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Home , Chain complex, Cone (category theory), Homological algebra, Kernel (category theory), Mapping cone (topology), Mapping cylinder

HOMOLOGICAL

Our goal will be to get far enough to develop some fluency working with spectral . We will follow Weibel. You should get errata for Weibel’s book because it has many errors.

Contents 1. Complexes of modules 1 2. Abelian categories 2 3. Algebraic constructions 5 4. Topologically-motivated constructions 7 5. Derived 10

1. Complexes of modules Let R be a (perhaps non-commutative).

Definition 1.1. A of R-modules is C = C• = ({Ci}, {di})i∈Z with {Ci}i a set of R-modules and di : Ci → Ci−1 satisfying di ◦ di−1 = 0. Remark 1.2. It is common in to omit indices when they are clear from context. For example, we will usually write the condition that di ◦di−1 = 0 as “d2 = 0”.

Definition 1.3. The cycles of C are Zi = Zi(C) := ker di, the boundaries of C are Bi := Bi(C) = im(di+1), and the of C is Hi(C) := Zi/Bi.

Example 1.4. Set Ci = Z/ h8i for i ≥ 0 and Ci = 0 for i < 0. Set di : n 7→ 4n for all i. One can check that  / h2i , i ≥ 1 Z Hi(C) = Z/ h4i , i = 0 0, else ♦ Definition 1.5. A of complexes C → D is a set of R-linear maps fi : Ci → Di such that the following commutes:

di Ci Ci−1

fi fi−1

di Di Di−1

A morphism is an if fi is an isomorphism for all i. We will denote the of chain complexes of R-modules by Ch(R). 1 2 HOMOLOGICAL ALGEBRA

Exercise 1.6. Show that a morphism of complexes sends cycles to cycles and bound- aries to boundaries and therefore induces a on homology. Show that each homology is a Hi : Ch(R) → Mod(R). Definition 1.7. A morphism of chain complexes is a quasi-isomorphism if it induces on all homology groups. Exercise 1.8. Check that 2 2 2 ··· Z/4 Z/4 Z/4 0 ···

··· 0 0 Z/2 0 ··· is a quasi-isomorphism.

Definition 1.9. If B,C ∈ Ch(R), then B is a subcomplex of C if Bn is a submodule of Cn for each n ∈ Z, and the differential of B is the restriction of the differential of C. If B is a subcomplex of C, then the C/B is the complex with (C/B)n = Cn/Bn with differential given by the induced maps on the .. Definition 1.10. If f : C → C0 is a morphism of complexes, then the ker(f) is the subcomplex (ker f)n = ker(fn) with differential the restriction of that of C. The and complexes, im(f) and coker(f), are defined similarly. • i i i Definition 1.11. A cochain complex is C = C = ({C }, {d })i∈Z with C R-modules and di : Ci → Ci+1 such that di+1di = 0 for all i. Definition 1.12. Let C be a cochain complex. The cocycles of C are Zi = ker di, the coboundaries are Bi = im di−1, and the is Hi = Zi/Bi.

Chain complexes and complexes are equivalent: given a chain complex C•, we i • obtain a cochain complex by setting C := C−i, and given a cochain complex C , −i we obtain a chain complex C• by Ci = C .

2. Abelian categories Abelian categories are the natural setting for homological algebra. Let Ab be the category of abelian groups. Definition 2.1. A category A is enriched in Ab if for all objects X,Y ∈ A, Hom(X,Y ) is equipped with an structure such that composition is Z-bilinear. Example 2.2. Both Mod(R) and Ch(R) are enriched in Ab (addition of in Ch(R) is done degree-wise). ♦ Definition 2.3. An additive functor is F : A → B between categories enriched in Ab is a functor whose induced maps on morphism sets are group . Definition 2.4. An is an Ab- A with a zero object that is both initial and terminal, and a product X × Y ∈ A for any objects X,Y ∈ A. Exercise 2.5. Show that initial and terminal objects are unique up to unique iso- morphism. HOMOLOGICAL ALGEBRA 3

Example 2.6. The zero 0 ∈ Mod(R) is both initial and terminal. ♦ Example 2.7. Both Mod(R) and Ch(R) are additive. ♦ While the definition only requires an additive category to have finite products, additive categories also have finite . Proposition 2.8. In an additive category, finite products and coproducts exist and are the same.

