Algebraic Topology Flashcards

Question 1

Q

Define: deformation retraction

Answer

A

A deformation retraction of a space X onto a subspace A is a family of maps f_t:X –> X for t in [0,1] s.t. f0 = I_X f1(X) is in A and f_t restricted to A is identity in A for all t and f_t jointly continuous.

1st semester pg 74

Question 2

Q

Examples of deformation retracts?

Answer

A

R^n to 0, R^n - {0} to S^n-1, Mobius band onto circle, Disk D^2 with 2 open disks removed pgs 75-76

Question 3

Q

Define: Homotopy, homotopic maps

Answer

A

a family of maps F_t: X –> Y st F:X x I –> Y is continuous. Say f0 and f1 are homotopic maps pg 77

Question 4

Q

Define: Homotopy relative to A

Describe def retraction in this language

Answer

A

If A a subspace of X, a homotopy relative to A is a homotopy F_t : X –> Y st f_t|A does not depend on t - i.e. points of A don’t move

A deformation retraction of X onto A is a homotopy rel A from identity on X to a map r: X –> A

Question 5

Q

Define: homotopy equivalence

Answer

A

A map f:X –> Y is a homotopy equivalence if exists a map g: Y –> X st gof = I_X and fog = I_Y

Generalizes homeomorphism

Question 6

Q

Describe how deformation retraction gives homotopy equivelence

Question 7

Q

Define: homotopy type

Compare to homeomorphism

Answer

A

If spaces X and Y are homotopy equivalent, we say they have the same homotopy type

looser than homeomorphism - R^2 and R not homeo but both def retract to a point so homotopy equiv. Measures in some sense the connectedness properties of spaces

Question 8

Q

Define: contractible

Answer

A

X is contractible if X has the same homotopy type as a one-point space

Question 9

Q

Prove: X contractible iff I_X nullhomotopic

Answer

A

nullhomotopic = homotopic to a constant map. pg 80

Question 10

Q

Define: CW Complex, attaching maps, characteristic maps

Answer

A

pg 81 weak topology Hatcher 5,7

Question 11

Q

What is a CW complex graph?

Answer

A

Just a 1-dim CW complex

Question 12

Q

Discuss cell structures for S^n

Answer

A

pg 84,

S^n = D^n / boundary(D^n) collapse boundary to a point. Two cells: one zero cell, one n-cell. phi maps boundary to 0-cell
Two hemispheres attached along S^n-1. Inductive. Two cells of each dimension
Iterated suspension of 0 sphere pg 92

Note 2 and 3 same

Question 13

Q

Discuss constructions of RP^n

Answer

A

Space of all lines through the origin in R^n+1 - modulii space of 1-dim subspaces - generalized by Grassmanian
= R^n+1 - {0} / x ~ lamda*x for all lamda in R - {0}
S^n/ x ~ -x
D^n / x ~ -x for all x on boundary of D

Viewing RP^n as D^n/~ suggests a simple cell structure. The boundary of D^n is a copy of RP^n-1 so we RP^n has a cell in each dimension:

85 - 88

Question 14

Q

What is Closed Map Lemma? Proof?

Answer

A

Mapping from compact space to Hausdorff space –> always closed map

Question 15

Q

Define: subcomplex of CW complex, CW pair

Examples?

Answer

A

A closed subset this is a union of open cells (Hatcher just calls these open cells e^n_alpha cells) - notice a subcomplex is a CW complex in its own right

If X = CW and A = subcomplex, (X,A) is a CW pair

n-skeleton best example

Question 16

Q

Topological properties of CW complexes?

Answer

A

Finite CW complex is compact
Every compact subset lies in a finite subcomplex
A set is open [closed] if its intersection with the closure of each cell is open [closed] - i.e. weak/coherent topology
CW complexes are normal
Locally path-connected ( so connected iff path-connected)

pg 89-90

Question 17

Q

List basic topological operations on spaces.

