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

251= (ζ ′ ⊗ A C) ◦ (p ⊗ A C) ◦ ρ N = (ζ ′ ⊗ A C) ◦ ρ Coker(f) ◦ pand since p is an epimorphism we get that ζ ′ ∈ (Mod-A) C .Lemma A.1<strong>1.</strong> Let C be an A-coring and assume that A C is flat. Then the category(Mod-A) C has kernels. Moreover if U : (Mod-A) C → (Mod-A) is the forgetfulfunctor we haveU (( Ker (f) , ρ Ker(f))) = Ker (f) .Proof. Since by Lemma A.10 the preadditive category (Mod-A) C has coproducts,it also has products. Now, let f : ( M, ρ M) → ( N, ρ N) a morphism in (Mod-A) C .Then in Mod-A we can consider the exact sequence0 → Ker (f)k−→ Mand, since A C is flat, we get the exact sequence0 Ker (f)ρ Ker(f)kMf−→ N0 Ker (f) ⊗ A C k⊗ AC M ⊗ A C f⊗ AC N ⊗ A CWe have0 = ρ N ◦ f ◦ k = (f ⊗ A C) ◦ ρ M ◦ k.By the properties of the kernel of f there exists a unique morphism ρ Ker(f) : Ker (f) →Ker (f) ⊗ A C such thatρ M ◦ k = (k ⊗ A C) ◦ ρ Ker(f) .We have to prove that ( Ker (f) , ρ Ker(f)) ∈ (Mod-A) C . Let us compute(k ⊗ A C ⊗ A C) ◦ ( ρ Ker(f) ⊗ A C ) ◦ ρ Ker(f) = ( ρ M ⊗ A C ) ◦ (k ⊗ A C) ◦ ρ Ker(f)= ( ρ M ⊗ A C ) ◦ ρ M ◦ k = ( M ⊗ A ∆ C) ◦ ρ M ◦ k = ( M ⊗ A ∆ C) ◦ (k ⊗ A C) ◦ ρ Ker(f)andρ M= (k ⊗ A C ⊗ A C) ◦ ( Ker (f) ⊗ A ∆ C) ◦ ρ Ker(f)k ◦ r Ker(f) ◦ ( Ker (f) ⊗ A ε C) ◦ ρ Ker(f) = r M ◦ (k ⊗ A C) ◦ ( Ker (f) ⊗ A ε C) ◦ ρ Ker(f)= r M ◦ ( M ⊗ A ε C) ◦ (k ⊗ A C) ◦ ρ Ker(f) = r M ◦ ( M ⊗ A ε C) ◦ ρ M ◦ k = k.Since k is mono we conclude. Let now ζ : ( Z, ρ Z) → ( M, ρ M) be a morphism in(Mod-A) C such that f ◦ ζ = 0. Then by the universal property of the kernel of f inMod-A there exists a unique morphism ζ ′ : Z → Ker (f) such thatk ◦ ζ ′ = ζ.We want to prove that ζ ′ ∈ (Mod-A) C . Let us compute(k ⊗ A C) ◦ ρ Ker(f) ◦ ζ ′ = ρ M ◦ k ◦ ζ ′ = ρ M ◦ ζ = (ζ ⊗ A C) ◦ ρ Z= (k ⊗ A C) ◦ (ζ ′ ⊗ A C) ◦ ρ Zand since k⊗ A C is mono we conclude that ζ ′ ∈ (Mod-A) C and U (( Ker (f) , ρ Ker(f))) =Ker (f) .□fNρ N□