pooya-git.github.io · 2. CW COMPLEXES 5 2. CWcomplexes Definition 2. f∶A—→XhasthestructureofarelativeweakCWcomplexif X(0)=A andwehavepushouts N ∈I(i)S n( ) / X(i) N ∈I(i)D

Algebraic TopologyMath 528, Spring 2011

Professor Donald Stanley

Notes by: Pooya Ronagh

Department of mathematics, University of British ColumbiaRoom 121 - 1984 Mathematics Road, BC, Canada V6T 1Z2

E-mail address: [email protected]

Chapter 1. Homotopy Theory 31. Cofibrations 32. CW complexes 53. Cellular and CW approximation 94. Hurewicz homomorphism 145. Moore spaces 166. Eilenberg-Maclane spaces/Postnikov approximations 18

Chapter 2. Fibrations 201. Fiber bundles and fibrations 20

Chapter 3. Spectral Sequences 251. Construction of spectral sequences 27

Chapter 4. H-spaces and algebras 341. H-spaces 342. (co)-Algebras 363. Hopf algebras 37

Chapter 5. Localization (rationalization) 421. Cohomology Serre spectral sequence 47

Chapter 6. Loop suspensions 501. Another homotopy computation 57

Chapter 7. Classifying spaces 601. Category of G-bundles 602. Fiber bundles 623. Milnor’s construction 64

Contents

CHAPTER 1

Homotopy Theory

1. Cofibrations

Last time talked about cofibrations in CG, the space of compactly generated, Hausdorffspaces (see appendix of Hatcher for more information).

Definition 1. X ∈ CG when this holds: C ⊆ X is closed iff for all K compact, C ∩K isclosed.

In particular all CW complexes are in CG.

Proposition 1. CG is closed under pushouts, coproducts, direct limits (if

A(0) ⊆ A(1) ⊆ A(2) ⊆ ⋯

each A(i)Ð→ A(i + 1) is a cofibration, then A(0)Ð→ limÐ→A(i) = ∪A(i) is a cofibration).

Proposition 2. In any category consider

A //

B

// C

D // E // F

(1) Both squares pushout then rectangle is a pushout.

(2) Left square and rectangle pushout then right square is a pushout.

(3) Right square and rectangle being pushout then it is not true that the left square isa pushout.

Proof. If

A //

f

C

g

B // D

3

1. COFIBRATIONS 4

is a pushout thenA × I ∪B × 0

// C × I ∪D × 0

B × I // D × Iis a pushout. Really the top right is the pushout of

C × 0 //

D × 0

C × I

.

If C ⊆D then it is the subspace C × I ∪D × 0 ⊆D × I. Hence if f is a cofibration then gis a cofibration.

InA //

C //

C × I

B // D // C × I ∪C Dleft is a pushout by hypothesis, right is a pushout by defintion. So rectanlge is a pushout.

A //

A × I //

C × I

B // A × I ∪A B // C × I ∪C Dleft is a pushout be defintion, rectangle is a pushout by last step. So right is a pushout.

A × I //

A × I ∪B //

B × I

C × I // C × I ∪D // D × ILeft is a pushout by last step. Rectangle is a pushout since in CG. So right is a pushoutas desired.

Next thing to show is that

Proposition 3. Sn Ð→Dn+1 is a cofibration.

Proof. Dn+1 × I ⊆ Dn+1 × R projecting from (0,2) gives retraction Dn+1 × I Ð→ Sn ×I ∪Dn+1 × 0.

fig(1).

Example 1.1. For all X, ∅Ð→X is a cofibration.

2. CW COMPLEXES 5

2. CW complexes

Definition 2. f ∶ A Ð→ X has the structure of a relative weak CW complex if X(0) = Aand we have pushouts

∐α∈I(i) Sn(α) //

X(i)

∐α∈I(i)Dn(α)+1 // X(i + 1)

with the indexing: n(α) ∈ −1,0,⋯, and S−1 = ∅ and D0 = ∗. And

X =X(∞) = ∪X(i)

is a (standard) relative CW complex if for every α ∈ I(i), n(α) = i − 1. And if A = ∅ thenX is a (weak) CW complex. In the (standard) CW case then Xi =X(i) is called i-skeletonof X.

Remark. In fact we need to define (weak) relative CW complexes up to an isomorphismof maps: f is a relative CW complex if f ≅ g and g can be given the structure of a relativeCW complex, where the isomorphism of f and g mean there exists commutative diagrams

X

f

// X ′

g

// X

f

Y // Y ′ // Y

with horizontal arrows isomorphism.

Exercise 1. Show that the (relative) weak CW complexes are the smallest class of mapscontaining ∅Ð→ ∗ and Sn Ð→ Dn+1 for all n ≥ 0 and closed under pushouts, coproducts and(countable) direct limits; i.e if

A //

f

B

g

C // D

is a pushout then f is relative weak CW complex then so is g.

Proposition 4. If f ∶ AÐ→X is a relative CW complex then f is a cofibraiton.

Proof. Given any diagram below we can complete the diagram:

A × I ∪X //

Z

X × I

ϕ::t

tt

tt

2. CW COMPLEXES 6

Fix X(0) = A and show that X(i) Ð→ X(i + 1) is a cofibration. Then X = colimX(i)so A Ð→ X is a cofibration by previous proposition. For every α, Sn(α) Ð→ Dn(α)+1 is acofibration. ∐α∈I(i) S

n(α) Ð→∐α∈I(i)Dn(α)+1 is a cofibration. But then we have a pushout

∐α∈I(i) Sn(α)

i

// X(i)

j

∐α∈I(i)Dn(α)+1 // X(i + 1)

i cofibation ⇒j cofibration. Then direct limit of cofibrations is a cofibration ⇒AÐ→X is acofibration.

For f ∶X Ð→ Y the mapping cylinder of f is the pushout

Xf

//

x↦(x,0)

Y

i

X × I // M(f)

Y Ð→ M(f) is a cofibration. i has strong deformation retraction. And in particular theyare homotopy equivalent.

If X is a CW complex then Y Ð→ M(f) is relative (weak) CW complex. If Y is a (weak)CW complex then X Ð→M(f) via x↦ (x,1) is a (weak) relative CW complex. Recall thatDn × I can be given the structure of a CW complex. There is a pushout

X //

Y

X∐X //

Y ∐X

X × I // M(f)

is a pushout so if X∐X Ð→X × I is a relative (weak) CW structure so is Y ∐X Ð→M(f).Also X Ð→X∐Y is relative CW if Y is. Y Ð→X∐Y is a relative CW complex if X is. Sobasically you have to show that X∐X Ð→X × I is a relative CW complex.

∐Si × I ∪Di+1 ∪Di+1 //

X∐X ∪X(i) × I

∐Di+1 × I // X∐X ∪X(i + 1) × I

And you can more generally write α(i) for any i.

2. CW COMPLEXES 7

Definition 3. f ∶ X Ð→ Y is a weak homotopy equivalence if for any x0 ∈ X and any nπn(f) ∶ πn(X,x0)Ð→ πn(Y, f(x0)) is an isomorphism.

Theorem 2.1 (Whitehead’s). If X,Y are path-connected (weak) CW complexes and f ∶X Ð→ Y is a weak homotopy equivalence then f is a homotopy equivalence. And in particularif f is a subcomplex then f has a strong deformation retraction.

Lemma 1. Let (X,A) be a relative (weak) CW complex and (Y,B) be just any pair andB ≠ ∅. Suppose for evey n such that X −A has n-cells (have the vanishing of obstructiontheory given by) πn(Y,B, y0) = 0 for any y0 ∈ B. Then for every map f ∶ (X,A) Ð→ (Y,B)of pairs, there exists g such that im(g) ⊆ B and f ≅ g rel A.

Proof. Recall the compression lemma: πn(Y,B, y0) = 0 iff for every f ∶ (Dn, Sn−1,∗)Ð→(Y,B, y0) there exists g such that f ≅ g rel Sn−1 and im(g) ⊆ B. Suppose

A =X(0)X(1) ⋯X(n) ⋯X(∞) =X

and we have a sequence of homotopies Hk ∶ fk ≅ fk+1 rel X(k − 1) with fk+1(X(k)) ⊆ Bfor all k < n. We will construct the next one: Denote by fn = fn∣X(n) the restriction. Webasically want to complete

X(n − 1)f ∣X(n−1)

//

B

X(n)fn

//

;;ww

ww

wY

But we also have a pushout

∐Sn(α)∐h(α)

//

X(n − 1)f ∣X(n−1)

//

B

∐Dn(α)+1∐h(α)

// X(n)fn

//

;;ww

ww

wY

So by the compression lemma for

Sn(α)

fnh(α)// B

Dn(α)+1fnh′(α)

// Y

2. CW COMPLEXES 8

there is

Sn(α) × I

constant // B

Dn(α)+1 × I H// Y

with H(−,0) = fnh′(α) and H(−,1)(Dn(α)+1) ⊆ B. So get by taking coproducts Hn as theleft square below is a pushout:

∐Sn(α) × Ih(α)

//

X(n − 1) × I

const. homotopy// B

∐Dn(α)+1 × Ih(α)

// X(n) × IHn

//__________ Y

X(n)X is a cofibration so in

X(n) × I ∪XHn∪fn

//

Y

X × I∃Hn

66llllllllllllllll

call fn+1(x) ∶=Hn(x,1) andHn(x,0) = fn(x), Hn ∶ fn ≅ fn+1 rel X(n−1) and fn+1(X(n)) ⊆B.

Last step is to put all these homotopies together to get

H(x, t) =Hk(x,2k+1t − (2k+1 − 1)), t ∈ [1 − 1/2k,1 − 1/2k+1].

These glue together to give a homotopy basically since Hk(x,1) = fk+1(x) =Hk+1(x,0). His continuous since H ∣X(k)×I is for k. Since H in eventually stationary on each X(k) (justcomposing finitely many homotopies on each X(k)).

Proof of Whitehead’s theorem. Suppose i ∶ X Y is (weak) relative CW com-plex. From the long exact sequence on πk

⋯Ð→ πnX Ð→ πxY Ð→ πn(Y,X) ∂Ð→ πn−1(X)Ð→ ⋯

implies when πnX ≅ πnY that πn(Y,X) = 0 for all n. So idY ≅ g rel X and g(Y ) ⊆ X. Sog is the homotopy inverse and hence X ≅ Y . More generally for any map f ∶X Ð→ Y , where

3. CELLULAR AND CW APPROXIMATION 9

Y ≅M(f) we have that i ∶X Ð→M(f) via x↦ (x,1) making the following

Xi //

f""EEEEEEEEE M(f)p

Y

commute. Since the down arrows are weak homotopy equivalences (f by hypothesis and psince homotopy equivalence implies weak homotopy equivalence), “2 out of 3” implies thati is a weak homotopy equivalence!

Also i is relative (weak) CW complex hence homotopy equivalence. So f = ip is homotopyequivalence.

3. Cellular and CW approximation

Suppose X,Y are have given CW structures then a map f ∶ X Ð→ Y is called cellular if forevery n, f(Xn) ⊆ Yn.

Lemma 2. Given diagramSn //

Y

Dn+1f// Y ∪Dk

with k > n + 1 there exists g such that f ≅ g rel Sn and g(Dn+1) ⊆ Y .

Lemma 3. Given diagramSn //

Yn+1

Dn+1f

// Y

where Y has CW structure, there exists g ∶Dn+1 Ð→ Y such that f ≅ g rel Sn and g(Dn+1) ⊆Yn+1.

Proof. By compactness of Dn+1, f factors through f ′ below; i.e. there exists a com-mutative diagram

Sn //

Yn+1

Dn+1

f##HHHHHHHHH f ′// Y ′

_

Y


where Y ′ − Yn+1 has finitely many cells all of dimension > n + 1 so apply lemma 2 finitelymany times.

Lemma 4. Given diagramYn

Xn

;;wwwwwwwww//

Yn+1

Xn+1f

//

g;;w

ww

ww

Y

where X,Y have CW structures, there exists a homotopy H ∶X × I Ð→ Y such that

H ∣Xn+1×I ∶ f ≅ g rel Xn and g(Xn+1) ≅ Yn+1.

Proof. As in Whitehead’s theorem! First construct H ∶Xn+1 × I Ð→ Y then extend toH using the fact that Xn+1 Ð→Xn is cofibration.

Theorem 3.1 (Cellular Approximation). If X,Y are given CW structures and f ∶ X Ð→ Yis any map then f ≅ g with g cellular (relative version is also true).

Proof. Use lemma 4 and same argument as in Whitehead’s theorem (gluing homo-topies).

What remains is

Proof of lemma 2. Given

Sn //

X

Dn+1f// X ∪ ek

where k > n + 1. Consider ek ≅ Rk. Since f is a map on a compact set, it is uniformlycontinuous. Hence there exist ε such that

∣x − y∣ < ε⇒ ∣f(x) − f(y)∣ < 12.

Cut Dn+1 ≅ In+1 into cubes with diameters < ε.

figure(2)

Let U = f−1B(0,1/2), V = f−1B(0,1) and let K1 be the union of all cubes intersecting U .Then K1 ⊆ V and f(K1) ⊆ B(0,1) and let now K2 be the union of all cubes which intersectK1. Note K1 is closed, ∂K2 is closed and hence ∂K2 ∩K1 = ∅ and also f(K2) ⊆ B(0,11

2) ⊆


Rk. So there exists some ψ ∶ K20,1Ð→ and ψ(∂K2) = 0 and ψ(K1) = 1 by normality of K2.

Triangulate K2 refining the cubes.fig(3)

Then let g ∶K2 Ð→ Rk ≅ ek be the linear map agreeing with f on vertices. Now we just givea homotopy:

H(x, t) = (1 − tψ(x))f(x) + tψ(x)g(y)Let h =H(−,1). SinceH is a homotopy relative to ∂K2 we can glue with constant homotopyto get

H ∶Dn+1 Ð→ Y ∪Dk ∶ f ≅ h′ rel Sn

where h′∣K2 = h. Also h′K1= gK1 . But gK1 is linear on simplices of dimension n+1 and they

are only finitely many mapping into Rk, k > n+ 1. So im(g) is in the union of finitely manyhyperplanes in Rk so is not surjective. Thus h′ ∶ Dn+1 Ð→ Y ∪Dk factors through Dk witha point removed:

Dn+1 h′ //

&&MMMMMMMMMM Y ∪Dk

Y ∪Dk − uSo h′ ≅ f rel Sn via

Sn

// Y

Dn+1 h′ // Y ∪Dk − u

r

\\

the deformation retraction r.