Answer

A

Suspension: SX = X x [0,1]/ ~ where X x {0} is collapsed to a point and X x {1} is collapsed to a point
Cone: CX = X x [0,1]/~ where X x {0} collapsed to a point
Join: X * Y = X x Y x [0,1]/~ where we collapse X x Y x {0} to X and collapse X x Y x {1} to Y. X on one end of space, Y on other end. Cone, suspension are special cases
Wedge sum: Glue two spaces at a point X v Y

Question 18

Q

Discuss criteria for homotopy equivalence

Examples?

Answer

A

Collapsing Subspaces - If (X,A) a CW pair st A contractible, the quotient X –> X/A is a homotopy equivalence
- Graphs - maximal tree
- S^2 v S^1 = S^2/S^0
Attaching Spaces - If (X1,A) a CW pair and f:A –> X0, g:A –>X0 are two attaching maps, then if f homotopy equiv to g, the spaces attained by attaching X1 to X0 are homotopy equivalent
- Dunce Cap
- S^2 v S^1

Question 19

Q

What is the homotopy extension property?

Equivalent property?

Answer

A

A pair of top spaces (X,A) satisfies the hom extension property if given any map f0: X –> Y and a homotopy f_t : A –> Y, there always exists an extension to a homotopy f_t: X –> Y

A pair (X,A) has the h.e.p. <=> X x {0} U A x I is a retract of X x I.

Question 20

Q

Prove: Every CW pair (X,A) has the Homotopy Extension Property

Question 21

Q

Prove: If (X,A) has the H.E.P. and A is contractible, then q: X –> X/A is a homotopy equivalence.

Question 22

Q

Prove: If (X,A) is a CW pair and attaching maps f,g: A –> X0 are homotopic, then the two spaces obtained by gluing are homotopy equivalent [rel X0]

Answer

A

pg 103. Do an example with simple spaces

Question 23

Q

Discuss the classification of surfaces, CW structure, proof idea

Answer

A

pg 105-111. Key points: Compact no boundary. Orientable vs. Nonorientable.

Rado - Every surface has a triangulation

Part 1 - Show every compact surface without boundary is homeo to one of the following S^2, M1, M2, M3, … N1, N2, N3, ….
-Triangulate and manipulate into nice form - polygon by glueing - cut and paste all justified since compact Hausdorff

Part 2 - These surfaces are all in fact topologically distinct (different homology groups)

Question 24

Q

Basic approach to showing two spaces are homeomorphic? not homeomorphic?

Answer

A

Homeomorphic: Exhibit explicit homeomorphism between the spaces - cts bijection with cts inverse

Not Homeomorphic: Harder - need topological invariants

Question 25

Q

Show B(0,1) is homeo to R^2

Question 26

Q

Discuss how to visualize R, R^2, R^3 not homeomorphic

Answer

A

lasso, disconnected, bubble pgs 2- 3

Question 27

Q

What is the main aim of algebraic topology

Answer

A

Convert topology to algebra. Generate algebraic invariants of a space. Functors from Category of top spaces and cts maps to some algebraic category. Here we will primarily be going to groups and homs

Question 28

Q

What are the basic types of homology? pros/cons

Answer

A

Simplicial - combinatorial
- space made of simplices
- simplest defs
Singular - maps of simplex onto space. Most abstract - easy invariance
Cellular - ‘polygons glued together’ easy examples

All agree where defined

Question 29

Q

Define: standard n-simplex, barycentric coordinates, vertices, dimension, face, boundary, open simplex

Discuss properties

Answer

A

pg 6 : let ai = (0, … , 0 , 1 , 0 , … 0) where 1 is in the ith spot. The standard n-simplex is the set of all points x = sum tiai s.t. ti sum to 1 and each ti >= 0

convex hull of {a0, …. , an}

barycentric coordinates are the ti

ai are the vertices, they span the simplex. A face of an n-simplex is a simplex spanned by a proper nonempty subset of {a0,…an}

The boundary of a simplex is the union of all proper faces

An open simplex or interior is just simplex - its boundary. All points of sigma st t_i(x) >0 for all i

n-simplex is compact, n dimensional, homeomorpic to closed n-ball

Question 30

Q

Define: simplicial complex, abstract simplicial complex K, dimension of simplex, dim(K), n-skeleton, vertices

Example?

