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

244Lemma A.4. Let A be an abelian category and let U ∈ A. Then, for every exactsequence in Athe sequence0 → Hom A (U, K) Hom A(U,k)−→0 → K k−→ Xis an exact sequence of abelian groups.f−→ Y,Hom A (U, X) Hom A(U,f)−→ Hom A (U, Y )Proof. Let h ∈ Hom A (U, K). Then Hom A (U, k) (h) = kh. Since k is a monomorphismit follows that kh = 0 if and only if h = 0. Hence Hom A (U, k) is also amonomorphism. Also (Hom A (U, f) ◦ Hom A (U, k)) (h) = fkh = 0. This impliesthat Im (Hom A (U, k)) ⊆ Ker (Hom A (U, f)). Let now g ∈ Hom A (U, X) and assumethat Hom A (U, f) (g) = 0 i.e. fg = 0. Since (K, k) = Ker (f) there exists a morphismg ′ : U → K such that g = kg ′ = Hom A (U, k) (g ′ ) ∈ Im (Hom A (U, k)). Thereforewe get that Ker (Hom A (U, f)) ⊆ Im (Hom A (U, k)) and hence Ker (Hom A (U, f)) =Im (Hom A (U, k)).□Lemma A.5. In the assumptions and notations of A.1, let (T, H) be an adjunctionwhere T : B → A and H : A → B and let f : X → Y be a morphism in A. Then fis a monomorphism (resp. epimorphism) if and only if T H (f) is a monomorphism(resp. epimorphism).Proof. First of all, for every X ∈ A let ɛX : T H (X) → X be the counit of theadjunction (T, H). Then, in view of Proposition A.3 and Proposition <strong>2.</strong>32, ɛX is anisomorphism and for every morphism f : X → Y in A we havef ◦ ɛX = ɛY ◦ T H (f) .Thus f is mono (resp. epi) if and only if T H (f) is mono (resp. epi).Lemma A.6. In the assumptions and notations of A.1, let m, n ∈ N, m, n ≥ 1,let f : B m → B n be a morphism of right B-modules, let X = Coim ( f ) and letj : X → B n be the canonical injection. Let T : Mod-B → A be a left adjoint of thefunctor Hom A (U, −) : A → Mod-B. Then T (j) is a monomorphism.Proof. Let m, n ∈ N, m, n ≥ 1, let f : U m → U n be a morphism in A. Let usconsider the diagram□0 Ker (f)k U m fU n iCoim (f)00where p is the canonical epimorphism and i is the canonical monomorphism. Byapplying to it the functor H = Hom A (U, −), in view of Lemma A.4, we obtain thep