Remark. From the proof it follows that if f is already cellular on Xn then we can assumef ≅ g rel Xn.

Corollary 1. If Y = X ∪Sn Dn+1 and n > 1, i ∶ X Ð→ Y , then π∗(i) is isomorphism for∗ < n and is surjective for ∗ = n.

Proof. For surjectivity let α = [f] ∈ πi(Y ) for any i ≤ n. f ∶ Si Ð→ Y so f ≅ g,g ∶ Si Ð→ Y that factors though X by lemma 2. Suppose α = [g] ∈ πk(X) for k < n such thati∗(α) = 0 then we get diagram

Skg

//

X

Dk+1 h // Y =X ∪Dn+1

since k + 1 < n + 1 again by lemma 2, h ≅ f rel Sk for f ∶ Dk+1 Ð→ X hence [g] = 0 inπk(X).


Definition 4. (X,x0) is n-connected if πi(X,x0) = 0,∀i ≤ n.

Hence 0-connected is the same as path-connected and 1-connected is the same as simplyconnected.

Definition 5. Similary (X,A) is n-connected, if πn(X,A) = 0 for every i ≤ n.

We are going to assume (as usual) that X and A are path connected. From the long exactsequence of pair (X,A)

⋯Ð→ πiAÐ→ πi(X)Ð→ πi(X,A)Ð→ πi−1AÐ→ πi−1X Ð→ ⋯we get

Proposition 5. (X,A) is n-connected if and only if

πi(A)Ð→ πi(X)is an isomorphism for i < n and is surjective if i = n.

Example 3.1. (X ∪Sn Dn+1,X) is n-connected from corollary 1.

Proposition 6 (CW approximation). Say X is path connected. Then there is a (weak)CW-complex Y and weak homotopy equivalence ηX ∶ Y Ð→X. (It probably works if X is notpath-connected?)

Remark. CW (−) is a functor and η ∶ CW (−) Ð→ id is a natural transformation. In theabove statement CW (X) = Y .

Before giving a proof we state for the record the following trivial statement:

Proposition 7. (Z, z0) ↦ (Z∐W,z0) induces an isomorphism on πi, i ≥ 1 and injectionon π0.

Proof of proposition 6. For step one consider

Y (1) =∐i≥0

∐f∈Hom(Si,X)

Si.

Note that this is functorial. And let f(1) ∶ Y (1)Ð→X be the obvious map.

Claim: f(1) is surjection on π∗. Precisely speaking, we mean that for any x0 ∈ X there isy0 ∈ Y (1) such that for any α ∈ π∗(X,x0) there exists β ∈ π∗(Y (1), y0) satisfying f(1)∗(β) =α. 1

In fact let α = [g ∶ Si Ð→X] then the composition Sig Ð→ Y (1)Ð→X is g and πi(g)([idSig]) =[g]. Then apply the previous proposition.

1A way to get around with the technicality here will be considering the wedge product rather thandisjoint union and consider only the pointed maps.


Assuming we have defined Y (i) and f(i) ∶ Y (i) Ð→ X for all i ≤ n, compatible with theinclusions ki,i+1 ∶ Y (i)Ð→ Y (i + 1), we will construct Y (n + 1) as the pushout

∐i∐(f,H) Si //

Y (n)

kn,n+1

f(n)

""FFFFFFFFFF

∐i∐(f,H)Di+1 // Y (n + 1) //___ X

where f ∶ Si Ð→ Y (n) and H ∶Di+1 Ð→X make

Sif//

Y (n)

f(n)

Di+1 H // X

commute. Define fn+1 in the obvious way. This is compatible with other maps and isfunctorial.

Claim: Suppose α ∈ πi(Y (n)) and f(n)(α) = 0 as element of πi(X) (slightly different forπ0). Then kn,n+1(α) = 0.

Say [α] = α. f(n)(α) = 0 if and only if there is an extension

Si

f(n)(α)// X

Di+1

==

which gives a commutative diagram

Si

α // Y (n)

f(n)

Di+1 H // X

and hence a homotopy of α to 0 in Y (n + 1).

As the last step of the construction, let

Y = limÐ→Y (n), f = limÐ→ f(n).

Note that Y and f are both functorial.

Claim f is a weak homotopy equivalence.

4. HUREWICZ HOMOMORPHISM 14

In factY (1) //

f(1)

Y

fzzzzzzzzz

X

commutes so f(1) is surjective. Consequently f is surjective. For injectivity (π0 will bedifferent but,) let α = [g] ∈ πi(Y ) and suppose f∗(α) = 0. By compactness of Sn

Hom(Sn, limÐ→Y (i)) = limÐ→Hom(Sn, Y (i)).

So g ∶ Si Ð→ Y must factor through Y (n) for some n. So we get a commutative diagram

Yn

j

kn,n+1

""DDDDDDDD

Si

g>>

g AAAAAAAA Yn+1

j′

||yyyyyyyy

Y

.

But (kn,n+1)∗[g] = 0. So j∗[g] = 0, thus [g] = 0. So f ∶ Y Ð→ X is a weak homotopyequivalence.

Remark. Another construction would be to let adding only the n-cells in the n-th step.Using

∨Sn //

Y (n)

∨Dn+1 // Y (n + 1)get functorial CW complex (not weak), hence also functorial cellular chains.

4. Hurewicz homomorphism

Consider [f] ∈ πn(X,A,x0) = πn(X,A) (or just ∈ πn(X)) where f ∶ (Dn, ∂Dn) Ð→ (X,A)or equivalently f ∶ Sn Ð→X. Get

f∗ ∶Hn(Dn, ∂Dn)Ð→Hn(X,A) ( or f∗ ∶Hn(Sn)Ð→Hn(X)).Pick an isomorphism Z ≅Hn(Dn, ∂Dn) which is really just a choice of ±1. Get the Hurewiczhomomorphism

h ∶ πn(X,A)Ð→Hn(X,A)via α = [f]↦ h(α) ∶=H∗(f)(1).

Remark. (1) This is well defined since f ≃ g implies H∗f =H∗g.

(2) The isomorphism Hn(Dn, ∂Dn) ≅ Z is needed in the definition so fix that.

4. HUREWICZ HOMOMORPHISM 15

Proposition 8. (a) h is compatible with exact sequences of pairs (for π∗ and H∗).

(b) h is a homomorphism for n > 1.

Remark. n = 1 is the absolute case is done (abelianization of fundamental group).

Proof. (a) Straightforward, check it basically for the connecting homomorphism. (b)In the absolute case recall that concatenation corresponds to addition of the abelian groupπn. Alternatively f + g is the composition of the pinch map ψ ∶ Sn Ð→ Sn ∨Sn (mod out the

equator in the first variable) to Sn ∨ Snf∨gÐÐ→X ∨X foldÐ→X. The folding is explicitly

fold(x1, x2) =⎧⎪⎪⎪⎨⎪⎪⎪⎩

x1 x2 = ∗x2 x1 = ∗∗ x1 = x2 = ∗

where the wedge product is represented as X ∨X = (x1, x2) ∶ x1 = ∗ or x2 = ∗. Now have

h(f + g) = (f + g)∗(1) = ((fold)(f ∨ g)ψ)∗(1)1= fold∗f∗(1) + g∗(1)

2= f∗(1) + g∗(1) = hf + hg.

Here the equality (1) follows from ψ(1) = (1,1) and the identity Hn(Sn ∨ Sn) ≅ Hn(Sn)⊕Hn(Sn). In fact the following

HnSn ⊕Hn(Sn)

≅

f∗⊕g∗// Hn(X)⊕Hn(X)

≅

Hn(S)

77oooooooooooo ψ

1↦(1,1)// Hn(Sn ∨ Sn) (f∨g)∗

// Hn(X ∨X)

commutes since the square commutes on each factor. And (2) follows since (fold)∗ ∶Hn(X∨X)Ð→H∗(X) is just (f, g)↦ f + g. For the relative case it is similar.

Proposition 9 (Hurewicz’s isomorphism). If (X,A) is (n−1)-connected (n ≥ 2), and X,Aare simply connected and A ≠ ∅ then

h ∶ πn(X,A)Ð→Hn(X,A)

(where the first nontrivial groups happen) is isomorphism and Hi(X,A) = 0 if i < n.

We will prove this later using spectral sequences (or see Hatcher).

Note 4.1. Note that if n = 1 and we are in the absolute case (A = ∗) then

π1(X)Ð→H1(X)

is the abelianization. So we actually need n ≥ 2 condition.

5. MOORE SPACES 16

Example 4.2. In the n = 2 case with A not simply connected, say A = S1 and S1 Ð→S1 ∨ S2 =X then from the long exact sequence of pairs

0 = π2(S1)Ð→ π2(S1 ∨ S2)Ð→ π2(S1 ∨ S2, S1)Ð→ π1(S1)

but the second term is isomorphic to π2(S1 ∨ S2), i.e. π2 of the covering which is homotopyequivalent to ⋁i∈Z S2. Thus this term is infinitely generated but H2(S1∨S2, S1) ≅ H2(S2) ≅Z. So we don’t get an isomorphism between π2 and H2 of (S1 ∨ S2, S1).

Example 4.3.πn(Sn) ≅ Z

andπn(Sn ∨ Sn) ≅ πn(Sn)⊕ πn(Sn) ≅ Z⊕Z.

Proposition 10. If X,Y are simply connected CW complexes and f ∶ X Ð→ Y , which ishomology isomorphism then it is a weak homotopy equivalence (and consequently homotopyequivalence).

Proof. We compare long exact sequences: Assume X Ð→ Y is a CW-pair (else considerthe mapping cylinder and use M(f) ≃ Y ). Consider the long exact sequence on π∗ and H∗,

πnX //

πnY //

πn(Y,X) //

πn−1(X) //

πn−1(Y )

Hn(X) // Hn(Y ) // Hn(Y,X) // Hn−1(X) // Hn−1(Y )

We know that Hn(Y,X) = 0 for all n, so by induction on n and from Hurewicz’s ho-momorphism it follows that πn(Y,X) = 0 for all n (note that we have assumed simplyconnectedness). Now finish using Whitehead’s theorem.

5. Moore spaces

Given an abelian group G, and n ≥ 2, there exists (uniquely up to homotopy equivalence)a CW-complex M(G,n), called a Moore space, such that

HkM(G,n) = G k = n0 else and πkM(G,n) = G k = n

0 i < n .

Notation: Pn(p) =M(Z/p,n − 1).

A quick example: M(Z, n) = Sn.

5. MOORE SPACES 17

We now show the existence of the Moore space for G = Z/`: Either consider the mappingcone of Sn ×`Ð→ Sn

Sn×` //

%%KKKKKKKKKKK Sn //

≃

C(×`)

M ∶=M(×`) // M(×`)/Sn

or the pushout

Sn×` //

Sn

Dn+1 // M(G,n)as definition. Check that these two definitions are the same. Get long exact sequence forthe pair (M,Sn) from the diagram above:

Hi(Sn)Ð→Hi(M)Ð→Hi(M,Sn) = Hi(M/Sn)Ð→Hi−1(Sn)For i ≠ 0, n, n+1 the vanishing ofHi(M) is immediate. For i = 0 and i = 1 deriveHi(M,Sn) =0 from the long exact sequence again, and for i = n

0Ð→ Z ×`Ð→ ZÐ→?Ð→ 0Ð→ 0

is exact and get ? = Z/`Z.

For uniqueness suppose there is another simply connected CW-complex M ′ with

Hk(M ′) = Z/` k = n0 else

By induction and Hurewicz M ′ is (n-1)-connected. By Hurewicz homomorphism again

πk(M ′) ≅ Z/` so there is SnfÐ→ M ′ for which the induced πn(Sn) Ð→ πn(M ′) via 1 ↦ [1]

is surjection on πn. Hence by Hurewicz it is also sujrective on Hn. On the other hand0 = `[f] ∈ πk(M ′), meaning we have a homotopy H ∶ `f ≃ 0 completing the diagram

Sn×` //

Sn

f

Dn+1H// M ′

.

But M(Z/`, n) is a pushout so there is ϕ ∶M(Z, `) Ð→M ′ arising from the pushout. Fromthe triangle

Sn

f

%%KKKKKKKKKK

M(Z/`, n) ϕ// M ′

6. EILENBERG-MACLANE SPACES/POSTNIKOV APPROXIMATIONS 18

we getZ

f∗

!! !!DDDDDDDD

Z/` ϕ∗// Z/`

on the n-th homology level. So ϕ∗ is a surjection and thus an isomorphism. This completesthe proof that ϕ∗ is an isomorphism on H∗. Now apply the previous proposition.

6. Eilenberg-Maclane spaces/Postnikov approximations

Proposition 11. For all X (path-connected as always!) and for all n, there is Y =∶ Pn(X)with f ∶X Ð→ Y such that

πi(Y ) = πi(X) i ≤ n0 i > n

and πi(f) is an isomorphism.

Note again that this construction can be made functorial.

Sketch of the proof. Let Yn =X. Assuming Yk is defined, then Yk+1 is the pushout

⋁g∈Hom(Sk,Yk) Sk //

Yk

⋁g∈Hom(Sk,Yk)Dk+1 // Yk+1

observe that πi(Yk) = πiX for i ≤ n and 0 if k > i > n. And let Y = Y∞ = limÐ→Yk andf ∶X Ð→ Y is the inclusion.

Proposition 12. X Ð→ PnX is relative CW complex and for all X Ð→W such that πi(W ) =0 if i > n then there is a unique (up to homotopy) dashed extension

X //

""DDDDDDDD PnX

W

so we can think of the functor Pn as a localization.

Proof. Exercise.

Proposition 13 (Eilenberg-Mac Lane spaces). For all n ≥ 1, and abelian group G thereexists a unique (up to homotopy) CW complex K(G,n) such that

πi K(G,n) = G i = n0 else

6. EILENBERG-MACLANE SPACES/POSTNIKOV APPROXIMATIONS 19

Proof. n = 1 is handled with K(π,1) = Bπ spaces (and G need not be abelian for thatmatter). So let n > 1 the existence is easy: let

K(G,n) = Pn+1M(G,n)and this is OK since πnM(G,n) ≅ HnM(G,n) ≅ G. For uniqueness the idea is to use thelast proposition: Suppose

πiK′ = G i = n

0 elseand construct the mapping M(G,n) Ð→ K ′ as in proof of existence of M(G,n). Thenthere is ϕ ∶ Pn+1(M(G,n)) Ð→ K ′ through which the former map factors and which is π∗equivalence and thus by Whitehead’s theorem a homotopy equivalence.

Terminology: A product of EM spaces is called a GEM (generalized Eilenberg- Mac Lane)space.

Example 6.1. πi(∏i(K(G,n))) = ⊕πi(K(Gi, i)) = Gi so we can construct spaces withany homotopy groups we want.

Example 6.2. K(Z,1) = S1 and K(Z,2) = CP∞.

CHAPTER 2

Fibrations

1. Fiber bundles and fibrations

Definition 6. Recall p ∶ E Ð→ B has the homotopy lifting property with respect to Xif for any solid diagram below there is a dashed extension ϕ completing the commutativediagram:

X //

E

X × Iϕ

<<xx

xx

x// B

Definition 7. p ∶ E Ð→ B is a (Hurewicz) fibration if p has HLP with respect to all X.

Definition 8. p ∶ E Ð→ B is called a Serre fibration if p has HLP with respect to all disksDn, n ≥ 0.

Remark. (Dn × I,Dn × 0∪Sn−1 × I) is homeomorphic to (Dn × I,Dn × 0). Thereforep is a Serre fibration if and only if for every solid arrow diagram below there exists a mapcompleting the commutative diagram:

Dn × 0 ∪ Sn−1 × I //

E

Dn × I //

ϕ

77ooooooooooooooB

Remark. Covering space⇒Hurewicz fibration⇒Serre fibration. Note that covering spacesare Hurewicz fibrations where homotopies lift uniquely.

Idea of fibration: Consider the pullback (B path-connected)

p−1(b) //

E

p

b // B.

If p is fibration, all fibers (i.e. all p−1(b)’s) should be homotopy equivalent and “varycontinuously”. This comes historically from the notion of fiber bundles:

20

1. FIBER BUNDLES AND FIBRATIONS 21

Definition 9. A fiber bundle structure on E with fiber F is a map p ∶ E Ð→ B such thatfor all b ∈ B there is a neighborhood U of b and homeomorphism h ∶ p−1(U) Ð→ U × F overB. h is called a local trivialization (and there are not unique). B is the base space of thebundle and E is the total space of the bundle and F is the fiber. We will sometimes denotea fiber bundle with these data as F Ð→ E

pÐ→ B

Example 1.1. p1 ∶X × Y Ð→X gives a fiber bundle structure.

Example 1.2. E = I × [−1,1]/ ≅ with the relation (0, v) ≅ (1,−v). The p ∶ E Ð→ S1 is afiber bundle induced by I × [−1,1]Ð→ (I Ð→ S1) and fibers are [−1,1].

Example 1.3 (Projective spaces).

C∗ Ð→ Cn+1 − 0pÐ→ Cn+1 − 0/C∗

with the standard trivialization Ui = (xi ≠ 0). One can create analogous spaces using R orH (the field of quaternions). The unit vectors versions will respectively be

S1 Ð→ S2n+1 Ð→ CP(n)S3 Ð→ S4n+3 Ð→ HP(n)S0 Ð→ Sn Ð→ RP(n).

Theorem 1.1. Suppose p ∶ E Ð→ B is a Serre fibration. Choosing b0 ∈ B,x0 ∈ F ∶= p−1(b0)then

p∗ ∶ πn(E,F,x0)Ð→ πn(B, b0)

coming from the map of triples (E,F,x0)pÐ→ (B, b0, b0) is an isomorphism for all n ≥ 1 and

so there is a long exact sequence

⋯Ð→ πn(F,x0)Ð→ πn(E,x0)p∗Ð→ πn(B, b0)

∂Ð→ πn−1(F,x0)Ð→ ⋯

Note: ∂ has to be calculated via the isomorphism p∗

πn(B, b0) // πn−1(F,x0)

πn(E,F,x0)

≅ p∗

OO 77nnnnnnnnnnnn

.

Proof. We prove the surjectivity only: let α = [f] ∈ πn(B) for f ∶ (In, ∂In)Ð→ (B, b0).Let J = ∂In − sn = 0 as we had denoted before. We want f ∶ (In, ∂In, J) Ð→ (E,F,x0).Define

f ∣J ∶ J Ð→ E via β ↦ x0.


Note that pf ∣J = f ∣J . E begin a Serre fibration, the following diagram completes

Dn−1 × 1 ∪ Sn−2 × I = J

f ∣J// E

p

Dn = Inf

//

33ggggggggggggggB.

So pf(∂In) = b0 and f(∂In) ⊆ p−1(b0) = F and so [f] ∈ πn(E,F,x0) since f(J) = x0 andalso p∗[f] = [f].

Proposition 14. p ∶ E Ð→ B is a Serre fibration if and only if p has the HLP with respectto all weak CW pairs.

Proof. Sufficiency is obvious. For necessity given a Serre fibration p ∶ E Ð→ B and arelative weak CW complex X Y we want to complete

X × I ∪ Y × 0 //

E

Y × I //

88qq

qq

qq

B

to a commutative diagram. Look at the rectangle below with the left square being a pushout

∐α∈I(i) Sn(α) × I ∪∐Dn(α)+1 //

X(i) × I ∪X(i + 1) × 0 //

E

∐D × I //

22eeeeeeeeeeeeeeeeeeeeeeeX(i + 1) × I // B

So we get lifts ϕ(i) ∶X(i + 1) × I Ð→ E and then take the direct limit of these lifts.

Proposition 15. A fiber bundle p ∶ E Ð→ B is a Serre fibration.

Remark. If B is paracomplact then p is actually a fibration.

Proof. We want to complete the following

In //

E

In × IH

//

;;xx

xx

xB

to a commutative diagram. Break In×I into smaller boxes Iα,j ∶= cα×[tj , tj+1] and supposeα ∈ J where J is some nicely ordered indexing set. Let this partition be fine enough sothat H(Iα,j) ⊆ Uα,j ⊆ B so that p∣Uα,j is trivial and say hα,j ∶ p−1(Uα,j) Ð→ Uα,j × F is


trivialization. Inductively assume that the lift has been defined (compatibly) on each cβfor β < α:

cβ

⊆// In // E

cβ × I

55kkkkkkkkkkkkkkkkkk ⊆// In × I

H// B

To extend to cα consider the diagram

Dn × 0

ρ′

++≅ //

(∪β<αcβ ∩ cα) × [tj , tj+1] ∪ cα × tj

// p−1(Uα,j)≅h//

Uα,j × Fπ

xxrrrrrrrrrrr

Dn × Iρ 44

22ddddddddddddddddddddddd ≅ // cα × [tj , tj+1] // Uα,j

also assume ϕα ∶ cα×I Ð→ E has been defined for t ≤ tj so constructing ψ will define ϕ∣α(−, t)for t ≤ tj+1 putting them together gives ϕα gluing together for all α gives lift ϕ ∶ In×I Ð→ E.One can check that

ψ(d, t) = h−1(ρ(d, t), π2ρ′(d))

works. Note that

hψ((d,0)) = (ρ(d,0), π2ρ′(d)) ,= (π1ρ

′(d), π2ρ′(d))

where , equality is because the solid arrow diagram commutes.

Example 1.4. Compute π3(S2) ≅ Z using the fibration S1 Ð→ S3 ηÐ→ CP (1) = S2. So η

is a fiber bundle hence is a Serre fibration. So we look at the long exact sequence on π∗πe(S1)Ð→ π3(S3)Ð→ π3(S2)Ð→ π2(S1)Ð→ π2(S3)

Since the unviersal cover of S1 is R, the R Ð→ S1 induces an isomorphism on πi for i ≥ 2(since it is a convering map) so πi(S1) = 0 if i ≥ 2. Hence π3(S3) π3Ð→ π3(S2) is isomorphism.By Hurewicz π3(S3) ≅ Z so π3(S2) ≅ Z. In fact πi(S3) ≅ πi(S2) for i≥ 3.

Example 1.5. Similarly consider the fiber bundle S3 Ð→ S7 Ð→ S4. Gives exact sequence

πn(S3)Ð→ πn(S7)Ð→ πn(S4)Ð→ πn−1(S3)but the first map factors through zero since S3 Ð→ S7 ≃ ∗ by cellular approximation so weget short exact sequences

0Ð→ πn(S7)Ð→ πn(S4)Ð→ πn−1(S3)Ð→ 0.

In fact all of these short exact sequences are split. So πn(S4) ≅ πn(S67) ⊕ πn−1(S3). Infact ΩS4 ≅ ΩS7 × S3 which we will turn back to later!


Let X be a CW complex. Let f ∈ πnX. Then we can get Σf ∈ πn(ΣX) where ΣX =X × I/(x,1) ≃ (x,0) ≃ (∗, t) is the reduced suspension. Σ is a homotopy preserving functor(i.e. f ≃ g⇒ Σf ≃ Σg

Σ ∶ Top∗ Ð→ Top∗.

So α = [g] ∈ πn(S) induces Σα ∶= [Σf] ∈ πn+1(ΣX).Theorem 1.2 (Freudenthal suspension). Suppose the connectivity of X is at least n, conn X ≥n.Then Σ ∶ πn(X)Ð→ πn+1(ΣX) is an isomorphism for i ≥ 2n and surjective for i = 2n + 1.

The functor Ω ∶ CG∗ Ð→ CG∗ is the functor: ΩX = Map∗(S1,X) with compact opentopology. Then Σ is the left adjoint to Ω:

Θ ∶ HomCG∗(ΣX,Y ) ≅ HomCG∗

(X,ΩY )in fact this correspondence is a hemeomorphism. Let us denote Θf by fa. Then

idaΣX = Θ(idΣX) ∈ Hom(X,ΩΣX)where [f] ∈ πn(X) = [ΣSn−1,X] and [fa] ∈ πn−1(ΩX). Theta in fact induces an isomor-phism

πnX Ð→ πn−1(ΩX)via [f]↦ [idaΣX .f] ∈ πn(ΩΣX) ≅ πn+1(ΣX) which is just Σf .

Recall that f ∶X Ð→ Y can be turned into a cofibration

X

""EEEEEEEEE // M(f)

Y

where the injection X M(f) is a cofibraiton andM(f)↠ Y is a fibration. Given XfÐ→ Y .

LetEf = (x, p) ∶ f(x) = p(0) ⊆X × Y I

be the pullbackX × Y I //

Y I

ev(0)

X // Y

Now p(1) ∶ Ef Ð→ Y via (x, p)↦ p(1) is a fibration. So we get a commutative diagram

Efp(1)

AAAAAAA

Xf

//

∗>>

Y

where ∗ is a strong deformation retract. Then p(1)−1(y) is a homotopy fiber of f . If Y ispath connected the homotopy type of p(1)−1(y) is independent of y.

CHAPTER 3

Spectral Sequences

Consider a sequence of bigraded modules (vector spaces) Erpq r ≥ n ≥ 0 for some startingpoint n (usually 0, 1 or 2), together with differentials dr with degree (−r, r − 1),

dr ∶ Erpq Ð→ Erp−r,q+r−1

satisfying dr dr = 0 and isomorphisms Er+1pq ≅ Hpq(Er, dr) = kerdrpq/ imdrp+r,q−r+1. The

sequence E+ = Erpq, drpqr is called a spectral sequence. These form a category wherearrows are sequences of bigraded maps Erpq Ð→ Er

′

pr compatible with the differentials.

E+ is a first quadrant spectral sequence if Erpq ≠ 0⇒ p ≥ 0 and q ≥ 0.

d0

d1oo

d2

ggOOOOOOOOOOOOOO

//

OO

p

q

Total degree of α ∈ Erpq is p + q. Thus the total degree of dr is always −1. Suppose E+

is first quadrant. Then for all p, q, Erpq stabilizes, i.e. given p, q there is k such that,r ≥ k⇒ Erpq ≅ Ekpq. We call these terms E∞

pq .

We will consider the first quadrant case from now on. If there exists a finite filtration0 = F−1Hn ⊆ F0Hn ⊆ ⋯ ⊆ FtHn =Hn then

E∞pq = FpHp+q/Fp−1Hp+q

is a spectral sequence for which Erpq ⇒Hpq.

Example 0.6 (Leray-Serre spectral sequence). Let F iÐ→ EπÐ→ B be a Serre fibration.

We want B to be simply connected. Then there exists a first quadrant spectral sequence[Serre’s PhD] with

E2pq =Hp(B,Hq(F ))⇒Hp+q(E)

25

3. SPECTRAL SEQUENCES 26

i.e. ⊕p+q=nE∞pq = Hn(E) up to extensions (with is equality if we work over a field). Also

note thatHp(B,Hq(F )) ≅HpB ⊗Hq(F )⊕TorZ

1 (Hp−1B,HqF )where the Tor term is zero if we are over a field.

Consider turning ∗Ð→X into a fibration, i.e.

PX = ∗ ×X XI = ϕ ∶ I Ð→X ∶ ϕ(0) = ∗where π = ev1 ∶ PX Ð→ X is the evaluation map at 1, ϕ ∶↦ ϕ(1) which is a fibration. Andπ−1(∗) = ΩX. So we get the path-loop fibration

ΩX Ð→ PX Ð→X.

Suppose X is path connected in which case PX is contractible, PX ≃ ∗.

Now we apply the Leray-Serre spectral sequence when X = Sn:E2pq ⇒Hp+q(E) = 0 unless n = q = 0.

Here E2pq = HpB ⊗HqF since HpB is free and H0F ≅ Z from the long exact sequence of

homotopy. So the second page looks like this:

2 ? 0 0 ?

1 ? 0 0 ?

0 Z 0 0 Z

0 1 n − 1 n

//

OO

p

q

We know that E∞n0 = 0 so there is r such that drn0(1) ≠ 0 for some r and is fact is an

isomorphism. SoE2

0,n−1 ≅ ZBut that means that Z ⊗Hn−1F ≅ H0B ⊗Hn−1F ≅ E2

0,n−1 ≅ Z so Hn−1F ≅ Z. Also derivethat E2

∗,i = 0 if 0 < i < n − 1.

By same argument H2n−2F ≅ Z and in general

HiF ≅ Z n − 1 ∣ i, i ≥ 00 else .

Similarly if ΩX ≃ S1 get path loop fibration S1 Ð→ PX Ð→X. A spectral sequence argumentshows that

Hi(X) ≅ Z 2 ∣ i, i ≥ 00 else .

1. CONSTRUCTION OF SPECTRAL SEQUENCES 27

1. Construction of spectral sequences

There are generally two ways of getting spectral sequences: filtrations and exzact cou-ples.

1.1. From exact couples. We will talk about exact couples here: An exact couple isa (non-commuting) diagram of modules

E ∶ D i // D

j~~

Ek

``AAAAA

which consists of two modules and three maps and is exact at each vertex, e.g. ker j = im i.Let d = jk and we get d2 = 0. Then we get the derived couple

E1 ∶ iD i′=i // iD

j′=ji−1yytttttt

H(E,d)k=k′

eeJJJJJJ

that is the maps are determined by, i′ = i∣D, j′(i(d)) = [j(d)], k′[e] = ke. Check that theseare well-defined and E1 is an exact couple. One can iterate this procedure and get the r-thderive couple

Er ∶ Dr = ir(D) i // Dr

j(r)=ji−r

xxrrrrrrrrrrr

Er =H(Er−1)

khhPPPPPPPPPPPP

i.e j(r)ir(d) = [j(d)]. Suppose now that D,E are bigraded Dpq,Epq, such that i has bi-degree (1,−1); i ∶ Dpq Ð→ Dp+1,q−1, j has bidegree (0,0) and k has bidegree (−1,0). Thenj(r) has bidegree (−r, r), dr has bidegree (−r−1, r). We have a spectral sequence (Er−1

pq , dr)

as above.

Example 1.1 (Bockstein Spectral Sequence (BSS), due to Browder). Consider the short

exact sequence 0 Ð→ Z ×`Ð→ Z πÐ→ Z/` Ð→ 0 where ` is a prime. Tensor this with C∗X (or anyfree chain complex) and get short exact seqeunce of complexes

0Ð→ C∗(X,Z)Ð→ C∗(X,Z)Ð→ C∗(X,Z/`)Ð→ 0.


The long exact sequence on H∗ gives an exact couple

H∗(X,Z) ×` // H∗(X,Z)

π∗wwppppppppppp

H∗(X,Z/`)

jggOOOOOOOOOOO

.

The associated spectral sequence is the BSS.

Exercise 2. Calculate BSS replacing C∗X with the chain complex

⋯Ð→ 0Ð→M1 = Z `kÐ→M0 = ZÐ→ 0Ð→ ⋯

given any integer k ≥ 1. We do the first cases here for illustration: when k = 1 the exactcouple is

H∗Ci // H∗C

jxxqqqqqqqqqq

H∗(C ⊗Z/`)k

ffMMMMMMMMMM

where k is the connecting homomorphism. In degree 0 we get

Z/` 0 // Z/`≅j

Z/`k

aaCCCCCC

and in degree 10 // 0

~~

Z/`

``AAAAAA

.

Thus the BSS is given by

E10 = Z/`

E11 = Z/`, d = kj ∶ E1

1 ≅ Z/` ≅Ð→ E10 ≅ Z/`

E2∗ = 0.

In k = 2 case,degree 1 degree 0

0 // 0

Z/`2 ×` // Z/`2

πzzzzzzzz

Z/`

__???????? δ

77oooooooZ/`

δ

aaDDDDDDDD


The differential is

d1 = jk = πδ ∶ E11 ≅ Z/`Ð→ Z/` ≅ E1

0

[1]↦ πδ[1] = π[`] = 0.

Now passing to the derived couple is given by

0 // 0

Z/` ×` // Z/`

ji−1

Z/`

__???????? [δ]

77oooooooZ/`

aaCCCCCCCC

.

with differential of the spectral sequence being

d2 = j′k′ = (π1`)[δ] ∶ E2

1 ≅ Z/`Ð→ Z/` ≅ E20

[1]↦ π1`[`] = π[1] = [1]

which is an isomorphism and therefore E3∗ = E∞

∗ = 0.

1.2. From filtrations on chain complexes. Let C be a chain complex. A filtrationon C is a sequence of sub-chain-complexes Fi = Fi(C)

0 = F−1 ⊆ F0 ⊆ ⋯ ⊆ Fp ⊆ ⋯ ⊆ C.

The filtration is exhaustive (or co-omplete ) if C = ∪FpC (complete corresponds to the limitbeing 0). Also assume C is non-negatively graded so that we get a first quadrant spectralsequence. The zero page is

E0pq = (FpC)p+q/(Fp−1C)p+q.

and d0 is induced by the differentials on quotient.

2

1

0

F0 F1/F0 F2/F1 F3/F2

//

OO

p

q

So in first page E1p,q =Hp+q(E0

p,∗, d0). Now let

Arp = c ∈ FpC ∶ d(c) ∈ Fp−rC


i.e. c ∈ Fp is a cycle mod Fp−rC; you may call is an approximate cycle. As a piece ofnotation also define

ηp ∶ FpC Ð→ FpC/Fp−1C = E0p

(and we are dropping q’s from our notations). Then

Zrp = ηp(Arp) ⊆ E0p

Br+1p−r = ηp−rd(Arp) ⊆ E0

p−rZ∞p = ∩∞r=1Z

rp

B∞p = ∪∞r=1B

rp

Observe that we have filtration

0 ⊆ B0p ⊆ B1

p ⊆ ⋯ ⊆ B∞p

⊆ Z∞p ⊆ ⋯ ⊆ Z0

p = E0p

The r-th page would be Zrp/Brp:

Arp ∩ Fp−1C = Ar−1p−1

Zrp ≅ Arp/Ar−1p−1

Erp = Zrp/Brp ≅

Arp + Fp−1C

dAr−1p+r−1 + Fp−1(C)

≅Arp

d(Ar−1p+1) +Ar−1

p−1

and the most important part defining the differential:

drp ∶ Erp

//___ Erp−r

Zrp

OOOO

d //___ Zrp−r

OOOO

Arp

OOOO

d// Arp−r

OOOO

defined since d2 = 0

d lands in here since d2 = 0

drp is unique since the vertical maps are surjections. To see that drp exists (making thediagram commute) it suffices that d′(Br

p) = 0. But since d im(dAr−1p+r−1) = 0 this is the case.

Clearly (dr)2 = 0. Finally check that Er+1 ≅ H∗(Er) to complete the construction of ourspectral sequence.

Example 1.2. Let our chain complex be concentrated only in degrees 3 and 4, where

C4 ≅ Ze31 ⊕Ze22, C3 = Ze21 ⊕Ze12

and we have chosen the indices this way since if the differential d ∶ C4 Ð→ C3 is given via(a, b)↦ (a + kb, b) then

F1C = Ze21, F2C = Ze12 ⊕Ze22 ⊕Ze21, F3C = C


E0 ∶ 0 Z Z 0

0 0 Z Z

0 0 0 0//

OO

p

q

E1 ∶ 0 Z 0 0

0 0 Z/k Z

0 0 0 0//

OO

p

q

E2 ∶ 0 Z 0 0

0 0 Z/k Z

0 0 0 0//

OO

p

q

E3 ∶ 0 Z 0 0

0 0 0 kZ

ggOOOOOOOOOOOO

0 0 0 0//

OO

p

q

Exercise 3. Compute the differentials in the previous example.

In a filtration of positively graded chain complex C, that is co-complete we have the con-vergence Er ⇒ FpHp+q/Fp+1Hp+q and one can check that this is the same as the spectralsequence associated to the exact couple

H∗Fishift // ⊕iH∗Fi+1

vvmmmmmmmmm

H∗(Fi+1/Fi)

ggNNNNNNNN

.

Let π ∶ E Ð→ B be a fibration and assume that B has a CW structure. Let f ∶ B′ Ð→ B bea weak homotopy equivalence. The by long exact sequence on π∗, f ′ ∶ π−1B′ Ð→ B′ is also aweak homotopy equivalence. Let Bn be the n-skeleton of B. Consider the filtration

Fn = imC∗(π−1Bn)

where π−1Bn Ð→ E is the pullback. Then F−1 = 0 and filtration is co-complete. Since

Dr //

$$HH

HH

H E

π−1(Bn)?

OO


factors for some n through π−1Bn since Dn is compact. So the associated spectral sequenceis a Serre spectral sequence. As always

E0p = Fp/Fp−1

E1p =H∗(Fp/Fp−1).

Note 1.3 (Recall Cellular homology). Let Ccelln = ⊕α∈I(n)Zα be the freely gener-

ated Z-module generated by n-cells in a (not weak) CW structure. To define a mapCcelln X Ð→ Ccell

n−1. We have a mapping that is the composition

Sn−1α

f(α)ÐÐ→Xn−1

qÐ→Xn−1/Xn−2 ≅ ⋁

β∈I(n−1)Sn−1β

pβÐ→ Sn01β .

Let pβ ⊕ZÐ→ Z be the projection. And consider deg(pβqf(α)1β). (Notice Sn−1 is compactso this will be a finite sum.) This induces a differential dn ∶ Ccell

n Ð→ Ccelln−1.

The following squares are pullbacks:

π−1(Bn−1) //

π−1Bn//

E

π

Bn−1 // Bn // B

and we have short exact sequence 0Ð→ Fp−1 Ð→ Fp Ð→ Fp/Fp−1 Ð→ 0 in the form

0Ð→ C∗(π−1Bn−1)Ð→ C∗(π−1Bn)Ð→ C∗(π−1Bm, π−1Bn−1)Ð→ 0.

So H∗(Fp/Fp−1) ≅H∗(π−1Bn, π−1Bn−1). But by excision

H∗(Bn,Bn−1) ≅Hn( ∐α∈I(n)

Dn, ∐α∈I(n)

Sn−1)

H∗(π−1Bn, π−1Bn−1) ≅Hn( ∐

α∈I(n)π−1Dn, ∐

α∈I(n)π−1Sn−1).

Consideri∗Dn ∶ π−1Dn //

E

π

Dn i // B

i ≃ ∗ =∶ 0 so by proposition 4.62 of Hatcher, i∗Dn is fiber homotopy equivalent to 0∗Dn, i.e.there are maps f, g making the following diagram commute.

i∗Dn

π ##GGGGGGG f 550∗Dn

guu

πwwwwwww

Dn


where by homotopies over Dn, gfH≃ idi∗Dn (for some homotopy H) and fg ≃ id0∗Dn . Thus

this induces an equivalence of pairs

(i∗Dn, i∗Sn−1) ≅ (0∗Dn,0∗Sn−1) ≅ (Dn × F,Sn−1 × F )

where F = π−1(∗).

Now

H∗(Fn, Fn−1) ≅H∗(∐π−1Dn,∏π−1Sn−1)≅ ⊕α∈I(n)H∗(π−1Dn, π−1Sn−1)≅ ⊕H∗(Dn × F,Sn−1 × F )

≅ Ccelln (B)⊗H∗F.

The last equality above holds because,

Zα ⊗H∗(F )in degree n

⋯ // H∗(Sn−1 × F )Kunneth

// H∗(Dn × F ) 0 // H∗(Dn × F,Sn−1 × F ) // ⋯

0 // H∗(Sn−1)⊗H∗(F ) // H∗(Sn−1)⊗H∗(F ) // H∗(F ) // 0

Note 1.4. If the base is not simply connected the above argument does not work.(Then say at n = 0, in Sn−1 ×F we can have disjoint copies of F and the kernel is not

the above one??)

So E1pq = Ccell

p (B)⊗Hq(F ) and d1 = dcell ⊗ id so

E2 ≅Hcell∗ B ⊗H∗F ⊕Tor(H∗B,H∗F )

ignoring the degrees.

Exercise 4. Work out the “fibration” S1 ∨ S2 Ð→ S1: You may turn this into a fibrationF Ð→ π−1(∗)Ð→ PS1, π ∶ PS1 Ð→ S1. Here F is homotopy equivalent to the covering space ofS1 ∨ S2.

CHAPTER 4

H-spaces and algebras

1. H-spaces

Definition 10. An H-space (H,ϕ) is a topological spaceH with base point e and a productmap

ϕ ∶H ×H Ð→H

(x, y)↦ x.y

making

H ∨H

fold

##GGGGGGGGG

H ×H ϕ// H

commute relative to e and only up to homotopy.

Definition 11. H is homotopy associative if

H ×H ×HidH×ϕ

ϕ×idH// H ×H

ϕ

H ×H ϕ// H

commutes relative to (e, e, e) and up to homotopy.

Definition 12. H has homotopy inverse . ∶H Ð→H if e = e and

H ×H id×. // H ×Hϕ

$$HHHHHHH

H

∆::vvvvvvv

e

44 H

commutes relative to e and up to homotopy.

34

1. H-SPACES 35

Example 1.1. Topological groups are homotopy associative H spaces with homotopyinverse (take all homotopies to be trivial). So are loop spaces ΩX: ΩX = Map∗(S1,X)with compact open topology. The box product is

ϕ ∶ ΩX ×ΩX Ð→ ΩX

(α,β)↦ α

traversing the first and then the second path with twice speed and

. ∶ ΩX Ð→ ΩX

α ↦ α ∶ t↦ α(t − 1).

Definition 13. A homotopy associative H-space with homotopy inverse is called an H-group.

One can dualize all the above notions:

Definition 14. A co-H-space is a pair (C,ψ) with C a topological space with base pointe, and coproduct ψ ∶ C Ð→ C ∨C making

C ∨C

C

ψ;;xxxxxxxxx

∆// C ×C

commute up to homotopy.

Homotopy co-associativity and homotopy co-inverses are defined likewise. A co-H-group isa homotopy co-associative co-H-space with homotopy co-inverse.

Example 1.2. ΣX (the reduced suspension) is a co-H-group, where the coproduct isthe pinching morphism along the equator:

ψ ∶

////////

////////

Ð→

////

////

////

////

Proposition 16. (1) If Y is an H-group and X is any space, then [X,Y ]∗ is a group.

(2) If X is a co-H-group and Y is any space, then [X,Y ]∗ is a group.

(3) If Y is an H-group and X is a co-H-group then [X,Y ]∗ is abelian (and both mul-tiplications in the above cases give the same result).