Proof. Let A be an additive category and X,Y ∈ A. Let πX : X × Y → X and πY : X × Y → Y be the projections from the product. We have a commuting diagram

X X × Y πY Y πX

ιX Id X 0 where ιX exists by the of the product. You should now check that ιX and the similarly-defined map ιY make X ×Y satisfy the universal property of the .  Exercise 2.9. Show that arbitrary products and finite coproducts commute with taking homology in Ch(R). Definition 2.10. A kernel of a morphism f : B → C in an additive category A is a map i : A → B such that (i) f ◦ i = 0 (ii) i is universal with respect to this property; i.e. for all morphisms i0 : A0 → B such that f ◦i0 = 0, there exists a unique g : A0 → A such that i0 = i◦g. Dually, we have Definition 2.11. A cokernel of a morphism f : B → C is a map π : C → D such that π ◦ f = 0 and π is universal with respect to this property. We will sometimes be sloppy with notation and write only the object rather than the map for a kernel or cokernel. Example 2.12. Kernals and in Mod(R) and Ch(R) are are categorical kernels and cokernels. ♦ Definition 2.13. A morphism i : A → B in an additive category is monic if for all g : A0 → A, ig = 0 =⇒ g = 0. Dually, π : B → C is epic if for all h : C → C0, hπ = 0 =⇒ h = 0. Exercise 2.14 (Tedious but healthy exercise). In an additive category, kernels are the same as monics, and cokernels are the same as epics. Definition 2.15. An additive category A is abelian if (i) Every map has both a kernel and a cokernel (ii) Every monic is the kernel of its cokernel (iii) Every epic is the cokernel of its kernel. Example 2.16. Mod(R) is abelian ♦ 4 HOMOLOGICAL ALGEBRA

In fact, Mod(R) is in some sense the only example. Theorem 2.17 (Freyd-Mitchell embedding). If A is a small , then there is a fully faithful embedding of Mod(R) into Mod(R) for some R that sends exact sequences to exact sequences. Definition 2.18. If A is an abelian category and f : A → B is a morphism, then the image is im(f) := ker coker f.

f g Definition 2.19. A A → B → C in an abelian category is exact if ker g = im f. The category of complexes in an abelian category A is defined as it is for the category of modules and is denoted by Ch(A). The proof of the following appears in Weibel as Theorem 1.2.3. We omit its proof here. Proposition 2.20. If A is an abelian category, then Ch(A) is abelian. Definition 2.21. Let A be an abelian category. A bicomplex or double com- h plex in A is a family {Cp,q}p,q∈Z of objects in A with maps d : Cp,q → Cp−1,q and v d : Cp,q → Cp,q−1 satisfying (i)( dh)2 = 0 and (dv)2 = 0 (ii) dvdh + dhdv = 0 The superscripts h and v on the differentials stand for “‘horizontal” and “verti- cal”. We will usually depict a bicomplex by a diagram with horizontal arrows going left and vertical arrows going down. . . . .

dv dv

··· Cp−1,q+1 Cp,q+1 ··· dh dh dh dv dv

··· Cp−1,q Cp,q ··· dh dh dh Remark 2.22. A bicomplex is not a complex in Ch(A), since the squares in a bicomplex must anticommute, rather than commute. Suppose that A has infinite products and coproducts. Definition 2.23. Given a bicomplex C in A, we can define two complexes: TotΠ(C) and Tot⊕(C), both called total complexes and defined by Π Y ⊕ M Tot (C)n = Cp,q Tot (C)n = Cp,q p+q=n p+q=n with differential dv + dh. Definition 2.24. A bicomplex C is bounded if C has only finitely many nonzero terms along each diagonal p + q = n. When C is bounded, TotΠ(C) = Tot⊕(C). Exercise 2.25. Let C be a bounded bicomplex with exact rows (or exact columns). Prove that TotΠ(C) = Tot⊕(C) is exact. HOMOLOGICAL ALGEBRA 5

3. Algebraic constructions 3.1. Truncations. There are two common ways to convert an unbounded chain complex into a bounded one.

Definition 3.1. Let C be a chain complex. Fix n ∈ Z and let σ≥nC be the subcomplex ( Ci, i ≥ n (σ≥nC)i := 0, else

Define σ≥nC similarly. The complexes σ≥nC and σ≤nC are the brutal trunca- tions of C. Unfortunately, brutal truncation doesn’t play well with homology: Exercise 3.2. Check that for C a chain complex,  H (C), i > n  i Hi(σ≥nC) = Cn/Bn, i = n 0, i < n and  0, i > n  Hi(σ≤nC) = Zn, i = n  Hi(C), i < n A better way to truncate is the following:

Definition 3.3. The good truncations of C are τ≥nC and τ≤nC, defined by  0, i < n  (τ≥nC)i = Zi, i = n and τ≤nC := C/τ≥n+1C  Ci, i > n

Note that τ≥nC is a subcomplex of C, while τ≤nC is a quotient. It is important to remember that C is a chain complex, as this situation is reversed if C is a cochain complex. Exercise 3.4. Show that ( ( 0, i < n Hi(C), i ≤ n Hi(τ≥nC) = and Hi(τ≤nC) = Hi(C), i ≥ n 0, i > n 3.2. Shifts.

