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9. Derived Functors a Functor F MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 27 9. Derived functors AfunctorF : between abelian categories , is called left exact if it is an A!B A B j p additive functor which takes every exact sequence 0 X Y Z 0in to a left exact sequence ! ! ! ! A Fj Fp 0 FX FY FZ ! ! ! in .Forexample,F (X) = Hom (A, X)isaleftexactfunctor B A Hom (A0, ): Mod-Z. A − A! We would like a formula for the cokernel of Fp : FY FZ. The “right derived functors” R1F, R2F, are the! answer to this problem. This sequence of functors complete the left exact sequence··· to a canonical long exact sequence: 0 FX FY FZ R1FX R1FY R1FZ R2FX R2FY ! ! ! ! ! ! ! ! !··· The formula for RiF also gives R0F = F . Contravariant left exact functors also give right derived functors. For example F = Hom ( ,B0)issuchafunctor.Thesearefunctorswhichtakeeachshortexactsequence 0 AX − Y Z 0toleftexactsequences ! ! ! ! 0 FZ FY FX ! ! ! and there right derived functors RiF extend this to a long exact sequence 0 FZ FY FX R1FZ R1FY R1FX R2FZ ! ! ! ! ! ! ! !··· Analogously, covariant and contravariant right exact functors (such as R B)give left derived functors (extending the sequence to the left): ⌦ L FZ L FX L FY L FZ FX FY FZ 0 ···! 2 ! 1 ! 1 ! 1 ! ! ! ! In this section we construct right derived functors and verify their properties. 9.1. Construction of RiF . Assume the category is abelian with enough injectives. Then each object X of has an injective coresolutionA X Q unique up to chain homotopy equivalence. A ! ⇤ Definition 9.1. The right derived functors RiF of F are defined to be the ith cohomology of the cochain complex F (Q ): ⇤ 0 F (Q ) F (Q ) F (Q ) ! 0 ! 1 ! 2 !··· In the case F = Hom (A, ), the right derived functors are the Ext functors: R − i i i ExtR(A, X):=R F (X)=H (HomR(A, Q )) ⇤ Note that the derived functors are only well-defined up to isomorphism. If there is another choice of injective coresolutions Q0 then Q Q0 which implies that F (Q ) ⇤ ⇤ ' ⇤ ⇤ ' F (Q0 )whichimpliesthat ⇤ i i H (F (Q )) = H (F (Q0 )) ⇤ ⇠ ⇤ By definition of F (Q )wetakeonlytheinjectiveobjects.ThetermF (X)isdeliber- ately excluded. But F is⇤ left exact by assumption. So we have an exact sequence 0 F (X) F (Q ) F (Q ) ! ! 0 ! 1 28 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA Thus Theorem 9.2. The zero-th derived functor R0F is canonically isomorphic to F .In particular, 0 ExtR(A, X) ⇠= HomR(A, X) 1 Example 9.3. Compute ExtZ(Z3, Z2). Take an injective coresolution of Z2: j 0 Z2 Q/2Z Q/Z 0 ! ! ! ! This gives a left exact sequence j] 0 Hom(Z3, Z2) Hom(Z3, Q/2Z) Hom(Z3, Q/Z) ! ! ! 1 Then, by definition, ExtZ(Z3, Z2)isthecokernelofj] (which is a monomorphism since Hom(Z3, Z2)=0). 1 Claim: This map is an isomorphism and, therefore, Ext (Z3, Z2)=0. Proof: A homomorphism Z/3 Q/Z is given by its value on the generator 1 + 3Z ! of Z/3Z. This must be a coset a/b + Z so that 3a/b Z.Inotherwordsb =3and 2 a =0, 1or2.Similarly,ahomomorphismZ/3 Q/2Z sends the generator of Z/3toa ! coset a/b +2Z so that 3a/b 2Z.So,b =3anda =0, 2or4.So,bothoftheregroups 2 have exactly three elements. We know j] is a monomorphism. So, it is a bijection. 9.2. connecting morphisms. One of the basic properties of the derived functors is that they fit into a long exact sequence where some maps are induced from the given maps ↵,β in the short exact sequence and every third map is a mysterious morphism called the “connecting map”. The main technical detail is to construct this map. There are several ways to do this. We can “chase the diagram” (the traditional method) or we can “rotate” the exact sequence (a conceptual method which is more difficult but makes some sense) Theorem 9.4. Given any short exact sequence β 0 A ↵ B C 0 ! ! ! ! there is a sequence of morphisms (the connecting morphisms) δ : RnF (C) Rn+1F (A) n ! making the following sequence exact: 0 F (A) F (B) F (C) δ0 R1F (A) R1F (B) ! ! ! ! ! ! R1F (C) δ1 R2F (A) R2F (B) R2F (C) δ2 R3F (A) ! ! ! ! !