Solutions and Elaborations for “An Introduction to Homological Algebra” by Charles Weibel

SOLUTIONS AND ELABORATIONS FOR “AN INTRODUCTION TO HOMOLOGICAL ALGEBRA” BY CHARLES WEIBEL

SEBASTIAN BOZLEE

Contents 1. Introduction 2 2. Missing Details 2 2.1. Chapter 1 2 2.2. Chapter 2 2 2.3. Chapter 3 2 2.4. Chapter 4 13 2.5. Chapter 5 15 2.6. Chapter 6 15 2.7. Chapter 7 15 2.8. Chapter 8 15 3. Solutions to Exercises 17 3.1. Chapter 1 17 3.2. Chapter 2 17 3.3. Chapter 3 17 3.4. Chapter 4 23 3.5. Chapter 5 28 3.6. Chapter 6 28 3.7. Chapter 7 28 3.8. Chapter 8 28 4. Additional Math 33 4.1. Short Exact Sequences 33 References 35

1 2 SEBASTIAN BOZLEE

1. Introduction Weibel’s homological algebra is a text with a lot of content but also a lot left to the reader. This document is intended to cover what’s left to the reader: I try to ﬁll in gaps in proofs, perform checks, make corrections, and do the exercises. It is very much in progress, covering only chapters 3 and 4 at the moment. I will assume roughly the same pre-requisites as Weibel: knowledge of some category theory and graduate courses in algebra covering modules, tensor products, and localizations of rings. I will assume less mathematical maturity.

2. Missing Details 2.1. Chapter 1. 2.1.1. Section 1.6. Example 1.6.8. It is not quite true that AU op = Presheaves(X), since there are functors that do not take the empty set to 0. In fact, some sources do not require that presheaves take the empty set to zero. With or without this requirement, the resulting category of presheaves is an abelian category. 2.2. Chapter 2. 2.2.1. Section 2.1. 2.2.2. Section 2.2. 2.2.3. Section 2.3. 2.2.4. Section 2.4. Theorem 2.4.6. It is asserted that the image of a split exact sequence under an additive functor is exact. See Theorem 4 in “Additional Math” below for a (lengthy, but I think neat) proof. 2.3. Chapter 3. 2.3.1. Section 3.1. Proposition 3.1.2. This is the way that a group is a direct limit of its finitely generated subgroups. Let A be an abelian group. First, we set up a diagram D whose objects are the finitely generated subgroups of A, and whose arrows are the inclusion maps ϕ : G → H whenever G ≤ H for finitely generated subgroups G, H of A. This diagram is indexed by a filtered category, since for any pair G, H of finitely generated subgroups of A, GH is also a finitely generated group, so G → GH and H → GH are arrows in the diagram (i.e. we satisfy axiom 1 of a filtered category in Definition 2.6.13) and there are no parallel arrows (i.e. we satisfy axiom 2 of a filtered category vacuously.) The claim is that the colimit of this diagram is A. To see this, first of all, we have a map ϕG : G → A for each f.g. subgroup G defined by the inclusion of G into A. This commutes with all maps in the diagram since the inclusion of G into H followed by the inclusion of H into A is just the inclusion of G into A. SOLUTIONS AND ELABORATIONS 3

It remains to show that A is a universal. Select another family of maps D → B. We will define a map ψ : A → B commuting with everything. Let g ∈ A. The element g is contained in some f.g. subgroup G of A. We define ψ(g) to be the image of g under the map G → B. This is the only way to define ψ so that G

ψ A / B commutes, so if ψ deﬁnes a map that commutes with the maps D → A and D → B, it is unique. On the other hand, if H is another f.g. subgroup of A containing g, then since

G / GH o H

! B } commutes, the image of g under H → B must be the same as its image under the map G → B. This shows that ψ is well defined. Finally, ψ is a group homomorphism, since for any g, h ∈ A, we may select a f.g. subgroup G so that g, h ∈ G. Let ϕ be the map G → B. Then ψ(g + h) = ϕ(g + h) = ϕ(g) + ϕ(h) = ψ(g) + ψ(h) since ϕ : G → B is a group homomorphism. So there exists a unique group homomorphism A → B so that everything commutes, as desired, and A is the colimit. Intuitively, the direct limit is a fancy way to express a union, and of course A is a union of its finitely generated subgroups, since each element of A is contained in some finitely generated subgroup. The argument above shows that the formal definition agrees with this intuition.

Z m m Proposition 3.1.4. The reason that Torn(Z ,B) = 0 is that Z is a free Z-module, hence projective, and Tor vanishes on projective modules. 2.3.2. Section 3.2.

Definition 3.2.1. The target category of − ⊗R B is intended to be Ab. If B is an R-S bimodule, we can also view − ⊗R B as a functor from mod-R to mod-S. Again, this is always right exact. It is equivalent for − ⊗ B to be exact as a functor to Ab and as a functor to mod-S, since in each case exactness is the same as taking every injection A0 → A of R-modules to an injective map A0 ⊗ B → A ⊗ B: we just view the map as a group homomorphism if Ab is the target and as an S-module homomorphism if mod-S is the target. This means that flatness of B does not depend on which category we take to be the target of − ⊗R B, although it does depend on R. Theorem 3.2.2. For an alternate proof without colimits, see [2, Proposition 2.5]. We show that the colimit indicated in the text is the localization, as claimed. Our tech- nique will be to use an explicit description of filtered colimits. Namely, if I is a filtered category, and F : I → R-mod is an additive functor, the colimit of F is the module C whose additive group has underlying set ! G C = F (i) / ∼ i∈I 4 SEBASTIAN BOZLEE

where a ∈ F (i) and b ∈ F (j) are equivalent a ∼ b if there exist arrows ϕ : i → k and ψ : j → k so that F (ϕ)(a) = F (ψ)(b). The addition of a ∈ F (i) and b ∈ F (j) is defined by taking a pair of arrows ϕ : i → k and ψ : j → k (whose existence is assured since I is a filtered category) and letting a + b be the sum F (ϕ)(a) + F (ψ)(b) in k. For r ∈ R we take the scalar product of r and a ∈ F (i) to be ra ∈ F (i). In our particular case, a ∈ F (s1) and b ∈ F (s2) are equivalent if and only if there exists a b −1 a number c ∈ S so that cs2a = cs1b. Next, note that a ∼ b if and only if = in S R. s1 s2 −1 a Therefore the map ϕ : C → S R defined by a ∈ F (s1) 7→ is a well defined bijection. s1 Next, the sum of a ∈ F (s1) and b ∈ F (s2) is s2a + s1b ∈ F (s1s2), so ϕ preserves addition. Finally, the scalar product of r ∈ R and a ∈ F (s1) is ra ∈ F (s1), so ϕ preserves scalar multiples. Altogether, the colimit C is isomorphic to the localization S−1R, as claimed.

∗ ∗ Lemma 3.2.6. When M = R, the map σ is from A ⊗R R → HomR(R,A) , with σ deﬁned by σ(f ⊗ r): h 7→ f(h(r)). There is an isomorphism A∗ → A∗ ⊗ R deﬁned by f 7→ f ⊗ 1 and there is an isomorphism

∗ ∗ HomR(R,A) = HomAb(HomR(R,A), Q/Z) → HomAb(A, Q/Z) = A

induced by the isomorphism A → HomR(R,A) that takes a to the map ϕa : R → A ∗ ∗ determined by ϕa(1) = a. More explicitly, the map HomR(R,A) → A is deﬁned by 0 h 7→ h : A → Q/Z a 7→ h ◦ ϕa

where ϕa is the element of Hom(R,A) so that ϕa(1) = a. Composing with these isomor- ∗ ∗ phisms, we may regard σ as a map σ : A → A deﬁned by σ(f): m 7→ f(ϕm(1)) = f(m). That is, σ takes f ∈ A∗ to f ∈ A∗, so σ is an isomorphism, as claimed. The use of the Five Lemma is with this diagram:

A∗ ⊗ Rm / A∗ ⊗ Rn / A∗ ⊗ M / 0 / 0

σ =∼ σ =∼ σ 0 0 α∗ Hom(Rm,A)∗ / Hom(Rn,A)∗ / Hom(M,A)∗ / 0 / 0

Proposition 3.2.9. We will prove that if T is R-ﬂat and P is a projective R-module, then P ⊗R T is a projective T -module. Recall that an R-module P is projective if and only if P is the direct summand of a free R-module (Proposition 2.2.1). So let F = P ⊕ Q be a free module with direct summand P . ∼ L Since F is free, we also have that F = s∈S R for some set S. This gives us the split exact sequence 0 / P / F / Q / 0.

Since T is R-ﬂat, − ⊗R T is exact as a functor from mod-R to mod-T . It therefore takes split exact sequences to split exact sequences (see comment on Theorem 2.4.6), so P ⊗ T is a direct summand of F ⊗ T . Also, − ⊗R T as a functor from mod-R to mod-T has the right adjoint Hom(T, −) (Proposition 2.6.3), so − ⊗R T preserves colimits (Theorem 2.6.10). L L L In particular, F ⊗ T = ( s∈S R) ⊗ T = s∈S(R ⊗ T ) = s∈S T , so − ⊗R T takes free R-modules to free T -modules. Together, P ⊗ T is a direct summand of a free T -module, so P ⊗ T is a projective T -module. SOLUTIONS AND ELABORATIONS 5

Lemma 3.2.11. To see why we require that r is central, let µi : Pi → Pi be the map p 7→ rp. Then µi is an R-module homomorphism if and only if for all s ∈ R, p ∈ P srp = sµ(p) = µ(sp) = rsp.

This holds with no conditions on Pi if sr = rs for all s ∈ R, that is, if r is central. Had we not required that r were central, we would then have to impose conditions on the projective resolution P → A to maintain thatµ ˜ : P → P is a chain map.

