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

Moreover (Q, C ρ Q ) is a left C-comodule functor.Let θ l = ( σ B P Q ) ◦ (P τ) and θ r = u B P Q : P Q → BP Q. Set(74) (D, j) = Equ Fun(θ l , θ r) .There exists a functorial morphism ρ D Q : Q → QD such that(75) (Qj) ◦ ρ D Q = τ.There exist functorial morphisms ∆ D : D → DD and ε D : D → B such thatD = ( D, ∆ D , ε D) is a comonad over B and D preserves equalizers. The functorialmorphisms ∆ D and ε D are uniquely determined by(76) (P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Moreover ( Q, ρ D Q)is a right D-comodule functor.Finally ( Q, C ρ Q , ρ D Q)is a C-D-bicomodule functor.Proof. See the dual Theorem 6.29.Theorem 6.6. Let Ξ = ( P, Q, A, B, σ A , σ B , u A , u B , ) be a regular preformal dualstructure on categories A and B such that the functors P, Q, A, B preserve equalizersand let τ : Q → QP Q be a pretorsor for Ξ. Assume that A and B aremonads, ( ) ( )P, B µ P is a left B-module functor and P, µAP is a right A-module functor.Moreover assume that the functorial morphism σ A is right A-linear, that isσ A ◦ ( )Qµ A P = mA ◦ ( σ A A ) and the functorial morphism σ B is left B-linear that isσ B ◦ (B µ P Q ) = m B ◦ ( Bσ B) and that they are compatible in the sense that(77)B µ P ◦ ( σ B P ) = µ A P ◦ ( P σ A) .Then there exists a comonad C = ( C, ∆ C , ε C) on the category A together with afunctorial morphism C ρ Q : Q → CQ such that ( Q, C ρ Q)is a left C-comodule functorand a comonad D = ( D, ∆ D , ε D) together with a functorial morphism ρ D Q : Q → QDsuch that ( Q, ρ D Q)is a right D-comodule functor. The underlying functors are definedas follows(C, i) = Equ Fun((QP σA ) ◦ (τP ) , QP u A)and(D, j) = Equ Fun((σ B P Q ) ◦ (P τ) , u B P Q ) .satisfying(iQ) ◦ C ρ Q = τ and (Qj) ◦ ρ D Q = τ.Furthermore1) The morphism can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA is an isomorphism.2) The morphism can 1 := ( σ B D ) ◦ ( P ρ D Q): P Q → BD is an isomorphisms.3)(QP σA ) ◦ (τP ) = (iA) ◦ can 1 and ( σ B P Q ) ◦ (P τ) = (Bj) ◦ can 191□(78)(79)4)i = can −11 ◦ (Cu A )j = (can 1 ) −1 ◦ (u B D)