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MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 1 1. Additive categories From Lang III 3: Abelian categories we take only the first two axioms: AB1, AB2. § Definition 1.1. An additive category is a category for which every hom set Hom (X, Y )isanadditivegroupand C C AB1a composition is biadditive,i.e.,(f + f ) g = f g + f g and f (g + g )= 1 2 ◦ 1 ◦ 2 ◦ ◦ 1 2 f g1 + f g2. AB1b ◦has a zero◦ object 0 having the property that Hom (X, 0) = 0 = Hom (0,X) forC all objects X. C C AB2 has finite products and coproducts (sums). C Theorem 1.2. In an additive category, finite products and finite sums are isomorphic, i.e., the canonical morphism A A is an isomorphism. i ! i Proof. By induction it sufficesL to considerQ sums and products of pairs. We will show A B = A B.Considerthefollowingdiagram. ⊕ ⇠ ⇥ idA (1.1) A / A ; jA pA jA # f g # A B / A B / A B ; ⊕ ⇥ ; ⊕ jB pB jB idB # B / B where f : A B A B is the unique morphism so that pA f jA = idA,pB f jB = id and the⊕ other! two⇥ compositions are zero: p f j ◦=0=◦ p f ◦j ◦.Let B B ◦ ◦ A A ◦ ◦ B g : A B A B be the sum g = jA pA + jB pB.Then,itiseasytoseethatf,g are inverse⇥ ! morphisms:⊕ ◦ ◦ (1) g f j = j p f j + j p f j = j id +0 = j and, ◦ ◦ A A ◦ A ◦ ◦ A B ◦ B ◦ ◦ A A ◦ A A similarly, g f jB = jB.Bydefinitionofcoproduct,thereisauniquemorphism h : A B ◦ A◦ B satisfying h j = j and h j = j .Sinceh = g f and ⊕ ! ⊕ ◦ A A ◦ B B ◦ h = idA B are two solutions, we have g f = idA B. ⊕ ◦ ⊕ (2) pA f g = pA f jA pA + pA f jB pB = pA and, similarly, pB f g = pB. So,◦f ◦g is the◦ identity◦ ◦ on A B◦ . ◦ ◦ ◦ ◦ ◦ ⇥ Therefore, A B and A B are canonically isomorphic in any additive category. ⊕ ⇥ ⇤ Example 1.3. Some examples of additive categories: (1) The category Mod-R of all right R-modules. (2) The category mod-R of finitely generated right R-modules and R-linear homo- morphisms. E.g., mod-Z is the category of finitely generated abelian groups. (3) The category tor-Z of finite abelian groups. (4) The category (Z) of finitely generated free abelian groups (groups isomorphic P to Zn for some finite n). The first and third categories are “abelian” but the other two are not. I don’t want to spend time formally defining what is an abelian category. But I will point out the problems with the second and fourth categories and how to fix them. 2MATH131B:ALGEBRAIIPARTA:HOMOLOGICALALGEBRA 2. Kernels and cokernels In a general additive category, a morphism might not have a kernel or cokernel. However, if they do exist, they are unique. Definition 2.1. The kernel of a morphism f : A B is an object K with a morphism j : K A so that ! ! (1) f j =0:K B (2) For◦ any other! ob ject X and morphism g : X A so that f g =0thereexists aunique˜g : X K so that g = j g˜ (such a!g ˜ is called a lifting◦ of g to K). ! ◦ Since this is a universal property, the kernel is unique (up to isomorphism) if it exists. X !˜g 0 9 g ~ j ✏ f K / A / B Theorem 2.2. K is the kernel of f : A B if and only if ! j f 0 Hom (X, K) ] Hom (X, A) ] Hom (X, B) ! C ! C ! C is exact for any object X. Here f] is left composition with f. Recall that a morphism f : A B is a monomorphism if, for any two maps ! g = h : C A, f g = f h,i,e,f] : Hom (X, A) Hom (X, B) is a monomorphism of6 groups. ! ◦ 6 ◦ C ! C Corollary 2.3. A morphism f : A B is a monomorphism i↵ ker f =0. ! f Notation 2.4. We say that 0 K A B is exact if K =kerf. ! ! ! Remark 2.5. A non-example:Thecategorymod-R of f.g. right R-modules does not have kernels if R is not right Noetherian (right ideals are finitely