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

153(81)= λ ′ ◦ σ B ◦ (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ)(173)= π Z ◦ (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) = π Z ◦ (hQ) (167)= ν Z ◦ y ′ .Since y ′ is an epimorphism, we getλ ′ ◦ ν B = ν Zso that we deduce that ν B : B ′ → B is an isomorphism and thus an isomorphism ofmonads.3) If P preserves equalizers and P C = Q → κ′Q ′ = DP then ν A and ν B areisomorphisms of monads.By assumption we have that q = Id P C and q ′ = Id DP and κ and κ ′ are isomorphisms.Then we can rewrite the initial diagram as followsDP CQDhQDP Q(P Cχ)◦(δD P CQ)(ε D P CQ)(P µ S Q )◦(jS)◦(Dσ S)(ε D P Q)P CQ y′ hQ P Q π Z ZS ′ ν Z((Since (C, i) = Equ ) )Fun QP σA((◦ (τP ) , QP u A , by Lemma <strong>2.</strong>10 we also have(CQ, iQ) = Equ Fun QP σ A Q ) ◦ (τP Q) , QP u A Q ) . We compute(QP σ A Q ) ◦ (τP Q) ◦ τ (68)= ( QP σ A Q ) ◦ (QP τ) ◦ τ (69)= (QP u A Q) ◦ τso that there exists a unique functorial morphism C ρ Q : Q → CQ such that (61)holds i.e.(iQ) ◦ C ρ Q = τas constructed in Proposition 6.<strong>1.</strong> Moreover ( Q, C ρ Q)is a left C-comodule by Proposition6.<strong>1.</strong> We also get(P iQ) ◦ ( P C ρ Q)= P τ.Let us computeand(hQ) ◦ ( P C ρ Q)=( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( P C ρ Q)q=Id P C=( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ ( P C ρ Q)(61)= (B µ P Q ) ◦ ( σ B P Q ) ◦ (P τ) (82)= ( µ A P Q ) ◦ ( P σ A Q ) ◦ (P τ)(69)= ( µ A P Q ) ◦ (P u A Q) P module= P Q(P C ρ Q)◦ (hQ) =(P C ρ Q)◦( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ)(82)= ( ) (P C ρ Q ◦ µAP Q ) ◦ ( P σ A Q ) ◦ (P iQ)(63)= ( ) (P C ρ Q ◦ µAP Q ) ◦ (P u A Q) ◦ ( P ε C Q )P module= ( P C ρ Q)◦(P ε C Q ) Qcomodule= P Q.