Rp Corollary 3.2.13. The equality Rp ⊗R M = Torn (Ap,Bp) follows by ﬁrst noting

R Rp Rp ⊗R M = Rp ⊗R Torn (A, B) = Torn (A ⊗R Rp,Rp ⊗R B) from Corollary 3.2.10, and then using that A ⊗ Rp and Rp ⊗ B are equal to the localizations of A and B at p, respectively. 2.3.3. Section 3.3. 0 Example 3.3.3. ExtZ(A, Z) is zero since we have assumed that A is a torsion group: torsion elements must be sent to torsion elements, so the only homomorphism A → Z is the zero homomorphism. 1 n The comment that Ext ( ∞ , ) = lim( /p ) is proved in Application 3.5.10. Z Zp Z ←− Z Example 3.3.5. ∼ L∞ (1) This is because A = i=1 Z/p, so ∞ ∞ 1 Y 1 Y ExtZ/m(A, Z/p) = ExtZ/m(Z/p, Z/p) = Z/p. i=1 i=1 where the ﬁrst equality follows from Proposition 3.3.4 part 1 and the second follows from Exercise 3.3.2. The last is a vector space of uncountably inﬁnite dimension. (2) The displayed equation follows from Proposition 3.3.4 part 2 and Calculation 3.3.2. 1 (ExtZ is intended.) The comment on divisibility follows since A is divisible if and only if A = pA for all p ≥ 2 if and only if A/pA = 0 for all p ≥ 2. Heuristically, there seems to be a duality between being torsionfree and divisible. Z 1 Q∞ Tor1 (−, Q/Z) tests for torsion and ExtZ(−, p=2 Z/p) tests for divisibility.

Lemma 3.3.6. Weibel asserts that HomAb(A, B) is a (left) R-module when R is commutative. Here is why. We deﬁne rf : A → B for r ∈ R and f ∈ HomAb(A, B) by (rf)(b) = f(rb). In order for this action to make HomAb(A, B) an R-module, we need rf to be an R-module homomorphism for all r. Now rf is an R-module homomorphism if and only if for all s ∈ R and b ∈ B, f(srb) = sf(rb) = s(rf)(b) = (rf)(sb) = f(rsb). If we assume R is commutative, then srb = rsb, so we get the equality above for free. Otherwise, we would have to impose conditions on A and B to maintain that HomAb(A, B) is an R-module. The assertion that multiplication by r gives a chain map when r is central is addressed in my comment on Lemma 3.2.11. Lemma 3.3.8. Let’s check the isomorphism for A = R. Then −1 −1 ∼ −1 S HomR(A, B) = S HomR(R,B) = S B. 6 SEBASTIAN BOZLEE

On the other hand, −1 −1 ∼ −1 HomS−1R(S A, S B) = S B. So both are equal to S−1B as R-modules, as promised. The implication that the same holds for Rm is due to the fact that additive functors preserve direct sums. This is a corollary of Theorem 4 and Proposition 3 in “Additional Math” above. 2.3.4. Section 3.4.

Theorem 3.4.3. According to the errata, the theorem should be stated with Θ : ξ 7→ ∂(idB), instead of idA. We will see why as we ﬁll in the details of this proof. When Weibel writes that X → A is the map induced by the maps B →0 A and P → A, he is referring to the universal property of pushouts: since

j M / P

β 0 B / A commutes (we get zero when composing M → P → A by exactness of 0 → M → P → A → 0) there is a unique map X → A so that the following diagram commutes:

j (1) M / P

β σ i ϕ B / X ϕ0

0 + A That map is the map we are to use in the sequence 0 → B → X → A → 0. More explicitly, if ϕ ϕ is the map P → A, then the dotted map, ϕ0 : X → A, is deﬁned by taking the equivalence class of (p, b) in X to the element ϕ(p) in A. This is well deﬁned since if (p, b) and (p0, b0) are two representatives of the same element of X, then ϕ(p) − ϕ(p0) = ϕ(p − p0) = ϕ(j(m)) j ϕ for some m ∈ M. By exactness of 0 → M → P → A → 0, ϕ(j(m)) = 0.

Now we show that the bottom row of the diagram

j ϕ (2) 0 / M / P / A / 0

β σ i ϕ0 0 / B / X / A / 0 is exact. (The left square commutes by definition of pushout. The right square commutes since the right triangle of (1) commutes.) Since ϕ is surjective, and ϕ = ϕ0 ◦ σ, it follows ϕ0 is surjective. This shows exactness at A. The map i : B → X is defined by b 7→ [(0, b)]. Suppose that [(0, b)] = [(0, 0)]. Then, by the definition of X, (0, b) = (0, 0) + (j(m), −β(m)) for some m ∈ M. Since j is injective, we must have that m = 0, so [(0, b)] = [(0, 0)]. Therefore i is injective. This shows exactness at B. SOLUTIONS AND ELABORATIONS 7

Finally, suppose (p, b) is the equivalence class of some element of the kernel of ϕ0. Then ϕ(p) = 0, so by exactness of the ﬁrst row, p = j(m) for some m ∈ M. Therefore, [(p, b)] = [(j(m), b)] = [(0, b+β(m))] is in the image of i. This shows exactness at X, and we are done.

The reference to naturality of ∂ refers to the fact that this diagram, induced from the morphism of short exact sequences (2) by the contravariant cohomological δ-functor Ext∗(−,B), commutes:

∂ Hom(M,B) / Ext1(A, B) O O β∗ id ∂ Hom(B,B) / Ext1(A, B)

∗ By commutativity of the diagram, ∂(idB) = ∂(β (idB)) = ∂(β) = x. This gives surjectivity of Θ.

Now we will check that the maps i : B → X and σ + i ◦ f : P → X induce an isomorphism X0 ∼= X and an equivalence between ξ0 and ξ. Recall that we have picked two lifts, β and β0 of some element x ∈ Ext1(A, B) under the map ∂ in the diagram

j∗ ∂ Hom(P,B) / Hom(M,B) / Ext1(A, B) / 0,

which we arrived at by taking the long exact sequence associated to 0 → M → P → A → 0 under Ext∗(−,B). By exactness, the maps β and β0 diﬀer by an element of the image of j∗, so we write β0 = β + f ◦ j, where f : P → B. First, we check that the diagram

j (3) M / P

β0 σ+i◦f i B / X

commutes. It does:

i ◦ β0 = i ◦ (β + f ◦ j) = i ◦ β + i ◦ f ◦ j = σ ◦ j + i ◦ f ◦ j = (σ + i ◦ f) ◦ j.

To get the third equaliity, we used the commutativity of the diagram

j M / P

β σ i B / X 8 SEBASTIAN BOZLEE

j Now P is a cocone of the diagram M / P so by the universal property

σ+i◦f β0 i B / X B of pushouts, there exists a unique map so that the following diagram commutes:

j (4) M / P

β0 σ0 i0 σ+i◦f B / X0

i + X By a symmetric argument, there exists a unique map so that

j M / P

β σ i σ0−i0◦f B / X

i0 + X0 commutes. Adding f ◦ j to β, i ◦ f to σ, and i0 ◦ f to σ0, we obtain the commutative diagram

j (5) M / P

β0 σ+i◦f i σ0 B / X

i0 + X0 with the same arrow from X → X0. Now, composing the arrows X0 → X and X → X0 in diagrams (4) and (5), we get a commutative diagram

j M / P

β0 σ0 i0 σ0 B / X0

i0 ! + X0 By the uniqueness in universal property of pushouts, the dashed arrow must be the identity on X0. By a symmetric argument, the other composition X → X0 → X is the identity on X. Therefore X and X0 are isomorphic. SOLUTIONS AND ELABORATIONS 9

Finally, we show that the isomorphism X0 → X induces an equivalence of extensions. Consider the diagram

i0 ξ0 : 0 / B / X0 / A / 0

i ξ : 0 / B / X / A / 0

where the middle arrow is the isomorphism X0 → X. The left square commutes since the the lower triangle of (4) commutes. The right square commutes because the right squares of

j ϕ 0 / M / P / A / 0

β σ i 0 / B / X / A / 0

and

j ϕ 0 / M / P / A / 0

β0 σ0 i0 0 / B / X / A / 0 commute. Therefore, ξ and ξ0 are equivalent extensions, as claimed.

Deﬁnition 3.4.4. I think it is helpful to compute the Baer sum of a simple example. Let us take the Baer sum of the extensions

p 1 ξ : 0 / Z/p / Z/p2 / Z/p / 0,

p 2 ξ0 : 0 / Z/p / Z/p2 / Z/p / 0.

1 2 Now the pullback X00 of Z/p2 → Z/p and Z/p2 → Z/p is the submodule {(x, y) ∈ Z/p2 × Z/p2 | x ≡ 2y (mod p)} of Z/p2 × Z/p2. The pullback diagram is

π2 X00 / Z/p2

π1 2 1 Z/p2 / Z/p

We let Y be X00/{(px, −px) | x ∈ Z/p}. Then the Baer sum of ξ and ξ0is the sequence

f g 0 / Z/p / Y / Z/p / 0

where f takes x to [(px, 0)] = [(0, px)] and g takes [(x, y)] to x (mod p) or equivalently to 2y (mod p). 10 SEBASTIAN BOZLEE

2 If p 6= 3, this is equivalent with the extension associated to i = 3 (mod p), via the following equivalence:

f g 0 / Z/p / Y / Z/p / 0

σ 2 p 3 0 / Z/p / Z/p2 / Z/p / 0

where σ : [(x, y)] 7→ x + y (mod p2). This is well deﬁned, since any other representative of (x, y) is of the form (x − pz, y + pz), and (x − pz, y + pz) 7→ (x − pz) + y + pz = x + y. To see 2 −1 g that the bottom right map is 3 , i.e. multiplication by 2 · 3 (mod p), note that (2, 1) 7→ 2, while (2, 1) 7→σ 3 (mod p2), and a morphism between cyclic groups must be multiplication by some number.

Corollary 3.4.5. When Weibel refers to using the notation of the previous theorem, he means that we have constructed the following diagrams from the extensions ξ and ξ0:

j (6) 0 / M / P / A / 0

γ τ i ξ : 0 / B / X / A / 0

and

j (7) 0 / M / P / A / 0

γ0 τ 0 i0 ξ0 : 0 / B / X0 / A / 0.

We construct them by ﬁrst using the lifting property of the projective module P to obtain a map τ : P → X and a map τ 0 : P → X0. Then by commutativity of the right squares, τ(j(M)) ⊆ ker{X → A} and τ 0(j(M)) ⊆ ker{X0 → A}. The exactness of the bottom rows allows us to lift τ and τ 0 to maps γ = i−1 ◦ τ ◦ j and γ0 = (i0)−1 ◦ τ ◦ j respectively. Recall that X00 is the pullback of the diagram

X00 / X

X0 / A.

