New PDF release: Lectures on Derived Categories

Algebraic Geometry

By Dragan Milicic

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5. Lemma. Let ϕ : M −→ N be a morphism in A[S −1 ] represented by a right roof . L }b •ddd f }} s d } d ∼ dd }} }} N M Then the following conditions are equivalent: (i) ϕ = 0; (ii) There exists t ∈ S such that t ◦ f = 0; (iii) There exists t ∈ S such that f ◦ t = 0. 6. Corollary. Let f : M −→ N be a morphism in A. Then the following conditions are equivalent: (i) Q(f ) = 0; (ii) There exists t ∈ S such that t ◦ f = 0; (iii) There exists t ∈ S such that f ◦ t = 0. Proof. The morphism Q(f ) is represented by the left roof || || | ∼ | |} | idM M Mf ff f ff ff f2 .

Hence, M is in B. It follows that B is a thick subcategory of A. Let B be a thick subcategory of A. Let SB be the class of all morphisms f : M −→ N in A such that ker f and coker f are in B. 5. Lemma. The class SB of morphisms in A is a localizing class. Proof. Clearly, if Aopp is the opposite category of A, the full subcategory of Aopp consisting of all objects in B is isomorphic to the opposite category of B, therefore we can denote it by B opp . Clearly, going from A to Aopp identifies SB with SBopp .

The morphism M −→ 0 has kernel equal to M and cokernel 0. Therefore, this morphism is in S, and M is isomorphic to 0 in A[SB−1 ]. On the other hand, if M is isomorphic to 0 in A[SB−1 ], the identity morphism on M represented by the left roof || || | | ∼ |} | idM M Mf ff ffidM ff f3 M has to be equal to the zero morphism represented by the left roof | || ||∼ | |} | idM M Mf ff ff0 ff f3 . M Therefore, there exists u : U −→ M such that the diagram M | y ff || u fffidM | ff || ∼ f3 |} | M M —f U ff |a ff∼ u ||| f || idM ff  || 0 M idM commutes and u ∈ SB .

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