Motivation?

Answer

A

A simplicial complex K in R^N is a collection of simoplicies in R^N st

(1) Every face of a simplex of K is in K
(2) The intersection of any two simplexes of K is a face of each of them

An abstract simplicial complex K consists of. collection S of finite nonempty sets called simplices st every nonempty subset of a simplex is also a simplex

dim(simplex) = cardinality - 1
dim(K) = sup dim(simplices in K) - could be infinite
dim(empty set) = -1

n-skeleton K^(n) = all p-simplices for p <= n

vertices are just the single elements

pg 8-9

Geometric approach gets very messy - all the essential data is in the combinatorial structure of the vertices

Question 31

Q

Define: simplicial map

example?

Answer

A

A simplicial map f: K1 –> K2 is a function defined on vertices V(K1) –> V(K2) st if x is a simplex, then f(x) is a simplex

ie a map defined on vertices sending simplices to simplices

Question 32

Q

Define: geometric realization of an abstract simplicial complex

Example?

Answer

A

pg 11 - 12

First, the geometric realization of an n-simplex is a standard geometric n-simplex . View as a set of functions from vertices –> R.

The geometric realization of an abstract simplicial complex K with vertex set V is denoted |K| inside R^V and defined |K| = {all functions a:V –>R} s.t.

For all a in |K|, {v in V : a(v) != 0} is a simplex of K
For all a in |K|, sum over V of a is 1
For all a, v, a(v) is in [0,1]

Question 33

Q

How can the geometric realization of an abstract simplicial complex be turned into a top space?

What choice do we make?

Answer

A

The metric topology using l_2 metric on R^V - not very natural
The weak/coherent topology on |K|. Each geometric simplex lies in a copy of R^n+1. Give the simplex the standard topology from R^n+1 (metric). Define a subset A in |K| to be open [closed] if its intersection with each simplex is open [closed]. “paste together subspace topologies”

We use weak topology!

Question 34

Q

When is a function f: |K| –> X continuous? where K is an abstract simplicial complex

Is |K| Hausdorff?

Answer

A

iff its restriction to every simplex is cts

yes, can pull back a separation of points from |K| –> |K|_d

Question 35

Q

Prove: Any subset A of |K| contains a discrete subset with exactly one point from each open simplex meeting A

Every compact subset of |K| meets only finitely many open simplices

Question 36

Q

Define: Locally finite (in context of simplicial complexes)

Answer

A

Locally finite means each vertex is in only finitely many simplices - in this case the metric topology on |K| matches the weak topology

Question 37

Q

Define: triangulated space

Examples of triangulations? Can all topological spaces be triangulated?

Answer

A

A top space is triangulated if X is homeomorphic to |K| for some abstract simplicial complex K

Polyhedra - built from the basic building blocks of lines, triangles, tetrahedra, n-simplices

The triangulation conjecture—first formulated by Kneser in 1924—claimed that every manifold was triangulable. The conjecture turned out to be false in general, although it is true for manifolds of dimension up to 3, and also for all differentiable manifolds

Floer homology - Manolescue 2016

Question 38

Q

Discuss calc 3 motivation to homology

Answer

A

pgs 17-18

Question 39

Q

When are two orderings of vertices in a simplex equivalent? What are equivalence classes called?

Answer

A

Equivalent if they differ by an even permutation - called orientations [v0, v1, … , vn]

pg 18

Question 40

Q

Define: group of (oriented) p-chains on a simplicial complex K

Answer

A

C_p(K) = abelian group gen by all oriented p-simplices with relations o = -o’ if o and o’ are opposite orientations of same simplex - “integer linear combinations of a bunch of simplices of order p”

free abelian with basis given by choosing an orientation for each p-simplex

Question 41

Q

Define: boundary map for simplicial complex K

Most important property? Proof?

Examples?