Definition 3.5. If (C, d) is a chain complex and p ∈ Z, C[p]i := Ci+p with differ- ential (−1)pd. If (C, d) is a cochain complex, then C[p]i = Ci−p with differential (−1)pd. The shifts [p] give functors Ch(A) → Ch(A). 3.3. product. Definition 3.6. If C and D are chain complexes, then C ⊗ D := Tot⊕(B) h v p where B is the bicomplex {Cq ⊗Dq} with differential d = d⊗1 and d = (−1) ⊗d. 6 HOMOLOGICAL ALGEBRA

3.4. Connecting . Theorem 3.7. Let f g 0 → A → B → C → 0 be an of complexes in an abelian category. There are canonical connecting morphisms ∂ : Hn(C) → Hn−1(A) such that

∂ f g ∂ · · · → Hn+1(C) → Hn(A) → Hn(B) → Hn(C) → Hn−1(A) → · · · is exact. Before proving this theorem, we introduce a useful lemma Lemma 3.8 (). Suppose that we have a

0 p0 A0 i B0 C0 0

f g h p 0 A i B C of R-modules over a ring R with exact rows. Then there is a natural map ∂ : ker(h) → coker(f) such that

ker f → ker g → ker h →∂ coker f → coker g → coker h satisfies (i) If A0 → B0 is injective, then ker(f) → ker(g) is injective. (ii) If B → C is surject, then coker(g) → coker(h) is surjective. Proof. We won’t write out the full proof because it is a length diagram chase, but some bits are below, and you should be able to complete it. • We define ∂ as follows: let c0 ∈ ker h. Choose b0 ∈ B0 such that p0(b0) = c0. Since hp0(b0) = 0, there is a unique a ∈ A such that i(a) = g(b0). Finally, we define ∂(c0) = a ∈ coker(f). One should now check that this map is well-defined. • Exactness at ker(h): Suppose that ∂(c0) ∈ im(f). Choose b0 such that p0(b0) = c0. Then there is a0 ∈ A0 such that if(a0) = g(b0). Now, p0(b0 − i0(a0)) = c0 and g(b0 − i0(a0)) = 0.  Remark 3.9. The Snake Lemma holds in all abelian categories by the Freyd-Mitchell embedding, applied to the smallest abelian containing the six objects in the diagram that appears in the Snake Lemma, which is always a small category. The following justifies the word “natural” used to describe ∂ in the Snake Lemma. It is stated very roughly to avoid lots of notation. Proposition 3.10. Let D and D0 be diagrams such as the one that appears in the hypotheses of the Snake Lemma. If there is a map D → D0, then it induces a map between the sequences of modules int he conclusion of the Snake Lemma.

Proof. Exercise  HOMOLOGICAL ALGEBRA 7

Proof of Theorem 3.7. Suppose that f g 0 → A → B → C → 0 is an exact sequence of complex in an abelian category. For each n, we have a commutative diagram

f g An/dAn+1 Bn/dBn+1 Cn/dCn+1 0

d d d f g 0 Zn−1(A) Zn−1(B) Zn−1(C) In order to the Snake lemma to this iagram, we need to knwot hat the rows are exact. At Cn/dn+1 and Zn−1(A), exactness is clear. For exactness at Bn/dBn+1: suppose that g(b) ∈ dCn+1. Choose c ∈ Cn+1 such 0 0 0 that d(c) = g(b) and b ∈ Bn+1 such that g(b ) = c. Then g(b − d(b )) = 0. Choose 0 a ∈ An such that f(a) = b − d(b ). Then f(a) = b mod dBn+1. For exactness at Zn−1(A): Suppose that g(b) = 0. Choose a ∈ An−1 such that f(a) = b. Then a ∈ Zn−1(A), since d(a) = 0 ⇐⇒ f(d(a)) = 0 ⇐⇒ df(a) = d(b) = 0. The Snake lemma gives a sequence ∂ Hn(A) → Hn(B) → Hn(C) → Hn−1(A) → Hn−1(B) → Hn−1(C) and gluing all of these together gives a long exact sequence.  Proposition 3.11. A morphism of exact sequences of complexes induces a mor- phism of long exact sequences of homology.

Proof. This follows from the naturality of the connecting homomorphism.  4. Topologically-motivated constructions 4.1. Chain . To motivate the notion of chain homotopy, let k be a 0 field and C ∈ Ch(k). For all integers n, choose decompositions Cn = Zn ⊕ Bn and 0 Zn = Bn ⊕ Hn, so that 0 ∼ ∼ 0 ∼ Bn = Cn/Zn = im(dn) = Bn−1 and Hn = Zn/Bn = Hn(C) Define maps

∼ 0 ∼ 0 ∼= 0 0 ∼ Cn = Zn ⊕ Bn Zn = Bn ⊕ Hn Bn Bn+1 Zn+1 ⊕ Bn+1 = Cn+1

sn

0 Observe the following: Note that ds and sd are projections onto Bn and Bn, respec- tively, so dsd = d, and ds + sd is an endomorphism of C with kernel and cokernel isomorphic to H∗(C). In particular, C is exact if and only if ds + sd = Id. Over a general ring R, this doesn’t all work because we can’t necessarily choose complements the way that we can over a field. Chain homotopy is the situation in which this does work. 8 HOMOLOGICAL ALGEBRA

Definition 4.1. A complex of R-modules is called split if there are maps sn : Cn → Cn+1 such that d = dsd. The maps sn are splitting maps. If in addition, C is exact, we say that C is split exact.

