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We Can Immediately Check the Following Properties, Hence, We Will Skip the Proof We can immediately check the following properties, hence, we will skip the proof. Proposition 133. Let A be an abelian category with enough injectives, and let F : A ! B be a covariant left exact functor to another abelian category B. Then (1) For each i 0, the functor RiF is an additive functor from A to B. It is independent ≥ of the choices of injective resolutions. (2) There is a natural isomorphism R0F F . ' (3) For each short exact sequence 0 A A A 0 and for each i 0, there ! 0 ! ! 00 ! ≥ is a natural morphism δi : RiF (A ) Ri+1F (A ), such that we have a long exact 00 ! 0 sequence i i i i δ i+1 R F (A0) R F (A) R F (A00) R F (A0) . ···! ! ! ! !··· (4) Given a morphism of the exact sequence of (c) to another 0 B B B 0, ! 0 ! ! 00 ! the δ’s give a commutative diagram i i δ i+1 R F (A00) / R F (A0) ✏ i ✏ i δ i+1 R F (B00) / R F (B0) (5) For each injective object I of A, we have RiF (I)=0for every i>0. Definition 134. Let F : A B be as above. An object J A is acyclic if RiF (J)=0 ! 2 for every i>0. We may use an acyclic resolution instead of an injective resolution to compute the derived functor: Proposition 135. Let F : A B be as above. Let A A be an object such that it ! 2 admits a resolution 0 A J 0 J 1 ! ! ! !··· i where each J is acyclic for F (we say J • is an F -acyclic resolution of A). Then for each i 0, there is a natural isomorphism RiF (A) hi(F (J )). ≥ ' • We will see that this definition of derived functors is very natural, by showing a universal property. We need to generalize our notion slightly. Definition 136. Let A, B be abelian categories. A covariant δ-functor from A to B is i i i i+1 a collection of functors T • =(T )i 0, together with a morphism δ : T (A00) T (A0) ≥ ! for each short exact sequence 0 A A A 0 and each i 0 such that ! 0 ! ! 00 ! ≥ (i) For each short exact sequence above, we have a long exact sequence i i i i δ i+1 T (A0) T (A) T (A00) T (A0) . ···! ! ! ! !··· 53 Topic 3 – Derived functors and cohomology (ii) For each morphism of a short exact sequence to another short exact sequence 0 B B B 0, we have a commutative diagram ! 0 ! ! 00 ! i i δ i+1 T (A00) / T F (A0) ✏ i ✏ i δ i+1 T (B00) / T F (B0) i A δ-functor T =(T )isuniversal if, given any other δ-functor T 0, and any given mor- phism of functors f 0 : T 0 T 0, there is a unique sequence of morphisms f i : T i T i ! 0 ! 0 for each i 0, starting with the given f 0, which commute with the δi for each short ≥ exact sequence. Proposition 137. Assume that A is an abelian category with enough injectives. For any i left exact functor F : A B, the derived functors (R F )i 0 : A B form a universal ! ≥ ! δ-functor with F R0F . Conversely, if T =(T i) is any universal δ-functor from A to ' B, then T 0 is left exact, and the T i are isomorphic to RiT 0 for every i 0. ≥ Hence, the right derived functor is a unique way to extend naturally a left exact functor. To conclude the construction, it only remains to show that the category of sheaves of abelian groups has enough injectives. Proposition 138. Let X be any topological space. The category Ab(X) of sheaves of abelian groups on X has enough injectives. Proof. We will first prove that the caegory Mod(X) of -modules has enough injectives OX when (X, ) is a ringed space. Let be an -module. For each point x X,the OX F OX 2 stalk is an -module. Since every module is a submodule of an injective module, Fx OX,x there is an injection , I where I is an injective -module. Let j : x , X Fx ! x x OX,x x { } ! be the inclusion map. The sheaf := x X (jx) (Ix), considered Ix as a (constant) sheaf I ⇤ on the one point set x , becomes injective:2 { } Q Let be any X -module. We have Hom( , )= x X Hom( , (jx) (Ix)) G O G I G ⇤ since the direct product commutes with the Hom functor.2 Note that each Q Hom( , (jx) (Ix)) is isomorphic to a module homomorphism group Hom ( x,Ix). G ⇤ OX,x G Since each I is an injective -module, the functor Hom( , )isadirect x OX,x − I product of exact functors, which