Now

τ P / X

τ 0 X0 / A SOLUTIONS AND ELABORATIONS 11

commutes, since each composition is the map P → A, by the diagrams (6) and (7) above. Accordingly, by the universal property of pullbacks, we get a unique map P → X00 so that P τ

$ X00 / X τ 0

X0 / A commutes. This is the induced map τ 00 : P → X00. More explicitly, τ 00 is the map P → X00 ⊆ X × X0 deﬁned by τ 00(p) = (τ(p), τ 0(p)). The map τ : P → Y is obtained from τ 00 by composition with the quotient map X00 → Y . Next, Weibel claims that

j 0 / M / P / A / 0

γ+γ0 τ i00 ξ + ξ0 : 0 / B / Y / A / 0. commutes. To see this, we first check that the left square commutes. Recall that the map i00 : B → Y is defined by b 7→ [(i(b), 0)] = [(0, i0(b))]. Then composing i00 ◦ (γ + γ0), we get (i00 ◦ (γ + γ0))(m) = i00(γ(m) + γ0(m)) = [(i(γ(m)), 0)] + [(0, i0(γ0(m)))] = [(τ(j(m)), 0)] + [(0, τ 0(j(m)))] = [(τ(j(m)), τ 0(j(m)))] = τ(j(m)), so the square commutes. To do: the right square commutes because. . . 2.3.5. Section 3.5. Definition 3.5.1. We check that the kernel of ∆ is lim A . To do ←− i Lemma 3.5.2. The snake lemma mentioned only gives the map lim C →∂ lim1 A . ←− i ←− i To see why we may take the higher derived functors of lim to vanish, note that the ←− induced map on cokernels lim1 B → Q lim1 C is onto, since the map Q B → Q C is onto. ←− i ←− i i i Accordingly, the sequence 0 → lim A → lim B → lim C →∂ lim 1A → lim 1B → lim 1C → 0 ←− i ←− i ←− i ←− i ←− i ←− i is exact. Example 3.5.5. Applying the δ-functor lim∗ − to the short exact sequence 0 → {pi } → ←− Z {Z} → {Z/pi} → 0, we get the long exact sequence lim pi lim lim /pi ∂ lim1 pi lim1 . ←− Z / ←− Z / ←− Z Z / ←− Z / ←− Z This simplifies to 0 ˆ ∂ lim1 pi 0. / Z / Zp / ←− Z / 12 SEBASTIAN BOZLEE

To see this, note that since each map pi → pi−1 is an inclusion, lim pi = T pi = 0. Z Z ←− Z i Z It is clear that lim = . Finally, lim1 = 0 by Lemma 3.5.3. By exactness, we conclude ←− Z Z ←− Z that lim1 pi = ˆ / . (To do: explain why the image of is the usual subgroup ←− Z Zp Z Z identified with Z?) Theorem 3.5.8. It is helpful to know that C = lim C is the complex whose jth degree ←− i component is C = lim C and whose differentials C →d C are induced by the collection j ←− i,j j j−1 i d(i) of maps Ci,j → Ci,j−1.(To do? Prove this statement.) We apply the left exact functor lim to 0 → {Z } → {C } →d {C [−1]}, and obtain that ←− i i i 0 → lim Z → C →d C[−1] ←− i (i) is exact. The last map is that induced by the collection of maps d : Ci,j → Ci,j−1, so its maps in each degree are the differentials of C. Then by exactness, the degree j part of lim Z ←− i is the kernel of the degree j part of C →d C[−1], so it really consists of the cycles of C, as claimed. d Next, from the exact sequence 0 → {Zi} → {Ci} → {Bi[−1]} → 0, we get the long exact sequence

0 Z C d lim B [−1] lim1 Z 0 lim1 B [−1] 0 / / / ←− i / ←− i / / ←− i / where we have lim1 C = 0 by the Mittag-Leﬄer condition. By exactness, lim1 B [−1] = 0, ←− i ←− i so the same holds of its shift, lim1 B = 0, as claimed. Also by exactness, the kernel of ←− i lim1 B [−1] → lim1 Z is the image of C →d lim B , which is B. This gives us that the ←− i ←− i ←− i sequence 0 B lim B [−1] lim1 Z 0 / / ←− i / ←− i /

is exact, as claimed. Next, the long exact sequence associated to 0 → {Bi} → {Zi} → {H∗(Ci)} → 0 is

0 lim B Z lim H (C ) 0 lim1 Z lim1 H (C ) 0. / ←− i / / ←− ∗ i / / ←− i / ←− ∗ i / since lim1 B = 0. This gives the exact sequence 0 → lim B → Z → lim H (C ) → 0 and the ←− i ←− i ←− ∗ i isomorphism lim1 Z ∼ lim1 H (C ). The chain of inclusions ←− i = ←− ∗ i 0 ⊆ B ⊆ lim 1B ⊆ Z ⊆ C ←− follows from the beginnings of the various exact sequences we have produced. The quotients of these come from those same exact sequences. The ﬁnal statement follows from the third isomorphism theorem: H (C) = Z/B has the subgroup lim1 H (C )[1] = lim B /B, and ∗ ←− ∗ i ←− i the quotient of these two is Z/ lim B = lim H (C ). Evaluating the resulting short exact ←− i ←− ∗ i sequence 0 lim 1H (C )[1] H (C) lim H (C ) 0 / ←− ∗ i / ∗ / ←− ∗ i / at a given degree gives the result. SOLUTIONS AND ELABORATIONS 13

Application 3.5.10. These towers of cochain complexes can be a bit diﬃcult to visualize at ﬁrst, so we go ahead and write out some diagrams. We have the chain of R-modules

A0 / ··· / Ai−1 / Ai / Ai+1 / ··· where each arrow is an inclusion map. The colimit of this chain of modules is the module A. For some R-module B we select an injective resolution

0 / B / E0 / E1 / E2 / ···

Applying the functors Hom(Ai, −) to the cocomplex E•, we get the following commutative diagram:

0 / Hom(A0,E0) / Hom(A0,E1) / Hom(A0,E2) / ··· O O O

...... O O O

0 / Hom(Ai−1,E0) / Hom(Ai−1,E1) / Hom(Ai−1,E2) / ··· O O O

0 / Hom(Ai,E0) / Hom(Ai,E1) / Hom(Ai,E2) / ··· O O O

0 / Hom(Ai+1,E0) / Hom(Ai+1,E1) / Hom(Ai+1,E2) / ··· O O O

......

The vertical arrows have reversed since Hom(−,Ei) is contravariant. Each row is a cochain complex and each collection of arrows from row to row forms a chain map, so this is actually a tower of cochain complexes,

Hom(Ai+1,E) / Hom(Ai,E) / ··· / Hom(A0,E). 1 ∼ ˆ i To get ExtZ (Zp∞ , Z) = Zp, we substitute Ai = Z/p , A = Zp∞ , and q = 1 to the ﬁrst sequence:

1 i 1 1 i 0 / lim Hom ( /p , ) / Ext ( ∞ , ) / lim Ext ( /p , ) / 0 ←− Z Z Z Z Zp Z ←− Z Z Z which simpliﬁes to

1 i 0 / 0 / Ext ( ∞ , ) / lim /p / 0 Z Zp Z ←− Z i 1 i i i 1 ∼ ∼ ∞ ∼ since HomZ(Z/p , Z) = 0 and ExtZ(Z/p , Z) = Hom(Z/p , Q/Z) = Z/p . Therefore ExtZ(Zp , Z) = lim /pi = ˆ . ←− Z Zp Q Application 3.5.11. Tot (C) is intended. 2.4. Chapter 4. 2.4.1. Section 4.1. 14 SEBASTIAN BOZLEE

2.4.2. Section 4.2.

Proposition 4.2.6. R and Homk(R, k) have the same dimension since Homk(R, k) is the dual space to R, and R is ﬁnite dimensional. Vista 4.2.7. Not every quasi-Frobenius ring is a Gorenstein ring since some quasi-Frobenius rings are non-commutative. For example, k[G], for any non-abelian group G. Lemma 4.2.8. The equality bR = abR + (1 − a)bR follows from the following result: if e ∈ R is idempotent, and M is an R-module, then M = eM ⊕ (1 − e)M. Proof: Every element x ∈ M may be written as ex + (1 − e)x, so eM + (1 − e)M = M. To see that the sum is direct, suppose ex = (1 − e)y ∈ eM ∩ (1 − e)M. Then, multiplying by e, ex = e2x = e(1 − e)y = (e − e2)y = (e − e)y = 0. Since eM + (1 − e)M = M and eM ∩ (1 − e)M = 0, we have that M = eM ⊕ (1 − e)M. 2.4.3. Section 4.3. Theorem 4.3.3. To show that A cannot be projective, note that A would then be a direct L L summand of I R, but x acts as a nonzerodivisor on I R, contradicting that x annihilates A. To justify the second sentence, notice ﬁrst that x 0 / R / R / R/x / 0 is a projective resolution of R-modules, so R/x has projective dimension less than or equal to 1. On the other hand, x(R/x) = 0, so R/x cannot be a projective module, for the same reason that A cannot, so R/x has projective dimension 1. Then by the general change of rings theorem 4.3.1,

0 < pdR(A) ≤ pdR/x(A) + pdR(R/x) = 1, so pdR(A) = 1. We have that pdR(A) = pdR/x(M) + 1 by dimension shifting, and the induction is on pdR/x(A). The next sentence comes from Exercise 4.1.2, plus applying the general change of rings theorem to show pdR(P ) ≤ pdR/x(P ) + pdR(R/x) = 1.

Theorem 4.3.7. It is claimed that pdT (M) ≤ 1 + pdT (T ⊗R U(M)). This follows from exercise 4.1.2, as follows:

pdT (T ⊗R U(M)) ≤ max{pdT (T ⊗R U(M)), pdT (M)} with equality unless pdT (M) = pdT (T ⊗R U(M)) + 1. If so, then trivially

pdT (M) ≤ pdT (T ⊗R U(M)) + 1.

