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Zvi Rosen Algebraic Topology Notes Kate Poirier

1. Friday, August 24, 2012

1.1. Course Information. Website: math.berkeley.edu/∼poirierE-mail: poirier@mathOffice: 813 EvansCourse Rubric: 25% HW every two weeks, 25% Oral Presentation, 50% Take-home Final.First Homework Due September 17

1.2. A Soft Introduction to Algebraic Topology.

(1) What is algebraic topology (Math 215A)?(2) Classification of Surfaces.(3) Rough introduction to categories and functors.

One goal: Classify topological spaces (up to homeomorphism) algebraically.

Question 1.1. Given spaces X and Y , does there exist a homeomorphism f : X → Y ? Recall ahomeomorphism is a continuous map with continuous inverse.

If there exists such an f then X and Y are topologically equivalent. Often, it is easier to answerthis question in the negative. For example, if X and Y don’t share some property preserved underhomeomorphism, e.g. connectedness or compactness.

Definition 1.2. Let |π0(X)| be the number of connected components of X.

If X and Y are homeomorphic, then |π0(X)| = |π0(Y )|. Failure to be equal would thus implyfailure of X and Y to be homeomorphic.

Certainly, there are X and Y such that |π0(X)| = |π0(Y )| but X and Y are not homeomorphic,e.g. the point and the line. This means that the numerical homeomorphism invariant |π0(X)| isjust okay.

In this class, we will examine three algebraic invariants:

(1) π1(X), the fundamental group.(2) H•(X), the homology.(3) H•(X), the cohomology.

Along the way, these invariants help us prove “purely topological” statements. The followingtheorem is an example.

Theorem 1.3 (Brouwer fixed point theorem). Let Bn = (x1, . . . , xn) ∈ Rn :∑x2i ≤ 1 be the

standard n-ball. Let f : Bn → Bn be a continuous map. Then f has a fixed point, i.e. ∃x ∈ Bn

s.t. f(x) = x.

Definition 1.4 (Imprecise). An n-dimensional manifold is a space that locally looks like Rn.

Example 1.5. • Rn• Sn = (x1, . . . , xn+1) ∈ Rn+1 :

∑x2i = 1, since locally it looks like Rn. (Fig1)

• Tn = S1 × · · · × S1︸︷︷︸n times

. (T 2 is often called T ) (Fig2)

• Bn = Dn the disk or ball. (with or without boundary)• A disk with small disks removed.

“Classification of 0-Manifolds” – a collection of points. There is only one zero manifold, up tohomeomorphsim.

Theorem 1.6 (Classification of Compact, Connected 1-Manifolds, with or without boundary). Upto homeomorphism, any 1-manifold is S1 or I = [0, 1], i.e. compact, connected 1-manifolds arecompletely determined by whether or not they have boundary.

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Theorem 1.7 (Classification of Surfaces). A surface is completely determined up to homeomor-phism by (1) χ, the Euler characteristic, (2) the number of boundary components, and (3) ori-entability.

2. Monday, August 27, 2012

Recall the following:

Definition 2.1 (Imprecise). A manifold is a space which looks locally Euclidean. A surface is a2-dimensional manifold (with boundary).

Theorem 2.2 (Classification of Surfaces). A compact, connected surface is completely determinedup to homeomorphism by

• χ, the Euler characteristic,• the number of boundary components• orientability/non-orientability.

Example 2.3. S2 = Sphere, Cylinder Segment, M = the Mobius strip, P = pair of pants, T =T 2 = S1 × S1 the torus, D = B2 the disk.

See Figure 3 for some constructions of surfaces. In the construction of RP2, the surface has oneface, one edge, and one vertex; we end up double wrapping that one edge + vertex around the face.

Definition 2.4 (Imprecise). A Connect Sum (#) is given in Figure 4. The identity of the monoiddefined by this operation is the sphere.

Definition 2.5 (Imprecise). A surface is nonorientable if it contains a Mobius strip (otherwise, itis orientable).

Definition 2.6 (Imprecise). A triangulation of a surface looks like Figure 5. Using this triangula-tion,

χ(Σ) = V − E + F,

where V = # vertices, E = # edges, and F = # faces (triangles).

Theorem 2.7. χ(Σ) is independent of the triangulation. For example Σ(S2) = 2.

Table 1. Orientable Surfaces

Theorem 2.8.

χ 2 1 0 -1 -2#∂0 S2 T 2 T 2#T 2

1 D T 2 −D2 S1 × S1

Table 2. Non-orientable Surfaces

Theorem 2.9. ∑orientable⇒ χ(Σ) = 2− 2g −#∂ comps.∑

non-orientable⇒ χ(Σ) = 2− g −#∂ comps.

where g = genus is defined roughly as the number of handles.

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3. Wednesday, August 29, 2012

I came 25 minutes late today, so my notes are incomplete. So far, Prof. Poirier has definedcategories, and given examples, including groups, topological spaces, and posets.

Definition 3.1 (Imprecise). A functor CF→ D is a morphism in the “category of categories.” It

maps the objects and morphisms of C to those of D respecting the associative structure.

In this course, we have three important functors:

(1) π1 : Top∗ → Group.(2) H∗ : Top→ grGroups, the category of graded groups.(3) H∗ : Top→ grRings, the category of graded rings.

Remark 3.2. These functors take homeomorphism in Top to isomorphism. But in these examples,homotopy equivalence (weaker than homeomorphism) also give isomorphism. So the invariants arejust okay.

Good news: they are often computable or comparable (even if they are not computable, you canfind some difference between them).

By the Poincare conjecture, π1 distinguishes S3 among all 3-manifolds. i.e. if π1(M3) ∼= π1(S3),then M3 is homeomorphic to S3. In fact, it generalizes to higher dimensions.

Remark 3.3. The Smooth Poincare Conjecture is false e.g. in dimension 7. Milnor’s exotic spheresare homeomorphic but (not?) diffeomorphic to S7.

4. Friday, August 31

Today: Homotopy of maps, and CW-complexes.

4.1. Homotopy. Fix topological spaces X and Y . Consider the space of continuous maps fromX to Y , f : X → Y . We use the notation:

Maps(X,Y ) = f : X → Y .Remark 4.1. Maps(X,Y ) has the compact-open topology.

What is a path in Maps(X,Y )?

Definition 4.2. Let Xf0⇒f1

Y be two continuous maps from X to Y . If there exists a continuous

map X × I F⇒ Y such that F (x, 0) = f0(x) and F (x, 1) = f1(x), then F is called a homotopybetween f0 and f1 and f0 and f1 are called homotopic. [Apparently, product with X and Mapsfrom X are adjoint functors: X × Z → Y ↔ Z →Maps(X,Y ).]

We think of a homotopy F as a continuous family of maps X → Y . Ft = F (•, t) : X × t ∼=X → Y . If f0 is homotopic to f1, write f0

∼= f1.

Exercise 4.3. Show that homotopy induces an equivalence relation on Maps(X,Y ).

Definition 4.4. Given

(1) Xf→ Y , Y

g→ X is called a homotopy inverse for f if f g ∼= idY and g f ∼= idX .

(2) If Xf→ Y has a homotopy inverse then f is called a homotopy equivalence.

(3) If there exists a homotopy equivalence Xf→ Y then X and Y are called homotopy equivalent

, write X ∼= Y .

Exercise 4.5. Show that homotopy equivalence induces an equivalence relation on spaces.3


Special case: A ⊂ X.

Definition 4.6. (1) A retraction from X to A is a map r : X → A that keeps A fixed.(2) A map r : X → A is called a deformation retraction if i r = idX .(3) If F : X × I → X is a homotopy between i r and idX and F (a, t) = a, for all a ∈ A, then

r is a strong deformation retraction.

Usually, when one says “retraction”, one means (2) or (3). (Please specify if you mean (1).)

Exercise 4.7. Write down an explicit formula for a strong deformation retraction

(1) R2 → (0, 0).(2) R2 − (0, 0) → S1.

⇒ R2 and pt have the same homotopy type. Similarly for (2).

Definition 4.8. If X has the homotopy type of a point, then X is called contractible.

Remark 4.9. Homotopy equivalence is coarse in that non-homeomorphic spaces may be homotopy-equivalent, e.g. it does not recognize dimension (Rn is contractible for all n).

4.2. CW-Complexes. It is possible to impose a CW structure on a surface.

Example 4.10. Consider the torus T 2, in its construction via rectangle. In this form, we havevertices, edges, and faces identified in such a way that they uniquely determine the surface.

Alternatively, construct T 2 by taking 1 point, 2 edges, and 1 face. Map the boundaries of the 2edges (i.e. vertices) onto the vertex and map the boundaries of the face (i.e. edges) and map themto the edges.

Definition 4.11. An n-cell is a space homeomorphic to Dn. A CW-complex is a space builtinductively by cells and attachments.

“0-skeleton” X0 = discrete set of points. “n-skeleton” Xn = Xn−1 tϕαDnα, where ϕα : ∂Dn

α →Xn−1, and α is in some indexing set.

The inclusion X0 ⊂ X1 ⊂ X2 ⊂ · · · defines X. If X = Xn then we call X an n-dimensionalCW-complex.

Exercise 4.12. Determine a CW decomposition for:

(1) S2.(2) S2 ∨ S1 ∨ S1.

5. Wednesday, September 5, 2012

Recall: A homotopy of maps Xf0→f1Y is a map F : X × I → Y such that F (x, 0) = f0(x) and

F (x, 1) = f1(x) for all x ∈ X.

Definition 5.1. Let Ai→ X be a subspace. We say f0, f1 : X → Y are homotopic relative

to A if f0(a) = f1(a) for all a ∈ A, and there exists a homotopy F : X × I → Y such thatF (a, t) = f0(a) = f1(a) for all a ∈ A.

(Think: e.g. strong deformation retraction r : X → A such that idX ∼ i r and homotopy fixesA pointwise.)

4


5.1. The Functor π1 : Top∗ → Groups. Preliminary definitions:

Definition 5.2. (1) A path in X is a map γ : I → X.(2) Its endpoints are γ(0) and γ(1).(3) If γ′ is a path in X such that γ′(o) = γ(1), then γ and γ′ may be concatenated by

(γ · γ′)(s) =

γ(2s) 0 ≤ s ≤ 1

2

γ′(2s− 1) 12 ≤ s ≤ 1

(4) A loop based at x0 ∈ X is a path such that γ(0) = γ(1) = x0.

Remark 5.3. Any two loops based at x0 may be concatenated (“composed”).

Definition 5.4. Let Ω(X,x0) be the set of loops in X based at x0. (This is really a space, calledthe based loop space of X.

Let · : Ω(X,x0)× Ω(X,x0)→ Ω(X,x0) be given by concatenation: (γ, γ′) = γ · γ′.Question 5.5. What kind of operation is ·? (What properties does it have?)

Associativity? Is (γ · γ′) · γ′′ = γ · (γ′ · γ′′)?No, but there is a natural homotopy between these two. See Figure 1. Therefore, · is associative

up to homotopy.

γ′γ γ′′

γ′γ γ′′

Figure 1. Homotopy from (γ · γ′) · γ′′ to γ · (γ′ · γ′′)

Remark 5.6. There are many homotopies between (γ · γ′) · γ′′ and γ · (γ′ · γ′′). In fact: anytwo homotopies are homotopic; we can stack layers of homotopies. This leads to the study of A∞algebras.

Definition 5.7. Let π1(X,x0) = Ω(X,x0)/ ∼, where ∼ is equivalence up to homotopy del x0. Inother words homotopy classes of loops in X based at x0.

For [γ] and [γ′] ∈ π1(X,x0), let [γ] · [γ′] = [γ · γ′].Exercise 5.8. Check that this is: 1) well defined on Ω(X,x0), and 2) associative.

Does this have identity?Candidate for identity in Ω(X,x0). Let e be the constant loop at x0 is γ · e = e · γ = γ?Yes.

5


γ γ−1

stay put

go forward move back

Figure 2. Homotopy from (γ · γ−1) the Constant Loop.

6. Friday, September 12, 2012

For the inverse element, for each γ, just take γ in the reverse orientation, i.e. γ−1(s) = γ(1− s).The homotopy is demonstrated in Figure 2.

Definition 6.1. (π1(X,x0), ·) is the fundamental group of X based at x0.

Example 6.2. (X,x0) = (Rn, 0). Let γ = Ω(Rn, 0). Let γ ∈ Ω(Rn, 0). Then γ is nullhomotopic.Proof: Let F : I × I → X. F (s, t) = (1− t)γ(s).

Therefore π1(Rn, 0) is the trivial group.

Question 6.3. Is the fundamental group dependent on the base point?

We assume that X is path-connected. Let h be a path in X from x1 to x0 (exists by assumption).Then (h · γ) · h−1 is a loop based at x1.

Proposition 6.4.Ω(X,x0) −→ Ω(X,x1). γ 7→ (h · γ) · h−1

induces a well-defined isomorphism π1(X,x0)βh→ π1(X,x1).

Therefore, π1(X) is well-defined up to isomorphism (though the isomorphism is not canonical).

Definition 6.5. X is simply-connected (1-connected) if

(0) X is connected.(1) π1(X) is trivial.

Example 6.6. Rn is simply connected for all n.

Remember that π1 is defined as a functor π1 : Top∗ → Group. We explained how to map theobjects, but how does π1 map the morphsims?

Given π1(X,x0) and a continuous map f : (X,x0) → (Y, y0), what is π1(f) = f∗ : π1(X,x0) →π1(Y, y0).

Notice:

(I, ∂I)γ→ (X,x0)

f→ (Y, y0),

we just take the composition as a loop in Y .

Definition 6.7. Ω(X,x0)Ωf→ Ω(Y, y0), sends γ 7→ f γ.

Proposition 6.8. (1) Ωf induces a well-defined map π1(X,x0)f∗→ π1(Y, y0).

6


(2) f∗ is a group homomorphism. (π1 takes morphisms to morphisms).

(3) (X,x0)f→ (Y, y0)

g→ (Z, z0)⇒ (g f)∗ = g∗ f∗.(4) (X,x0)

id→ (X,x0)⇒ id∗ = id : π1(X,x0)→ π1(X,x0).Therefore, π1 is a bona fide functor.

Proposition 6.9. If f0 : (X,x0) → (Y, y0) and f1 : (X,x0) → (Y, y0) are homotopic, then (f0)∗ :π1(X,x0) → π1(Y, y0) and (f1)∗ : π1(X,x0) → π1(Y, y0) are equal. (There exists an analogousstatement, even if f0(x0) 6= f1(x0)).

Sketch of proof: if γ ∈ Ω(X,x0) then (Ωf0)(γ) ∼ (Ωf1)(γ).

7. Monday, September 10, 2012

Remark 7.1. Recall that π1 : Top∗ → Group is a functor taking continuous maps to grouphomomorphisms.

(X,x0)

f0**

f1

44↓ (Y, y0)

g

(Z, z0)

−→π1(X,x0)

(f0)∗=(f1)∗// π1(Y, y0)

g∗

π1(Z, z0)

Specifically, it takes pointed spaces to groups, maps of pointed spaces to group homomorphism,homotopic maps of spaces to the same group homomorphism.

Remark 7.2. Recall also that Rn is simply connected (connected and π1(Rn) is trivial.)

Proposition 7.3. If f : X → Y is a homotopy equivalence, then f∗ : π1(X,x0) → π1(Y, f(x0)) isan isomorphism.

Proof. Let g : Y → X be a homotopy inverse for f . f g ∼ idY and g f ∼ idX . Usually thismeans that g∗ f∗ = “id” (where we need to be careful about basepoints).

π1(X,x0)

id++

f∗// π1(Y, y0)

g∗// π1(X, g(f(x0)))

π1(X,x0)

βh∼=

OO

⇒ g∗ f∗ is an isomorphism⇒ f∗ is injective. ⇒ f∗ g∗ is an isomorphism⇒ f∗ is surjective.

Corollary 7.4. If i : A → X is a deformation retract, x0 ∈ A. Then π1(X,x0) ∼= π1(A, x0).

Proposition 7.5. If r : X → A is any retraction (not necessarily deformation retraction), theni∗ : π1(A, x0)→ π1(X,x0) is injective.

The proof uses the fact that r i = idA...

Proposition 7.6. X path-connected. Then, π1(X) is trivial if and only if ∀x0, x1 ∈ X any twopaths from x0 to x1 are homotopic.

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Theorem 7.7. π1(S1, 1) ∼= Z. Specifically, if we define

ωn : I → S1, by s 7→ (cos 2πns, sin 2πns),

thenΦ : Z→ π1(S1, 1) n 7→ [ωn]

is an isomorphism.

Corollary 7.8 (Brouwer’s Fixed Point Theorem for n = 2). Any continuous map f : D2 → D2

has a fixed point.

Proof. Assume f(x) 6= x for all x ∈ D2, then x, f(x) determine a line for all x ∈ D2. Let r : D2 → S1

map x to the point where the ray from f(x) to x hits the circle.Check: r is continuous, fixes S1.So, r is a retraction ⇒ i : S1 → D2 induces i∗ : π1(S1, 1) → π1(D2, 1). But Z cannot inject into

the trivial group, contradiction! Therefore, there must be a fixed point where r is not defined.

Ingredient for the proof of Theorem 7.7:

Definition 7.9. The map (π : X → X) has the homotopy lifting property if given

(1) F : Y × I → X

(2) a lift f of f = F (· · · , 0).

there exists a unique lift F of F inducing f .

X

π

Yf//

f??

X

⇒X

π

Y × I F //

∃!F<<

X

Example 7.10. (1) Y = pt, then (π : X → X) has the path lifting property: given any

path γ ∈ X and a lift γ(0) of γ(0), then there exists a unique lift of the whole path γ to Xagreeing with γ(0)

(2) Y = I, then (π : X → X) has the “homotopy of paths” lifting property: given a homotopy

of paths in X and a lift of one end to X, then there exists a unique lift f the whole homotopyto X.

The map we use is π : X → X where x is R3 and we project onto the plane. If we map theinterval by s 7→ (cos 2πns, sin 2πns, s), then we have the homotopy lifting property, and we knowsome things about this space that we can use.


I arrived late to class. Prof. Poirier is proving the isomorphism of fundamental groups assertedin the last class. She went in three steps: homomorphism, surjectivity, and injectivity, using paths“upstairs” in the preimage spiral, and paths “downstairs” on the circle.

Theorem 8.1. Let X,Y be connected spaces, and let the product have projections as shown. Then,

π1(X × Y, (x0, y0)) −→ π1(X,x0)× π1(Y, y0), [γ] 7→ ((p1)∗[γ], (p2)∗[γ])

is an isomorphism.

Exercise 8.2. γ is completely determined by p1(γ), p2(γ).

8


X × Y p2//

p1

Y

Y

Example 8.3. The n-torus,

Tn := S1 × · · · × S1

n times⇒ π1(Tn) ∼= Z⊕ · · · ⊕ Z.

The homotopy class of the curve for T 2 is completely determined by two integers.


[Absent for Monday, September 17 due to Rosh Hashanah]

Theorem 9.1 (Van Kampen Theorem). X = U ∪ V , such that U, V are open, path connected,U ∩ V path connected, and x0 ∈ U ∩ V . There are canonical maps

U ∩ Vj

##

i

U

i′ ##

V

j′

U ∪ V

(1) π1(X) is the pushout of:

π1(U ∩ V )j∗

%%

i∗

yy

π1(U) π1(V )

(2) π1(X) ∼= [π1(U) ∗ π1(V )]/N , where N is the normal subgroup generated by i∗[γ] · (j∗[γ])−1,for γ ∈ π(U ∩ V ).

Example 9.2. Take X = RP 2. Let

U = RP 2 − pt ∼= S1.

V = nbhd of pt ∼= pt .

U ∩ V = ( nbhd of pt )− pt ∼= S1.

π1(U) ∼= Z, so take a generator a. Then the map of spaces induces the following map of fundamentalgroups:

π1(U ∩ V ) ∼= Z

((

c 7→2a

vv

π1(U) ∼= Z π1(V ) ∼= 1

So the pushout will satisfy π1(X) ∼= [〈a〉 ∗ 1]/〈2a · 1〉 ∼= Z/2Z.

9


π1(U ∩ V )j∗

&&

i∗

xx

π1(U)

%%

i′∗

π1(V )

yy

j′∗

π1(U)∗π1(V )N

∃!Φ

π1(X)

Idea of Proof. Want to show that Φ is an isomorphism, in the following diagram: Method:

• Define a map Φ : π1(U) ∗ π1(V )→ π1(X) a surjective homomorphism.• Show that ker Φ = N .• By the first isomorphism theorem Φ induces the desired isomorphism.• π1(U) ∗ π1(V ) is generated by [γ]u and [γ]v, i.e. γ is a loop in U and [γ]u is its homotopy

class in U (V resp).• Define Φ on generators by Φ([γ]u) = [γ]X , resp. V . Then extend Φ as a homomorphism.• This Φ is surjective: Take any γ ∈ X, look at where it crosses back and forth between U

and V . Define base points in each transition section. Then these will be base points for asequence of loops entirely in U or entirely in V .• N ≤ ker Φ is easy. i∗[γ]U∩V = [γ]U , and j∗[γ]U∩V = [γ]V . Then Φ(i∗[γ](j∗[γ])−1) =

[γ]X([γ]X)−1 = 1 ∈ π1(X).• Ker Φ ≤ N is trickier.

10. Friday, September 21, 2012

Proof Continued. To show that the ker Φ = N , we have the following idea:

• We have N < ker Φ < π1(U) ∗ π1(V ).• Consider the cosets of N and kerΦ.• The coset Φ−1([γ]) = all possible factorizations of [γ].

Strategy:

(1) Define equivalence relation ∼ on Φ−1([γ]) so that if two factorizations are equivalent, theylie in the same N coset.

(2) Show that any two factors are equivalent.

Definition 10.1. Let the following generate ∼ on Φ−1([γ])

(1) [γ1][γ2] · · · [γi][γi+1] · · · [γN ] ∼ [γ1][γ2] · · · [γi · γi+1] · · · [γN ] where [γi] and [γi+1] are both inU or V .

(2) [γ1][γ2] · · · [γi]U · · · [γN ] ∼ [γ1][γ2] · · · [γi]V · · · [γN ] where [γi] and [γi+1] are both in U or V .

11. Monday, September 24, 2012

11.1. Covering Spaces. Recall that in proving the fundamental group of S1, we had a map fromR to S1 via P .

10


R

S1

P

Figure 3. Examples of Covering Spaces

We used covering spaces and covering maps to compute π1(S1, 1) and π1(S1 ∨ S1, ∗) in anotherexample.

Definition 11.1. Let X be a topological space. A covering space for X is a space X, together witha covering map p : X → X such that ∀x ∈ X there is a neighborhood Ux of x such that p−1(Ux) ishomeomorphic to Ux × S, where S is some discrete set.

In other words, there is a finite set of pre-image neighborhoods. (If S = F , some topologicalspace, see “fiber bundle.”)

S1

S1S1

tS1

Figure 4. Examples (1) and (2) of covers of the circle.

In Example (2), the map sends z to zn, where n is the number of pre-image neighborhoods. InExample (3), the map sends z on the boundary of the Mobius strip to z2.

Exercise 11.2. If p : X → X and p′ : X ′ → X ′ are two covering spaces then so is p× p′ : X × X ′.Corollary 11.3. Since R is a covering space of S1, R2 = R× R is also a cover of S1 × S1 = T 2.

Recall:

Definition 11.4. The map p : X → X satisfies the homotopy lifting property if for all Y , and forall maps F : Y × I → X and for all lifts f0 : Y → X of f0 := F (•, 0) : Y → X, there exists a unique

lift F : Y × I → X of F that completes the diagram in Figure 6.

Recall:

(1) If Y = pt. This is called the path lifting property.11


Figure 5. Examples (3) and (4) of covering spaces.

Y × 0 f0 //

X

p

Y × IF

//

F

;;

X

Figure 6. Homotopy Lifting Property

(2) If Y = I, we called this the “homotopy of paths” – lifting property.

We used these two to prove that π1(S1, 1) ∼= Z.This was okay because the first map in Figure 3 is a covering space.

Proposition 11.5. Covering spaces and covering maps satisfy the homotopy lifting property.

We will prove that covering spaces satisfy the path lifting property in Figure 7.

0 γ(0)//

X

p

Y × I γ//

γ<<

X

Figure 7. Path Lifting Property

Intuition: the neighborhood of our base point has multiple pre-images. We choose one of them:by definition, the neighborhood we have chosen is homeomorphic to the neighborhood in the image,so we can “invert” the map into this neighborhood, and get the pre-image of the path in the imagespace.

We observe that⋃x∈Im γ γ

−1(Ux) covers I.I is compact, therefore there is a finite subcover to any cover. So there is a partition of I into

∪Ni=1Ii such that γ(I1) lies in one of the Ux’s.Consider the component Vγ(0) of p−1(Uγ(0)) containing γ(0).

p|Vγ(0) : Vγ(0) → Uγ(0).

Lift γ|I1 by γ|I1 = p−1 γ|I1 .Proceed analogously.

12


Exercise 11.6. (1) Check that the definition of γ doesn’t change if you choose another finitesub cover of I and partition.

(2) The “same” method works to prove the general homotopy lifting property for coveringspaces.

12. Friday, October 5, 2012

[Three classes missed, for seminar and the holiday of Sukkot.]Today, we are finishing up covering spaces, and discussing deck transformations (see Hatcher).

Example 12.1. Consider the covering space R × R → S1 × S1 = T 2. The fundamental groupπ1(T 2, x0) = Z⊕ Z = 〈a, b|[a, b]〉, where a and b are the generating cycles.

Specifically π1(T 2, x0) acts on R2 by a, b, which send x 7→ x+ma+ nb. In words, generator [a]of π1(T 2, x0) acts by translation by a.

isomorphism classes of covering spaces ⇔ subgroups of π1(X,x0)

Definition 12.2. The (isomorphism class of) simply connected cover(s) is called the “universalcover of X.”

A simply connected cover is unique up to isomorphism and the path-space construction of Xgives us a model for the universal cover.

12.1. Homology Cartoon. We want to generalize π1 to higher dimensions.

Example 12.3 (Higher Homotopy Groups). Consider In = [0, 1]n. Let πn(X,x0) = set of homo-topy class of (In, ∂In)→ (X,x0).

n = 0. I0 = pt , and ∂I0 = ∅. Therefore, π0(X,x0) = set of path components of X.

n ≥ 1. We introduce a group operation (the analogue of concatenation of loops):

(f + g)(s1, . . . , sn) =

f(2s1, s2, . . . , sn) s1 ∈ [0, 1/2]

g(2s− 1, s2, . . . , sn) s1 ∈ [1/2, 1].

This is well-defined on homotopy classes.

X

f

Figure 8. Higher Homotopy Concatenation.

In fact, πn(X,x0) for n ≥ 2 is abelian. See Figure 9 for a visual proof.

Theorem 12.4 (Whitehead’s Theorem). Let X,Y be connected CW complexes. Let f : X → Ybe a continuous map such that f∗ : πn(X)→ πn(Y ) is an isomorphism for all n. (i.e. f is a weakhomotopy), then f is a homotopy equivalence.

13


ff f

fgg g

gconst.

const. const.

const.∼= ∼= ∼=

Figure 9. Visual Proof that Higher Homotopy Groups are Abelian.

Problem: πn are notoriously hard to compute, even for simple spaces. (Covered in Math 215B).

Another idea: (Co)Bordism. For πn, Sn is a manifold. Homotopy of Sn → X is Sn × I → X.

Sn × I is an n+ 1-dimensional manifold with boundary.We generalize by replacing Sn withMn an n-dimensional manifold, andWn+1 is n+1-dimensional

manifold with boundary.Two n-dimensional manifolds M1 and M2 are (co)bordant if ∃Wn+1 such that

∂Wn+1 = Mn1 tMn

2 .

Example 12.5. The pair of pants is the cobordism of two copies of S1 and one copy of S1. Ahemisphere is a cobordism of S1 and ∅.

13. Monday, October 15, 2012

Definition 13.1. A homology theory is a (covariant) functor

(pairs of topological spaces A ⊂ X) → (graded Abelian groups)

satisfying four “Eilenberg-Steenrod” axioms.

(0) (Dimension)

H∗(point) =

“coefficients” ∈ Z ∗ = 0.

0 ∗ 6= 0.

(1) (Homotopy) Let f, g be homotopic maps from (X,A) → (Y,B). Then, H∗(f) = H∗(g) asmaps from H∗(X,A)→ H∗(Y,B).

(2) (Exactness) There exists a map δ such that the inclusion i : A → X induces a long exactsequence:

Hi+1(A) Hi+1(X) Hi+1(X,A)

Hi(A) Hi(X) Hi(X,A)

· · ·

δ

δ

(3) (Excision) If Z ⊂ A ⊂ X and Z ⊂ int(A), then i : (X − Z,A − Z) → (X,A) induces an

isomorphism H∗(i) : H∗(X − Z,A− Z)∼→ H∗(X,A).

Theorem 13.2 (Eilenberg-Steenrod). There is only one homology theory.

Definition 13.3. An exotic homology theory is a functor satisfying (1),(2),(3).

Definition 13.4. A functor is covariant if it preserves the directions of arrows. A functor iscontravariant if it reverses directions of arrows.

Definition 13.5. A contravariant functor satisfying versions of (0), (1), (2), and (3) is called acohomology theory.

14


13.1. Simplicial Homology of Simplicial Complexes.

Definition 13.6. An n-dimensional simplex [v0, . . . , vn] is the convex hull of n+ 1 ordered pointsv0, . . . , vn in Rm that do not lie in an affine subspace of dim < n.

Definition 13.7. The standard n-simplex ∆n is the convex hull of the endpoints of the standardbasis vectors in Rn+1:

(1, 0, . . . , 0), . . . , (0, . . . , 0, 1).

In “barycentric coordinates”,

∆n = (t0, . . . , tn) :∑

ti = 1, 0 ≤ ti ≤ 1.

Example 13.8. ∆0 ⊂ R1 is a point.∆1 ⊂ R2 is a line segment.∆2 ⊂ R3 is a triangle.

Remark 13.9. There exists a canonical linear homeomorphism from any n-simplex to ∆n.

Remark 13.10. Sometimes we only care about the combinatorial structure of an n-simplex, i.e.the vertices. Other times we’ll care about the underlying space.

Definition 13.11. A face of [v0, . . . , vn] is the sub simplex given by any (nonempty) (ordered)subset of the vi.

Remark 13.12. Sometimes empty is okay. Hatcher only uses the n− 1 dimensional faces.

Example 13.13. Consider the 3-simplex. [123] is a 2-dimensional face of the 3-simplex [0123].[23] is a 1-dimensional face of [0123].

14. Wednesday, October 17, 2012

14.1. Simplicial Complexes. Roughly, a simplicial complex is a space formed from a collectionof simplifies identified along faces.

Definition 14.1. A simplicial complex K is a collection of simplifies in Rn such that

(1) K is closed under taking faces.(2) Simplices intersect only in faces.

Remark 14.2. A space may have a number of simplicial decompositions. For example, ∆2 can bedivided using barycentric subdivision.

Figure 10. Barycentric Subdivision of the 2-simplex.

Remark 14.3. There is a purely combinatorial definition - the abstract simplicial complex:A (nonempty) collection of subsets of 1, . . . , n which is closed under taking subsets.

15


Remark 14.4. In a simplicial complex (as defined), a simplex is completely determined by itsvertices.

Hatcher uses a more general notion: ∆-complexes where faces of simplices may be identified.

Example 14.5. Spaces which are ∆-complexes, but not simplicial complexes:

• A vertex with a loop.• A triangle with two edges identified, i.e. a cone.• A pair of triangles with the same set of vertices, embedded in a cone.

14.2. Simplicial Homology of a Simplicial Complex. Let K be a ∆-complex.

Definition 14.6. The ∆-chain complex of K,

C∆∗ (K), ∂)

is given by:

• C∆n (K) is the free abelian group generated by the n-simplices of K.

• ∂ : C∆n (K)→ C∆

n−1(K) is defined on generators [v0, . . . , vn] by

∂n[v0, . . . , vn] =

n∑i=0

(−1)i[v0, v1, . . . , vi, . . . , vn].

i.e. ∂ is given by taking a signed sum of codimension 1 faces.

Lemma 14.7.(∂n−1 ∂n) = 0.

Proof.

(∂n−1 ∂n)[v0, . . . , vn] = ∂n−1

(n∑i=0

(−1)i[v0, v1, . . . , vi, . . . , vn]

).

=

n∑i=0

(−1)i∂n−1 ([v0, v1, . . . , vi, . . . , vn])

=∑j<i

(−1)j(−1)i[v0, v1, . . . , vj , . . . , vi, . . . , vn]

.


Recall:

• Simplicial chains of a ∆-complex K: (C∆∗ (K), ∂).

• C∆n (K) = the free abelian group generated by the n-simplices of K.

• ∂n : C∆n → C∆

n−1(K), acting like ∂n[v0, . . . , vn] =∑

(−1)i[v0, . . . , vi, . . . , vn].

Recall: “Definition”: A ∆-complex K is a space |K| together with maps σα : ∆n → |K| suchthat

(1) σα|int(∆n) is injective and each point of |K| is in exactly one “open simplex.”

(2) σα|face of ∆n is one of the σβ.

(3) A ⊂ |K| is open if and only if σ−1α (A) is open in ∆n∀α.

In this setting, the σα : ∆n → |K| generate C∆n (K) and

∂n(σnα) =∑

(−1)iσα|[v0,...,vi,...,vn].

16


Definition 15.1.

H∆∗ (K) := H∗(C

∆∗ (K), ∂)

is the simplicial homology of K.

Example 15.2. Let K = pt.

C∗(pt) = · · · → 0∂1→ Z ∂0→ 0.

All ∂i are 0. H∆0 (pt) = ker ∂0/Im ∂1 = Z/0 = Z.

So H∆∗ (pt) =

Z ∗ = 0

0 ∗ > 0.

Example 15.3. Let K = the 2-simplex. The faces are:

0 [v0], [v1], [v2]1 [v0, v1], [v0, v2], [v1, v2]2 [v0, v1, v2]

n(n > 2) ∅

We know already that C∆n (K) = 0, for n > 2⇒ H∆

n (K) = 0, for n > 2.

∂2[v0, v1, v2] = [v1, v2]− [v0, v2] + [v0, v1] 6= 0.⇒ ∂2 =

1− 11

∂1[v0, v1] = [v1]− [v0]. ∂1[v0, v2] = [v2]− [v0]. ∂1[v1, v2] = [v2]− [v1].

⇒ ∂1 =

− 1 −1 01 0 − 10 1 1

Now to compute homology groups:

H∆2 (K) = ker ∂2/Im ∂3 = 0/0 = 0.

H∆1 (K) = ker ∂1/Im ∂2 = 〈[v1, v2]− [v0, v2] + [v0, v1]〉/〈[v1, v2]− [v0, v2] + [v0, v1]〉 = 0.

H∆0 (K) = ker ∂0/Im ∂1 = 〈[v0], [v1], [v2]〉/〈[v1]− [v0], [v2]− [v0]〉

= 〈[v0], [v1]− [v0], [v2]− [v0]〉/〈[v1]− [v0], [v2]− [v0]〉 = 〈[v0]〉 ∼= Z.Therefore,

H∆∗ (K) =

Z ∗ = 0

0 ∗ 6= 0.

Example 15.4. Consider the torus as a ∆-complex as in Figure 11.

C∆∗ (T ) = · · · → 0

∂3→ ZA ⊕ ZB∂2→ Za ⊕ Zb ⊕ Zc

∂1→ Zv∂0→ 0.

∂2(A) = a− c+ b, ∂2(B) = b− c+ a⇒ ker ∂2 = 〈A−B〉, Im ∂2 = 〈a+ b− c〉.

H∆∗ (T ) =

Z ∗ = 0

Z⊕ Z ∗ = 1

Z ∗ = 2

0 ∗ > 2, ∗ < 0

.

17


a

a

b bc

v

v v

v

A

B

Figure 11. Delta Complex for the Torus.


Recall: Simplicial homology of a ∆-complex–

• H∆∗ (K) := H∗(C

∆∗ (K), ∂).

• H∆∗ (pt) =

Z ∗ = 0

0 ∗ 6= 0.

• H∆∗ (T ) =

Z ∗ = 0

Z⊕ Z ∗ = 1

Z ∗ = 2

0 ∗ > 2, ∗ < 0.

.

Remark 16.1. Note that H1 is the abelianization of π1. (The homology groups have to be abelian.)

Question 16.2. (1) Does H∆∗ (K) depend on the ∆-decomposition of |K|?

(2) What about homology for a general topological space X?

16.1. Singular Homology of a Space X. Program:

(1) Build a complex S(X) from X, called the “singular complex.”(2) Define C∗(X) := C∆

∗ (S(X)), the “simplicial chain complex.”(3) Define H∗(X) := H∗(C∗(X)) = H∆

∗ (S(X)), “singular homology.”

Definition 16.3. S(X) has one n-simplex for each continuous map

σ : ∆n → X (“singular simplex of X).

The (n− 1)-faces of σ are given by

σ|[v0,...,vi,...,vn] : ∆n−1 → X.

Equivalently, we can start by saying:

Definition 16.4. C∗(X) is generated by all continuous σn : ∆n → X. ∂ is defined on the generatorsby

∂(σn) =

n∑i=0

(−1)iσn|[v1,...,vi,...,vn].

Example 16.5. Let X be a point. There is one continuous map σn : ∆n → X for each n.Therefore, S(X) has one simplex in each dimension.

⇒ C∗(X) = · · · → Z→ Z→ Z→ 0→ · · · .18


What is the differential?

∂(σn) =∑

(−1)iσn|[...,vi,...] =∑

(−1)iσn−1 =

0 n odd

σn−1 n even > 0

0 n = 0

.

ker ∂n =

Z n odd

0 n even > 0

Z n = 0

.

Im ∂n+1 =

Z n odd

0 n even > 0

0 n = 0

.

⇒ Hn(pt) =

Z/Z = 0 n odd

0/0 = 0 n even > 0

Z/0 = Z n = 0

.

16.2. Relative Homology / Chains.

Definition 16.6. If f : X → Y is a continuous map, then we define f# : C∗(X) → C∗(Y ) with(σ : ∆n → X) 7→ (f σ : ∆n → Y ).

Exercise 16.7. Check that f# is a chain map, i.e. it commutes with ∂.

Denote the map induced by f# on the homology groups by

f∗ : H∗(X)→ H∗(Y ).

Definition 16.8. Let i : A → X be a subspace. Then i# : C∗(A) → C∗(X) is an injection. Denotethe quotient C∗(X)/C∗(A) by C∗(X,A).


Recall:

H∗ : Pairs of Top. Spaces (X,A), A ⊂ X. → graded abelian groups.

satisfying

(0) (Dimension)

H∗(point) =

“coefficients” ∈ Z ∗ = 0.

0 ∗ 6= 0.

(1) (Homotopy) Let f, g be homotopic maps from (X,A) → (Y,B). Then, H∗(f) = H∗(g) asmaps from H∗(X,A)→ H∗(Y,B).

(2) (Exactness) There exists a map δ such that the inclusion i : A → X induces a long exactsequence:

Hi+1(A) Hi+1(X) Hi+1(X,A)

Hi(A) Hi(X) Hi(X,A)

· · ·

δ

δ

19


(3) Excision Theorem/ Axiom:Let Z ⊂ A ⊂ X be a subspace such that Z ⊂ int(A). Then,

(X − Z,A− Z)i→ (X,A)

induces an isomorphism:

H∗(X − Z,A− Z)i∗→ H∗(X,A)

Proof. The proof is in two steps:

Definition 17.1. Let U be a collection of sets whose interiors cover X.Let CUn (X) ⊂ Cn(X) be generated by σ : ∆n → Ui, for Ui ∈ U . Call those U-small chains.

Easy Claim 1:

∂ : CUn (X)→ CUn−1(X).

so (CU∗ (X), ∂) is a chain complex with homology HU∗ (X).Easy Claim 2:

CU∗ (X)I→ C∗(X).

is a chain map by inclusion σ 7→ σ.

Proposition 17.2.

CU∗ (X)I→ C∗(X)

is a quasi-isomorphism, i.e. it induces an isomorphism

I∗ : HU∗ (X)∼→ H∗(X).

A picture of the previous proof is given in Figure 12. This allows us to get pieces of a subspace

Figure 12. Iterated Barycentric Subdivision.

small enough so that each piece fits in an open set in the cover.Why does the proposition prove the theorem?Let U = A,X−Z. Write: CU∗ (X) = C∗(A)+C∗(X−Z), i.e. the subgroup of C∗(X) generated

by C∗(A) and C∗(X − Z).

Definition 17.3. CU∗ (X,A) is defined as CU∗ (X)/CU∗ (A).

Consider the maps

Cn(X − Z,A− Z) //

((

Cn(X,A)

CUn (X,A)

88

20


The left map is induced by inclusion. The right map is quasi-isomorphism by the proposition.We want to show that the top map is a quasi-isomorphism.

Cn(X − Z,A− Z) = Cn(X − Z)/Cn(A− Z) = Cn(X − Z)/(Cn(X − Z) ∩ Cn(A).

CUn (X,A) := CUn (X)/CUn (A) = (Cn(A) + Cn(X − Z))/Cn(A).

is an isomorphism for all n by the second isomorphism theorem for groups. This forces the topmap in terms of the homology to also be an isomorphism.


We are still discussing the proof of the excision theorem for singular homology.

Remark 18.1. Suppose instead we were talking about simplicial homology, with X an A-complex,and A,Z subcomplexes of X. Consider C∆

∗ (X − Z,A − Z) and C∆∗ (X,A). These are given by

simplicial complexes that are already broken up into pieces.With the singular picture on the other hand, we have to define these subcomplexes by hand.

We use barycentric subdivision as pictured above. The barycenter of ∆n ⊆ Rn+1 can be definedas the point ( 1

n+1 , . . . ,1

n+1).Our goal right now is to build a chain homotopy inverse ρ to I.

CU∗ (X) ρI C∗(X).

Point of Barycentric Subdivision: n simplices in barycentric subdivisions are strictly smallerthan [v0, . . . , vn]. Iterate barycentric subdivision until the diameter of an n-simplex approaches 0.

Idea: Subdivision map:

S : C∗(X) // C∗(X)

(σ : ∆n → X) // (Subdivide ∆ and take sum of restrictions.)

This is a chain map, is chain homotopic to

id : C∗(X)→ C∗(X),

via chain homotopy

T : C∗(X)→ C∗+1(X).

Therefore, S induces an isomorphism H∗(X)→ H∗(X).Given σ : ∆n → X, there is m(σ) ∈ Z≥0 such that

Sm(σ)(σ) ∈ CUn (X).

We would like to define

ρ : Cn(X)→ CUn (X)

by ρ(σ) = Sm(σ)(σ).Problem: This might not be a chain map. For example, let τ be a face of σ and m(τ) < m(σ)

(we always have m(τ) ≤ m(σ)).For example, see Figure 13The solution to this problem is to add a correction term. (Capturing the failure of ρ to be a

chain map.) This is illustrated in Figure 14.Notation:

Dm =∑

0≤i≤mTsi : C∗(X)→ C∗+1(X)

is a chain homotopy between Sm and id.I missed some details of this proof.

21


τ

σX

Figure 13. Failure to be a Chain Map.

Figure 14. Correction to the Chain Map.

19. Friday, November 2, 2012

[Arrived late to class.]For CW complexes X, we want a simpler way of computing homology – cellular homology. We

will

(1) define HCW∗ (X), and

(2) show HCW∗ (X) ∼= H∗(X).

The definition uses a lot of things.

(1) First,

H∗(Sn) =

Z ∗ = 0, n

0 else.

H0(X) is generated by the path components of X. We will use LES of (Dn, ∂Dn) byinduction on n. For n = 1,

C∆∗ (S1) = (0→ Z 0→ Z→ 0.

⇒ H∗(S1) =

Z ∗ = 0, 1

0 else.

For n > 1 we have the exact sequence:

→ Hk+1(Dn)→ Hk+1(Dn, ∂Dn)→ Hk(∂Dn)→ · · ·

which is equivalent to:

0→ Hk+1(Sn)∼→ Hk(S

n−1)→ 0→ Hk(Sn)→ · · · .

22


Therefore, by induction and the argument for k = 1, we have Hk(Sn) ∼= Hk−1(Sn−1) by

induction and argument for k = 1.(2) Let X = ∨αXα. If xα ∈ Xα corresponding to the wedge point gives a good pair (Xα, xα),

then

H∗(X) ∼=⊕α

H∗(Xα).

Proof. Idea: (tαXα, xα) is a good pair and X = tαXα/ tα xα. We use the idea that : ifX = tαXα, then

C∗(X) =⊕α

C∗(Xα)⇒ H∗(X) =⊕

H∗(Xα).

H∗(∧Sn) =

Z# copies of Sn ∗ = n

Z ∗ = 0

0 else

(3) Let X be a CW complex with Xn its n-skeleton, then (Xn, Xn−1) is a good pair and

Xn/Xn−1 ∼= ∨# of n-cellsSn.

(4) H∗(Xn) = 0 if ∗ > n. Proof uses LES of (Xn, Xn−1) and 1.

(5) Xn → X induces an isomorphism Hk(Xn)∼→ Hk(X) if k < n.

Definition 19.1 (Cellular Chain Complex). Let X be a CW complex.

CCWn (X) := Hn(Xn, Xn−1).

(i.e. generated by the n-cells of X.)The differential dn : CCWn (X)→ CCWn−1(X), i.e. Hn(Xn, Xn−1)→ HCW

n−1(Xn−1, Xn−2).

20. Monday, November 4, 2012

Theorem 20.1.

HCW∗ (X) ∼= H∗(X).

Proof. Recall: H∗(Xn) = 0 if ∗ > n.

Xn+1/X induces an isomorphism H∗(Xn+1)→ H∗(X) if ∗ < n+ 1.

This allows us to construct the diagram:

0 = Hn(Xn+1)

**

Hn(Xn+1) ∼= Hn(X)jn// Hn(Xn+1, Xn) = 0

Hn(Xn)

in 44

jn**

Hn+1(Xn+1, Xn)dn+1

// Hn(Xn, Xn+1)

δn++

dn // Hn−1(Xn−1, Xn−2)

Hn−1(Xn−1)jn−1

33

0

22

Example 20.2. Let X = CPn be a cell structure with cells in dimension 0, 2, 4, . . . , 2n. Inductively,

• n = 1. CP1 ∼ S2.• n > 1. CPn ∼ CPn−1 ∪ e2n.

23


⇒ CCW∗ (CPn) =

(· · · → 0→

2nZ →

2n−10 →

2n−2Z → 0→ · · · →

10→

0Z→ 0→ · · ·

)This means dn = 0 for all n.

⇒ H∗(CPn) =

Z ∗ = 2k, k = 0, . . . , n.

0 otherwise..

Remark 20.3. In general, dn is hard to work with as defined. We want a more geometric definition,one that uses the degree of a map from Sn → Sn.

Definition 20.4. Given a continuous map f : Sn → Sn, f induces

f∗ : Hn(Sn)→ Hn(Sn),

i.e. a homomorphism Z ·d→ Z, sending α 7→ dα. d is called the degree of f .

Example 20.5. The degree of id is 1, since the induced homomorphism of groups is the identity.If f is not surjective then f factors through Dn → Sn (the 1-point compactification of Dn). This

implies that f is nullhomotopic, which means that the degree f .

Definition 20.6. Cellular Boundary Formula.

dn : Hn(Xn, Xn−1) −→ Hn−1(Xn−1, Xn−2)(generated by n-cells) (generated by (n− 1)-cells)

is defined by

dn(enα) =∑β

dαβen−1β

where dαβ is the degree of the map

Snϕα→ Xn−1 → Xn−1/(Xn−1 − en−1

β ) ∼= Sn−1.

21. Wednesday, November 7, 2012

Example 21.1. Let X = RPn. X has a cell structure with a cell in each dimension up to n.

RPn ∼= Sn/ antipodal map.

So, we use the cell structure on Sn with 2 cells in each dimension up to n. After quotienting, RPninherits the cell structure.

⇒ CCW∗ (RPn) =

(· · · → 0→

nZ→

n−1Z →

n−2Z → Z→ · · · →

1Z→

0Z→ 0→ · · ·

)Sk−1 ϕα→ Xk−1 → Xk−1/(Xk−1 − ek−1

β ) ∼= Sn.

ϕα is a two-sheeted cover.

Exercise 21.2. Show: q ϕα has degree

2 if k is even

0 if k is odd. Read about local degree.

⇒ CCW∗ (RPn) =

0→nZ ×2→

n−1Z ×0→

n−2Z ×2→ Z ×0→ · · · ×2→

1Z ×0→

0Z→ 0→ · · · n even

0→nZ ×0→

n−1Z ×2→

n−2Z ×0→ Z ×2→ · · · ×2→

1Z ×0→

0Z→ 0→ · · · n odd

24


Now we can compute the homology:

⇒ HCW∗ (RPn) =

Z ∗ = 0

Z ∗ = n odd

Z/2Z ∗odd, ∗ < n

0 otherwise

.

21.1. Mayer-Vietoris Sequence. (Homology analogue of Van Kampen’s Theorem for π1)The Van Kampen’s Theorem told us that if we have a decomposition for a space, we can compute

the fundamental group in terms of the groups of each space.

(1) Let A,B ⊂ X be subsets such that int(A) ∪ int(B) = X.Let C∗(A+B) = CU∗ (X) where U = A,B.[We saw that CU∗ (X) → C∗(X) is a quasi-isomorphism.]

(2) Inclusions A ∩BiA

iB

##

A

jA ##

B

jB

X

induce a short exact sequence (exercise) of chain com-

plexes:

0 −→ C∗(A ∩B)→ C∗(A)⊕ C∗(B)→ C∗(A+B)→ 0.x 7→ (iA∗(x),−iB∗(x)) (x, y) 7→ jA∗(x) + jB∗(y)

(3) The short exact sequence of chain complexes induces a long exact sequence of homology

Hn(A ∩B) Hn(A)⊕Hn(B) Hn(X)

Hn−1(A ∩B) Hn−1(A)⊕Hn−1(B) Hn−1(X)

· · ·

ϕ∗ ψ∗

δ

δ

and we have a similar statement for reduced / relative homology. This is called theMayer-Vietoris Sequence.

22. Wednesday, November 14, 2012

[Class missed on Friday 11/9 due to travel.]

22.1. Homology with Coefficients. Fix an abelian group G.

Definition 22.1. Singular chains with coefficients in G are written as:

Cn(X;G) =∑

giσi | σi : ∆n → X, gi ∈ G.

∂(∑

giσi

)=∑

gi(∂σi) =∑

gi(±σi|v0···vi···vn).

Remark 22.2.

(C∗(X;Z), ∂) = (C∗(X), ∂).

Again ∂2 = 0⇒ (C∗(X;G), ∂) is a chain complex with homology H∗(X;G).

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Definition 22.3 (Informal). Analogous definitions, for A ⊂ X,

C∗(X,A;G)→ H∗(X,A;G).

similar for C∆∗ (X;G) and CCW∗ (X;G), and the same proof shows

H∆∗ (X;G) ∼= H∗(X;G) ∼= HCW

∗ (X;G).

Remark 22.4.Cn(X;Z) ∼=

⊕singular n-simplices of X

Z.

Cn(X;G) ∼=⊕

G.

