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

Let( (CQ ) ()D, ι Q) C = Equ Fun ρ DC C QU, C Q C Uγ D . Then we have( CQ ) D=C ( Q D) : D B → C A.55Proof. Since Q is endowed with a left C-comodule structure, by Lemma 4.28 thereexists a unique functor C Q : B → C A such that C U C Q = Q and C Uγ C C Q = C ρ Q .Note that, since Q is a C-D-bicomodule functor, in particular the compatibilitycondition(Cρ D Q) ◦ C ρ Q = (C ρ Q D ) ◦ ρ D Qholds, that is ρ D Q : Q = C U C Q → QD = C U C QD is a morphism in C A. Thus, thereexists a functorial morphism ρ DC Q : C Q → C QD such thatC Uρ DC Q = ρD Q.By the coassociativity and counitality properties of ρ( )D Q we get that also ρDC Q iscoassociative and counital, so thatC Q, ρ DC Qis a right D-comodule functor. Thuswe can consider the equalizer(38)( CQ ) D ι C Q C Q D Uρ D C Q D UC Q D Uγ D C QD D Uso that we get a functor (C Q ) D: D B → C A. Since C preserves equalizers, by Lemma4.20 also C U preserves equalizers. Then, by applying the functor C U to (38) we stillget an equalizerC U (C Q ) DC Uι C Q C U C Q D UC Uρ D C Q D UC U C Q D Uγ D C U C QD D Uthat isC U (C Q ) DC Uι C Q Q D Uρ D Q D UQ D Uγ D QD D UBy Proposition 4.29 ( Q D , ι Q) (= Equ Fun ρD DQ U, Q D Uγ D) , then we haveC U (C Q ) D= Q D and C Uι CQ = ι Q .MoreoverC Uγ C ( CQ ) D: C U (C Q ) D= Q D → C C U (C Q ) D= CQDso that, using Proposition 4.29 where we prove that ( Q D , C ρ Q D)is a left C-comodulefunctor and that C Uγ C C ( Q D) = C ρ Q D, we getC Uγ C ( CQ ) D= C ρ Q D = C Uγ C C ( Q D)i.e.( CQ ) D=C ( Q D) .□