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

∐Hom B (T P, T (M i )). Then we have the following commutative diagrami∈I(Hom A P, ∐ )M ii∈Iφ ` P, M ii∈I∇(Hom A (P,ε i )) i∈I(Hom B T P, ∐ ∇(HomB (T P,T (ε i ))) i∈I(T M i ))i∈I∐Hom A (P, M i )i∈I`φ P,Mii∈I∐Hom B (T P, T (M i ))i∈I259where we observed that the first row is an isomorphism and also the φ’s are isomorphism.Then we deduce that ∇ (Hom B (T P, T (ε i ))) i∈Iis an isomorphism, so thatT P preserves coproducts, i.e. T P is finite.4) Let {Q i } i∈Ibe a direct family of subobjects of T P such that T P = ∑ i∈IQ i .Then, since T is an equivalence, there exists a direct family {P i } i∈Iof subobjects ofP such that T P i = Q i for every i ∈ I and P = ∑ i∈IP i . Since P is finitely generated,there exists an index i 0 ∈ I such that P = P i0 and then T P = T P i0 = Q i0 , i.e. T Pis finitely generated.5) By Proposition A.18, P is finite and thus by 3) we deduce that T P is alsofinite. Since T P is moreover projective, by Proposition A.18 we conclude that T Pis finitely generated.6) By Proposition A.18, since P is finite projective, P is finitely generated andprojective. Then we conclude by 5).□Definition A.20. Let A be an abelian category. A finite projective generator P inA is called progenerator.Corollary A.2<strong>1.</strong> Let A be a Grothendieck category. There exists an equivalenceF : A → Mod-B, where B is a ring, if and only if A contains a progenerator P .Moreover• If P is a progenerator of A, then Hom A (P, −) : A → Mod-T where T =Hom A (P, P ) .• If F is an equivalence, there exists a progenerator P in A such thatHom A (P, P ) ≃ B and F ≃ Hom A (P, −).Proof. Assume first that A contains a progenerator P . Let B = Hom A (P, P ) andconsider the functor Hom A (P, −) : A → Mod-B. Since P is a generator and A is aGrothendieck category, by Proposition A.3 we deduce that Hom A (P, −) is full andfaithful. Hence we only have to prove that Hom A (P, −) is surjective on the objects.Let M ∈ Mod-B. Then we have the following exact sequence in Mod-BB (X) −→ B (M) −→ M → 0.Since B = Hom A (P, P ) we can rewrite is asHom A (P, P ) (X) −→ Hom A (P, P ) (M) −→ M → 0.