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

where p : B (M) → M is the usual epimorphism of right B-modules and (P, f ′ , p ′ ) isthe pullback of (p, f). Recall thatP = { (x, y) ∈ B (M) × L | p (x) = f (y) }and f ′ : P → B (M) is defined by setting f ′ ((x, y)) = x while p ′ : P → L is defined bysetting p ′ ((x, y)) = y. Moreover i ′ : Ker (p) → P is defined by setting i ′ (x) = (x, 0).Since f is a monomorphism we have that also f ′ is a monomorphism and since p isan epimorphism, also p ′ is an epimorphism. Then, by the foregoing, both T (f ′ ) andT (i) are monomorphism. Since T (i) is a monomorphism we get that T (i ′ ) is alsoa monomorphism so that (T (Ker (p)) , T (i ′ )) is a kernel of T (p ′ ). Since T (f ′ ) isa monomorphism and T (f ′ ) T (i ′ ) = T (i) we get that (T (Ker (p)) , T (i ′ )) is also akernel of T (p) T (f ′ ). In fact T (p) T (f ′ ) T (i ′ ) = T (p) T (i) = 0 and if ζ : Z → T (P )is a morphism such that T (p) T (f ′ ) ζ = 0 there exists a unique morphism ζ ′ : Z →T (Ker (p)) such that T (f ′ ) ζ = T (i) ζ ′ so that T (f ′ ) ζ = T (i) ζ ′ = T (f ′ ) T (i ′ ) ζ ′and since T (f ′ ) is mono we get that ζ = T (i ′ ) ζ ′ . Since T (p) T (f ′ ) = T (f) T (p ′ )we deduce that (T (Ker (p)) , T (i ′ )) is a kernel of T (f) T (p ′ ). Therefore we obtainthatKer (T (f) T (p ′ )) = Ker (T (p ′ )) .Since T is right exact we know that T (p ′ ) is an epimorphism and hence, in view ofLemma A.8, we deduce that T (f) is a monomorphism.02490 T (Ker (p)) T (i′ )T (P )T (p ′ )T (L)0Id T (Ker(p))T (f ′ )0 T (Ker (p)) T (i) T ( B (M)) T (p)T (f) T (M) 0Lemma A.10. Let C be an A-coring. Then the category (Mod-A) C has coproductsand cokernels so that it is cocomplete. Moreover if U : (Mod-A) C → (Mod-A) isthe forgetful functor we have((∐ ))U (Mi , ρ i ) i∈I, ρ M = ⊕ M i and U (( Coker (f) , ρ Coker(f))) = Coker (f) .i∈I□Proof. Let (M ι , ρ i ) i∈Ibe a family of right C-comodules and let ε i : M i → M = ⊕ M ii∈Ibe the canonical injection and π i : M → M i the canonical projection. Since π i ε i =Id Mi the mapε i ⊗ A C : M i ⊗ A C → M ⊗ A Cis injective and hence also the mapρ i = (ε i ⊗ A C) ◦ ρ i : M i → M ⊗ A Cis injective. Letρ M : M → M ⊗ A C