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

50Since we already observed that the columns are equalizers and also the second andthe third row are equalizers by Proposition 4.13, in view of Lemma <strong>2.</strong>13 also thefirst row is an equalizer, so that (E, e) has a left C-comodule structure, i.e. thereexists (Ξ, ξ) ∈ C A such that (Ξ, ξ) = EquC A (f, g) and C U (Ξ, ξ) = (E, e) . □Lemma 4.20. Let C = ( C, ∆ C , ε C) be a comonad on a category A with equalizers andlet (C U, C F ) be the adjunction associated. The following statements are equivalent:(i) C : A → A preserves equalizers(ii) CC : A → A preserves equalizers(iii) C A has equalizers and they are preserved by C U : C A → A(iv) C U : C A → A preserves equalizers.Proof. (i) ⇒ (ii) and (iii) ⇒ (iv) are clear.(ii) ⇒ (iii) follows by Lemma 4.19.(iv) ⇒ (i) Note that C F is a right adjoint, so that in particular it preserves equalizers.Then C U C F = C also preserves equalizers.□Lemma 4.2<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A and assumethat C preserves coequalizers. Then C F preserves coequalizers where (C U, C F ) isthe adjunction associated to the comonad.Proof. Dual to proof of Lemma 3.2<strong>2.</strong> LetXfgbe a coequalizer in A. Let us consider the fork obtained by applying the functor C Fto the coequalizeri.e.C F XC F fC F g Y C F Yk KC F k C F K(CX, ∆ C X ) Cf (CY, ∆ C Y ) Ck ( CK, ∆ C K )CgNow, let ( Z, C ρ Z)∈ C A and z : ( CY, ∆ C Y ) → ( Z, C ρ Z)be a morphism in C A suchthat z ◦ (Cf) = z ◦ (Cg). Since C preserves coequalizers, we know that (CK, Ck) =Coequ A (Cf, Cg) . By the universal property of the coequalizer (CK, Ck) in A, thereexists a unique morphism z ′ : CK → Z in A such that z ′ ◦ (Ck) = z. We now wantto prove that z ′ is a morphism in C A, i.e. that (Cz ′ ) ◦ ( ∆ C K ) = C ρ Z ◦ z ′ . Since zis a morphism in C A we have that(Cz) ◦ ( ∆ C Y ) = C ρ Z ◦ zand since also Ck is a morphism in C A we have thatThen we have(CCk) ◦ ( ∆ C Y ) = ( ∆ C K ) ◦ (Ck) .(Cz ′ ) ◦ ( ∆ C K ) ◦ (Ck) = (Cz ′ ) ◦ (CCk) ◦ ( ∆ C Y )propz= (Cz) ◦ ( ∆ C Y ) = C ρ Z ◦ z propz= C ρ Z ◦ z ′ ◦ (Ck)