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

Then we have(P Cx) ◦ ( ρ C P X ) ◦ l ′ = ( ρ C P Y ) ◦ (P x) ◦ l ′ = ( ρ C P Y ) ◦ h= ( ρ C P Y ) ◦ (P x) ◦ l = (P Cx) ◦ ( ρ C P X ) ◦ l.Since P C preserves equalizers, we have that P Cx is a monomorphism. Since ρ C P X isalso a monomorphism, we deduce that l = l ′ . Therefore we obtain that (P X, P x) =Equ B (P f, P g) . The second statement can be proved similarly.□Lemma 4.19. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let f, g :(X, C ρ X)→(Y, C ρ Y)be morphisms in C A. Assume that there exists (E, e) =Equ A( CUf, C Ug ) and assume that CC preserves equalizers. Then there exists (Ξ, ξ) =EquC A (f, g) and C U (Ξ, ξ) = (E, e) .Proof. Since CC preserves equalizers and ( C, ∆ C) is a right C-comodule functor,also C preserves equalizers by Lemma 4.18, in particular, C preserves (E, e) . Since(C C Uf ) ◦ C ρ X ◦ e f∈C A= C ρ Y ◦ (C Uf ) ◦ eeequ= C ρ Y ◦ (C Ug ) ◦ e g∈C A= ( C C Ug ) ◦ C ρ X ◦ eby the universal property of the equalizer (CE, Ce) there exists a unique morphismC ρ E : E → CE such that(Ce) ◦ C ρ E = C ρ X ◦ e.Moreover, by composing with ε C X the first term of this equality we get(ε C X ) ◦ (Ce) ◦ C ρ Eε C = e ◦ ( ε C E ) ◦ C ρ Ewhereas the second term becomes(ε C X ) ◦ C ρ X ◦ e = eso that we obtain the following equalitye ◦ ( ε C E ) ◦ C ρ E = e.Since e is a monomorphism we deduce that(ε C E ) ◦ C ρ E = E.Now, consider the following serially commutative diagram49EC ρ ECE∆ C EC C ρ ECCEeCeCCeXC ρ XCX∆ C XC C ρ XCCXC UfC UgC C UfC C UgCC C UfCC C UgYC ρ Y CY∆ C YC C ρ Y CCY