Otherwise, pdT (M) ≤ pdT (T ⊗R U(M)), so pdT (M) ≤ pdT (T ⊗R U(M)) + 1. Proposition 4.3.11. Rn/mRn = kn for obvious reasons, and P/mP ∼= kn by the hypothesis that P/mP is generated by n elements. Now P R P ⊕ ker P ker ∼= kn = = = ⊕ mP mR m(P ⊕ ker mP m ker so ker /m ker = 0 (this works since kn is a ﬁnite dimensional vector space), whence ker ⊆ mR. SOLUTIONS AND ELABORATIONS 15

2.4.4. Section 4.4. Proposition 4.4.5. The Krull intersection theorem (see [2, Corollary 5.4]) tells us that ∩xnR = 0. This is useful since 0 is a prime ideal if and only if R is a domain. There are ﬁnitely many prime ideals of R by a theorem of Noether (see [2, Exercise 1.2]). We include a statement and a proof since I should do the exercise anyway: If R is Noetherian, and I ⊆ R is an ideal then among the primes of R containing I, there are only ﬁnitely many that are minimal with respect to inclusion. Write P (I) for the set of minimal primes containing an ideal I of R. Select an ideal I maximal among those so that |P (I)| = ∞. I is not prime, since if it were, we would have P (I) = {I}. Therefore there exist f, g ∈ R so that f, g 6∈ I but fg ∈ I. Consider a prime p ∈ P (I). Since fg ∈ I ⊆ p, either f ∈ p so (I, f) ⊆ p or else g ∈ p so (I, g) ⊆ p. Since p was a minimal prime containing I, p is also a minimal prime containing (I, f) or (I, g). Then P (I) ⊆ P ((I, f)) ∪ P ((I, g)), so either |P ((I, f))| = ∞ or |P ((I, g))| = ∞, contradicting the maximality of I. The result follows.

The implication that m ⊆ Pi for some i is from “prime avoidance;” see [2, Lemma 3.3]. Since m is maximal, we actually have that m = Pi, and since we have assumed that m contains no prime ideals, this shows that there are no proper inclusions of prime ideals. It follows R has dimension 0. Standard Facts 4.4.6. An associated prime ideal of R is a prime ideal of the form AnnR(M) for a prime R-module M.A prime R-module M is one so that AnnR(M) = AnnR(N) for any nonzero submodule N of M. Corollary 4.4.12. The inequality sup{pd(R/I)} ≤ fd(R/m) comes from the previous proposition. Each module R/I is ﬁnitely generated (actually generated by one element) R so the lemma tells us that pd(R/I) is the maximum integer d so that Tord (R/I, k) 6= 0. Thus each pd(R/I) ≤ fd(k) = fd(R/m), giving the inequality. 2.4.5. Section 4.5.

2.5. Chapter 5.

2.6. Chapter 6.

2.7. Chapter 7.

2.8. Chapter 8.

2.8.1. Section 8.1. Although there are upper and lower indices used throughout this section, it does not appear that there is a convention for which go with simplicial objects and which go with cosimplicial objects. This appears to agree with other sources.

Lemma 8.1.2. 0 ≤ is ≤ · · · ≤ i1 ≤ m should be replaced by 0 ≤ is < ··· < i1 ≤ m in the statement of the lemma. To do: explain why the factorization is equal to α This expression for α is unique since a different factorization into face maps and degenera- cies would imply that different elements in the image are skipped or that different elements in the domain are mapped to the same place. 16 SEBASTIAN BOZLEE

Example 8.1.8. The words “combinational” and “combinatorial” are used interchangeably here and later. According to the errata, the word “combinatorial” was intended, so we will use the name combinatorial in these notes.

2.8.2. Section 8.2. We check that d2 = 0:

n−1 ! n ! 2 X i X j d = (−1) ∂i (−1) ∂j i=0 j=0 n−1 n X X i+j = (−1) ∂i∂j i=0 j=0 n−1 i n n X X i+j X X i+j = (−1) ∂i∂j + (−1) ∂i∂j i=0 j=0 i j=i+1 n−1 i n n X X i+j X X i+j = (−1) ∂i∂j + (−1) ∂j−1∂i i=0 j=0 i j=i+1 n−1 i n n−1 X X i+j X X i+j = (−1) ∂i∂j − (−1) ∂j∂i i=0 j=0 i j=i n−1 i X X i+j i+j = (−1) ∂i∂j − (−1) ∂i∂j i=0 j=0 = 0.

Lemma 8.2.8. We verify a few equalities asserted in the course of the lemma. Recall that we assume inductively that we have found g = gr−1 ∈ Gn+1 so that ∂i(g) = xi −1 for all i ≤ r − 1, i 6= k. In the case that r 6= k, we set u = xr ∂r(g). Then ∂i(u) = 1 for i < r since

−1 ∂i(u) = ∂i(xr) ∂i(∂rg) −1 = ∂i(xr) ∂r−1(∂i(g)) (simplicial identities) −1 = ∂i(xr) ∂r−1(xi) (induction hypothesis) −1 = ∂i(xr) ∂i(xr) (hypothesis) = 1.

This implies ∂i(σru) = 1 since ∂i(σru) = σr−1(∂i(u)) = σr−1(1) = 1, using the simplicial −1 −1 identities. The formula gr = g(σku) should be gr = g(σru) . This satisﬁes the induction hypothesis since, for i < r, i 6= k,

−1 ∂i(gr) = ∂i(g(σr(u)) ) −1 = xi∂i(σr(u))

= xi. SOLUTIONS AND ELABORATIONS 17

And for i = r, −1 ∂r(gr) = ∂r(g(σr(u)) ) −1 = ∂r(g)∂r(σr(u)) −1 = ∂r(g)u (simplicial identities) −1 −1 = ∂r(g)(xr ∂r(g))

= xr. as required. 2.8.3. Section 8.3.

0 00 Lemma 8.2.9. We check that y = ∂n+2z is a homotopy from x to x . First, for i < n,

∂iy = ∂i(∂n+2z) = ∂n+1(∂iz) = ∂n+1(∗) = ∗. Next, 0 0 ∂n(y) = ∂n(∂n+2z) = ∂n+1(∂nz) = ∂n+1y = x . and ﬁnally, 00 00 ∂n+1(y) = ∂n+1(∂n+2z) = ∂n+1(∂n+1z) = ∂n+1y = x .

3. Solutions to Exercises 3.1. Chapter 1. 3.2. Chapter 2. 3.2.1. Section 2.1. 3.2.2. Section 2.2. 3.2.3. Section 2.3. 3.2.4. Section 2.4. Exercise 2.4.5. 3.3. Chapter 3. 3.3.1. Section 3.1. 3.3.2. Section 3.2. Exercise 3.2.1. Suppose that B is a ﬂat R module and that A is any R-module. Let P• → A → 0 be a projective resolution of A. Then P• ⊗ B is exact except at P0 ⊗ B, since R − ⊗ B is an exact functor and P• is exact except at P0. Then Torn (A, B) = Hn(P• ⊗ B) = 0 for n 6= 0. This gives that 1 =⇒ 2. The implication 2 =⇒ 3 is trivial. R 0 00 Finally, suppose Tor1 (A, B) = 0 for all A. Let 0 → C → C → C → 0 be any exact R sequence. Then, from the long exact sequence for Tor∗ (−,B), we get that

R 0 ∂ 0 00 Tor1 (C ,B) / C ⊗ B / C ⊗ B / C ⊗ B / 0 18 SEBASTIAN BOZLEE

R 0 is exact. By hypothesis, Tor1 (C ,B) = 0, so in fact

0 / C0 ⊗ B / C ⊗ B / C00 ⊗ B / 0 is exact. Since 0 → C0 → C → C00 → 0 was arbitrary, this shows − ⊗ B is exact, so B is ﬂat. This gives the implication 3 =⇒ 1. Exercise 3.2.2. Suppose that 0 → A → B → C → 0 is an exact sequence of R-modules R and A and C are ﬂat R-modules. We take the long exact sequence associated to Tor∗ (−,D) for an R-module D and obtain that

∂ R R R ··· / Tor1 (A, D) / Tor1 (B,D) / Tor1 (C,D) / ···

R R is exact. Since A and C are ﬂat, Tor1 (A, D) = 0 and Tor1 (C,D) = 0. By exactness, R T or1 (B,D) = 0. Since D was arbitrary, B is ﬂat, by the implication 3 =⇒ 1 in the preceding exercise. Exercise 3.2.3. As hinted, we use the projective resolution of k given by

−y [ x ] [ x y ] 0 / R / R2 / R / k → 0. From the projective resolution, we deduce that k is intended to be an R module by having x and y act on k trivially, and the constants act by the usual multiplication. R Now we compute Tor2 (k, k) as the cohomology at the second location of the complex −y [ x ]⊗1 [ x y ]⊗1 0 / R ⊗ k / R2 ⊗ k / R ⊗ k / 0, which may be rewritten −y [ x ] [ x y ] 0 / k / k2 / k / 0. −y This is the same as the kernel of the map [ x ]. But x and y both act as zero on k, so this R kernel is k. Thus Tor2 (k, k) = k. R ∼ R Next, we establish the isomorphism Tor1 (I, k) = Tor2 (k, k) by a long exact sequence argument. Consider the short exact sequence

0 / I / R / k / 0. R A portion of the associated long exact sequence under the homological δ-functor Tor∗ (−, k) is R R R R Tor2 (R, k) / Tor2 (k, k) / Tor1 (I, k) / Tor1 (R, k). R ∼ Since R is projective, each of the modules on the ends are zero. By exactness, Tor1 (I, k) = R R Tor2 (k, k). Combining this with the previous paragraph Tor1 (I, k) = k, so I is not ﬂat, despite being torsionfree.

Exercise 3.2.4. Let A → B → C be an exact sequence of R-modules. Then since Q/Z is an injective abelian group (since Q/Z is divisible: see Corollary 2.3.2), HomAb(−, Q/Z) is an exact functor from Ab to Ab. Since the forgetful functor F from R-mod to Ab is also exact, the composition HomAb(F (−), Q/Z) is an exact functor. The image of A → B → C under this composition is A∗ → B∗ → C∗, so A∗ → B∗ → C∗ is exact. Conversely, suppose To do SOLUTIONS AND ELABORATIONS 19

3.3.3. Section 3.3.

∗ Exercise 3.3.1. We use the long exact sequence of the functor ExtZ(−, Z) associated to the short exact sequence 1 ∞ 0 / Z / Z[ p ] / Zp / 0, then hijack the calculation given in Example 3.3.3. Since Hom is contravariant in its ﬁrst input, Ext is a contravariant cohomological δ-functor in its ﬁrst input, which means that the order of the parts of the short exact sequence is reversed in the long exact sequence and indices on Ext are increasing. The relevant portion of the long exact sequence is

1 1 1 1 1 ∞ Hom(Z[ p ], Z) / Hom(Z, Z) / ExtZ(Zp , Z) / ExtZ(Z[ p ], Z) / ExtZ(Z, Z).

1 1 First, we show Hom(Z[ p ], Z) = 0. Suppose 1 7→ a under a homomorphism ϕ : Z[ p ] → Z. 1 a 1 n Then ϕ( pn ) 7→ pn for all n; that is, pn is mapped to an element of Z whose p th multiple is 1 a. This is impossible unless a = 0, so Hom(Z[ p ], Z) = 0. ∼ Next, Hom(Z, Z) = Z. By the universal property of free abelian groups, any group homomorphism Z → Z is determined by the image of 1, and conversely, the image of 1 may be chosen freely. Then if ϕa : Z → Z denotes the group homomorphism such that 1 7→ a, the map a 7→ ϕa is clearly a group isomorphism Z → Hom(Z, Z). 1 ∞ ˆ Next, ExtZ(Zp , Z) = Zp by Exercise 3.3.3. 1 Next, ExtZ(Z, Z) = 0, since Z is projective. Our exact sequence now looks like

ˆ 1 1 0 / Z / Zp / Ext (Z[ p ], Z) / 0

1 1 ˆ So Ext (Z[ p ], Z) is the quotient of Zp by some subgroup isomorphic to Z. I have not checked that this subgroup is the is usual subgroup identiﬁed with Z. ˆ ∼ The next isomorphism we were given to prove, Zp/Z = Zp∞ , is false. This can be checked ˆ by noting that Zp/Z has an element of order m where m is relatively prime to p (namely 1 ∞ [ m ]) while Zp does not. The error is acknowledged in the errata online. Exercise 3.3.2. Let R = Z/m and B = Z/p with p | m, so that B is an R-module. We must show that

i p m/p p m/p 0 / Z/p / Z/m / Z/m / Z/m / Z/m / ··· is an inﬁnite periodic injective resolution of B. The ﬁrst arrow is the inclusion map, p denotes multiplication by p (mod m), and m/p denotes multiplication by m/p (mod m). Multiplication by p in Z/m vanishes precisely on Z/p = (m/p)Z/m and multiplication by m/p in Z/m vanishes precisely on pZ/m, so this is an exact sequence. Each of the modules Z/m are injective by Exercise 2.3.1, so this is actually an injective resolution of B. ∗ To compute ExtZ/m(A, Z/p) we take the cohomology of the image of the injective resolution above under the functor HomZ/m(A, −), namely

p∗ (m/p)∗ p∗ (m/p)∗ 0 / A∗ / A∗ / A∗ / A∗ / ··· 20 SEBASTIAN BOZLEE

∗ where A is shorthand for HomZ/m(A, Z/m) as suggested in the problem statement. Then we get that  A∗ if n = 0  n ∗ ∗ ExtZ/m(A, Z/m) = ker(m/p) /im p if n > 0 is odd ker p∗/im (m/p)∗ if n > 0 is even 2 ∗ Consider the case that A = Z/p and p | m. Then we may identify A = Hom /m(Z/p, Z/m) ∼ Z with (m/p)Z/m = Z/p via the isomorphism ϕ 7→ ϕ(1). This is injective since each homomorphism is determined by the image of 1 and surjective since each homomorphism must take 1 to (p/m)Z/m ⊆ Z/m by order considerations. Now note that p and m/p both act as zero on Z/p since both are divisible by p. It follows that each kernel in the chain complex is Z/p and each image is 0. Taking quotients, n ExtZ/m(Z/p, Z/m) = Z/p, for all n ≥ 0. 3.3.4. Section 3.4. Exercise 3.4.1. Suppose (8) 0 / Z/p / X / Z/p / 0 is an extension of Z/p by Z/p in Ab. We know from finite group theory that any extension of a group A by a group B has order |A||B|. In this case, the group X must have order p2. ∼ ∼ 2 By the fundamental theorem of abelian groups, X = Z/p ⊕ Z/p or X = Z/p . We will show that in each case the extension above is equivalent to one of the given extensions. Then we will show that each of the given extensions are distinct. ∼ Suppose first that X = Z/p ⊕ Z/p. Let (a, b) ∈ Z/p ⊕ Z/p be the image of 1 under the map Z/p → Z/p ⊕ Z/p and let (c, d) be a preimage of 1 under the map Z/p ⊕ Z/p → Z/p. Then (c, d) and (a, b) form a basis of Z/p ⊕ Z/p, viewing Z/p ⊕ Z/p as a Z/p vector space, since (c, d) 6∈ ker{Z/p ⊕ Z/p → Z/p} = Span {(a, b)} and Z/p ⊕ Z/p has dimension two. Let ϕ : Z/p ⊕ Z/p → Z/p ⊕ Z/p be the Z/p-linear (hence Z-linear) map that takes (a, b) to (1, 0) and (c, d) to (0, 1). This is an isomophism since we take a basis to a basis. Then the following diagram commutes, where the first row is the extension (8) and the second row is the usual split extension: 0 / Z/p / Z/p ⊕ Z/p / Z/p / 0 ϕ

i π 0 / Z/p / Z/p ⊕ Z/p / Z/p / 0 ∼ 2 2 2 Next suppose that X = Z/p . The image of Z/p → Z/p must be pZ/p , as it is the unique subgroup of Z/p2 of order p. Then 1 is mapped by Z/p → Z/p2 to some element of the form ap. Let b be the image of 1 under the map Z/p2 → Z/p. Let ϕ : Z/p2 → Z/p2 be the map taking 1 to a−1. This is an isomorphism since a is relatively prime to p. Then the diagram 0 / Z/p / Z/p2 / Z/p / 0

p ab 0 / Z/p / Z/p2 / Z/p / 0 SOLUTIONS AND ELABORATIONS 21

commutes, where the top row is extension (8) and the bottom row is one of the given extensions, where i = ab. Now we have to show that the given extensions are distinct. Let

p 2 i ξi : 0 / Z/p / Z/p / Z/p / 0. ∼ 2 Since Z/p ⊗ Z/p 6= Z/p , we really only have to show that the ξi are distinct. Suppose p i 0 / Z/p / Z/p2 / Z/p / 0

ϕ p j 0 / Z/p / Z/p2 / Z/p / 0 is an equivalence of extensions. In order that the right square commute, we must have that 1 ∈ Z/p2 is mapped by ϕ to some k mod p2 so that k ≡ ij−1 (mod p). We must also have that px is mapped by ϕ to px for all x ∈ Z/p so that the left square commutes. So px = kpx (mod p) for all x. This implies that k ≡ 1 (mod p), so ij−1 ≡ 1 (mod p), so i ≡ j (mod p), which completes the proof. 3.3.5. Section 3.5.

Exercise 3.5.1. Suppose that {Ai} is a tower of abelian groups with injective maps Ai+1 → Ai. We get a topology on A = A0 out of the sets a + Ai (a ∈ A, i ≥ 0) by using these a basis for a topology. To see this is a basis, let a + Ai and b + Aj be basic open sets. Then  a + A a ∈ b + A , i ≥ j  i j (a + Ai) ∩ (b + Aj) = b + Aj b ∈ a + Ai, j ≥ i ∅ else So we have a basis. Suppose first that A is Hausdorff. Let a be any nonzero element of A. Since A is Hausdorff, 0 there exists an open set U containing 0 but not containing A. Since the sets a + Ai form a 0 0 0 basis, there is a set a +Ai ⊆ U containing 0. Since 0 ∈ a +Ai, a +Ai = Ai. Altogether, there is, for any nonzero element a ∈ A, an integer i so that a 6∈ A . Therefore lim A = T A = 0. i ←− i i i Suppose conversely that lim A = T A = 0. Then, given any two distinct points a, b, ←− i i i choose i so that a − b 6∈ Ai. Then a + Ai and b + Ai are disjoint open neighborhoods of a and b respectively, so A is a Hausdorff topological space. For the second part, we need a definition of completeness that makes sense in a topological group. We say A is complete if every Cauchy sequence in A converges, where by a Cauchy sequence (aj) we mean a sequence such that for all i ≥ 0, there exists an integer Ni so that for all j, k ≥ Ni, aj − ak ∈ Ai. (Thanks to Sebastian Casalaina-Martin for the definition.) We take the long exact sequence associated to the short exact sequence of towers 0 → {Ai} → {A} → {A/Ai} → 0, which includes lim A A lim A/A lim1 A 0 ←− i / / ←− i / ←− i / where we have used that lim A = A and that lim1 A = 0, since the maps A → A are ←− ←− onto (Lemma 3.5.3). Now A → lim A/A is an isomorphism if and only if lim A = 0 and ←− i ←− i lim1 A = 0. This only shows that if A is Hausdorff, then A is complete if and only if ←− i lim1 A = 0. ←− i 22 SEBASTIAN BOZLEE

To complete the proof, we show that if A is Hausdorﬀ, A is complete if and only if the induced map A → lim A/A is an isomorphism. (The problem statement is adjusted in the ←− i errata.) First, suppose (aj) is a Cauchy sequence. For each i, let Ni be as in the deﬁnition of

Cauchy sequence. Note that then for all i, j ≥ Ni, aj ≡ ai (mod Ai). Let ϕ((aj))i = aNi (mod A ). Then ϕ((a )) is an element of lim A/A . Now suppose that ϕ((a )) = ϕ((b )) for i j ←− i j j two Cauchy sequences (aj) and (bj). Then (aj) and (bj) are equivalent Cauchy sequences. Conversely, given (b ) ∈ lim A/A , we can ﬁnd a Cauchy sequence (a ) so that ϕ((a )) = (b ) j ←− i j j j by choosing each a to be a lift of b in A. This tells us that lim A/A is the completion of A; j j ←− i that is, it is the space of Cauchy sequences in A modulo equivalence of Cauchy sequences. Now the induced map lim A → lim A/A is deﬁned by (a) 7→ ([a]). That is, its image is the ←− ←− i space of equivalence classes of Cauchy sequences that are equivalent to a constant sequence. This map is an isomorphism if and only if all Cauchy sequences converge if and only if A is complete.

Exercise 3.5.2. Suppose that {Ai} is a tower of finite abelian groups. Denote by ϕj,k the map Aj → Ak. Then for any k, the sequence of subgroups ϕj,k(Aj) is decreasing in j (i.e. ϕj+1,k(Aj+1) ⊆ ϕj,k(Aj)). Since the collection of subgroups is finite, ϕj,k(Aj) must eventually be constant; that is, {Ai} must satisfy the Mittag-Leffler condition. By Proposition 3.5.7, lim1 A = 0. ←− i Similarly, suppose that {Ai} is a tower of finite-dimensional vector spaces. Then for any k, the sequence ϕj,k(Aj) is decreasing in j. In particular, the dimension of ϕj,k(Aj) is decreasing. Since the dimension is strictly positive, this implies that the dimension of ϕj,k(Aj) is eventually constant. Then since ϕj,k(Aj) is also decreasing, this implies that ϕj,k(Aj) is also eventually constant. Therefore {Ai} satisfies the Mittag-Leffler condition, so by the proposition lim1 A = 0. ←− i Exercise 3.5.5. Suppose

ϕ Ay / Az ψ is an element of AI . So that we can deal with elements, I will let A = R-mod. The only nonzero terms of the associated cochain complex are C0 and C1, so the only differential we have to worry about is the differential d : C0 → C1. By definition, the object C0 is Ax×Ay ×Az, with the terms corresponding to the chains x, y, and z respectively, and the object C1 is Az × Az with the first term of the product corresponding to the chain x < z and 0 the other term coresponding to the chain y < z. Following definitions again, d : C0 → C1 1 is defined by (a, b, c) 7→ (ϕ(a), ψ(b)) and d : C0 → C1 is defined by (a, b, c) 7→ (c, c). All p 0 1 other d are zero, so the differential d : C0 → C1 has the formula d − d which takes (a, b, c) 7→ (ϕ(a) − c, ψ(b) − c). The kernel of d is then

{(a, b, c) ∈ C0 | ϕ(a) = ϕ(b) = c} which is isomorphic to

{(a, b) ∈ Ax × Ay | ϕ(a) = ϕ(b)}, SOLUTIONS AND ELABORATIONS 23

the traditional concrete form given to the pullback of Ax and Ay over Az. This gives that 0 H (C∗) is the pullback, as desired. 1 Next, since d : C1 → C2 is zero, H (C∗) is the cokernel of d : C0 → C1. Since ∆ = {(c, c): c ∈ Az} = d(0×0×Az) is contained in im d, we may quotient out by it ﬁrst. More explicitly, we have the diagram with exact rows

d 1 C0 / C1 / H (C∗) / 0

i j

d 1 Ax × Ay / Az / H (C∗) / 0 where i :(a, b, c) 7→ (a, b), j :(c, c0) 7→ c − c0 and d :(a, b) 7→ (ϕ(a) − ψ(b)). This commutes 1 since ker i = 0 × 0 × Az, ker j = ∆ = d(ker i), and ∆ ⊆ ker{C1 → H (C∗). This makes 1 H (C∗) the cokernel of the diﬀerence map, as claimed. 3.4. Chapter 4. 3.4.1. Section 4.1. Exercise 4.1.1. Exercise 4.1.2. The proof is by taking long exact sequences. We’ll tackle the projective dimension part, then claim that the other parts have analogous proofs. Suppose that d > pd(A) and d > pd(B). Then for any R-module D,

Extd(C,D) / Extd(B,D) / Extd(A, D) is exact, and Extd(C,D) = Extd(A, D) = 0, so Extd(B,D) = 0. By the pd Lemma, pd(B) ≤ max{pd(A), pd(C)} as desired. For the other inequality, suppose ﬁrst that ∞ > pd(C) > pd(A) + 1. By the pd Lemma, there is some R-module D so that Extpd(C)(C,D) 6= 0. The long exact sequence associated to Ext∗(−,D) contains the sequence

Extpd(C)−1(A, D) / Extpd(C)(C,D) / Extpd(C)(B,D) / Extpd(C)(A, D). By hypothesis, the terms on the ends are zero, so Extpd(C)(B,D) 6= 0, which proves pd(B) ≥ pd(C) = max{pd(A), pd(C)}. Next, suppose pd(C) < pd(A) + 1 < ∞. There is an R- module D so that Extpd(A)(A, D) 6= 0. This gives us the exact sequence

Extpd(A)(B,D) / Extpd(A)(A, D) / Extpd(A)+1(C,D) By assumption, the term on the right vanishes, so by exactness, Extpd(A)(B,D) 6= 0. This proves pd(B) ≥ pd(A) = max{pd(A), pd(C)}, when both pd(A) and pd(C) are finite. When pd(A) is infinite but pd(C) is not, the long exact sequence associated to Ext∗(−,D) for any D eventually becomes a sequence of isomorphisms of Extp(B,D) with Homp(A, D). p By choosing Dp so that Hom (A, Dp) is nonzero for each p, we get that pd(B) = ∞, as desired. The same argument works in the case that pd(A) is finite while pd(C) is infinite, which completes the proof. When pd(A) = pd(C) = ∞ (in which case pd(C) = pd(A) + 1 could be interpreted as true) we need not have equality, as is shown by the short exact sequence 0 → Z/p → Z/p2 → Z/p → 0 in the category of Z/p2 modules. Here pd(Z/p) = ∞ (see Calculation 3.1.6) while pd(Z/p2) = 0, since Z/p2 is projective. 24 SEBASTIAN BOZLEE

It’s interesting to see that the result cannot be strengthened to include pd(C) = pd(A)+1 p even with ﬁnite projective dimensions, since 0 → Z → Z → Z/p → 0 is a counterexample over Z-modules. More generally, for any non-projective C, we can ﬁnd a short exact sequence 0 → M → F → C → 0 where F is free, so we have a counterexample for every non-projective object C. Parts 2 and 3 have exactly analogous proofs. For the second part, we take long exact sequences with respect to Ext∗(D, −) instead of Ext∗(−,D), and in the third part we take ∗ long exact sequences with respect to Tor (−,D). The short exact sequence 0 → Z → Q → Q/Z → 0 is an example that shows the second part’s result cannot be strengthened to include the case that id(A) = id(C) + 1. Exercise 4.1.3.

(1) First, choose a projective resolution Pi → Ai of minimal length for each i. Then L L L L i Pi → i Ai is a projective resolution of i Ai. The length of Pi is the given supremum, so M pd( Ai) ≤ sup{pd(Ai)}. To do: other inequality. (2) (3) Choose a sequence of R-modules Ai such that pd(Ai) ≥ i for each i ∈ N. Then by L part 1, A = i Ai is a module with inﬁnite projective dimension. 3.4.2. Section 4.2.

Exercise 4.2.1. Z/m is Noetherian because it is finite. To show that Z/m is injective, we use Baer’s criterion (2.3.1). Each ideal of Z/m is of the form (n). By order considerations, any map (n) → Z/m must take n 7→ k · n, and since (n) is cyclic this determines the map. This extends to the map Z/m → Z/m determined by 1 7→ k. Therefore Z/m is also injective as a module over itself, and Z/m is quasi-Frobenius. Exercise 4.2.5. Let a ∈ R. We want to show that there exists x ∈ R so that axa = a. By hypothesis, aR = eR for some idempotent e. From this equality, there is some x ∈ R so that ax = e and some y ∈ R so that a = ey. Then axa = eey = ey = a, as desired. Exercise 4.2.6. 3.4.3. Section 4.3. Exercise 4.3.1. (Thanks to Jonathan Wise for the suggestion to use the local criterion for flatness.) By the first change of rings theorem,

pdk[[x1,...,xn]](k) = 1 + pdk[[x1,...,xn−1]](k).

By induction and noting that pdk(k) = 0, pd(k) = n, so the global dimension of R is greater than or equal to n. Since R is Noetherian, the flat and global dimensions of R coincide, so we now show that the flat dimension of M over R is less than n for all R-modules M. By the fd lemma, it is enough to show that Torn+1(M,N) = 0 for all R-modules N. Assume first that M is SOLUTIONS AND ELABORATIONS 25

finitely generated. Then Torn+1(M, k) = 0, since fd(k) = pd(k) = n. By the local criterion for flatness [2, Theorem 6.8], M is flat. If M is not finitely generated, write M as a colimit of its finitely generated submodules, lim M . Then Tor (M,N) = lim Tor (M ,N) = 0. −→ I n+1 −→ n+1 I Therefore R has global dimension n. Exercise 4.3.2. By the first change of rings theorem,

pdR(A/xA) = 1 + pdR/x(A/xA).

We need to relate pdR(A) with pdR(A/xA), so we consider the short exact sequence x 0 / A / A / A/xA / 0

By Exercise 4.1.2, pdR(A) ≤ max{pdR(A), pdR(A/xA)}, with equality unless pdR(A/xA) = 1 + pdR(A). If pdR(A) + 1 = pdR(A/xA), then pdR(A) = pdR/x(A/xA), and the theorem holds. Otherwise, we have that pdR(A) ≥ pdR(A/xA), so pdR(A) ≥ pdR/x(A/xA) + 1, and the theorem holds. Exercise 4.3.3. 3.4.4. Section 4.4. Exercise 4.4.1. (Thanks again to Jonathan Wise for noting that the remark from the “Standard Facts” was applicable.) 2 Since R is a regular local ring, dimk(m/m ) = dim(R) = d. By the fourth isomorphism theorem for rings, R/(x1, . . . , xi)R is local with maximal ideal m/(x1, . . . , xi)R. Now 2 2 (m/(x1, . . . , xi)) = m /(x1, . . . , xi) so the quotient in question is 2 m m ∼ m ∼ m (x1, . . . , xi) / = 2 = 2 / 2 (x1, . . . , xi) (x1, . . . , xi) m + (x1, . . . , xi) m m which has dimension d − i. It remains to check that the Krull dimension of R/(x1, . . . , xd)R is as claimed. By the remark in the “Standard Facts” following this exercise, if x is a nonzerodivisor on R, then R/x has dimension dim R − 1. Inductively, R/(x1, . . . , xd) has dimension d − i as desired. Exercise 4.4.2. 3.4.5. Section 4.5. Exercise 4.5.1.

(1) We deﬁne the product of [a] ∈ ker dp/im dp+1 and [b] ∈ ker dq/im dq+1 to be the 0 element [ab] ∈ ker dp+q/im dp+q+1. To show this is well deﬁned, suppose a = a + d(α) 0 for some α ∈ Kp+1 and b = b + d(β) for some β ∈ Kq+1. Now ab − a0b0 = d(α)b + ad(β) + d(α)d(β). I claim the right side is equal to d(αb + aβ + αd(β)): d(αb + aβ + αd(β)) = d(α)b + d(α)d(b) + d(a)β + ad(β) + d(α)d(β) + αd2(β) = d(α)b + ad(β) + d(α)d(β), 26 SEBASTIAN BOZLEE

where the second equality follows from the assumption that a, b ∈ ker d and d2 = 0. This shows that the products of different representatives differ by an element of im dp+q+1, so the product is well defined. Associativity, distributivity, and graded commutativity come directly from the corresponding properties of K after reduction modulo the image of d. (2) We will need to assume that R is commutative. To see d2 = 0, it suffices to check that this holds on basis vectors. Suppose i1 < . . . < in. Then n ! 2 X k+1 d (ei1 ∧ · · · ∧ ein ) = d (−1) xkei1 ∧ · · · ∧ eîk ∧ · · · ∧ ein k=1 X j+k+1 = (−1) xij xik ei1 ∧ · · · eîj ··· eîk · · · ∧ ein 1≤j

Noting that the 3rd and 4th lines have additional minus signs and swapping eij+1 and

eij on the 2nd and 5th lines, we get the desired equality. We may now check the Leibnitz rule. By linearity, it suﬃces to check on basis

vectors a = ei1 ∧ · · · ∧ eip and b = ej1 ∧ · · · ∧ ejq , where i1 < ··· < ip and j1 < ··· < jq. Then

d(a · b) = d(ei1 ∧ · · · ∧ eip ∧ ej1 ∧ · · · ∧ ejq ) p ! X k+1 = (−1) xik eip ∧ · · · eˆik · · · ∧ eip ∧ ej1 ∧ · · · ∧ ejq k=1 q ! X p+k+1 + ei1 ∧ · · · ∧ eip ∧ (−1) xjk ej1 ∧ · · · eˆjk · · · ∧ ejq k=1 = d(a)b + (−1)pad(b). Next, the wedge product functions as a product which respects the grading and is graded-commutative. We have now shown that the Koszul complex K(x) is a graded-commutative DG-algebra. The desired external product is deﬁned by the bilinear map

(a(ei1 ∧ · · · ∧ eip ), b(ej1 ∧ · · · ∧ ejq )) 7→ a ⊗ b(ei1 ∧ · · · ∧ eip ∧ ej1 ∧ · · · ∧ ejq ). The Koszul homology is a graded-commutative R-algebra by the part 1. (3) We note that since each xi ∈ I acts as zero, each map in the Koszul complex is the zero map. It follows that the homology is isomorphic to the Koszul complex itself, Λ∗(An). Exercise 4.5.2. We reproduce the argument from page 13 in this context to show that {Hq(x, −)} is a homological δ-functor. If we can show that for each short exact sequence 0 → A → B → C → 0 the diagram

ΛqAn/dΛq+1An / ΛqBn/dΛq+1Bn / ΛqCn/dq+1Cn / 0

d d d 0 / Zq−1Λ∗An / Zq−1Λ∗Bn / Zq−1Λ∗Cn has exact rows, then the Snake Lemma will give the required connecting homomorphisms. The top row can be obtained from the short exact sequence by tensoring over R with the R-module ΛqRn/dΛq+1Rn. Thus the top row is exact by the right exactness of the tensor product. For exactness of the bottom row, consider the diagram

0 / Zq−1Λ∗An / Zq−1Λ∗Bn / Zq−1Λ∗Cn

0 / ΛqAn / ΛqBn / ΛnCn / 0 The bottom row of this diagram is exact since it is obtained by tensoring the short exact sequence with ΛqRn, which is a free and therefore ﬂat R-module. The map Zq−1Λ∗An → Zq−1Λ∗Bn is injective since it is the restriction of an injective map. We have exactness at Zq−1Λ∗Bn by a diagram chase. This completes the proof of the existence of the connecting homomorphism. 28 SEBASTIAN BOZLEE

For naturality of the connecting homomorphism, we can repeat the argument of Proposi- tion 1.3.4, noting that Hq(x, −) is a functor, since it is the composition of the homology and ∗ n Λ R ⊗ − functors. This completes the proof that {Hq(x, −)} is a homological δ-functor. To do: {Hq(x, −)}. Exercise 4.5.3. Exercise 4.5.4. By Exercise 1.4.3, it suffices to show that there is a chain contraction s of the identity map. Suppose without loss of generality that x1 is the unit. I claim that the maps {sq} defined by −1 sq : ei1 ∧ · · · ∧ eiq 7→ x1 e1 ∧ ei1 ∧ · · · ∧ eik is such a chain contraction. We check: q −1 X k+1 (ds + sd)(ei1 ∧ · · · ∧ eiq ) = d(x1 e1 ∧ ei1 ∧ · · · ∧ eiq ) + s( (−1) xkei1 ∧ · · · eîk · · · ∧ eiq ) k=1 q X k −1 = ei1 ∧ · · · ∧ eik + (−1) xkx1 e1 ∧ ei1 ∧ · · · eîk · · · ∧ eiq k=1 q X k+1 −1 + (−1) x1 xke1 ∧ ei1 ∧ · · · eîk · · · ∧ eiq k=1

= ei1 ∧ · · · ∧ eik . So id = ds + sd, as desired. ∗ The groups H∗(x,A) = H (x,A) = 0 are zero since split exact sequences are preserved by additive functors, namely − ⊗ A and Hom(−,A). 3.5. Chapter 5.

3.6. Chapter 6.

3.7. Chapter 7.

3.8. Chapter 8.

3.8.1. Section 8.1.

Exercise 8.1.1. Recall that i :[n − 1] → [n] and ηi :[n + 1] → [n] are deﬁned by ( j if j i To get the idea of what’s going on, it helps to draw a picture of each simplicial identity for n = 4 or so. Of course, this won’t do for a proof, so now we plug and chug. Fix an integer SOLUTIONS AND ELABORATIONS 29 n greater than or equal to zero, and assume i, j are integers so that 0 ≤ i < j ≤ n. Then ( j(k) if k < i j(i(k)) = j(k + 1) if k ≥ i k if k < i and k < j  k + 1 if k < i and k ≥ j = k + 1 if i ≤ k and k + 1 < j  k + 2 if i ≤ k and k + 1 ≥ j  k if k < i  = k + 1 if i ≤ k ≤ j k + 2 if j < k while ( i(k) if k < j − 1 i(j−1(k)) = i(k + 1) if k ≥ j − 1 k if k < j − 1 and k < i  k + 1 if k < j − 1 and k ≥ i = k + 1 if k ≥ j − 1 and k + 1 < i  k + 2 if k ≥ j − 1 and k + 1 ≥ i  k if k < i  = k + 1 if i ≤ k ≤ j k + 2 if j < k so ji = ij−1. Intutively, ij−1 inserts a space at the j − 1st spot then the ith spot of [n]. If we were to do this in the opposite order, we would have to take into account that the j − 1st element of [n] is moves one spot to the right after inserting a space at the ith spot, so ji = ij−1. Next, with the same hypotheses, ( ηj(k) if k ≤ i ηj(ηi(k)) = ηj(k − 1) if k > i k if k ≤ i and k ≤ j  k − 1 if k ≤ i and k > j = k − 1 if k > i and k − 1 ≤ j  k − 2 if k > i and k − 1 > j  k if k ≤ i  = k − 1 if i < k ≤ j + 1 k − 2 if j + 1 < k 30 SEBASTIAN BOZLEE while ( ηi(k) if k ≤ j + 1 ηi(ηj+1(k)) = ηi(k − 1) if k > j + 1 k if k ≤ j + 1 and k ≤ i  k − 1 if k ≤ j + 1 and k > i = k − 1 if k > j + 1 and k − 1 ≤ i  k − 2 if k > j + 1 and k − 1 > i  k if k ≤ i  = k − 1 if i < k ≤ j + 1 k − 2 if j + 1 < k so ηjηi = ηiηj+1. If 0 ≤ i < j ≤ n, then ( ηj(k) if k j  k if k j − 1  k if k j − 1  k if k < i  = k − 1 if i ≤ k < j k if k ≥ j so ηji = iηj−1 in this case. If i = j or i + 1 = j, then ( ηj(k) if k < i ηj(i(k)) = ηj(k + 1) if k ≥ i = k SOLUTIONS AND ELABORATIONS 31

Finally, if j < i, ( ηj(k) if k j  k if k ≤ j  = k − 1 if k > j and k − 1 < i − 1 k if k − 1 ≥ i − 1  k if k ≤ j  = k − 1 if j < k < i k if i ≤ k

so ηji = i−1ηj in this case, as required. Exercise 8.1.2. Suppose that K is an ordered combinatorial simplicial complex consisting of subsets of V . Then given any subset W = {v0, . . . , vn} ∈ K, with v0 < v1 < ··· < vn, we get a corresponding element (v0, . . . , vn) ∈ SSn(K). This cannot be degenerate, as any degenerate n-simplex has a repeated index, while this has none. Conversely, if (v0, . . . , vn) ∈ SSn(K) is nondegenerate, then {v0, . . . , vn} is in K, by deﬁ- nition. This establishes the one-to-one correspondence.

Exercise 8.1.3. Note that each (v0, . . . , vk) in SSk(K) may be interpreted as a morphism v :[k] → [n] in ∆ defined by i 7→ vi. This is a one-to-one correspondence. Recall that given α :[l] → [k], we have defined α∗(v0, . . . , vk) = (vα(0), . . . , vα(l)). The associated morphism takes i 7→ vα(i), so α∗ acts by precomposition on associated morphisms. That is, SS(K) is isomorphic to the functor Hom∆(−, [n]), and we will regard them as interchangeable from here on out. ∗ k Now, to each k and α ∈ Hom∆([k], [n]), we have an associated continuous map α : ∆ → ∆n (from Example 8.1.5, although there it is written with a lower star). These induce, ` k n by the universal property of coproducts, a map ϕ : k SS(K) × ∆ → ∆ defined by ∗ n ` k (α, x) 7→ α (x). Going the other direction, we have the map ψ : ∆ → k SS(K) × ∆ that n n includes ∆ into the copy of ∆ corresponding to id[n]. Clearly, ϕ ◦ ψ = id∆n . k l Next, I claim that (α, x) ∈ SSk(K) × ∆ is equivalent (see 8.1.6) to (β, y) ∈ SSl(K) × ∆ if and only if ϕ(α, x) = ϕ(β, y). Suppose first that ϕ(α, x) = ϕ(β, y). Then α∗(x) = β∗(y), so ∗ ∗ (α, x) = (α ◦ id[n], x) ∼ (id[n], α (x)) = (id[n], β (y)) ∼ (β ◦ id[n], y) = (β, y). It is clear that ϕ(α, x) = ϕ(β, y) is an equivalence relation, so for the other implication it suffices to show that if there is γ :[l] → [k] so that (α ◦ γ, x) = (β, γ∗(y)), then ϕ(α, x) = 32 SEBASTIAN BOZLEE