Exercise 4.2. Let C be a chain complex. Prove that C is split if Zn is a summand of Cn and Bn is a summand of Zn for all n.

Example 4.3. The complex

2 2 2 ··· → Z/4 → Z/4 → Z/4 → · · · is exact but not split exact, as Z/4 6= Z/2 ⊕ Z/2. ♦

Definition 4.4. A complex C if contractible if IdC = ds + sd for some sn : Ci → Ci+1.

Exercise 4.5. A complex is split exact if and only if it is contractible.

Definition 4.6. A morphism f : C → D of complexes is nullhomotopic if there are maps sn : Cn → Dn+1 such that f = ds + sd. Maps f, g : C → D are homotopic if f − g is nullhomotopic. The maps s are a homotopy.

Remark 4.7. There is a topological motivation for this definition. Consider the as a cell complex with two 0-cells a0 and a1 and a 1-cell a. Let I be the complex

· · · → 0 → 0 → Za → Za0 ⊕ Za1 → 0

a 7→ (a0, −a1) with Za in homological degree 1 and Za0 ⊕ Za1 in homological degree 0. Consider Z as a chain complex with Z in degree 0 and the 0 module elsewhere. We have maps i0, i1 : Z → I given by ij : n 7→ naj for j = 0, 1. With this set-up, if f, g : C → D are morphisms, a homotopy from f to g is equivalent to a morphism h : I ⊗ C → D such that h ◦ (i0 ⊗ IdC ) = f and h ◦ (i1 ⊗ IdC ) = g. This looks a lot like a topological homotopy!

Exercise 4.8. Explicitly, a map h : I ⊗ C → D looks like

(f g s) C ⊕ C ⊕ C[−1] D because (I ⊗ C)n = Cn ⊗ I0 ⊕ Cn−1 ⊗ I1. Show that s is a homotopy between f and g.

Proposition 4.9. If f : C → D is nullhomotopic, then it induces the zero map on homology.

Proof. If f is nullhomotopic, then there are maps sn : Cn → Dn+1 such that f = ds + sd. Let x ∈ Hn(C) and choose a preimagex ˜ ∈ Zn(C). Then f(˜x) = (ds + sd)˜x = ds(˜x) ∈ Bn(D). 

Corollary 4.10. Homotopic chain maps induces the same maps on homology. HOMOLOGICAL ALGEBRA 9

4.2. Cones and cylinders. Definition 4.11. Let f : B → C be a morphism of complexes. The of f is the complex cyl(f) defined by cyl(f)n = Bn ⊕ Bn−1 ⊕ Cn with differential d 1 0 0 −d 0 . 0 −f d

Definition 4.12. The mapping of f : B → C is given by cone(f)n = −d 0 B ⊕ C with differential . n−1 n −f d

Note that the is just the mapping cylinder modulo Bn. Remark 4.13. Again, these constructions are topologically motivated. If f : X → Y is a map of spaces, then the mapping cylinder is cyl(f) = (I × X) t Y/(0, x) ∼ f(y) and the mapping cone of f is cone(f) = cyl(f)/({1} × X). This suggests that the mapping cylinder of a map f : B → C of complexes should be given in degree n by

Za1 ⊗ Bn ⊕ Za0 ⊗ Bn ⊕ Za ⊗ Bn−1 ⊕ Cn Dn := a0 ⊗ b − f(b) with the canonical differential on I ⊗ B ⊕ C/ ∼. You should check: there is an isomorphism D − n → cyl(f)n = Bn ⊕ Bn−1 ⊕ Cn given by the matrix 1 0 0 0 0 0 1 0 0 f 0 1 Similarly, the mapping cone of the map of spaces f : X → Y is cyl(f)/({1}×X). This suggests that for complexes, we should have ∼ Za0 ⊗ Bn ⊕ Za ⊗ Bn−1 ⊕ Cn cone(f : B → C)n := Bn−1 ⊕ Cn = (a0 ⊗ b, 0, 0) − (0, 0, f(b)) Again, you can check that the isomorphism is given by 0 1 0 f 0 1 4.2.1. Applications of mapping cones. Let f : B → C be a map of complexes with induced map f∗ on homology. There is a SES of complexes g 0 → C → cone(f) →h B[−1] → 0 given by g(c) = (0, c) and h(b, c) = −b. This yields a LES

∂ · · · → Hn(C) → Hn(cone(f)) → Hn−1(B) → Hn−1(C) → · · ·

Lemma 4.14. ∂ = f∗

Proof. Recall that ∂ is defined on b ∈ Zn−1(B) by “the unique element of C that maps via g to dˆb, ∂(b) := ˆ where ∈ˆ cone(f)n satisfies h(b) = b” 10 HOMOLOGICAL ALGEBRA

ˆ For this complex, b ∈ Zn−1(B) lifts to b = (−b, 0) ∈ cone(f)n. We then have −d 0 d(−b, 0) = ˆb = (db, f(b) = (0, f(b)), and g(f(b)) = (0, f(b)), so ∂ = f as −f d desired.  Corollary 4.15. f : B → C is a quasi-isomophism if and only if cone(f) is exact. Remark 4.16. A topological statement that is somewhat (but not entirely) analo- gous to Corollary 4.15 is: if f : X → Y is a homotopy equivalence, then cone(f) is contractible.

