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Triangulated Categories Part III

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Triangulated Categories Part III Triangulated Categories Part III Daniel Murfet October 5, 2006 The aim of this note is to prove the Brown Representability theorem. This was originally proved by Neeman [Nee01] but our presentation follows recent simplifications due to Krause [Kra02]. For most of this note the only required background is our Triangulated Categories Part I notes. Contents 1 Brown Representability 1 1.1 Dual Notions ....................................... 9 2 Compactly Generated Triangulated Categories 10 3 Portly Considerations 12 4 Representability for Linear Categories 14 1 Brown Representability First we introduce a portly abelian category A(S) for any preadditive category S. This will be a certain full subcategory of the category of all contravariant additive functors S −→ Ab. Modulo the fact that S is not assumed to be small, this is precisely what we call a right module over a ringoid in our Rings With Several Objects (RSO) notes. For background on portly abelian categories, the reader is referred to (AC,Section 2.4). Definition 1. Given a preadditive category A, the objects of the portly abelian category ModA = (Aop, Ab) are called right A-modules. A sequence of right modules M 0 −→ M −→ M 00 is exact in ModA if and only if the following sequence is exact in Ab for every A ∈ A M 0(A) −→ M(A) −→ M 00(A) Similarly kernels, cokernels and images in ModA are computed pointwise. See (AC,Corollary 59) and the proof of (AC,Proposition 44) for details. A morphism φ : M −→ N in ModA is a monomorphism or epimorphism if and only if φA : M(A) −→ N(A) has this property for every A ∈ A. For any object A ∈ A we have the right module HA = Hom(−,A): A −→ Ab defined in the obvious way. Proposition 1 (Yoneda). If A is a preadditive category, then (i) For any object A ∈ A and right A-module T there is a canonical isomorphism of abelian groups HomA(HA,T ) −→ T (A) defined by γ 7→ γA(1). (ii) The functor A 7→ HA defines a full additive embedding A −→ ModA. (iii) The objects {HA}A∈A form a (large) generating family of projectives for ModA. 1 Proof. All three results are proved in the usual way. See (RSO,Lemma 1), (RSO,Lemma 2) and (RSO,Proposition 3). Observe that since the object class of A is not assumed to be small, the family {HA}A∈A is not indexed by a set, so is not a generating family in the usual sense. In particular we cannot take coproducts and conclude that ModA has a generator. Remark 1. Given the Yoneda embedding A −→ ModA it is natural to identify A with its image in ModA. So at least in our intuition, we will tend to confuse A with HA and identify a morphism α : A −→ B with its corresponding natural transformation HA −→ HB. Definition 2. Let S be a preadditive category. We say that a right S-module F is coherent if there exists an exact sequence in ModS of the following form HA −→ HB −→ F −→ 0 (1) That is, F is the cokernel of some morphism of S considered as a morphism of S-modules. Clearly any representable functor is coherent. We denote by A(S) the full replete subcategory of ModS consisting of the coherent modules. At the moment we only know that this is a preadditive portly category. Remark 2. Let S be a preadditive category. Given a morphism ϕ : M −→ N in ModS and presentations HA / HB / M / 0 ϕ HA0 / HB0 / N / 0 there exist by projectivity vertical morphisms making the above diagram commute. Lemma 2. Let S be a preadditive category. If a right S-module F is coherent, then as a covariant functor Sop −→ Ab it preserves products. Proof. Suppose that F has a presentation of the form (1). Given a family of objects {Xi}i∈I in L S and a coproduct i Xi we have a commutative diagram in which the rows are exact L L L Hom( i Xi,A) / Hom( i Xi,B) / F ( i Xi) / 0 Q Q Q i Hom(Xi,A) / i Hom(Xi,B) / i F (Xi) / 0 The first two vertical morphisms are clearly isomorphisms, and therefore so is the third. This shows that F sends coproducts to products, as required. Remark 3. Suppose we have a commutative diagram of abelian groups a A / B (2) c d A0 / B0 b Taking cokernels of the rows, we get a morphism B/A −→ B0/A0. The cokernel of this morphism is the quotient