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

(112)= (Cx) ◦ (Cδ C ) ◦ ∆ C ◦ ( ε C C ) (103)= (Cu A ) ◦ ( Cε C) ◦ ∆ C ◦ ( ε C C )Ccomonad= (Cu A ) ◦ ( ε C C )and since ε C C is an epimorphism we get thatΛ ◦ (u A C) = (Cu A ) .Finally we compute(ε C A ) ◦ Λ ◦ (xC) defλ= ( ε C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )119ε C = x ◦ ( ε C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= x ◦ (χP ) ◦ (QP δ C ) (102)= m A ◦ (xx) ◦ (QP δ C )x= m A ◦ (xA) ◦ (QP x) ◦ (QP δ C ) (103)= m A ◦ (xA) ◦ (QP u A ) ◦ ( QP ε C)x= m A ◦ (Au A ) ◦ x ◦ ( QP ε C) Acomonad,x=and since xC is an epimorphism we get that(ε C A ) ◦ Λ = Aε C .2) Consider the functorial morphism given by(AεC ) ◦ (xC)DP Q δ DP Q−→ P QP Q P χ−→ P Q P ρD Q−→ P QD yD−→ BDγ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .Recall that z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q and let us computethat isLet us computeγ ◦ ( Dz l) ? = γ ◦ (Dz r )(yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)?= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .(yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q) ◦ (Dδ D P Q)δ D ,(111)= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (P QδD P Q) ◦ (δ D DP Q)(113)= (yD) ◦ ( (P ρQ) D ◦ (P χ) ◦ P Qε D P Q ) ◦ (δ D DP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .Since (DB, Dy) = Coequ Fun(Dz l , Dz r) , there exists a functorial morphism Γ :DB → BD such thatΓ ◦ (Dy) = γ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .