Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

120We want to prove that Γ is an opposite mixed distributive law. We compute(m B D) ◦ (BΓ) ◦ (ΓB) ◦ (Dyy) y = (m B D) ◦ (BΓ) ◦ (ΓB) ◦ (DyB) ◦ (DP Qy)defγ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (δ D P QB) ◦ (DP Qy)δ D= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (P QP Qy) ◦ (δ D P QP Q)χ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P Qy) ◦ (P χP Q) ◦ (δ D P QP Q)ρ D Q= (m B D) ◦ (BΓ) ◦ (yDB) ◦ (P QDy) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (BΓ) ◦ (BDy) ◦ (yDP Q) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)defγ= (m B D) ◦ (ByD) ◦ ( )BP ρ D Q ◦ (BP χ) ◦ (BδD P Q) ◦ (yDP Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yDP Q) ◦ (P QyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(109)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(127)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P δC QP Q) ◦ ( P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)δ=C(yD) ◦ (P χD) ◦ (P δ C QD) ◦ ( (P CρQ) D ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)(112)= (yD) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)ε= C (yD) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (P χP Q) ◦ (δ D P QP Q)Qcomfun= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (δD P QP Q)(111)= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ (DP χ)defγ= Γ ◦ (Dy) ◦ (DP χ) (109)= Γ ◦ (Dm B ) ◦ (Dyy)and since Dyy is an epimorphism we deduce thatLet us compute(m B D) ◦ (BΓ) ◦ (ΓB) = Γ ◦ (Dm B ) .(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) ◦ (Dy) ∆D= (ΓD) ◦ (DΓ) ◦ (DDy) ◦ ( ∆ D P Q )