··· Furthermore, δn is natural in the sense that, given any commuting diagram with exact rows: ↵ β 0 / A / B / C / 0 f g h ✏ ✏ ✏ ↵0 β0 0 / A0 / B0 / C0 / 0 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 29 we get a commuting square: δn RnF (C) / Rn+1F (A) h f ⇤ ⇤ ✏ ✏ n δn n+1 R F (C0) / R F (A0) Here is one construction of these δ operators. First I needed the following lemmas. Lemma 9.5. If Q is injective then RnF (Q)=0for all n 1. ≥ Proof. The injective coresolution of Q has only one term: 0 Q Q 0 ! ! ! So, R⇤F (Q)isthecohomologyoftheexactsequencewithonlyoneterm: 0 F (Q) 0 ! ! and this is in degree 0. ⇤ Lemma 9.6. If 0 A Q K 0 is a short exact sequence where Q is injective, then we have an exact! sequence! ! ! 0 F (A) F (Q) F (K) ⇡ R1F (A) 0 ! ! ! ! ! and n n+1 R F (K) ⇠= R F (A) for all n 1. ≥ Proof. We can use Q = Q0 as the beginning of an injective coresolution of A: j j j j 0 A Q 0 Q 1 Q 2 Q ! ! 0 ! 1 ! 2 ! 3 !··· Since coker j ⇠= im j0 ⇠= ker j1 ⇠= K,wecanbreakthisupintotwoexactsequences: j 0 A Q K 0 ! ! 0 ! ! j j 0 K Q 1 Q 2 Q ! ! 1 ! 2 ! 3 !··· The second exact sequence shows that the injective coresolution of K is the same as that for A but shifted to the left with the first term deleted. So, n n+1 R F (K) ⇠= R F (A) for all n 1. When≥n =0wehave,byleftexactnessofF ,thefollowingexactsequence: (j1) 0 F (K) F (Q ) ⇤ F (Q ) ! ! 1 −−! 2 In other words, F (K) = ker(j1) .Theimageof(j0) : F (Q0) F (Q1) lands in ⇠ ⇤ ⇤ ! F (K)=ker(j1) .Thecokernelisbydefinitionthefirstcohomologyofthecochain complex F (Q )whichisequalto⇤ R1F (A). So, we get the exact sequence ⇤ F (Q ) F (K) ⇡ R1F (A) 0 0 ! ! ! We already know that the kernel of F (Q ) F (K)isF (A)sothisprovesthelemma. 0 ! ⇤ 30 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA The construction of the delta operator proceeds as follows. Start with any short exact sequence 0 A B C 0. Then choose an injective coresolution of A: ! ! ! ! j j j j 0 A Q 0 Q 1 Q 2 Q ! ! 0 ! 1 ! 2 ! 3 !··· Let K =kerj1 =imj0 =cokerj.SinceQ0 is injective, the map A Q0 extends to B and cokernels map to cokernels giving a commuting diagram: ! ↵ β (9.1) 0 / A / B / C / 0 idA f g ✏ j ✏ p ✏ 0 / A / Q0 / K / 0 The map g : C K induces a map g : RnF (C) RnF (K)andtheconnecting morphism δ for n! 1canbedefinedtobethecomposition:⇤ ! n ≥ n g n n+1 δ : R F (C) ⇤ R F (K) = R (A). n ! ⇠ When n =0,themapontherightisanepimorphism,notanisomorphism: g ⇡ 1 δ : F (C) ⇤ F (K) R (A). 0 ! ! We will skip the following proof that δ is independent of the choice of g : C K. ⇤ ! This follows from the fact that, for any other choice g0,thedi↵erenceg g0 lifts to Q0 since − n f f 0 : B Q0 is zero on A and therefore factors through C.So,g g0 = R F (g g0) − ! n ⇤ − ⇤ − factors through R F (Q0)=0sog = g0 . ⇤ ⇤ ↵ β 0 / A / B / C / 0 h 0 f f g g − 0 − 0 ✏ j ✏ { p ✏ 0 / A / Q0 / K / 0 h β = f f 0 p h = g g0 ◦ − ) ◦ − n 1 n ↵ n β n R − F (C) / R F (A) / R F (B) / R F (C) / 0 h 0 (f f 0) ⇤ g g0 =0 − ⇤ ⇤− ⇤ ✏ = if n 2 ✏ j ✏ w p ✏ n 1 ⇠ ≥ n n n R − F (K) / / R F (A) / R F (Q0)=0 / R F (K) / 0 In the case n =0,g g0 = F (g g0)stillfactorsthroughF (Q0), which is not zero in general. But this is⇤ OK− since⇤ F (Q− )goestothekernelof⇡ : F (K) R1F (A). So, 0 ! ⇡ (g g0 )=0makingδ0 = ⇡ g = ⇡ g0 . ⇤ ⇤ − ⇤ ⇤ ◦ ⇤ ⇤ ◦ ⇤.
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