Answer

A

del_p:C_p(K) –> C_p-1(K) is defined on each oriented simplex, then extended linearly

del_p[v0, … , vp] = sum_{i=0}^p (-1)^i [v0, …, vi hat, … vp]

key property: del^2 = 0

these alternating sums must have something to do with differential forms

pgs 20-22

Question 42

Q

Define: p-cycles, p-boundaries, pth homology group of K

K a simplicial complex

Answer

A

p-cycles: Z_p(K) = p-chains without boundary

p-boundaries: B_p(K) = p-chains that are a boundary of a p+1-chain

Z_p(K) = ker del_p

B_p(K) = im del_p+1

del^2 = 0 implies B_p(K) < Z_p(K) – every boundary is a cycle!

H_p(K) = Z_p(K) / B_p(K) = cycles/boundaries

pgs 23-24

Question 43

Q

Do examples of homology groups via simplicial homology for loop graph and filled in square

Answer

A

pgs 24-25

Question 44

Q

Define: homologous p-cycles, homologous to zero

chain c carried by a subcomplex L

Answer

A

p-cycles are homologous if they differ by a p-boundary. They represent the same element in quotient group H_p(K). i.e. c -c’ = boundary_p+1(d) for some p+1 chain d

homologous to 0 if boundary_p+1(d) = c. Also say “c bounds”

Question 45

Q

Compute homology of T^2 using simplicial complex

Answer

A

pg 29 - 31 Show we can push off to outer edge

H2 = Z, H1 = Z^2 H0 = Z

Munkres 34 - 36

Question 46

Q

Prove: The decomposition of simplicial complex |K| into components {K_alpha} gives an isomorphism H_p(K) = direct sum over alpha H_p(K_alpha)

What about singular homology?

Answer

A

Simplicial: pg 33

Singular: Look at path components of X.

Delta^n path connected, so image must be path connected so must be contained in path component of X. boundary also lies in the same path component

pg 114

Question 47

Q

Discuss H_0(K) in simplicial setting. Proof?

What about in singular homology?

Answer

A

If |K| connected, then H_0(K) = Z. In general free abelian with one factor for each component.

pgs 34-35

Munkres 41

Singular: In general have direct sum of Z’s, one for each path component

pg115

Question 48

Q

Define: Chain complex

Answer

A

A sequence of abelian groups with boundary homomorphisms satisfying boundary^2 = 0. Homology groups H_p(C) = im boundary_p=1 / ket boundary_p

pg35

Question 49

Q

Define: reduced homology

simplicial vs. singular?

Answer

A

We let epsilon: C_0(K) —> Z take sum n_i*v_i —> sum n_i
epsilon is called the “augmentation map” of C_0(K) and define the “augmented chain complex” which just has epsilon tacked on at the end.

Confirm epsilon after boundary = 0.

Reduced homology groups are just H tilda _p(K) = H_p(K) if k>0 and ker(epsilon)/im(bdy) if p =0

The only thing that changes is 0 dimensional homology group

singular on pg 115

Question 50

Q

What is the relationship between reduced homology and standard homology?

Answer

A

Only dim 0 changes. H_0(K) = H_0 tilda(K) + Z

H_0(K) = 0 when K is connected - closer to measuring 0 dim holes - disconnection

Question 51

Q

Define: join of simplicial complex

Examples?

Answer

A

Let K_1 and K_2 be simplicial complexes. The join K_1*K_2 is a simplicial complex with vertx set Vert(K_1) disjoint union Vert(K_2) with simplices sigma_1 disjoint union sigma_2 for sigma_i a simplex of K_i

pg 38-40

Cone and suspension

Question 52

Q

What are the homology groups of a cone? Proof?

Define: bracket operation

Answer

A

A cone has trivial reduced homology - i.e. is acyclic

pg 41-42

For bracket see Munkres 45

Question 53

Q

Compute the homology of (n-1) - sphere using boundary of simplex

Answer

A

pg 43 -44

Munkres 46

Question 54

Q

Define: pair of simplicial complexes

Answer

A

(K,L) K = simplicial complex, L = subcomplex

Question 55

Q

Define: relative chains for simplicial pair (K,L)

Answer

A

C_p(K,L) = C_p(K) / C_p(L) i.e. the free abelian group with basis elements sigma^p + C_p(L) where sigma^P is in K - L.

cosets are represented by chains not involving L

pg 45

Question 56

Q

Define: relative homology groups

Discuss also for singular homology

Answer

A

First, we define a relative chain complex by observing boundary maps are well-define on quotient groups and boundary^2 = 0. Define H_p(K/L) = ker(boundary_p)/ im(boundary_p+1) = Z_p(K,L) / B_p(K,L)

pg 46, 56

pg 127

Question 57

Q

Discuss relative cycles and relative boundaries - draw example

Answer

A

pgs 47 - 48

Question 58

Q

What is the Excision Theorem - simplicial setting - intuition?

Proof?

Now discuss in singular setting, what changes?

Answer

A

The main idea is that H_p(K,L) ignores everything inside L (treats it as 0) so then modifying K inside of L does not change H_p(K,L)

Excision Thm. Let A and L be subcomplexes of K with A union L = K. Let B = A intersect L. Inclusion induces an isomorphism H_p(A,B) —> H_p(K,L).

Proof. Show the two chain complexes are isomorphic so that their homology groups are isomorphic – pg 50

Singular: Much more complex…source of nearly all explicit computations of singular homology groups of topological spaces.

Suppose B < A < X with B closure < int A. Then (X - B, A - B) —> (X, A) induces an isomorphism of singular homology groups. pgs 129 - …

Strategy:

Define subdivision sd
sd is chain homotopic to identity
Iterated subdivision leads to arbitrarily small singular simplicies
Hi(X) is determined by the “small” simplices

With this in hand, the excision thm is easy to prove: pg148

Hatcher 117

Question 59

Q

How does a simplicial map induce a chain map? - simplicial

vs.

How does a continuous map induce a chain map? - singular

Answer

A

simplicial: F_sharp [v0, … , vp] = [f(v0) … f(vp)] where [f(v0) … f(vp)] = 0 if any of the vertices are repeated pg 51
singular: f_sharp Sn(X) –> Sn(Y) is composition of maps f after sigma pg 117

The key property of chain maps is

boundary f# = f# boundary

Question 60

Q

Prove: A simplicial map f: K —> L induces a homomorphism f_* : H_p(K) —> H_p(L)

Same for continuous map inducing a homomorphism

Answer

A

51 and 117

Question 61

Q

Discuss simplicial and singular homology categorically

Answer

A

Simplicial: Covariant functor from category of simplicial complexes and simplicial maps to category of abelian groups and homomorphisms

Singular: Covariant functor from category of topological spaces and continuous maps to category of abelian groups and homomorphisms

Question 62

Q

Define: chain homotopy simplicial vs. singular

Answer

A

A chain homotopy is a homomorphis between f_# and g_#

D:C_p(K) —>C_p+1(L) st boundary(D sigma) = g_#(sigma) - f_#(sigma) - D(boundary sigma)

singular: 120

Question 63

Q

Simplicial: When do 2 simplicial maps induce the same homology map? Proof?

Singular: When do 2 continuous maps induce the same homology map? Proof. Corollaries?

Answer

A

If f and g are chain homotopic. See book

If f and g are homotopic. So the homology maps depend only on the homotopy class of f. pg 119

Cor. If f: X –> Y is a homotopy equivalence, then f_* : Hn(X) —> Hn(Y) is an isomorphism - i.e singular homology depends only on homotopy type

Cor. If X is contractible, then H_n(X) tilda = 0 for all n - acyclic.