ϕ(β, y). For this, we note ϕ(β, y) = β∗(y) = (α ◦ γ)∗(y) = α∗(γ∗(y)) = α∗(x) = ϕ(α, x). Therefore ϕ respects the equivalence relation. The universal property of quotients then ` k tells us that ϕ factors uniquely through the quotient map π : k SSk(K) × ∆ → |SS(K)| as ϕ : |SS(K)| → ∆n. In fact, since points are equivalent if and only if their images under ϕ agree, ϕ is a bijection. It remains to see that its inverse function is continuous. For this, we note that

id∆n = ψ ◦ ϕ = ψ ◦ π ◦ ϕ, so ψ ◦ π is its ϕ’s inverse, and ψ ◦ π is continuous. Altogether, |SS(K)| ∼= ∆n. Exercise 8.1.4. Exercise 8.1.5. Exercise 8.1.6. Considering Lemma 8.1.2 in the case that α :[n] → [m] is injective, we get that every morphism in ∆s is a composition of face maps. Therefore a semi-simplicial object K : ∆s → A is determined by its values on objects, Kn = K([n]), and its values on face maps, ∂i : Kn → Kn−1 = K(i :[n − 1] → [n]), for i = 0, . . . , n. We know from the simplicial identities that ∂i∂j = ∂j−1∂i when i < j. On the other hand, given objects Kn and morphisms ∂i satisfying ∂i∂j = ∂j−1∂i for i < j, we get a semi-simplicial object since, the relations ∂i∂j = ∂j−1∂j determine all relations among the i in ∆s.(To Do: This is a bit handwavey. Why is this a complete set of relations?) 3.8.2. Section 8.2. Exercise 8.2.1. We did this in Exercise 8.1.3. Exercise 8.2.2. By Exercise 8.1.2, it is enough to enumerate the non-degenerate n-simplices of ∆[1] × ∆[1] and ∆[2] × ∆[1]. Writing the elements (v0, . . . , vk) × (w0, . . . , wk) of each as ((v0, w0), (v1, w1),..., (vk, wk)), we see that the non-degenerate simplices correspond to sequences of pairs of numbers, increasing in each coordinate, without repeating pairs of numbers. For ∆[1] × ∆[1], we have the non-degenerate 0-simplices ((0, 0)), ((0, 1)), ((1, 0)), and ((1, 1)), ﬁve non-degenerate 1-simplices, ((0, 0), (0, 1)), ((0, 0), (1, 0)), ((0, 1), (1, 1)), ((1, 0), (1, 1)), and ((0, 0), (1, 1)), and two non-degenerate 2-simplices, ((0, 0), (0, 1), (1, 1)) and ((0, 0), (1, 0), (1, 1)). The geometric realization of the corresponding combinatorial simplicial complex consists of two triangles meeting on one edge.

01 11

00 10 SOLUTIONS AND ELABORATIONS 33

This is a square, just like |∆[1]| × |∆[1]|. For ∆[2] × ∆[1], the list of non-degenerate simplices gets unwieldy. We will just draw a picture:

20 21

10 11

00 01 This is a triangular prism made up of three tetrahedrons, namely one with vertices 00, 01, 11, 21, another with vertices 00, 10, 11, 21, and ﬁnally one with vertices 00, 10, 20, 21. |∆[2]| × |∆[1]| is also a triangular prism, so |∆[2] × ∆[1]| = |∆[2]| × |∆[1]|, as desired.

Exercise 8.2.3. Let n > 0. Consider x0 = (0, 1), x1 = (1, 1) in ∆[n]1. We have that ∂0x1 = 1 = ∂0x0. In order that y ∈ ∆[n]2 satisfies ∂0y = x0 and ∂1y = x1, we would need y = (1, 0, 1). But this is not an increasing sequence, so ∆[n] is not fibrant. To Do: Prove that a non-constant fibrant simplicial set must have non-degenerate n-cells for all n. Exercise 8.2.4. Exercise 8.2.5. 3.8.3. Section 8.3.

n Exercise 8.2.6. Take G0 to act on G by g · h = σ0 (g)h for g ∈ G0, h ∈ Gn.

4. Additional Math 4.1. Short Exact Sequences. ([3, p.132] and [1] provided the necessary hints to complete this section.) Deﬁnition 1. A split exact sequence in an abelian category A is a sequence

f g 0 / A0 / A / A00 / 0 such that there exist maps ϕ : A00 → A and ψ : A → A0 so that

(1) g ◦ ϕ = idA00 , (2) ψ ◦ f = idA0 , (3) g ◦ f = 0, (4) ψ ◦ ϕ = 0, (5) f ◦ ψ + ϕ ◦ g = idA. (I don’t think anyone else deﬁnes a split exact sequence this way.) The next proposition shows that a split exact sequence can be thought of as one which is “exact when read forwards as well as backwards,” or as a consequence, as a sequence in which both A0 and A00 are subobjects as well as quotient objects of A. 34 SEBASTIAN BOZLEE

Proposition 2. If a sequence of R-modules

f g 0 / A0 / A / A00 / 0

f g is split exact, and we ﬁx associated maps ϕ and ψ, then both 0 → A0 → A → A00 → 0 and ϕ ψ 0 → A00 → A → A0 → 0 are exact.

00 Proof. The map g is an epimorphism since g ◦ ϕ = idA00 , so the first sequence is exact at A . 0 Similarly, f is a monomorphism since ψ ◦ f = idA0 , so the first sequence is exact at A . Next, ker g ⊇ im f, since g ◦ f = 0. For the other inclusion, let x ∈ ker g. Then x = (f ◦ ψ + ϕ ◦ g)(x) = f(ψ(x)) + ϕ(0) = f(ψ(x)). so x ∈ im f. This proves the first sequence is exact at A. The second sequence is exact by a symmetric argument, since the definition is symmetric under the exchange of g with ψ and f with ϕ. Other definitions of split exact sequences are equivalent to the one we have given by the following proposition:

f g Proposition 3. Let 0 → A0 → A → A00 → 0 be a short exact sequence of R-modules. The following are equivalent: f g (1)0 → A0 → A → A00 → 0 is a split exact sequence. 00 (2) There exists a map ϕ : A → A so that g ◦ ϕ = idA00 . 0 (3) There exists a map ψ : A → A so that ψ ◦ f = idA0 . (4) There exists an isomorphism τ : A → A0 ⊕ A00 so that τ ◦ f is the usual inclusion a0 7→ (a0, 0) and g ◦ τ −1 is the usual projection (a0, a00) 7→ a00. Proof. The implications (1) =⇒ (2) and (1) =⇒ (3) are trivial. We shall show (3) =⇒ (4), (4) =⇒ (1), and leave (2) =⇒ (4) as an adjustment of the proof that (3) =⇒ (4).

00 ((3) =⇒ (4)) Suppose that ψ : A → A is a map so that ψ ◦ f = idA0 . Note that for any x ∈ A, we have that x − f(ψ(x)) ∈ ker ψ, since ψ(x − f(ψ(x)) = ψ(x) − (ψ ◦ f)ψ(x) = ψ(x) − ψ(x) = 0. Thus, any x ∈ A can be written as a sum x = f(ψ(x)) + (x − f(ψ(x))). This implies A = im f + ker ψ. Suppose x ∈ im f and x ∈ ker ψ. Then x = f(y) for some y in A0. It follows x = f(y) = f(ψ(f(y))) = f(ψ(x)) = 0, so in fact A = im f ⊕ ker ψ. Let τ : A = im f ⊕ ker ψ → A0 ⊕ A00 be deﬁned by f(y) + w 7→ (y, g(w)), where w ∈ ker ϕ. We must show τ is an isomorphism. For injectivity suppose x = f(y) + w ∈ A and x0 = f(y0) + w0 ∈ A are elements of A so that τ(x) = τ(x0). Since f is an injection, y = y0. By exactness, w − w0 ∈ im f, so w − w0 = 0. For surjectivity, let (a0, a00) ∈ A0 ⊕ A00. Then for any lift w ∈ A so that g(w) = a00, f(a0) + (w − f(ψ(w))) is a preimage of (a0, a00). Clearly, τ ◦ f is the usual inclusion. Next, (g ◦ τ −1)(a0, a00) = g(f(a0) + (w − f(ψ(w)))) = g(w) = a00 is the usual projection, where w is any preimage of a00 under g. SOLUTIONS AND ELABORATIONS 35

((4) =⇒ (1)) Suppose we have such an isomorphism τ. Then we have the following commutative diagram:

f g 0 / A0 / A / A00 / 0

τ i p . 0 / A0 A0 ⊕ A00 + A00 / 0 j n ψ0 ϕ0 where i : a0 7→ (a0, 0), p :(a0, a00) 7→ a00, ψ0 :(a0, a00) 7→ a0, and ϕ0 : a00 7→ (0, a00). Let ψ = ψ0 ◦τ and let ϕ = τ −1 ◦ ϕ0. Then 0 ψ ◦ f = ψ ◦ τ ◦ f = idA0 and −1 0 g ◦ ϕ = g ◦ τ ◦ ϕ = idA00 by the commutativity of the diagram. This gives us equations (1) and (2) of the deﬁnition. The next equation, g ◦ f = 0, holds by the hypothesis that 0 → A0 → A → A00 → 0 is exact. Next, ψ ◦ ψ = ψ0 ◦ τ ◦ τ −1 ◦ ϕ0 = ψ0 ◦ ϕ0 = 0, so we have the fourth equation. Finally, f ◦ ψ + ϕ ◦ g = f ◦ ψ0 ◦ τ + τ −1 ◦ ϕ0 ◦ g = τ −1 ◦ (i ◦ ψ0 + ϕ0 ◦ p) ◦ τ −1 −1 = τ ◦ idA0⊕A00 ◦ τ

= idA.

0 f g 00 Therefore 0 → A → A → A → 0 is split exact. Our deﬁnition of short exact sequence makes the following theorem easy. Theorem 4. The image of a short exact sequence 0 → A0 → A → A00 → 0 under an additive functor to an abelian category is a short exact sequence.

f g Proof. Let 0 → A0 → A → A00 → 0 be a short exact sequence in an abelian category A, with associated maps ϕ : A00 → A and ψ : A → A0. Let F : A → B be an additive functor. Then F f F g 0 → F (A0) → F (A) → F (A00) → 0 is a short exact sequence since equations (1)-(5) of the deﬁnition are preserved by F . References [1] Dremodaris. Additive functors preserve split exact sequences. http://math.stackexchange.com/questions/133387/ additive-functors-preserve-split-exact-sequences, 2015. Online; accessed 19-September-2015. [2] David Eisenbud. Commutative Algebra with a View Toward Algebraic Geometry. Springer, New York, 2004. [3] Serge Lang. Algebra: Revised Third Edition. Springer, New York, 2002.