5. Derived functors Let A and B be abelian categories. Definition 5.1. A homological (resp. cohomological) δ-functor between A n and B is a collection of additive functors Tn : A → B (resp. T : A → B) for n ≥ 0 n n n+1 together with morphisms δn : Tn(C) → Tn−1(A) (resp. δ : T (C) → T (A)) for all SES 0 → A → B → C → 0 n n in A. Set T = Tn = 0 for n < 0. The functors {Tn} (resp. {T }) must satisfy (i) For any SES 0 → A → B → C → 0 in A, there are LES’s

δ · · · → Tn+1(C) → Tn(A) → Tn(B) → Tn(C) → · · · and · · · → T n−1(C) →δ T n(A) → T n(B) → T n(C) → · · · (ii) The LES’s are natural; that is, a morphism of SES induces a morphism of LES. Example 5.2. Homology (resp. cohomology) is a homological (resp. cohomological) ≥0 δ-functor Ch≥0(A) → A (resp. Ch (A) → A) where Ch≥0(A) is complexes with nonzero modules only in non-negative degree (and Ch≥0(A) is defined similarly). ♦ Definition 5.3. An additive functor F : A → B is left exact (resp. right exact) if for all SES 0 → A → B → C → 0 in A, the sequence 0 → F (A) → F (B) → F (C) (resp. F (A) → F (B) → F (C) → 0) is exact. A functor is exact if it is both left and right exact.

Proposition 5.4. If {Tn} is a homological δ-functor, then T0 is right exact. If {T n} is a cohomological δ-functor, then T 0 is left exact.

Definition 5.5. Let S, T be δ-functors. A morphism S → T is a family Sn → Tn (or Sn → T n) of natural transformations of functions that commute with the maps δ. HOMOLOGICAL ALGEBRA 11

Definition 5.6. A homological δ-functor T is universal if given a δ-functor S and a map f0 : S0 → T0, there exists a nique map S → T extending f0. A cohomological δ-functor T is universal if given a δ-functor S and a map f 0 : T 0 → S0, there exists a unique T → S extending f 0. Our next goal will be to show that there are conditions such that given a right (resp. left) F : A → B, there is a universal homological (resp. 0 cohomological) δ-functor T such that T0 = F (resp T = F ). To do this, we will need to introduce some machinery. 5.1. Projective resolutions. Definition 5.7. If A is an abelian category, P ∈ A is projective if for any π : B → C and any morphism γ : P → C, there existsγ ˜ : P → B such that γ = π ◦ γ˜. Remark 5.8. Beware! The morphismγ ˜ in Definition 5.7 is not necessarily unique! 5.1.1. Criteria for projectivity. Proposition 5.9. An R-module is projective if and only if it is a summand of a . Proof. Let M be an R-module. Suppose M is projective. Pick a generating set Λ ⊂ M. Then we have a SES 0 → ker(π) → RΛ →π M → 0. If M is projective, then there is a diagram M

Id γ˜ RΛ π M so the SES is split and there is an isomorphism RΛ =∼ M ⊕ ker(π). Conversely, suppose that F is a free module such that F =∼ M ⊕ N. Let ρ : F → M and ι : M → F be the canonical projection and inclusion, respectively. Let π : B → C be a surjection and let γ : M → C be any map. Let α = γρ and fix a basis {ei}i∈I of F and choose for each i ∈ I some bi ∈ B such that π(bi) = αιρ(ei). Define β : F → B by ei 7→ bi and letγ ˜ = β ◦ ι. You should check thatγ ˜ lifts γ.  Over some rings, all projectives are free. Some important cases of this are noe- therian local rings, Z, and polynomial rings over fields. However, this nice behavior does not always happen. Example 5.10. If S = R × R with R a ring, then R × {0} is a projective S-module that is not free. ♦ 2 2 Example 5.11. The ideal h1 + x, yi ⊂ R[x, y]/ x + y − 1 is projective but not free. ♦ Proposition 5.12. Let A be an abelian category and M ∈ A.