Question 64

Q

Define: exact sequence of abelian groups

Answer

A

A sequence of homomorphisms … –> A_n+1 —> A_n —> A_n-1 —> … is exact at A_n if im alpha_n+1 = ker alpha_n

Exact sequence means exact at each term

Answer 60

A

surjection
injection
isomorphism
s.e.s

Answer 61

A

Let 0–>A–>B–>C–>0 be a s.e.s. The following are equivalent:

The short exact sequence splits
Exists j:C–>B s.t, b after j = Id_C
B = A + C
Exists i:B–>A s.t. a after i = id_A

Z x2 map

Answer 62

A

Can easily define j:C –> B by mapping basis of C to preimage pg 54

Answer 63

A

pg 55 A sequence of homomorphisms making diagram of two chain complexes commute

-also called a chain map

Answer 64

A

If (K,L) is a simplicial pair, the following sequence is exact:

… –> H_p+1(K,L) –> H_p(L) –> H_p(K) –> H_p(K,L) –> H_p-1(L) –> …

Discuss in terms of short exact sequence of chain complexes.
Boundary is connecting homomorphism

Proof:

Define boundary*
Show boundary* well-defined
Show boundary* is homomorphism
Check exactness via diagram chase

pg 57 - 63

pg 127 same proof works for either…if we have good pair can prove something stronger - replace H(X,A) with H(X/A)

Answer 65

A

Same as Long Exact Sequence of Pair but we use augmented chain complexes
pg 64

Answer 66

A

pg 66

pg 123

Answer 67

A

Given a homomorphism from one ses of chain maps to another, we get a homomorphism of homology long exact sequences. pg69-71

Answer 68

A

5 maps, 4 known to be isomorphisms, then 5th is…

pg 72

Answer 69

A

If f: G –> H then cokernal is H/f(G)

Answer 70

A

An abelian group with a basis as a Z-module–> each g in G can be written uniquely as finite sum n_i g_i with n an integer

Answer 71

A

The set of all elements of finite order in G is the torsion subgroup. torsion free if torsion subgroup = 0.

If there is a collection of subgroups of G, G_i st each g in G can be written uniquely as a finite sum of gi, the G is internal direct sum of the G_i

Answer 72

A

Start with a space X0 and another space X1 that we wish to attach to X0 by identifying points in a subspace A in X1 with points of X0. The data needed is a map f:A –> X0 for then we can form the quotient space of the disjoint union of X0 and X1 by identifying each point of a in A with its image f(a) in X0. Hatcher 13

Let f: X –> Y, then the mapping cylinder M_f is the quotient space if the disjoint union (X x I) U Y obtained by identifying each (x,1) in X x I with f(x) in Y. We see this is a special case of the attaching construction…

Def retracts onto Y

Hatcher pg 2

Answer 73

A

No, consider def retract of X onto point

Answer 74

A

No. A space X always retracts onto any point x in X via constant map but a space that deformation retracts onto x must be path connected since def retract gives path joining each x in X to x_0

More generally, spaces with nontrivial fundamental group not homotopy equivalent to point - no def retract

Answer 75

A

<= easy
=> Take mapping cylinder of homotopy equivalence f:X –> Y. We know this def retracts onto Y. Need to show def retracts onto X. Hatcher 16-17

Answer 76

A

Hatcher 15. Let (X, A) be a pair.
A mapping cylinder neighborhood is a closed neighborhood N containing a subspace B, thought of as the boundary of N, with N - B an open neighborhood of A, s.t. there exists a map f: B –> A and a homeomorphism h : M_f –> N with h| A U B = identity

A pair (X,A) has the homotopy extension property if A has a mapping cylinder neighborhood in X

Example: Thick letters

Answer 77

A

A complex whose reduced homology vanishes in all dimensions is stb acyclic

Answer 78

A

If X is a topological space, a singular n-simplex in X is a map sigma: delta^n –> X

where delta^n is the standard n-simplex {(t0, …, tn) in R^n+1 | sum ti =1 and ti in [0,1]}

Only requirement is that sigma is cts. Need not be injective etc. Could be very complicate - 1-simplex is just [0,1] so all cts paths allowed - space filling curves etc

Answer 79

A

S_n(X) = free abelian group with basis all singular n-simplicies. Elements are finite sums ni sigma_i ni in Z, sigma a singular n-simplex

mirror simplicial bit everything singular now. Crazy uncountable basis

pg 112

Answer 80

A

del_n : Sn(X) —> Sn-1(X)

del_n(sigma) = sum_i=0^n (-1)^i sigma|[v0, …, vi hat , … vn]

we canonically identify the face with delta^n-1 by a linear map that preserves order of vertices

