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

257Since f i = 0 for every i ∈ I\F we have∑ε i ◦ f i = ∑i∈Ii∈Fso that it is sufficient to prove thatε i ◦ f if = ∑ i∈Fε i ◦ f iand by (267) and (268)t ◦ s ◦ f = ∑ i∈Ft ◦ h ◦ ε F i ◦ f i .Since t is mono we only need to prove thats ◦ f = ∑ i∈Fh ◦ ε F i◦ f iand thus, for every j ∈ F, thatπ j ◦ h −1 ◦ s ◦ f = π j ◦ h −1 ◦ ∑ i∈Fh ◦ ε F i ◦ f i .Let us computeOn the other handπ j ◦ h −1 ◦ s ◦ f (271)= q j ◦ h ◦ h −1 ◦ s ◦ f(266)= q j ◦ t ◦ h ◦ h −1 ◦ s ◦ f= q j ◦ t ◦ s ◦ f (267)= q j ◦ f (272)= f j .π j ◦ h −1 ◦ ∑ i∈Fh ◦ ε F iπ j ◦ h −1 ◦ h ◦ ε F iπ j ◦ ε F i◦ f i◦ f i = ∑ i∈F◦ f i = ∑ i∈F(269)= ∑ δ ij ◦ f i = f ji∈Fso that we conclude that ∐ (Hom A (P, B i ) ∇(Hom(P,ε i)) i∈I−→ HomA P, ∐ )B i is an epimorphism.Let now (f i ) i∈Ii∈Ii∈I∈ ∐ Hom A (P, B i ) where f i ’s are almost all zero, be( i∈I ) ∑such that ∇ (Hom (P, ε i )) i∈I (fi ) i∈I = ε i ◦ f i = 0. Since f i ’s are almost all zero,i∈Ilet F ⊆ I be a finite subset such that f i ≠ 0 for every i ∈ F and f i = 0 for everyi ∈ I\F . Then 0 = ∑ (265)ε i ◦ f i = ∑ h ◦ ε F i ◦ f i . Since h is mono, we also havei∈Ii∈F0 = ∑ (ε F i ◦ f i ∈ Hom A P, ∐ )B i so that, for every j ∈ F,i∈Fi∈F0 = π j ◦ ∑ i∈Iε F i◦ f i = ∑ i∈Iπ j ◦ ε F i ◦ f i(269)= f jand thus f i = 0 for every i ∈ I.□