)(id⊠([ξ1+(1)→ξ2−(−1)]⋅ξ~1([x→1]))⊠ξ~1([−1→x]))⋅(β⊠idξ1−(−1))d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\bigl([\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\boxtimes\tilde{\xi}_{1}([-1\to x])\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)}) =∑x∈ξ~1−1(U)(id⊠ξ~1([−1→x])⊠(αξ2−(−1)⋅[ξ1+(1)→ξ2−(−1)]⋅ξ~1([x→1])))⋅(β⊠idξ1−(−1))\displaystyle=\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\boxtimes\bigl(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)}) =f1∘u(α⊗β).\displaystyle=f_{1}\circ u(\alpha\otimes\beta). The lemma follows. ∎