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

61Proof. Consider the following diagramP D ιP P D Uf Df D UQ D ιQ Q D Uρ D P D UP D Uγ D ρ D Q D UQ D Uγ DP D D UfD D U QD D USince f is a functorial morphism and it is a functorial morphism of right D-comodulefunctors, the right square serially commutes. Note that(ρDQ D U ) ◦ ( f D U ) ◦ ι P = ( Q D Uγ D) ◦ ( f D U ) ◦ ι Pso that, by the universal property of the equalizer, there exists a unique morphismf D : P D → Q D such that((42)f D U ) ◦ ι P = ι Q ◦ f D .We now want to prove that f D is a functorial morphism of left C-comodule functor.In fact we have(CιQ ) ◦ C ρ Q D ◦ f D (33)= (C ρ Q D U ) ◦ ι Q ◦ f D(42)= (C ρ D Q U ) ◦ ( f D U ) ◦ ι P(Cf D U ) ◦ (C ρ D P U ) ◦ ι PfleftCcolin=(33)= ( Cf D U ) ◦ ( Cι P ) ◦ C ρ P D(42)= ( Cι Q) ◦ ( Cf D) ◦ C ρ P Dand since C preserves equalizers Cι Q is a monomorphism so that we getC ρ Q D ◦ f D = ( Cf D) ◦ C ρ P D.Then there exists a functorial morphism C f D : C P D → C Q D such thatC U C f D = f D .Corollary 4.4<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizersand assume that C preserves equalizers and let P, Q : B → A be C-bicomodulefunctors. Let f : P → Q be a functorial morphism of C-bicomodule functors. Thenthere exists a unique functorial morphism of left C-comodule functorssatisfyingThen we can considersuch thatf C : P C → Q Cι Q ◦ f C = ( f C U ) ◦ ι P .C f C : C P C → C Q CC U C f C = f C .□Proof. We can apply Proposition 4.40 to the case D = C and B = A.□