Cn(X;G) ∼= Cn(X)⊗G.Cn(X;G)

∂⊗id→ Cn−1(X;G).

We need not have Hn(X;G) ∼= Hn(X)⊗G! Universal Coefficient Theorem.

Sometimes calculations over G 6= Z reveal something new.

Example 22.5. Let X = RPn. Take G = F a field. Recall: CCW∗ (RPn) is:

· · · → 0→nZ ×0 n odd−→×2 n even

n−1Z → · · · ×2−→

1Z ×0−→

0Z→ 0.

which implies that:

H∗(RPn) =

Z ∗ = 0, n, n odd

Z/2Z 0 < ∗ < n odd

0 otherwise

Over F , CCW∗ (RPn;F ) is given by

· · · → 0→nF×0 n odd−→×2 n even

n−1F → · · · ×2−→

1F×0−→

0F → 0.

if char F = 2, then

H∗(RPn;F ) =

F 0 ≤ ∗ ≤ n0 otherwise

If char F 6= 2, then every ×2 is an isomorphism which means that

H∗(RPn;F ) =

F ∗ = 0, ∗ = n odd

0 otherwise

22.2. Cohomology.

Definition 22.6. A cochain complex (C∗, δ) is a sequence of abelian groups (R-modules/F -vectorspaces...) Ci together with δi+1 : Ci → Ci+1 such that ∂i+1 ∂i = 0.

· · · ← Ci+2 δ← Ci+1 δ← Ciδ← Ci−1 ← · · ·

Definition 22.7. The i-th cohomology group of a cochain complex (C∗, δ)

H i(C∗, δ) := ker δi+1/im δi.

Definition 22.8. Given a chain complex (C∗, ∂) and an abelian group G, one can build a cochaincomplex

Ci := Hom(Ci, G) = ϕ : Ci → G.δi : Hom(Ci, G)→ Hom(Ci+1, G), ϕ 7→ ϕ ∂.

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Proposition 22.9. Given a chain map

f# : C∗ → D∗,

there exists an induced cochain map

f# : D∗ → C∗.

Proposition 22.10. A cochain map f# : D∗ → C∗ induces a map on cohomology

f∗ : H∗(D∗)→ H∗(C∗).

23. Friday, November 16, 2012

Everything said last class about normal cohomology is true also for G-valued cohomology.

Definition 23.1. An ordinary cohomology theory is a contravariant functor.

(Pairs of topological spaces A ⊂ X, (X,A))H∗(−;G)

// (graded abelian groups)

satisfying the following four axioms:

(1) “dimension”.

H∗(pt;G) =

G ∗ = 0

0 ∗ 6= 0

(2) “homotopy”. If

(X,A)

f**

g 44(Y,B)

are homotopic, then

H∗(X,A;G) H∗(Y,B;G)f=g

oo

(3) “exactness”. Maps induce Long Exact Sequence

· · · Hn+1(X,A;G)oo

Hn(A;G)

δ11

Hn(X;G)oo Hn(X,A;G)oo

Hn−1(A;G)

δ11

· · ·oo

(4) “excision”. Given Z ⊂ A ⊂ X such that Z ⊂ int(A), then inclusion (X−Z,A−Z) → (X,A)induces an isomorphism

H∗(X − Z,A− Z;G) H∗(X,A;G)i∗∼oo

Definition 23.2. An extraordinary/exotic cohomology theory satisfies the last three axioms.

Theorem 23.3 (Eilenberg-Steenrod). Any two cohomology theories satisfying the four axioms areequivalent.

Theorem 23.4. If X has a ∆-structure or CW-structure, then

H∗(X;G) ∼= H∗∆(X;G) ∼= H∗CW (X;G).

Theorem 23.5. Singular cohomology satisfies the four axioms.

Theorem 23.6 (de Rham). If X is a smooth R-manifold, its differential forms form a cochaincomplex (δ = exterior derivative).

H∗dR(X;R) ∼= H∗(X;R).

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Remark 23.7. All “co-proofs” work. For example, M-V sequence for cohomology.

Remark 23.8. In general,Hn(X;G) 6∼= Hom(Hn(X);G),

but the universal coefficient theorem provides conditions when they are.

It looks like H∗(X;G) and H∗(X;G) carry the same info about X but if G = R a commutativering with 1, then ⊕

H i(X;G) = H∗(X;G)

can be given a ring structure.Plan:

(1) We will define Ck(X;R)× C l(X;R)∪→ Ck+l(X;R).

(2) We will prove an algebraic statement about when a product on a cohcain complex Ck×C l →Ck+l induces a product on the cohomology Hk ×H l → Hk+l.

(3) We will show (1) satisfies the property in (2).(4) We will see an example where this structure distinguishes two spaces with isomorphic ho-

mology groups.

24. Monday, November 19, 2012

We will begin with step (2) of the plan from last class. The algebraic statement will motivatethe definition of the cup product.

Question 24.1. Let (C∗, δ) be a cochain complex and let · : Ck × C l → Ck+l be an associativeproduct for all k, l (i.e. a map C∗ × C∗ → C∗).

What must (·, δ) satisfy to induce a well-defined product

Hk ×H l ·→ Hk+l ∀k, l,H∗ ×H∗ → H∗?

We want [ϕ] · [ψ] ∈ Hk+l for [ϕ] ∈ Hk, [ψ] ∈ H l for cocycles ϕ ∈ Ck, ψ ∈ C l.We want: ϕ · ψ cocyle to satisfy [ϕ] · [ψ] := [ϕ · ψ]. We need:

• product of cocycles = cocycle.• coboundary times cocycle = coboundary.• cocycle times coboundary = coboundary.

Lemma 24.2. If δ(ϕ · psi) = (δϕ) · ψ + (−1)kϕ · (δψ), for all ϕ,ψ ∈ C∗ where ϕ ∈ Ck, then thechain map induces a well-defined map on homology.

Proof. Let δϕ = δψ = 0. Then δ(ϕ · ψ) = 0.Assuem ϕ = δφ, and δψ = 0, then δ(φ · ψ) = δφ · ψ ± φ · δψ. Then δ(φ · ψ) = ϕ · ψ.

Definition 24.3. If δ satisfies the hypothesis of the Lemma, then it is called a derivation of ·.Now we return to step (1), defining the cup product for singular cochains of a space C∗(X;R),

with X a space, and R a commutative ring with 1.

Definition 24.4. We define the map on singular cochains

Ck(X;R)× C l(X;R)∪→ Ck+l(X;R)

by taking ϕ ∈ Ck(X;R) and ψ ∈ C l(X;R), and defining a function acting on k + l-chains.Let the k + l-simplex be denoted by ∆k+l = [v0, . . . , vk, vk+1, . . . , vk+1]. Then,

(ϕ ∪ ψ)(σ) = ϕ(σ|[v0,...,vk]) · ψ(σ|[vk,...,vk+l]).where · is the product in R.

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This defines a product ∪ on (C∗(X;R), δ).

Example 24.5. The following image has k = 2, l = 1, and shows which faces of the 3-simplex ϕand ψ act on.

v0

v1

v2

v3

ϕ

ψ

Exercise 24.6. Show that the cup product satisfies the lemma, so that ∪ induces a well-definedproduct on homology.

Now we proceed to computing an example.[Recall: C∆

∗ (X) → C∗(X) is a quasi-isomorphism. Likewise for C∗∆(X)← C∗(X).]Let X = T = S1 × S1, and Y = S1 ∨ S1 ∨ S2.We claim that the cup product for homology distinguishes these two spaces.

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