generated). To see this suppose that I is a right ideal which is not finitely generated. Then IR as R-module is not an object in the category. So, the morphism R R/I does not have a kernel in the category mod-R. ! Cokernel is defined analogously and satisfies the following theorem which can be used as the definition. Theorem 2.6. The cokernel of f : A B is an object C with a morphism p : B C so that ! ! p] f ] 0 Hom (C, Y ) Hom (B,Y ) Hom (A, Y ) ! C ! C ! C is exact for any object Y where f ] is right composition with f. The standard diagram defining coker f is the following. f p A / B / C g 0 !g ✏ 9 Y Again, letting C =0wegetthestatement: MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 3 Corollary 2.7. A morphism f is an epimorphism (i.e., f ] is a monomorphism) if and only if coker f =0. f Notation 2.8. We say A B C 0isexact if C =cokerf. f g ! ! ! We say 0 A B C 0isexact if A =kerg and C =cokerf. ! ! ! ! These two theorems can be summarized by the following statement. Corollary 2.9. For any additive category , Hom is left exact in each coordinate. C C Exercise 2.10. (1) Show that mod-R has cokernels. (2) In the category of f.g. free abelian groups, show that the morphism f : Z Z 2 ! given by multiplication by 2 is an epimorphism: 0 Z Z 0isexact!! ! ! ! 3. Projective and injective objects We will discuss the definition of projective and injective objects in any additive cat- egory. And it was easy to show that the category of R-modules has sufficiently many projectives. But the category of free R-modules does not. Definition 3.1. An object P of an additive category is called projective if for any epimorphism p : A B and any morphism g : P C B there exists a morphism g˜ : P A so that p !g˜ = g.Themap˜g is called a lifting! of g to A. ! ◦ A ? g˜ 9 p g ✏ P / B Theorem 3.2. P is projective if and only if Hom (P, ) is an exact functor. C − Proof. If 0 A B C 0 is exact then, by left exactness of Hom we get an exact sequence: ! ! ! ! 0 Hom (P, A) Hom (P, B) Hom (P, C) ! C ! C ! C By definition, P is projective if and only if the last map is always an epimorphism, i.e., i↵we get a short exact sequence 0 Hom (P, A) Hom (P, B) Hom (P, C) 0 ! C ! C ! C ! ⇤ Theorem 3.3. Any free R-module is projective. Proof. Suppose that F is free with generators x↵.TheneveryelementofF can be written uniquely as x↵r↵ where the coefficients r↵ R are almost all zero (only finitely many are nonzero). Suppose that g : F B is2 a homomorphism. Then, for P ! every index ↵,theelementg(x↵)comesfromsomeelementy↵ A.I.e.,g(x↵)=f(y↵). Then a liftingg ˜ of g is given by 2 (3.1)g ˜( xar↵)= y↵r↵ The verification that this is a liftingX is straightforward.X ⇤ 4MATH131B:ALGEBRAIIPARTA:HOMOLOGICALALGEBRA To do the last step in full detail, we use the universal property of free modules: A homomorphism from the free module F generated by a set S to another module M is uniquely determined by a set mapping S M which is arbitrary, i.e., (3.1) defines the ! unique homomorphism F A sending x↵ to y↵. For every R-module M !there is a free R-module which maps onto M, namely the free module F generated by symbols [x]forallx M and with projection map p : F M given by 2 ! p r↵[x↵] = r↵x↵ The notation [x]isusedtodistinguishbetweentheelement⇣X ⌘ X x M and the corresponding generator [x] F . Using the universal property of free modules,2 the homomorphism p is defined by the2 equation p[x]=x. For M afinitelygeneratedR-module, take a finite generating set S.Thenthefree R-module generated by S maps onto M..
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