(i) HomA(M, −): A → Ab is left exact. (ii) HomA(M, −) is exact if and only if M is projective. Proof. 12 HOMOLOGICAL ALGEBRA

(i) Let 0 → A → B → C → 0 be a SES in A. By Freyd-Mitchell, we assume without loss of generality that A is a module category. We show that

0 → HomA(M,A) → HomA(M,B) → HomA(M,C) is exact. Suppose that α : M → A and f ◦ α = 0. Since f is monic, α = 0. Next, suppose that β : M → B and g ◦ β = 0. Then im(β) ⊂ ker(g) = im(f). Since f is injective, we may define β˜ : M → A by m 7→ a, where ˜ β(m) = f(a). Then f∗(β) = β. γ (ii) Given a surjection π : X → Y and a map M → Y , we have an SES

0 → ker π → X →π Y → 0 which yields a (perhaps not exact) sequence

0 → Hom(M, ker π) → Hom(M,X) →π∗ Hom(M,Y ) → 0 The exactness of this sequence on the right is equivalent to the existence of a preimageγ ˜ ∈ Hom(M,X) for γ ∈ Hom(M,Y ), which is equivalent to the projectivity of M.  5.1.2. Resolutions by projectives. Definition 5.13. Let M be an object in an abelian category A.A projective of M is a complex P such that

(i) Pi = 0 for i < 0 (ii) Pi is projective for all i ( M, i = 0 (iii) Hi(P ) = 0, else

Remark 5.14. A projective resolution can equivalently be define as a complex of projective modules that is quasi-isomorphic to · · · → 0 → 0 → M → 0 → 0 → · · · where M is in homological degree 0. Definition 5.15. An abelian category A has enough projectives if for all A ∈ A there is a projective P with an epimorphism P  A. Example 5.16. The category of R-modules has enough projective for any ring R: if M M is an R-module, then take the R → M with em 7→ m for m ∈ M. ♦ Lemma 5.17. If A has enough projective, then all objects of A have a projective resolution.

Proof. Let M ∈ A and let π0 : P0 → M be an epimorphism with P0 projective. For i > 0, pick Pi so that there is an epimorphism πi : Pi → ker πi−1. You can check that the complex P has the desired properties.  HOMOLOGICAL ALGEBRA 13

Theorem 5.18. Suppose that

εP · · · → P2 → P1 → P0 → M → 0 → 0 → · · ·

εQ · · · → Q2 → Q1 → Q0 → N → 0 → 0 → · · · are complexes with Pi and Qi projective for all i and with the bottom complex exact. If f 0 : M → N, then there is a map f : P → Q, where

P = · · · → P2 → P1 → P0 → 0 → 0 → · · ·

Q = · · · → Q2 → Q1 → Q0 → 0 → 0 → · · · 0 such that εQf0 = f εP , and f is unique up to chain homotopy.

0 Proof. We prove the existence of f inductively. Set f−1 = f . Suppose that fi has been constructed for all i ≤ n such that the diagram

εP Pn ··· P0 M

0 fn f0 f

εQ Qn ··· Q0 N

commutes. By the projectivity of Pn+1, choose fn+1 : Pn+1 → Qn+1 completing the diagram

Qn+1

fn+1 d

Pn+1 Bn(Q) fnd

The bottom arrow of this diagram is well-defined because fn sends cycles to cycles and Zn(Q) = Bn(Q) by the exactness of Q. Next, we show that f is unique up to homotopy. Let g : P → Q be another map 0 lifting f . We inductively construct sn : Pn → Qn+1. For n < 0, sn = 0. Since εQf0 = εQg0, we may define s0 by

Q1

s0 d

P0 ker(εQ) f0−g0

Suppose by induction that we have si : Pi → Qi+1 for i < n such that fi − gi = dsi + si−1d. We can construct sn as

Qn+1 s n d

Pn Zn(Q) fn−gn−sn−1d where the bottom map’s codomain is Zn(Q) because

dsn−1 = fn−1 − gn−1 − sn−2d 14 HOMOLOGICAL ALGEBRA so

d(fn − gn − sn−1d) = d(fn − gn) − dsn−1d

= d(fn − gn) − (fn−1 − gn−1 − sn−2d)d

= d(fn − gn) − (fn−1 − gn−1)d = 0  Lemma 5.19 (Horseshoe). Suppose that there is a diagram 0

0 0 0 0 ε 0 ··· P2 P1 P0 A 0

ιA A

πA

0 00 00 00 ε 00 ··· P2 P1 P0 A 0

0

0 00 with the column exact and the rows projective resolutions. Set Pn = Pn ⊕Pn . Then the Pn assemple to give a projective resolution of A such that 0 → P 0 →ι P →π P 00 → 0 is exact, with ι and π the inclusion and projection, respectively.