2. (CO)-ALGEBRAS 36

Proof. In (1) the group operation is the composition

f.g ∶X ∆Ð→X ×Xf×gÐÐ→ Y × Y

ϕÐ→ Y

and in (2)

f.g ∶XψÐ→X ∨X

f∨gÐÐ→ Y ∨ Y foldÐÐ→ Y.

For (3) look at [Bredon, p. 442].

Remark. The question of studying pairs of spaces X and Y for which ΩX ≃ ΩY (de-looping!) is related to the above ideas. In fact, recall the fibration F = ΩX Ð→ PX

evÐ→ Xand observe that PX ≃ ∗. The long exact sequence on π∗ looks in part like

⋯Ð→ πn(ΩX)Ð→ πn(PX) = 0Ð→ πn(X) ≅Ð→ πn(ΩX)Ð→ ⋯thus πnX ≅ πn−1(ΩX). So it is hopeless to try to recover X and Y from the loop spaces.One need more structure on the spaces to tackle the problem.

2. (co)-Algebras

Fix R (or k), your favorite underlying commutative ring or field. An algebra is a triple(A,ϕ,µ) with multiplication ϕ ∶ A⊗AÐ→ A and unit µ ∶ R Ð→ A.

R⊗A∼

JJJJJJJJJJ

JJJJJJJJJJµ⊗id

// A⊗Aϕ

A⊗Rid⊗µoo

∼tttttttttt

tttttttttt

A

We may want to have a grading on A and want ϕ to respect the grading, ϕ(Ak⊗As) ⊆ Ak+s,im(µ) ⊆ A0. ϕ is said to be associative:

A⊗A⊗Aid⊗ϕ

ϕ⊗id// A⊗A

ϕ

A⊗A ϕ// A

(without the position of parentheses matter) and is (graded-)commutative if

A⊗A

""FFFFFFFFFT // A⊗A

ϕ||xxxxxxxxx

A

commutes such that T (α⊗ β) = (−1)∣α∣∣β∣β ⊗ α, where ∣.∣ denotes the degree.

A coalgebra is the dual object: a triple (C,∆, η) where C is an R-module, ∆ ∶ C Ð→ C ⊗Cand η ∶ C Ð→ R and everything is defined dually likewise! A (co)algebra is connected ifA0 = R and it is non-negatively graded.

3. HOPF ALGEBRAS 37

Remark. When the coalgebra C is connected then for α ∈ Cn

∆(α) = ∑i+j=n

α′i ⊗ α′′j ∈ (C ⊗C)n = ⊕i+j=nCi ⊗Cj

idC ⊗ η(∆(α)) = ∑i+j=n

α′i ⊗ η(α′′j )´¹¹¹¹¸¹¹¹¹¶=0if j≠0

= α′n ⊗ α′′0 = α⊗ 1

So we can change α′n, α′′0 so that α′n = α and α′′0 = 1. Similarly for the other factor. So wecan write

∆(α) = α⊗ 1 + 1⊗ α + ∑0<i<n

α′i ⊗ α′′n−i

and therefore we have a reduced diagonal ∆ given via

∆(α) = ∆(α) − (α⊗ 1 + 1⊗ α).

If ∆(α) = 0 then α is called primitive, and the R-module of such elements is denoted byP (C).

Example 2.1. H∗(X,R) is an associative commutative algebra and H∗(X,R) is aco-associative co-commutative coalgebra and are connected if X is path-connected.

3. Hopf algebras

Definition 15 (Hopf algebra). A Hopf algebra is a quintuple (A,ϕ,∆, µ, η) where (A,ϕ,µ)is an algebra and (A,∆, η) is a coalgebra and the structures are compatible: ϕ is a coalgebramap (or equivalently ∆ is an algebra map), i.e.

A⊗Aϕ

∆⊗∆//

∆(A⊗A)

''OOOOOOOOOOOO A⊗A⊗A⊗Aid⊗T⊗id

ϕA⊗A

tt

A

''OOOOOOOOOOOOO A⊗A⊗A⊗Aϕ⊗ϕ

A⊗A

commutes (but be aware that T introduces signs).

Remark. In a Hopft algebra A, P (A) is a Lie algebra with product

[α,β] = αβ − (−1)∣α∣∣β∣βα

3. HOPF ALGEBRAS 38

where the products on the right hand side are multiplication in A. For if α,β ∈ P (A)

∆[α,β] = ∆(α)∆(β) − (−1)∆βδα since ∆ is linear and an algebra map

= (α⊗ 1 + 1⊗ α)(β ⊗ 1 + 1⊗ β)− (−1)∣α∣∣β∣(β ⊗ 1 + 1⊗ β)(α⊗ 1 + 1⊗ α)

= (αβ − (−1)∣α∣∣β∣βα)⊗ 1 + 1⊗ (αβ − (−1)∣α∣∣β∣βα)

= [α,β]⊗ 1 + 1⊗ [α,β].

Example 3.1. To be safe (?) let k be an underlying field and (X,ϕ) be a path-connected H-space. So we know that H0(X,k) ≅ k. Then H∗(X,k) is a connected,commutative, Hopf algebra. If X = ΩY or is a topological group then H∗(X,k) will beco-associative: ∆ ∶X Ð→X ×X induces the isomorphism fitting into the following diagram:

H∗(X)⊗H∗(X) ≅ //

ϕ((RRRRRRRRRRRRRR

H8(X ×X)

H∗(∆)

H∗X

Since X is an H-space we also have X ×XϕÐ→X which induces

H∗(X)

∆

66

H∗(ϕ)// H∗(X ×X) // H∗(X)⊗H∗(X)

≅ss

completing the construction of our Hopf algebra structure. Similarly H∗(X,k) is a Hopfalgebra. If X = ΩY then this is associative, but typically not commutative. However ifX = Ω2Y then H∗(X,k) will be commutative. Here Ω2Y ∶= ΩΩY fitting in the adjunction

[Σ2X,Y ] ≅ [ΣX,ΩY ] ≅ [X,Ω2Y ].

Note 3.2. Note that Hom(A × B,Y ) ≅ Hom(B,Hom(A,Y )) true as sets or asmapping spaces if spaces satify basic topological properties! Similarly

Map∗(S1 ∧X,Y ) ≅ Map∗(X,Map∗(S1, Y ))

where X ∧ Y =X × Y /X ∨ Y is the smash product. Since ΣX ≅ S1 ∧X we have

Map∗(ΣX,Y ) ≅ Map∗(X,ΩY )

which is compatible with homotopies and therefore

[ΣX,Y ] ≅ [X,ΩY ]

i.e. Σ and Ω are adjoint functors (actually π0 Map∗(ΣX,Y ) = [ΣX,Y ]).

3. HOPF ALGEBRAS 39

Let R[α] be the polynomial algebra over R on one generator α with positive even degree∣α∣. Let α be primitive so that ∆α = 1⊗ α + α⊗ 1 and in particular

∆(αn) = (∆α)n = (α⊗ 1 + 1⊗ α)n

=n

∑i=0

(ni)αi ⊗ αn−i

Note that the coproduct on generators always determine the coproduct. And here sincethis algebra is free the above formula gives a well-defined coproduct.

Let E(α) = ∧R[α] be the exterior algebra on generator α with the degree of α being odd.And assume α is primitive. Then E(α) is a free graded commutative algebra on α. Thiswill imply that the formula for ∆ is well-defined.

Example 3.3. Say char(k) ≠ 2 and ∣α∣ = 2 contrary to the above assumption. We wantto have ∆(α) = 1⊗ α + α⊗ 1. Then

0 = ∆(α2) = (∆α)(∆α)= (1⊗ α + α⊗ 1)(1⊗ α + α⊗ 1)

= 1⊗ α2

=0+ α⊗ α + (−1)∣α∣∣α∣α⊗ α + α2 ⊗ 1

=0

= 2α⊗ α ≠ 0.

Therefore we do not get a Hopf algebra unless in characteristic 2. More generally k[α]/(αpk)with α primitive, determines a Hopf algebra if and only if char(k) = p.

If A and B are Hopf algebras then so is A⊗B in the obvious way.

Theorem 3.1. If char(k) = 0, A is a (graded-)commutative associative connected Hopfalgebra. Then as algebras (and not Hopf algebras) we have an isomorphism

A ≅ E(a1,⋯, an,⋯)⊗ k[b1,⋯, b`,⋯]

with ∣ai∣’s odd and ∣bj ∣’s even.

Notation: We denote the above by ⋀(a1,⋯, an,⋯, b1,⋯, b`,⋯) the free graded commutativealgebra on the ai and bj .

Example 3.4. We have seen using Serre’s spectral sequence thatHi(ΩSn,Z) ≅ Z n − 1∣i0 otherwise .

Thus

H i(ΩSn,Q) ≅ Q r − 1∣i0 otherwise.

The latter is a Hopf algebra so if n is odd we have H i(ΩSn,Q) ≅ Q[a] with ∣a∣ = n − 1.

3. HOPF ALGEBRAS 40

Working overR = Z let ∣a∣ be even and ∣cn∣ = n∣a∣ then the divided powers algebra

Γ(a) = ⊕n∈Z≥0cnZ

is defined via c0 = 1, c1 = a and cicj = (nj)cn. Then ∆(cn) = ∑i+j=n ci ⊗ cj . And note that

Γ(a) ≅ k[a] if chark = 0 (in fact a Hopf algebra isomorphism if the isomorphism is definedin a convenient way). Finally

Γ(a)∗ = HomZ(Γ(a),Z) ≅ R[a]

as Hopf algebras.

We work out the proof only when An is finite dimensional for all n, i.e. A is of finite type.A CW-complex X is called finite type if H∗X is finite type so this is a concrete case wherethe following prove goes through.

Proof. Say A has a generating set a1,⋯, an,⋯ with ∣ai∣ ≤ ∣ai+1∣. Set A(0) = k, andA(n) the subalgebra (not sub-Hopf algebra) generated by a1,⋯, an. It turns out that A(n)is also a sub-Hopf algebra:

∆(ai) = ai ⊗ 1 + 1⊗ ai + something in ⊕p+q=∣ai∣,p>0

Ap ⊗Aq

so the q’s above have to be ∣q∣ < ∣ai∣ therefore the extra terms are in

⊕Ap ⊗Aq ⊆ A(n − 1)⊗A(n − 1).

We conclude that ∆A(n) ⊆ A(n) ⊗ A(n). Assume A(n − 1) ≅ γ(a1,⋯, an−1) if ∣an∣ even.There is a surjection A(n− 1)⊗k[an]

fÐ→ A(n) (uses commutativity). It is injective as well:

say f(α) = ∑0≤i≤t αiain, αi ∈ A(n − 1).

∆fα = ∑0≤i≤t

∆(αiain)

= ∑0≤i≤t

(∆αi)(∆an)i since Delta is an algebra map.

But observe that

(∆αi)(∆an)i = (αi⊗1+A(n−1)⊗A(n−1)+1⊗αi)(an⊗1+A(n−1)⊗A(n−1)+1⊗an)i.

so ∆f(α) = ∑0≤i≤t(αi ⊗ 1)(an ⊗ 1 + 1⊗ an)i.

Let I ⊆ A(n) be the ideal generated by A(n − 1) and a2n. Consider the composition

A(n) ∆Ð→ A(n)⊗A(n)Ð→ A(n)⊗A(n)/I

The image of fα under composition is therefore

∑0≤i≤t

(αi ⊗ 1)(iai−1n ⊗ an) + ∑

0≤i≤t(α⊗ 1)(ain ⊗ 1)

3. HOPF ALGEBRAS 41

so if f(α) = 0 then ∆f(α) = 0 thus we should have

A(n)∈ ∑1≤i≤t

iαiai−1n = 0.

If we take α ∈ ker f of lowest non-zero degree the we have found an element of lower degreein the kernel which is a contradiction. So α has to be of degree zero. But f is injective ondegree zero and we are done with proving the injectivity.

We conclude that f is an isomorphism. The case of ∣an∣ being odd is similar, so takingunions (direct limits) we get that A ≅ γ(a1, c . . . , an).

Corollary 2. H∗i (ΩX; Q) ≅ ⋀(a1,⋯, an,⋯).

Remark. The quesiton of realizability of an algebra as the cohomology of a space or morespecifically the loop space of some topological space in rational coefficient will be done laterin this course.

∃X,H∗(ΩX,Q) ≅⋀(a1,⋯, an,⋯).The question is difficult integrally though. It is known however that any polynomial algebraZ[x1,⋯, xk] can be realized with a space H∗(X,Z) ≅ Z[x1,⋯, xk].

CHAPTER 5

Localization (rationalization)

There are two approaches: Serre classes (C classes) of 50’s and localization of 60’s. We willstudy the latter in what follows. All spaces will be simply connected, CW complexes.

Definition 16. For a topological space X, suppose there is a space X(0) and a mapη ∶ X Ð→ X(0) such that η induces an isomorphism on H∗(−,Q), H∗(X(0),Z) is a vectorspace over Q and for all f ∶X Ð→ Y such that H∗(Y,Z) is a Q-vector space, there is a unique ?map ϕ ∶X(0) Ð→ Y (up to homotopy), making

Xη//

f!!CCCCCCCCCX(0)

ϕ

Y

commute up to homotopy. Then X Ð→X(0) is called the rationalization of X.

Aside: There’s a construction to show that H∗(Sn,Z) is a Q-vector space. This is aconstruction by Sullivan to make a rational sphere by a telescope construction from mapsSn

×pÐ→ Sn for all primes appearing infinitely many times.

Proposition 17. There is a functorial localization ηX ∶X Ð→X(0) on the category of simplyconnected CW complexes.

Denote ηX(f) by f(0), then f(0) is homotopy equivalence if and only if H∗(f) ⊗ Q is anisomorphism if and only if (using simply connectedness) π∗(f)⊗Q is an isomorphism (forcertain tensor product). Localizations preserves fibrations (everything is simply connectedagain).

Recall K(Π, n) for abelian group Π:

πiK(Π, n) = Π i = n0 otherwise.

By this we mean a more general construction than what we presented above which coversadditive abelian groups of infinite generation; In gist, take a presentation of any abeliangroup G, 0 Ð→ ⊕Z Ð→ ⊕Z Ð→ G Ð→ 0. Consider a map ∨Sn Ð→ ∨Sn such that passing to

42

5. LOCALIZATION (RATIONALIZATION) 43

H∗ gives the injective homomorphism of the sequence. Construct a space K(G,n) by thegluing map.

If we assume K(Π, n) is a CW complex then it is determined up to homotopy. We want tocalculate the localizations of these spaces. Note that

(0.5) ΩK(Π, n) ∼K(Π, n − 1).

In fact this is a weak homotopy equivalence from what we know. But a result of Milnorimplies that loop spaces are CW-complexes and therefore by Whitehead’s theorem, thecomputation of π∗ as follows, results the existence of the homotopy equivalence 0.5 ofCW-complexes. Using the long exact sequence on π∗ for the fibration

ΩK(Π, n)Ð→ PK(Π, n)Ð→K(Π, n)

so π∗−1ΩK(Π, n) ≅ π∗K(Π, n).1

Example 0.6 (Localization of CP∞). The fibrations S1 Ð→ S2n+1 Ð→ CPn form com-mutative diagrams with compatible actions:

S1 // S2n+1 //

CPn

S1 // S2n+3 // CPn+1.

Taking the direct limit we get a fibration

S1 Ð→ S∞ Ð→ CP∞.

We know that π∗(limÐ→(Xi)) ≅ limÐ→π∗(Xi) since Sn is compact, and since π∗(Sn)Ð→ π∗(Sn+1)is the zero map we get that π∗(S∞) = 0 and therefore S∞ ∼ ∗ since S∞ has the a naturalCW-structure (or directly see that S∞ ∼ ∗). S1 ∼ K(Z,1) as we know.2 We conclude thatΩCP∞ ∼ S1 and

πi(CP∞) = Z i = 20 otherwise

thus CP∞ =K(Z,2).

Note also that π∗(X(0)) ≅ π∗X ⊗Q so CP∞(0) =K(Q,2). So ?

CP∞(0) ∼K(Q,2) ∼ ΩK(Q,3)

and in particular CP∞(0) is a loop space.

Before doing another example we state a general fact that will be lectured on later:

1It is more generally the case that π∗−1(ΩX) ≅ π∗(X) if X is simply connected.2And in fact S1

(0) ∼K(Q,1).


Theorem 0.2. K(Π, n) represents H∗(−,Π): One can think of [−,K(Π, n)] as a functorfrom the category of pointed homotopy classes of maps to groups. If Π is cyclic

[−,K(Π, n)] ≅H∗(−,Π) via f ↦ f∗(ιn),where ιn ∈HnK(Π, n) ≅ πnK(Π, n) ≅ Π is the generator.

Example 0.7 (Localization of ΩS3). We start with the observation that

H∗(CP∞,Z) ≅ Z[a], ∣a∣ = 2.

In fact H∗(CPn,Z) ≅ Z[a]/(an+1) and H∗(CPn,Z) Ð→ H∗(CPn−1,Z) maps a to a underthe isomorphism above.

We want to find H∗(CP∞,Q) ≅ ⋀(a1,⋯, an,⋯). Observe that

H0(CP∞,Q) = Q, H1(CP∞,Q) = 0, H2(CP∞,Q) ≅ Q.a, ⋯.Therefore there is an injection3

Q[a]Ð→H∗CP∞.

Checking dimensions implies that the map is also surjective:

dim(Q[a])k = 1 2∣k, k ≥ 00 otherwise ,dim(H∗CP∞)k = 1 2∣k, k ≥ 0

0 otherwise .

So any injective (graded) map Q[a]Ð→H∗CP∞ is sujrective. On the other hand

H∗(ΩS3; Q) ≅ Q[a′], ∣a′∣ = 2.

(Compute H∗(ΩS3,Q) as a ring with the Serre spectral sequence; Then take the map ofalgebras Q[a′]Ð→H∗(ΩS3,Q), ∣a′∣ = 2 via a′ ↦ ι2. This is injective, and by dimension countof the Q-vector spaces in all degrees it is surjective.)

By 0.2 there is a map f ∶ ΩS3 Ð→ CP∞, with f(a) = a′. Since f∗ is an algebra map itmust be an isomorphism, after tensoring with Q. Thus ΩS3

(0) ∼ K(Q,2). Really H∗(f,Q)is an isomorphism, hence H∗(f(0),Z) is isomorphism: This follows from the commutativediagram

H∗(X,Q) ≅ //

≅

H∗(X(0),Z) ≅ Q

H∗(Y,Q) ≅// H∗(Y(0),Z) ≅ Q

f(0) is a homotopy equivalence ΩS3(0)

f(0)ÐÐ→K(Q,2).

We will end this section with a generalization of the above example, that is the follow-ing

3When we omit the coefficient in cohomologies we are assuming rational coefficients unless statedotherwise.


Theorem 0.3. ΩS2n+1(0) ∼K(Q,2n) and S2n+1

(0) ∼K(Q,2n + 1) for all n ≥ 1.

In fact, the computation in previous example motivates

H∗(K(Q,2n)) ≅ Q[a], ∣a∣ = 2n,H∗(K(Q,2n + 1)) = E(a), ∣a∣ = 2n + 1,

where E(a) is the exterior algebra in one generator. We will be proving these in whatfollows.

Lemma 5. There exists g ∶ S2n+1(0) Ð→ K(Q,2n + 1) that is an isomorphism on π2n+1 and

hence on H2n+1.

Proof. π2n+1(K(Z,2n + 1)) ≅ Z with generator ι2n+1. Let g′ ∶ S2n+1 Ð→ K(Z,2n + 1)be representing ι. Thus π2n+1(g′) is an isomorphism. By π∗(f(0)) = π∗f ⊗Q, g = g′(0) is anisomorphism on π2n+1 ≅H2n+1.

Proposition 18. Assume H∗(K(Q,2n)) = Q[a], ∣a∣ = 2n. Then

(1) f = Ωg is a homotopy equivalence.

(2) g is a homotopy equivalence.

Proof. We have a map between fibration sequences

(ΩS2n+1)(0)

≅

// (PS2n+1)(0)

≅

// S2n+1(0)

Ω(S2n+1(0) ) //

PS2n+1(0)

// S2n+1(0)

ΩK(Q,2n + 1)≅K(Q,2n)

// PK(Q,2n + 1) // K(Q,2n + 1)

Deriving the long exact sequence of π∗ on the fibration sequence we get that f is anisomorphism on π2n and consequently on H2n and also on H2n. H∗(ΩS2n+1) ≅ Q[a′], ∣a′∣ =2n, hence H∗(f)(a′) = αa for some α ≠ 0. Hence H∗(f) is an isomorphism. Thus H∗(f)(which is an algebra morphism) is an isomorphism so f is a homotopy equivalence byWhitehead’s theorem.

For g use the same long exact sequence on π∗ and note that π∗(f) is an isomorphism in alldimensions.

Proposition 19. Assuming H∗(K(Q,2n + 1)) ≅ E(a), ∣a∣ = 2n + 1, we have

H∗(K(Q,2n + 2)) ≅ Q[b], ∣b∣ = 2n + 2.


Proof. By our assumption

Hi(K(Q,2n + 1)) = Q i = 0,2n + 10 otherwise .

Consider Serre spectral sequence for homotopy fibration sequence

K(Q,2n + 1) //

∼ΩK(Q,2n+2)

∗∼PK(Q,2n+2)

// K(Q,2n + 2)

in which

E2pq ≅Hp(K(Q,2n + 2))⊗Hq(K(Q,2n + 1))⇒H∗(∗) ≅ Q.

So we know that the spectral sequence looks like

E2 ∶ 2n+1 Q

2n 0 ⋯ 0 ⋯

⋮

1 0 ⋯ 0 ⋯

0 Q

0 2n+2//

OO

p

q

The only page at which E0,2n+1 may degenerate at 0 is r = 2n + 1. We also want E2n+2,0

to degenerate at 0 on the same page (otherwise it will never stabilize at 0). Therefore thedr-differential has to be the isomorphism shown below

E2 ∶ 2n+1 Q

2n 0 ⋯ 0 ⋯

⋮

1 0 ⋯ 0 ⋯

0 Q Q

≅

ffMMMMMMMMMMMMMMMMMMMMMMMMMMMM

0 2n+2//

OO

p

q

.

Likewise analysis shows that all E2k,0,1 ≤ k ≤ 2n + 1 should already be zero in the second

page.

1. COHOMOLOGY SERRE SPECTRAL SEQUENCE 47

Now from E22n+2,0 = Q we conclude that E2

2n+2,2n+1 ≅ Q also. We can now use the sameargument on the terms E2k,0,2n + 3 ≤ k ≤ 4n + 2. Inductively we conclude that

H i(K(Q,2n + 2)) = Q 2n + 2∣i, i ≥ 00 otherwise.

Hence as a commutative, connected, Hopf-algebra over a field of characteristic zero

H i(K(Q,2n + 2)) ≅ Q[b], ∣b∣ = 2n + 2

as desired.

Proof of theorem 0.3. This is the result of the induction starting with

H i(CP∞(0)) ≅ Q i even , i ≥ 0

0 i odd

and induction steps of propositions 18 and 19.

So we have computed π∗(S2n+1)⊗Q. What about the even dimensional spheres?

π∗(S2n)⊗Q =?

Consider f ∶ S2n(0) Ð→ K(Q,2n) such that π2n(f) is an isomorphism. This map exists since

K(Q,2n) represents H2n. Form the Serre spectral sequence associated to the fibrationsequence

F Ð→ S2n(0) Ð→K(Q,2n).

1. Cohomology Serre spectral sequence

Given a fibration sequence F Ð→ E Ð→ B where B is simply connected and all spaces arepath-connected, there exists a spectral sequence

Epq2 ⇒H∗(E)

with the E2-termsEpq2 ≅Hp(B)⊗Hq(F ).

This is also a spectral sequence of algebras. In particular the differentials satisfy the Leibnitzrule

(1.1) dr(ab) = (dra)b + (−1)∣a∣adrb,

where the product structure is induced by the product structure on the tensor product. Notethat since dr satisfies the Leibnitz rule ??, Er+1 has an induced algebra structure.


1.1. Edge homomorphism. In the Serre spectral sequence of the fibration

FiÐ→ E

pÐ→ B

as above (B simply-connected and F connected) we have the mappings

Hn(E)

Hn(p)

))//

⊂E∞n,0

//

≅E2n,0

// Hn(B)

FnH FnH/Fn−1H.

The composition E∞n,0 Ð→ E2

n,0 Ð→Hn(B) is called the edge homomorphism.

We wanted to find πi(S2n)⊗Q. Consider a fibration sequence

F Ð→ S2n(0)

fÐ→K(Q,2n)

so that f is π>n-isomorphism. Take the associated cohomology Serre spectral sequence:

Epq2 ≅HpK(Q,2n)⊗HqF.

We know that H0F ≅ Q so

Ep,02 ≅HpK(Q,2n) ≅ Q[a], ∣a∣ = 2n.

Also the edge homomorphism

HpB≅Ð→ Ep,02 ↠ Ep,0∞ Ð→Hp(S2n

(0))

tells us

Epq∞ ≅HpS2n(0) = Q p = 0,2n

0 otherwise .

So in the second page of the spectral sequence, since imdr ∩ (E2n,0r ) = 0

HqF = 0,0 < q < 4n − 1.

So there is α ∈ E0,4n−14n such that d4nα ≠ 0 and hence E0,4n−1

2 ≅ Q. Therefore

Ep,4n2 ≅ Q 2n∣p, p ≥ 00 otherwise.

So since this is a spectral sequence of algebras we can rewrite this computation as

E∗,4n−12 ≅ Q[a]⊗Q⟨b⟩, ∣a∣ = 2n, ∣b∣ = 4n − 1.

So since dr(a) = 0 = dr(b) then dr(E∗,4n−1r ) = 0 if r < 4n. Also since d4r(1⊗b) = a2⊗1

d4n(ak ⊗ b) = (d4rak).b + (−1)∣ak ∣ak.d4rb = ak+2 ⊗ 1.


So we get that

Epq∞ ≅ Q p = 0,2n, q ≤ 4n − 10 otherwise.

Consequently E0,q2 = 0, q ≥ 4n so

H iF = Q i = 0,4n − 10 otherwise.

If α ∈ HqF , the dr(α) = 0 for all r: The only interesting cases are d2n and d4n. If d2nα =c(a⊗ b), c ≠ 0 then we should have

d4n(a⊗ b) = 0

but this is a contradiction by the computation we did: d4n(a ⊗ b) = a3 ⊗ 1. So we shouldhave d2n(α) = 0. For the 4n-page just use the fact that d4n is a differential, so d4n(α) = 0.For all other r

dr(α) ∈ Er,q−rr = 0but E0,q

∞ = 0 if q ≠ 0 (by edge homomorphism).

We conclude that

Hi(F ) ≅H i(F ) ≅ Q i = 0,4n − 10 otherwise.

Claim F ∼ F(0): Because in the long exact sequence on π∗ for F Ð→ S2n(0) Ð→ B(0) we get

thatπ∗(F ) ≅ π∗F ⊗Q.

Then

H i(F,Z) ≅⎧⎪⎪⎪⎨⎪⎪⎪⎩

Z i = 0Q i = 4n − 10 otherwise.

So F ∼M(Q,4n − 1) = S4n−1(0) =K(Q,4n − 1).

Now use long exact sequence on π∗ for F Ð→ S2n(0) Ð→K(Q,2n) so

πi(S2n(0)) = Q i = 2n,4n − 1

0 otherwise.

This completes our comuptation.

CHAPTER 6

Loop suspensions

Recall we have adjoint functors Σ and Ω and unit of adjunction X Ð→ ΩΣX. Let X beany pointed space (with base point ∗). Let JX be the free monoid on X with identity ∗,i.e.

JX = x1⋯xk ∶ k ∈ N, xi ∈X/ ≃where the equivalence relation is x1⋯xi⋯xn = x1⋯xi⋯xn if xi = ∗. In fact let JnX ⊂ JXbe the set of all k-tuples for all k ≤ n of JX:

JnX =Xn/ ≃with the quotient topology. Then

JX = limÐ→JnX,

and it is a topological monoid by concatenation. This is called James’ construction. It is anassociative H-space, with a strict identity; i.e. ϕ(e, x) = x (not just upto homotopy).

