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

256We will prove that h : ∐ B i → ∑ ε i (B i ) is in fact an isomorphism. We define thei∈F i∈Ffamily (η j ) j∈Iby settingη j = 0 for every j ∈ I\F and η j = ε F j for every j ∈ F .Then we can take ∇ (η i ) i∈I: ∐ B i → ∐ B i . Let us compute for every j ∈ Fi∈I i∈F∇ (η i ) i∈I◦ h ◦ ε F jdefh= ∇ (η i ) i∈I◦ ∇ (ε i ) i∈F◦ ε F j= ε F j = Id ` B i◦ ε F ji∈F(265)= ∇ (η i ) i∈I◦ ε jand thus ∇ (η i ) i∈I◦ h = Id ` B i. Therefore we deduce that h is mono and then h isi∈Fan isomorphism. Let us consider (δ ij : B i → B j ) j∈Ithe family defined by settingδ ii = Id Bi and δ ij = 0 for every j ≠ i.Since ∐ B i is a finite coproduct, we can view it as a product and call π j : ∐ B i =i∈Fi∈F∏B i = × i∈F→ B j the projections for every j ∈ F satisfyingi∈F(269) π j ◦ ε F i = δ ij and∑ε F i π i = Id ×.i∈FNote that, by the universal property of the coproduct, there exist q j : ∐ i∈IB i → B jsuch thati∈F(270) q j ◦ ε i = δ ij .Let us compute, for every i ∈ F and for every j ∈ I,and thusq j ◦ h ◦ ε F i = q j ◦ ε i(270)= δ ij(269)= π j ◦ ε F i(271) q j ◦ h = π j .We define the family (f i ) i∈I∈ ∐ i∈IHom A (P, B i ) by settingNote thati.e.f i = π i ◦ h −1 ◦ s ◦ f for every i ∈ F and f i = 0 for every i ∈ I\F .f i = π i ◦ h −1 ◦ s ◦ f (271)= q i ◦ h ◦ h −1 ◦ s ◦ f(266)= q i ◦ t ◦ h ◦ h −1 ◦ s ◦ f = q i ◦ t ◦ s ◦ f (267)= q i ◦ f(272) f i = q i ◦ f.For such a family (f i ) i∈I, we have to prove thatf = ∑ i∈Iε i ◦ f i .