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

Since j G : Ker (λ G ) → U (G) there exists a unique morphism i ̂GF : Ker (λ F ) → Ker (λ G )such thati G F ◦ j F = j G ◦ i ̂GF .We want to prove that β G ◦ ̂ i G F = β F for every F, G finite subsets of Hom A (U, X) .Let us computeu ◦ β G ◦ i ̂GF= j ◦ h G ◦ i ̂GF = i G ◦ j G ◦ i ̂GF = i G ◦ i G F ◦ j F = i F ◦ j F= j ◦ h F = u ◦ β F .Since u is mono we conclude. Let us consider the exact sequence0 → Ker (λ F ) j F−→ U (F ) λ−→FXSince A is a Grothendieck category, we have that lim → are exact and hence we getthe exact sequence0 → lim Ker (λ F ) lim→ j F−→ lim U (F ) = U (Hom A(U,X)) lim→ λ F =λ−→ X.→ →It follows that Ker (λ) = lim → Ker (λ F ) and hence there exists a unique monomorphismβ = lim → β F : lim → Ker (λ F ) = Ker (λ) → Ker (µ) such thatβ ◦ h F = β F for every finite subset F of Hom A (U, X) .Since for every finite subset F of Hom A (U, X)243we get thatand henceu ◦ β ◦ h F = u ◦ β F = j ◦ h Fu ◦ β = jµ ◦ j = µ ◦ u ◦ β = 00 Ker (λ)j U (Hom A(U,X))λXtp∼ k∼Coker (j) Ker (χ)eµZλχ=0 Coker (λ)Therefore there exists a unique morphism ˜µ : Im (λ) ≃ Coker (j) → Z such that˜µ◦p = µ where p : U (Hom A(U,X)) → Coker (j) is the canonical projection. By LemmaA.2, we have that Im (λ) = X and then X = Im (λ) = Coker (j). Then there existsan isomorphism t : X → Coker (j) such that t ◦ λ = p. Set g = ˜µ ◦ t and for everyf ∈ Hom A (U, X), we computeg ◦ f = ˜µ ◦ t ◦ f = ˜µ ◦ t ◦ λ ◦ i f = ˜µ ◦ p ◦ i f = µ ◦ i f = ϕ (f) .This means that ϕ = Hom A (U, g).□