is also exact. There is a natural morphism of -modules , obtained from the local maps OX F !I , . It is obviously injective, hence, is isomorphic to a subsheaf of an injective Fx !Ix F -module I . We conclude that the category Mod(X) has enough injectives. OX Ab(X) also has enough injectives since Ab(X) coincides with Mod(X) when we assign a sheaf of rings X on a topological space X as the constant sheaf Z on X. O 54 2 Sheaf cohomology and its properties Now we are ready to define the sheaf cohomology as derived functors. We also define the higher direct image functor, since it follows from the exactly same argument. Definition 139. Let X be a topological space. We define the cohomology functors Hi(X, ) to be the right derived functor of the global section functor Γ(X, ). For any − − sheaf of abelian groups on X, the groups Hi(X, ) are the cohomology groups of . F F F If f :(X, ) (Y, ) is a morphism of ringed spaces, we define higher direct images OX ! OY Rif ( ) to be the right derived functor of the direct image functor f . ⇤ − ⇤ Remark 140. The category Mod(X) of X -modules does not have enough projectives 1 O in general. Let X = P be the projective line with the Zariski topology, and let X = Z k O be the constant sheaf associated to Z.IfMod(X) has enough projectives, then there would be a surjection ' : P from a projective sheaf P. In particular, the induced !OX map on a stalk ' is surjective for every x X. x 2 Let denote the sheaf on X obtained by restricting onto U and then extending by OU OX zero outside of U. Let U be an open neighborhood of x, and let V be a strictly smaller open neighborhood. Consider the surjection X x V X .SinceP is projective, O \{ } ⊕O !O the surjection ' lifts to P X x V . The map P(U) OX (U) factors through !O \{ } ⊕O ! X x (U) V (U) = 0, hence, any map P(U) X (U) is zero for any choice of open O \{ } ⊕O !O neighborhood U of x. We conclude that ' : P is zero since it gives a zero map !OX on each stalk. Next, we see that any injective -module on a ringed space is acyclic. In particular, an OX injective resolution is an example of an acyclic resolution for the global section functor. We will have a small detour; passing through the notion of flasque sheaves. Lemma 141. Let (X, ) be any ringed space. Any injective -module is flasque, OX OX I i.e., every restriction map (U) (V ) is surjective. I !I Proof. Let U X be an open subset. Let denote the sheaf j ,whichisthe ✓ OU !OX |U restriction of to U, extended by zero outside of U. If we have any inclusion of open OX sets V U, we have an inclusion 0 . Hence, for any injective -module ✓ !OV !OU OX , we have a surjection I Hom( , ) Hom( , ) 0. OU I ! OV I ! On the other hand, the functor Hom( , ) is same as the global section functor Γ(U, ) OU − − taking on U, in particular, the above surjection identifies with the restriction map (U) (V ). I !I Then it is easy to see that any flasque sheaf is acyclic: Proposition 142. Let be a flasque sheaf on a topological space X. Then Hi(X, )=0 F F for all i>0. 55 Topic 3 – Derived functors and cohomology Proof. Let be an injective sheaf in Ab(X) containing , and let be its quotient. We I F G have a short exact sequence 0 0. !F!I!G! Since is flasque, we have a short exact sequence on sections F 0 (U) (U) (U) 0 !F !I !G ! for every open set U X. Now is also flasque, one can immediately check that is ✓ I G also flasque. Since is injective, we have Hi(X, ) = 0 for all i>0. I I Therefore, from the long exact sequence of cohomology, we have H1(X, ) = 0 and F Hi+1(X, ) Hi(X, ) for every i 1. But is also flasque, hence, we get the desired F ' G ≥ G result by the induction on i. In particular, we can calculate the cohomology using a flasque resolution, which does not depend on the choice of the sheaf of rings . It means, there is no di↵erence between OX the right derived functor of Γ(X, ):Ab(X) Ab (= the sheaf cohomology functor − ! Hi(X, )) and the right derived functor of Γ(X, ):Mod(X) Ab. It does not depend − − ! on the choice of the ring structure on a topological space X.
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