of B0/A0 by the subobject (Im(d) + A0)/A0, so it is isomorphic to the quotient B0/(Im(d) + A0). In other words, it is the cokernel of the morphism B ⊕ A0 −→ B0 induced by the morphisms d, b. Now let S be a preadditive category, and suppose we are given a commutative diagram (2) in S. Mapping this diagram to ModS and taking cokernels of the rows yields a commutative diagram with exact rows HA / HB / M / 0 ϕ HA0 / HB0 / N / 0 2 There is an induced morphism ϕ : M −→ N. Let κ : HB0 −→ H be the cokernel of the morphism HB ⊕ HA0 −→ HB0 induced by the morphisms d, b. Then it is not difficult to check that the induced morphism N −→ H is the cokernel of ϕ in ModS. This observation motivates the proof of the next result. Lemma 3. Let S be an additive category. If ϕ : M −→ N is a morphism in A(S) then any cokernel of ϕ in ModS also belongs to A(S). Proof. In light of Remark 2 and Remark 3 we need only observe that the Yoneda embedding is ∼ additive, so HB ⊕ HA0 = HB⊕A0 and so H belongs to A(S). Definition 3. Let S be a preadditive category. A morphism u : X −→ Y is a weak kernel of v : Y −→ Z if vu = 0 and if any other morphism f : T −→ Y with vf = 0 factors through u (not necessarily uniquely). In other words, the induced sequence HX −→ HY −→ HZ is exact in ModS. If every morphism of S has a weak kernel, then we say that S has weak kernels. Lemma 4. Let S be an additive category with weak kernels. If ϕ : M −→ N is a morphism in A(S) then any kernel of ϕ in ModS also belongs to A(S). Proof. Choose presentations of M, N and construct a commutative diagram inducing ϕ as follows Ha HA / HB / M / 0 Hc Hd ϕ HA0 / HB0 / N / 0 Hb We have a commutative diagram in S a A / B c d A0 / B0 b Taking our cue from Remark 3 we try to construct a square sitting on top of this one, which induces a kernel of ϕ when we apply the Yoneda embedding and take cokernels. Take a weak kernel B00 −→ B ⊕ A0 of the morphism B ⊕ A0 −→ B0, and a weak kernel A00 −→ B00 ⊕ A of the morphism B00 ⊕ A −→ B. In other words, we have two weak kernels 0r1 @sA “d b” B00 / B ⊕ A0 / B0 0m1 @ n A “r a” A00 / B00 ⊕ A / B and a commutative diagram m A00 / B00 −n r A a / B Apply the Yoneda embedding to this diagram, and take a cokernel HB00 −→ K of Hm : HA00 −→ HB00 . We have an induced morphism ψ : K −→ M, which one checks is a pointwise kernel of ϕ : M −→ N. Since K is certainly coherent, the proof is complete. 3 Proposition 5. Let S be an additive category with weak kernels. Then A(S) is a portly abelian category. If S has coproducts then A(S) is cocomplete and the induced Yoneda functor H(−) : S −→ A(S) preserves coproducts. Proof. We know from Lemma 3 and Lemma 4 that A(S) has kernels and cokernels, which can be calculated in ModS. That is, the inclusion A(S) −→ ModS preserves kernels and cokernels. Since the Yoneda functor H(−) : S −→ ModS preserves finite coproducts, it is clear that A(S) is closed under finite products and coproducts in ModS, and is therefore portly abelian. Now suppose that S has coproducts, and that we are given a nonempty family {Fi}i∈I of objects of A(S). Choose presentations Hαi HAi / HBi / Fi / 0 and let H⊕iBi −→ F be the cokernel in ModS of H⊕iαi . We have a commutative diagram for each i ∈ I HAi / HBi / Fi / 0 H⊕iAi / H⊕iBi / F / 0 and therefore an induced morphism ui : Fi −→ F . Given Q ∈ A(S) and morphisms φi : Fi −→ Q it is easy to see that there is a morphism φ : F −→ Q with φui = φi. To prove uniqueness, suppose we have a morphism φ : F −→ Q with Q ∈ A(S) and φui = 0 for every i ∈ I. Lift the morphism φ to a morphism of presentations H⊕iAi // H⊕iBi / F / 0 φ HC / HD / Q / 0 Since φui = 0 for each i, we deduce that for each i ∈ I the composite HBi −→ H⊕iBi −→ HD vanishes on HD −→ Q, and therefore factors through HC −→ HD (using projectivity of HB). It is now not difficult to see that H⊕iBi −→ HD must factor through HC −→ HD, which implies immediately that the composite H⊕iBi −→ F −→ Q is zero. Since the first morphism is an epimorphism, we deduce that φ = 0 as required. This proves that A(S) is cocomplete, and as a L special case of the above construction we find that given a coproduct {ui : Ai −→ i Ai}i∈I in S, the morphisms Hui : HAi −→ H⊕iAi are a coproduct in A(S).
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