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

yields the canonical section h F : A (Z F ) → A (Z) of the canonical projection π F :A (Z) → A (Z F ) since π F (e z ) = e F z for every z ∈ F . Set i F = π F ◦ j ◦ j F . Then wehaveIm (j ◦ j F ) ⊆ ∑ z∈F e zA = ∑ z∈F (h F ◦ π F ) (e z ) Aand sincewe obtain that(h F ◦ π F ) (e z ) = e z for every z ∈ F(263) h F ◦ i F = h F ◦ π F ◦ j ◦ j F = j ◦ j F .Assume now that F, G ∈ P 0 (X) and that F ⊆ G. Then Z F ⊆ Z G so that wecan consider the canonical section h G F : A(Z F ) → A (ZG) of the canonical projectionπF G : A(ZG) → A (Z F ) . We have( ) ( )πFG eGz = eFz and h G F eFz = eGz for every z ∈ Z F .Moreoverh G ◦ h G F = h F .Let j G F : X F → X G be the canonical inclusion. Thenso that we getj G ◦ j G F = j F and π G ◦ h F = h G Fi G ◦ j G F = π G ◦ j ◦ j G ◦ j G F = π G ◦ j ◦ j F =(263)= π G ◦ h F ◦ i F = h G F ◦ i Fand henceh G ◦ i G ◦ jF G = h G ◦ h G F ◦ i F = h F ◦ i Fwhich proves that (h F ◦ i F ) F ∈P0 (X)is a compatible family of morphisms. Sincelim −→ (X F ) F ∈P0 (X)= X, to prove thatlim (h F ◦ i F ) = j−→it is enough to prove thatj ◦ j F = h F ◦ i Ffor every F ∈ P 0 (X) . This holds in view of (263).Lemma A.8. Let f : X → Y and p : W → X be morphisms in an abelian categoryA. Assume that p is an epimorphism and thatThen f is a monomorphism.Ker (f ◦ p) = Ker (p) .Proof. Since p is an epimorphism, we have that Coker (Ker (p)) = (X, p). It followsthat Coker (Ker (f ◦ p)) = (X, p). Let f ◦ p : Coker (Ker (f ◦ p)) → Ker (Coker (f ◦ p))be the isomorphism such that(264) f ◦ p = k ◦ f ◦ p ◦ p247□