There is an obvious inclusion η ∶ X Ð→ JX via x ↦ x. JX has the universal property thatgiven f ∶X Ð→H there is f ∶ JX Ð→H making

X

f !!CCCCCCCCηX// JX

f

H

commutes exactly. f is well-defined since e is strict (we define it by (x1⋯xn)↦ (x1(x2(⋯(xn)⋯)))or any other choice of parenthesis).

And if H is strictly associative then f is strictly multiplicative.

JX × JX //

f×f

JX

f

H ×Hϕ

// H.

In fact if H is strictly associative the choice of parenthesis does not matter since H isstrictly associative.

We will show that JX is a model for ΩΣX (i.e. it is homotopy equivalent to ΩΣX; in factmore it true, it is multiplicative homotopy equivalent to this space).

50

6. LOOP SUSPENSIONS 51

Definition 17 (Moore loops). Let (X,∗) be a based space. The More loops of X is

ΩmX = (η, s) ∶ η ∶ R≥0 Ð→X,s ∈ R≥0, η(t) = ∗ if t ≥ s or t = 0with subspace topology of Map(R≥0,X) ×R≥0.

ΩMX is associative H-space with strict identity. The structure is given as follows:

ΩMX ×ΩMX Ð→ ΩMX

(η, s) × (η′, s′)↦ (η ◻ η′, s + s′)where

η ◻ η′(t) = η(t) 0 ≤ t ≤ sη′(t − s) s ≤ t

and e = (e∗,0), i.e. e∗(t) = ∗ is the identity map.

Consider the continuous mapping

f ∶ ΩX Ð→ ΩMX

η ↦ (η,1)and the rescaling map

g ∶ ΩMX Ð→ ΩX

(η, s)↦ η′ ∶ η′(t) = η(st).

Proposition 20. g and f are homotopy multiplicative inverse homotopy equivalences.

Proof. gf = idΩX and fg ≃ idΩMX by scaling to length 1. Rescaling also gives ahomotopy

ΩX ×ΩX

f×f// ΩMX ×ΩMX

ΩXf

// ΩMX

by the top right composition one is mapping a path to some (∗,2) and by the bottom leftcomposition to (∗,1). For the other direction let r be the rescaling (η, s)↦ (η,1/2). Thenthe outer pentagon commutes exactly in

ΩMX ×ΩMXid

))SSSSSSSSSSS

ΩMX ×ΩMXg×g

//

r×r 55kkkkkkkkkkkΩMX ×ΩMX

g

ΩX ×ΩX // ΩX.


Recall the map

X Ð→ ΩΣX

x↦ ϕ,ϕ(t) = (x, t) ∈ ΣX =X × I/ ≅ .

This factors through X Ð→ ΩMΣX via x ↦ (ϕ,1). We need an extra hypothesis ∗ Ð→ X isa cofibration (one says X is well-pointed).

Proposition 21. Suppose (X,∗) is well-pointed and f ∶ X Ð→ ΩMΣX is the map above.Then f ≃ g such that g(∗) = e.

Proof. Consider

∗ × I ∪X × 0

H∪f// ΩMΣX

X × I.H

77nnnnnn

where H is a path in ΩMΣX from f(∗) to e: say f(∗) = (ϕ,1) where ϕ(t) = (∗, t). ThenH(s) = (ϕ,1 − s).

Now H making the diagram commute is our homotopy.

Notation: Let g ∶X Ð→ ΩMΣX be the mapping

g(x) = H(x,1)

and θ ∶ JX Ð→ ΩMΣX its multiplicative extension.

We will show that H∗(θ) is an isomorphism.

Theorem 0.1. Let k be any field. Then H∗(ΩΣX;k) ≅ TH∗(X;k) as algebras.

Recall that if V is a (graded) vector space, then TV = ⊕0≤n<∞V ⊗n, called the tensor algebraand has the obvious multiplicative structure.

In fact in case of X = Sn and chark = 0, we know that

H∗(ΩSn+1, k) = k[a]where a is primitive for degree reasons. So we knowH∗(ΩSn+1, l) as Hopf algebra and

(H∗(ΩSn+1, k))∗ ≅H∗(ΩSn+1, k)as Hopf algebras (universal coefficients theorem).

Recall that X ∧X =X ×X/X ∨X and from that

H∗(X ∧X) ≅ H∗(X)⊗ H∗(X).

Proposition 22. If X is connected, Σ(X ×X) ≅ Σ(X ∨X) ∨Σ(X ∧X).


Proof. Consider the cofibration sequence

X ∨X Ð→X ×X Ð→X ∧X. (∗)By this we mean a sequence A B Ð→ C where C is a quotient of a cofibration A B. Wehave maps

pi ∶X ×X Ð→X

(x1, x2)↦ xi.

Then (∗) gives a sequence

Σ(X ∨X)

id

// Σ(X ×X) //

pinch twice

Σ(X ∧X)

id

Σ(X ×X) ∨Σ(X ×X) ∨Σ(X ×X)

Σp1∨Σp2∨Σπ

ΣX ∨ΣX // ΣX ∨ΣX ∨Σ(X ∧X) // Σ(X ∧X)

The diagram commutes up to homotopy and the bottom row is cofibration sequence. Sousing long exact sequence on H∗ of bottom and top cofibration sequences and the fivelemma we get that

Σ(X ×X)Ð→ ΣX ∨ΣX ∨Σ(X ∧X)is homology isomorphism and hence homotopy isomorphism since the suspension of con-nected spaces are simply connected. Note that we also using the fact that the (weak)product of CW complexes are CW complexes and Σ of CW complexes are CW complexes.Now we apply Whitehead’s theorem.

Remark. The above is a space level version of the homology decomposition

H∗(X ×X) ≅H∗(X)⊗H∗(X)the last one being only linearly equivalent to

H∗(X)⊕ H∗(X)⊕ H∗(X)⊕ H∗(X ∧X).The idea is to start from a decomposition of algebraic invariants and prove a decompositionof objects with more structure.

RecallTnX = (x1,⋯, xn) ∈Xn ∶ xi = ∗ for some i

is the flat wedge of X.

Proposition 23. (1) Let X be a CW-complex, then the diagram

TnX //

Jn−1X

Xn // JnX


is a pushout.

(2) Xn/TnX ≅ JnX/Jn−1X.

(3) Also TnX Ð→Xn is a cofibration so Jn−1X Ð→ JnX is a cofibration.

Proof. (1) is obvious. (2) follows since if

A //

B

C // C

is a pushout, then

C/Af// D/B

ϕ

ZZ

is a homeomorphism. If ϕ exists such that the diagram commutes, ϕ induced by mapD Ð→ C/A determind by projection C Ð→ C/A and

A/A

""EEEEEEEE

B

>>||||||||// C/A

clearly agree on A so we get a map D/B Ð→ B/A. Check f and ϕ are inverses. We skip theproof of (3).

Proposition 24. TnX Ð→Xn Ð→X∧n is a cofibration sequence i.e. Xn/TnX ≅X∧n.

Proof. Look at X ∨ Y Ð→X × Y Ð→X ∧ Y =X × Y /X ∨ Y . So

X∧n =Xn/ ≅where the equivalence is generated by (x1,⋯, xn) ≅ ∗ if some xi = ∗. This proves theassertion.

Corollary 3. Jn−1X Ð→ JnX Ð→X∧n is a cofibration sequence.

Proof. Follows from two previous propositions.

Since J1X =X we getH∗X //

zzuuuuuuuuu

H∗JX

H∗X// TH∗X.

ϕ

99tt

tt

t

So take ϕ to be the unique algebra extension.


Proposition 25. If H∗X = H∗(X,R) is a free R-module then the map ϕ above is anisomorphism.

Proof. LetTn(H∗X) = ⊕0≤i≤n(H∗X)⊗i

be the tensors of lengths ≤ n.

Claims:

(1) ϕ(TnH∗X) ⊆H∗JnX.

(2) H∗(X)⊗n TnH∗(X)ϕÐ→H∗JnX Ð→H∗(X∧n) is an isomorphism.

Considering how ϕ is defined we have

(H∗X)⊗n // (H∗X)⊗n //

''NNNNNNNNNNN

>>>>>>>>>>>>>>>>>>>

H∗(J1X)⊗n //

H∗(JX)⊗n

multiply

H∗(X)⊗n

77ooooooooooooH∗(JX)

H∗(Xn)

77oooooooooooo// H∗(JnX)

OO

which proves claim (1).

For claim (2) look at

H∗(X)⊗n //

&&MMMMMMMMMMM

†

''

H∗(Xn)

// H∗(X∧n)

‡

H∗(Jn(X) // H∗(X∧n)† is an isomorphism by Kunneth formula and induction. ‡ is also an isomorphism sincecofibers are heomoemorphic.

So we get maps between short exact sequences

0 // Tn−1H∗(X) //

TnH∗X//

H∗(X)⊗n //

≅by claim (1)

0

0 // H∗(Jn−1X) // H∗(JnX) // H∗(X∧n) // 0

where the vertical maps are induced by ϕ. So by induction on n we get that TnH∗X Ð→H∗(JnX) is an isomorphism.


So we have a sequence of isomorphisms

TH∗X ≅ limÐ→TnH∗X

≅ limÐ→H∗(JnX)≅H∗ limÐ→JnX since limÐ→ commutes with H∗

≅H∗(JX) since JX ∶= limÐ→JnX

Therefore ϕ is an isomorphism.

Proposition 26. The map JX Ð→ ΩΣX from the last class is a homotopy equivalence.

Proof. We are assuming everything is of finite type. We have a homotopy commutingdiagram

X //

""FFFFFFFF JX

Θ

ΩΣX

where Θ is homotopy multiplcative. So taking H∗ we get multiplcative map H∗(Θ). Wehave a commutative diagram

T (H∗(X))

≅

H∗X //

99ssssssssss

%%LLLLLLLLLL

999999999999999999 H∗(JX)

H∗Θ

H∗(ΩΣX)

T (H∗(X))

≅ by Bott-Samuelson theorem

OO

so H∗(Theta, k) is an isomorphism for any field k. So H∗(Θ,Z) is an isomorphism. So Θis a homotopy equivalentce. Since X is a CW complex TX is a CW complex and ΩΣX ishomotopy equivalent to a CW-complex.

Proposition 27. ΣJX ≅ Σ(⋁1≤i<∞X∧i).

Proof. Consider the commutative diagram

0 // TnX //

Xn //

X∧n

// 0

0 // Jn−1X // JnX // X∧n // 0.

1. ANOTHER HOMOTOPY COMPUTATION 57

The top row splits after suspension. Therefore so does the bottom row. So we have

ΣJnX ≃ ΣJn−1X ∪ΣXn.

Now induction and passing to limÐ→ gives the proposition.

Theorem 0.2 (Freudenthal suspension theorem). Suppose the connectivity of X is ≥ n−1.Then

πiX Ð→ πi(ΩΣX) ≅ πi+1(ΣX)is an isomorphism for i ≥ 2n − 1 and is a surjection for i ≤ 2n − 1.

Proof. Look at Serre spectral sequence for fibration

F Ð→X Ð→ ΩΣX.

Then E2p,0 ≅ T (H∗(X)). Lowest degree homology is at least Hn(X) so the lowest degree

homology in the cokernel will at least be

HnX ⊗ HnX.

So connectivity of the fiber, F , is at least 2n−2. Then we use the long exact sequence onπ∗to complete the proof.

1. Another homotopy computation

Recall that for the fibration

FgÐ→ ΩS3 f

Ð→ CP∞ =K(Z,2)By Serre’s method, g induces an isomorphism on πi for i > 2 and πi(F ) = 0 for i ≤ 2, and finduces isomorphism on π2 and is zero on πi for i > 2. As a result of applying Hurewitcz’stheorem we had

π3F =H3(F,Z) = π3(ΩS3) ≅ π4S3.

We now want to compute H3(F,Z). As an algebra

H∗(ΩS3,Z) ≅ Z[a], ∣a∣ = 2.

This is a consequence of Bott-Samuelson theorem: we have ΩS3 = ΩΣS2 and hence

H∗(ΩS3) = T (H∗(S2,R)) = T (R.a) ≅ R[a]

since H∗(S2) is a free R-module.

Moreover as Hopf algebras we have an isomorphism

H∗(CP∞,Z) ≅ Z[a′]with a′ primitive (just work out the diagonal ∆(a′) to see this). Since

∆(a′)n =∑(i + ji

)(a′)i ⊗ (a′)j


the dual algebra is the divided powers algebra,

Γ(α) ∶= ⊕i≥0Zαiwhere αi = (ai)∗ and αiαj = (i+j

u)αi+j :

H∗(CP∞,Z) ≅ (H∗(CP∞,Z))∗ ≅ Γ(α) both as Hopf algebras.

In Z/p coefficients we have H∗(f,Z/p) is an isomorphism for ∗ < 2p and is zero if ∗ =2p.

H∗(ΩS3,Z/p) ≅ Z/p[a]Ð→ Γ(α)⊗Z/p ≅H ∗ (CP∞,Z/p)ap ↦ (α1)p = p!(αp) = 0

Note 1.1. In cases of both coefficients H∗(f) is algebra map, since f = Ωg. So f isa Hopf algebra map.

In the next step we look at Serre’s spectral sequence with coefficients in Z/p: in the secondpage we have

E`,0 ≅H`B ≅H`CP∞.

so

E2 ∶ 2p Z/p

2p-1 Z/p

0 Z/p Z/p Z/p

0 2 2p//

OO

p

q

The edge homomorphism is

H`(ΩS3)Ð→ E∞`,0 ≅ E

2`,0 Ð→H2(CP∞)

which is an isomorphism. HenceE∞

2p,0 = 0.So on second page we need to have

E20,2p−1 ≅ Z/p.

Also E20,2p ≅ Z/p since ⊕i+j=2pE

∞ij = Z/p ≅H2p(ΩS3) but we don’t need this.

So the spectral sequence yields

H2p−1(F,Z/p) = Z/p,Hi(F,Z(p)) = 0 if i < 2p − 1.


From the short exact sequence

0Ð→ Z(p) Ð→ Z(p) Ð→ Z/pÐ→ 0

we getH2p−1(F,Z(p)) = Z(p) or Zpn

but we can exclude Z/pn from knowing H∗(f)⊗Q is isomorphism so H∗(F )⊗Q = 0. Onthe other hand assuming finite type, and localization techniques we have

Hi(F,Z(p)) = 0, i < 2p − 1

(so F(p) is 2p − 2 connected). By Hurewicz’s theorem

π2p−1(F )⊗Z(p) ≠ 0.