00 00 Proof. We construct α : P0 → A using the fact that P0 is projective. Now, define 0  ε : P0 → A by the row matrix ιAε α to yield a diagram 0 0 0

0 0 0 ε 0 0 ker(ε ) P0 A 0

ιA

ε 0 ker(ε) P0 A 0 α πA

0 00 00 ε 00 0 ker(ε ) P0 A

0 0 0

By the snake lemma, there is an exact sequence 0 → ker(ε0) → ker(ε) → ker(ε00) → coker(ε0) → coker(ε) → coker(ε00) → 0 | {z } | {z } =0 =0 HOMOLOGICAL ALGEBRA 15 so all the rows and columns of the diagram are exact. We now iterate with the diagram 0

0 0 d 0 P1 ker(ε ) 0

ker(ε)

00 00 d 00 P1 ker(ε ) 0

0  5.2. Injective resolutions. Definition 5.20. An object I in an ablian category A is injective if for all moninc f : A → B and all maps α : A → I, there exists a β : B → I such that α = βf. Definition 5.21. An abelian category A has enough injectives if for all A ∈ A, there is a monic A → I with I injective. 5.2.1. Characterizing injectivity. Definition 5.22. If A is an abelian category, then the Aop has the same objects as A and HomA(A, B) =: HomAop (B,A). Lemma 5.23. Let I ∈ A with A an abelian category. The following are equivalent. (i) I injective in A (ii) I is projective in Aop op (iii) HomA(−,I): A → Ab is exact Proof. Exercise.  The following does not hold in a general abelian category, but is useful in module categories. Proposition 5.24 (Baer’s criterion). A right R-module E is injective if and only if for all right ideals J ⊂ R, every map J → E extends to a map R → E. Proof. =⇒ : clear. ⇐: Suppose we have a diagram B

E α A with A ⊂ B right R-modules. Let E be the poset of all extensions α0 : A0 → E of α to a submodule A ⊂ A0 ⊂ B, ordered by (α0 : A0 → E) ≤ (α00 : A00 → E) when 0 00 0 00 A ⊂ A and α = α |A0 . Every chain of this poset has an upper bound (union of the elements of the chain), so by Zorn’s lemma, there is a maximal α0 : A0 → E. 16 HOMOLOGICAL ALGEBRA

Suppose towards a contradiction that A0 6= B. Then there is b ∈ B \ A0, and we may form the right ideal J = {r ∈ R : br ∈ A0} ⊂ R. We now have a diagram R f

E J α0◦(b·) where the dashed arrow f exists by hypothesis. Set A00 = A0 + bR, and define α00 : A00 → E by α00(a + br) = α0(a) + f(r). You can check that this map is well-defined, so α00 is a nontrivial extension of α0, contrary to the maximality of 0 α .  Corollary 5.25. If R is a (commutative) PID, then an R-module M is injective if and only if it is divisible; that is, for all r ∈ R \{0} and m ∈ M, there is m0 ∈ M such that rm0 = m. Proof. ⇐=: Let hri be a nonzero ideal of R and let f : hri → M be an R-. Choose m0 ∈ M such that rm0 = f(r) and define f˜ : R → M by 1 7→ m0. ˜ You can now check that f|hri = f, so f extends to all of R and M is injective by Proposition 5.24. =⇒ : Suppose that M is injective an let r 6= 0 ∈ R and m ∈ M. Given f : hri → M, ˜ ˜ there is an extension f : R → M. We now hav rf(1) = m, so M is divisible.  5.2.2. Injective hulls.

Proposition 5.26. Let A be an abelian category and suppose that {Ii}i∈S is a Q family of injective objects in A. If i∈S Ii exists, then it is injective. f Q Proof. Let j : A,→ B be a monic in A and A → i∈S Ii a morphism. Since the Ii’s are injective, for each i ∈ S, we have B g i j Q Ii Ii A πi i∈S f Q so by the definition of the product, there is a unique g : B → i∈S Ii given by gi in each coordinate. Since πif = gij = πigj for all i, the uniqueness of g implies that gj = f.  Theorem 5.27. Mod(Z) has enough injectives. Proof. Let A ∈ Mod(Z). Set Y I = (Q/Z)α. α∈Hom (A, / ) Z Q Z Q/Z is injective by Corollary 5.25, so I is injective by Proposition 5.26. Define a map e : A → I by a 7→ (α(a))α. To show that e is injective, first suppose that the order of a 6= 0 is n < ∞, so there is an injection j : Z/n → A given by 1 7→ a. Define k : Z/n → Q/Z 1 by 1 7→ n . Since Q/Z is injective, there is f : A → Q/Z such that fj = k. In 1 particular, f(a) = k(1) = n 6= 0, so e(a) 6= 0 ∈ I. HOMOLOGICAL ALGEBRA 17

Otherwise, if a has infinite order, then there is an injection j : Z → A with 1 1 7→ a. Define a map k : Z → Q/Z by 1 7→ 2 . By the injectivity of Q/Z, there 1 is f : A → Q/Z with fj = k. Now, f(a) = k(1) = 2 6= 0, e(a) 6= 0 ∈ I and e is injective.  Definition 5.28. Functors L : A → B and R : B → A are adjoint if there are natural bijections for each X ∈ A and Y ∈ B ∼ τX,Y : HomB(L(X),Y ) ←→ HomA(X,R(Y )). By “natural”, we mean the following commutes:

L(f)∗ g Hom(L(X),Y ) Hom(L(X),Y ) ∗ Hom(L(X),Y 0)

∼= ∼= ∼= f ∗ R(g) Hom(X0,R(Y )) Hom(X,R(Y )) ∗ Hom(X,R(Y 0)) and L is the left adjoint and R is the right adjoint.

