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).