Again from H∗(f)⊗Q being an isomorphism, we have

pkH2p−1(F,Z(p)) ≅ π2p−1(F )⊗Z(p) = 0

so π2p−1(F )⊗Z(p) ≅ Z/pn.

Now we look at the Serre spectral sequence of

F(p) Ð→ ΩS3(p) Ð→ CP∞

(p)

with Z-coefficients.

E2 ∶ 2p-1 Z(p)

0 Z(p) Z(p) Z(p)

0 2 2p//

OO

p

q

By the edge homomorphism

H2p(ΩS3)Ð→ E∞2p,0 Ð→ E2

2p,0 ≅H2p(CP∞)

is multiplication by p, Z(p) Ð→ Z(p). So αp /∈ Z∞2p and pαp ∈ Z∞

2p aso there is β ∈ E20,2p−1 and

pβ = 0. And β mut generate E20,2p−1. So

E20,2p−1 = Z/p ≅H2p−1(F(p),Z(p)) ≅ π2p−1(F )⊗Z(p)

Also H`(F(p),Z) = 0, ` < 2p − 1. Since πi(F ) ≅ πi(ΩS3) ≅ πi+1(S3) for i > 2

πi(S3)⊗Z(p) = 0,3 < i < 2p

and getπ2p(S3)⊕Z(p) = Z/p.

CHAPTER 7

Classifying spaces

1. Category of G-bundles

Let G be a topological group, a space with a continuous right G-action is called a rightG-space. We will usually be working with right actions and shortening the above term byG-space. Morphisms of G-spaces are G-equivariant maps. We know that if X is a G-spacethen the is an open map π ∶X Ð→X/G to the quotient space with quotient topology.

Definition 18. A map ξ ∶XpÐ→ B is a G-bundle if it is isomorphic to X ′ Ð→X ′/G for some

G-space X.

Example 1.1. The action of GL(n,R) on Rn gives a quotient space Rn/GL(n,R)consisting of two points with one of them being open and the other one not open. It isobvious from this example that G-bundles may not be fibrations.

Definition 19. a G-space X is free is xs = x implies s = 1.

If X is free then we make a notation

X∗ ∶= (x,xs) ∶ x ∈X,s ∈ G ⊂X ×X.The map τ ∶ X∗ Ð→ G via (x,xs) ↦ s is well-defined since G is free and is called thetranslation function.

Definition 20. A G-space X is principal if X is free and τ ∶ X∗ Ð→ G is continuous. AG-bundle X

pÐ→ B is principal if X is a principal G-space.

Proposition 28. A principal G-bundle XpÐ→ B has fiber G (i.e. for all b ∈ B, p−1(b) is

homeomorphic to G).

Proof. Let b ∈ B pick x ∈ p−1(b) define u ∶ G Ð→ p−1(b) via x ↦ u(s) ∶= xs. The inverseis y ↦ τ(x, y).

Example 1.2. Let G = Z/2 act on Sn. Then Sn Ð→ Sn/G is principal. Clearly theaction is free and observe that τ from the space

(x,±x) ∶ x ∈ S ⊂ Sn × Sn

to 1,−1 is continuous. 60

1. CATEGORY OF G-BUNDLES 61

If X is a G-space with associated G-bundle XpÐ→ B then any map B′ f

Ð→ B can pullback pto a G-bundle on B′:

X ′

◻p′

f// X

p

B′f// B.

The G-action on X ′ is given via

(x, b′).s = (xs, b′).

It is clear that f is a map of G-spaces. Also if X is principal then X ′ is one as well.

Definition 21. For any map f ∶ Y Ð→ Y ′ of G-spaces we get a commutative diagram

Y //

Y ′

Y /G // Y ′/G

a map of G-bundles is something isomorphic to this (here meaning only consistent homeo-morphisms).

Let Bun[G] be the category of G-bundles and G−BunB is the groupoid of G-bundles overB. Any map f ∶ B′ Ð→ B gives a functor

f∗ ∶ BunB[G]Ð→ BunB′[G].The fact that this is a groupoid will be left unproved,

Proposition 29. If f ∈ HomBunB[G](X ′,X) then f is an isomorphism.

Proof. Let f be the equivariant map over B in the diagram

Xf

//

p

@@@@@@@@ X ′

p′~~

B

Injectivity: If f(x1) = f(x2) then p(x1) = p′(f(x)) = p′(f(x2)) = p(x2). So x1s = x2 forsome s ∈ G. Thus

f(x1)s = f(x1s) = f(x2) = f(x1).Therefore x = 1 since ξ′ is free. So x1 = x2.

Sujrectivity: Given x′ ∈ X ′, let x ∈ X be such that p(x) = p′(x′) (by surjectivity of p). Sop′(x′) = p(x) = p′(f(x)). Thus x′ = f(x)s for some x ∈ G. Therefore x′ = f(xs) so f issurjective. Now check that f−1 is continuous.

2. FIBER BUNDLES 62

2. Fiber bundles

Let ξ ∶ (XpÐ→ B) be a principal G-bundle. Let also F be a left G-space and we have a

diagonal right action of G on X × F via

(x, f).s = (xs, s−1f)

and the orbit space will be denoted by

XF ∶=X × F /G =X ×G F.

The map pF ∶XF Ð→ B is the unique factorization of

X × F Ð→X Ð→ B

through X ×G F . Then ξF ∶ (XFpFÐ→ B) is called a fiber bundle over B with fiber F (as a

G-space).

One Session missing here

Definition 22. A covering Ui of X is a numerable overing if there is a locally finitepartition of unity Uii∈S . In particular

Ui((0,1]) ⊆ Ui,∀i ∈ S.

For Hausdorff spaces and paracompact spaces all coverings are numerable. A G-bundle ξis numerable if it is trivial over a numerable covering.

Proposition 30. For a map of bases f ∶ B′ Ð→ B, numerable G-bundle ξ pulls back tonumerable f∗(ξ).

Proof. Use u′i = ui f and U ′i = f−1(Ui).

For convenience we write E(ξ) for the total space of a G-bundle ξ.

Proposition 31. Given r ∶ B × I Ð→ B × I via (b, t) ↦ (b,1) and ξ a numerable G-bundleover B×I then there is a G-morphism (g, r ∶ ξ Ð→ ξ such that there is a commutative diagram

E(ξ)g//

E(ξ)

B × I r// B × I.

Corollary 4. ξ and r∗ξ are isomorphic over B × I.

2. FIBER BUNDLES 63

Proof. There is a unique map making the diagram

E(ξ)

$$II

II

I g

$$

$$

E(r∗ξ) //

E(ξ)

B × I // B × I

commute. So check that it is a G-equivariant map and by last session it must be anisomorphism.

Theorem 2.1. Given f, g ∶ B′ Ð→ B and ξ a bundle over B, if f ≅ g that f ∗ (ξ) ≅ g∗(ξ).

Proof. Let H be a homotopy from f to g. Let E0 ∶ B′ Ð→ B;×I be b↦ (b,0). We haveHE0 = f , and HrE0 = g. Now we have

f∗(ξ) = (HE0)∗(ξ) = E∗0H

∗(ξ) ≅ E∗0 r

∗(H∗(ξ)) ≅ (HrE0)∗(ξ) = g∗(ξ).

Notation: kG(B) is the set of isomorphism classes of numerable principal G-bundles overB. And by isomorphism we mean isomorphisms over the identity map.

Corollary 5. For f ∈ [X,Y ] we get kG(f) ∶ kG(Y ) Ð→ kG(X). And kG(f) is a bijectionif f is a homotopy equivalence.

Definition 23. A principal G-bundle ω ∶ E0p0Ð→ B0 is a universal (numerable) G-bundle if

ω is numerable and[−,B0]Ð→ kG(−)

is an isomorphism of contravariant functors, i.e. it’s a natural transformation such that forall X,

[X,B0]Ð→ kG(X)is a bijection.

Note that [−,B0] is clearly a functor on the homotopy category Top≅ so is kG(0) morphismsgiven by pullback.

Proposition 32. A numerable principal G-bundle ω is universal if and only if

(1) ∀ξ principal numerable G-bundle over X, there is f ∶X Ð→ B0, ξ ≅ f∗(ω) over X.

(2) For any f, g ∶X Ð→ B0. f∗(ω) ≅ g∗(ω) implies f ≃ g.

3. MILNOR’S CONSTRUCTION 64

3. Milnor’s construction

Definition 24 (The join). A ∗B is a topological space A ×B × I/ ≅ with the relations

(a, b,1) = (a, b′,1), (a, b,0) = (a′, b,0).We can view it as having points (t0a, t1b) with t0 + t1 = 1 with relations

(0a, t1b) = (0a′, t1b).

A ∗B fits into a pushout

(I) A ∗B A ×CB

CA ×B A ∗B

where CX is the notation for cone on X. Consider the homotopy pushout

(II) A ∨B A ∨CB

CA ∨B CA ∨CB

the inclusion IIÐ→ I is a cofibration on each space in part.

colim(I/II) ≅ colim I/ colim II ≅ A ∗Bsince colim II ≃ ∗. But

I/II ∶ A ∧B //

A ∧CB(≃ ∗)

(∗ ≃)CA ∧B

and this diagram is “equivalent” to

A ∧B //

C(A ∧B)

∗ // Σ(A ∧B)

and we conclude that colim I/II ≃ Σ(A ∧B). And finally that

A ∗B ≃ ΣA ∧B.


We make the notation EnG = ∗nG = (∗n−1G) ∗G. And as a set we should view it as

(t0x0, t1x1,⋯, tn−1xn−1) ∶∑ ti = 1/ ≅with the relations that if ti = 0 then

(⋯, t2xi,⋯) ≅ (⋯, tix′i,⋯).Clearly EnGÐ→ En+1G via

(t0x0,⋯, tn−1xn−1)↦ (t0x0,⋯, tn−1xn−1,0).As an aside note that one can show that EnGÐ→ En+1G is null-homotopic.

We now make the final definition

EG ∶= ∗∞G ∶= colimn(EnG).

Proposition 33. EG ≃ ∗.

Proof. EnG ≃ En−1G ∗G ≃ Σ(En−1G ∧G). So it’s connected to all degrees.

EnG with diagonal action is a free G-space. In fact, since for any (t0g0,⋯, tn−1gn−1) wehave ∑ ti = 1, some ti ≠ 0 so the action is free. The action is continuous as well (check it!).So we can define

BnG ∶= EnG/G, BG ∶= EG/G.

Proposition 34. EnGpnÐ→ BnG and ωG ∶ EG

pÐ→ BG are locally trivial, numerable, princi-

pal G-bundles.

Proof. We will only work out the proofs of local triviality and numerability for theEG case. Let ti ∶ EGÐ→ I be given via

(t0x0,⋯, tixi,⋯)↦ ti.

There is a unique ui ∶ BG0,1Ð→ such that uip = ti. Letvi = u−1

i ((0,1]) = p(t−1i ((0,1])) ⊆ B.

To show that ωG is trivial over vi, we show that there is a section over vi,

si ∶ vi Ð→ t−1i (0,1] ⊆ EG.

So define s′i ∶ t−1i (0,1]Ð→ t−1

i (0,1] that is constant on G-orbits

s′i ∶ t−1i (0,1]Ð→ t−1

i (0,1]to be constant on G-orbits, via

s′i ∶ t−1i (0,1]Ð→ t−1

i (0,1](t0x0,⋯, tixi,⋯)↦ a(xi)−1.

s′i is well-defined since ti ≠ 0. And si(ag) = s′i(a) for all g ∈ G, hence s′i induces

si ∶ t−1i (0,1]/GÐ→ t−1

i (0,1].


It is easy to check that psi = idvi . So ωG is locally trivial, also numerable since the ui’s givea partition of unity.

Corollary 6. EGpÐ→ BF is a fibration with fiber ΩBG and so B is an inverse to Ω on

the homotopy category.

Proof. EG ≃ ∗ and EGpÐ→ BF a fibration, since ωG is locally trivial G-bundle over a

CW complex. Thereforep−1(∗) ≃ ΩBG.

The constructions we have for EnG, BnG, EG and BG are clearly functorial on grouphomomorphisms. So B is a functor

Top group Ð→ Top.(probably we need some topological restrictions on the category of Topological groups weare considering.) Recall that Ω can be considered as a functor

Top Ð→ Top group.This is due to Milnor. The first thing one has to show is an equivalence of categoriesbetween simplicial sets and topological spaces. Then he works on simplicial sets to createa fibration PX Ð→ X. PX turns out to be contractible so the fiber is the loop space ΩX.The way the fibration is constructed ΩX has a natural group structure on it as well.

From now on we work with CW complexes. So all coverings are numerable since CWcomplexes are paracompact.

Theorem 3.1. For a principal G-bundle, ξ, over B, there is f ∶ B Ð→ BG such that

ξ ≅ f∗(ωG).

Proof. Since ξ is numerable there is un over B and Un = u−1n (0,1], trivializing ξ

for all n ≥ 0. (Quesiton: Why can we assume Un is a countable collection?) There is atrivialization

hn ∶ Un ×G ≅ E(ξ∣Un) ⊆ Eξ).Let gn ∶ Un ×GÐ→ G be the projection and p ∶ E(ξ)Ð→ B. Define g ∶ E(ξ)Ð→ EG

z ↦ (u0(p(z)g0h−10 (z),⋯, un(p(z))gnh−1

n (z),⋯).If hn is not defined, then unp(z) = 0. We conclude that g is well-defined and that g(zs) =g(z)s for all s ∈ G since hn is G-equivariant. So g induces f ∶ B Ð→ BG and (g, f) ∶ ξ Ð→ ωGis a bundle map, i.e.

E(ξ)g//

EG

Bf// BG

commutes. Hence ξ ≅ f∗(ωG).


Theorem 3.2. For f and g are maps B Ð→ BG, and f∗(ωG) ≅ g∗(ωG) then f ≃ g.

This completes the proof of [B,BG]Ð→ kG(B) being a bijection.

pooya-git.github.io · 2. CW COMPLEXES 5 2. CWcomplexes Definition 2. f∶A—→XhasthestructureofarelativeweakCWcomplexif X(0)=A andwehavepushouts N ∈I(i)S n( ) / X(i) N ∈I(i)D

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