Example 5.29. If M ∈ Mod(R) with R commutative, then −⊗R M and Hom(M, −) are adjoint. To see this, define θ : HomR(A ⊗R M,B) → HomR(A, HomR(M,B)) −1 by θ(α): a 7→ (m 7→ α(a ⊗ m). The inverse is θ : HomR(A ⊗R M,B) ← −1 HomR(A, HomR(M,B)) given by θ : β 7→ (a⊗m 7→ β(a)m). Checking naturality is left as an exercise. More generally, if M is an (R,S) bimodule, then −→ − ⊗R M : RightMod(R) ←− RightMod(S) : HomS(M, −) are adjoint in the same way, with the action of r ∈ R on HomS(M,N) given by (α · r)(m) := α(rm). If we have a ring map f : R → S, and set M = S, then these are the extension and restriction of scalars of right modules, respectively. ♦ Example 5.30. If f : R → S is a ring map, then we have an adjunction of left modules ∗ −→ f : Mod(S) ←− Mod(R): f∗ ∗ ∗ with f∗(M) := HomR(f (S),M). If α ∈ HomR(f (S),M) and s ∈ S, then for t ∈ f ∗(S)S,(sα)(t) = α(ts). ♦ Proposition 5.31. If an additive functor R : B → A is right adjoint to an exact functor L : A → B and I is an injective object of B, then R(I) is injective in A. Dually, if an additive functor L : A → B is left adjoint to an exact functor and P is projective in A, then L(P ) is projective. op 0 Proof. We show that HomA(−,R(I)) : A → Ab is exact. Let f : A → A be a monic. It suffices to show 0 HomA(A ,R(I)) → HomA(A, R(I)) is epic. We have a diagram

∗ 0 f HomA(A ,R(I)) HomA(A, R(I))

∼= ∼= ∗ 0 L(f) HomB(L(A ),I) HomB(L(A),I) 18 HOMOLOGICAL ALGEBRA

Since L is exact, L(f) is monic, so by the injectivity of I, L(f)∗ is epic. It follows ∗ that f is epic as well.  Corollary 5.32. If f : R → S is a ring map and I is an injective R-module, then ∗ f∗(I) := HomR(f (S),I) is an injective S-module. ∗ Proof. f∗ is right adjoint to the restriction of scalars f , and restriction of scalars is exact.  Theorem 5.33. Mod(R) has enough injectives ∗ Proof. Let M ∈ Mod(R). Set I0 = HomZ(f (R), Q/Z) ∈ Mod(R) = f∗(Q/Z), where → R is the canonical map. Set I = Q (I ) , which is injective Z α∈HomR(M,I0) 0 α by Corollary 5.32 and the fact that products of injectives are injective. Define e : M → I by m 7→ (α(m))α. To see that e is injective,

HomR(M,I0) = HomR(M, HomZ(R, Q/Z)) ∼ = HomZ(M ⊗R R, Q/Z) ∼ = HomZ(M, Q/Z) and we can now refer back to our proof that Mod(Z) has enough injectives (Theorem 5.27).  5.3. Back to δ-functors. Let F : A → B be a right exact functor and sassume that A has enough projectives.

Definition 5.34. The left derived functors are LiF (A) := Hi(F (P )) with i ≥ 0 and P a projective resolution of A. ∼ Proposition 5.35. L0F (A) = F (A)

Proof. F is right exact, so we have a right exact sequence F1(P1) → F (F0) → F (A) → 0. 

Theorem 5.36. L∗F for a universal homological δ-functor. We will need some set-up before proving this theorem.

Lemma 5.37. If A ∈ A, then Lif(A) is well-defined up to canonical isomorphism. That is, if Q is another projective resolution of A, then there is a canonical iso- ∼ morphism H∗(F (P )) = H∗(F (Q)).

Proof. Choose a chain map f : P → Q lifting IdA. We get an induced map F (f): F (P ) → F (Q), which in turn induces F (f)∗ : H∗(F (P )) → H∗(F (Q)). Similarly, we obtain F (g)∗ : H∗(F (Q)) → H∗(F (P )) with g : Q → P a of IdA.

Now, gf and IdP are both lifts of IdA, so F (g)∗ ◦ F (f)∗ = IdH∗(F (P )). Similarly,

F (f)∗ ◦ F (g)∗) = IdH∗(F (Q). 

Corollary 5.38. If A is projective, then L∗F (A) = 0 for i > 0. Proof. A is a projective resolution of itself. 

Exercise 5.39. Show that the map H∗F (f) is independent of the lift f chosen in the preceding proof. (Hint: F (f) and F (f 0) are homotopic for any two lifts f and f 0)