Again, the key property is boundary^2 = 0

pg 113

Answer 81

A

H_n(X) = ker del_n / im del_n+1. Possible to define since d^2 = 0 i.e. im d_n+1 < ker d_n

pg 113

Answer 82

A

A pair (X,A) is a good pair if A has a nbd that def retracts onto A

Example: All CW complexes - go a little out into cell tofind neighborhood

Answer 83

A

Follows easily from functorial properties of homology

No retraction r : D^n –> S^n-1. Just plug into commutative diagram
pg 124

Answer 84

A

Thm: Every map D^n –> D^n has a fixed point

Pf. Existence of such a map would imply D^n retracts onto S^n-1

Hatcher 113, notes

Answer 85

A

Let A be an nxn matrix with all entries real and positive. Then A has a positive eigenvalie and a corresponding positive eigenvector (all entries real and postive

Pf. Define a cts map and appeal to Brower fixed point

pg 126

Answer 86

A

notice this is different then l.e.s. of pair - have quotient space, not relative homology.

This follows from the prop. If (X, A) is a good pair, then (X,A) –> (X/A, A/A) induces an isomorphism H_n(X, A) = H_n(X/A, A/A) = H_n(X/A) reduced
pg 149 -150

Moral: H_n(X,A) is the same as H_n(X/A) reduced as long as you have a good pair

Answer 87

A

Be sure to take reduced homology, then wedge sum of spaces yields direct sum of homology groups

Good pair lets us move from relative homology to quotient space

Answer 88

A

When both defined, they are equal

There is a canonical map from simplicial to singular homology groups taking a simplex to its inclusion map into X.

Thm. The canonical map is an isomorphism for any simplicial pair (K,L).

If it is known for single spaces, then 5 lemma gives for simplicial pairs.

pg 154

Answer 89

A

Recall: H_n(S^n) = Z for n >= 1. f: S^n –> S^n induces f_* : H_n(S^n) —> H_n(S^n) i.e from Z –> Z. Must be multiplication by some d in Z.

The degree of a map f: S^n –> S^n is the number d s.t. f_*: Z –> Z is multiplication by d. deg(f) = d

Examples
S^1 –> S^1

f(z) = z^d had degree d
f(z) = complex conjugation - degree -1

Triangulate…

pg 159-160

Answer 90

A

deg I = 1
deg f = 0 if f not surjective
If f, g homotopic, then deg(f) = deg(g)
deg(fg) = deg(f)deg(g) — multiplicative — implies homotopy equivalence S^n to S^n has degree +- 1
deg(f) = -1 if f is a reflection
The antipodal map f(x) = -x is a composition of n+1 reflections, hence has degree (-1)^n+1
If f has no fixed point, then deg(f) = (-1)^n+1 — homotopic to antipodal map

pg 161 - 162

Answer 91

A

Thm. S^n admits a continuous nonzero vector field <=> n is odd.

Note: Then S^n not parallelizable for even n, can’t be Lie group

If n is odd, consider unit sphere in C^K. Let v(z1, … zk) =

If n even, use vector field to define a homotopy from 1 to -1 –> a contradicition

pg162-163

Answer 92

A

The local homology of space X at a point x in X is H_n(X, X- {x}). Only depends on the shape of small neighborhood of x.

If U and V are open sets of R^m and R^n that are homeomorphic, then m = n.

This follows from excision:

Answer 93

A

Hatcher 114

Answer 94

A

Purely algebraic result. Hatcher 114 - 115

Diagram chasing

Answer 95

A

If X is a CW complex, the group of cellular n-chains is H_n(X^n, X^n-1) — the nth singular homology group of the pair (X^n, X^n-1)

Answer 96

A

if k = n, this is free abelian with basis the n-cells of X
if k != n, 0

Key: this is a good pair, so reduced homology is the same as X^n/X^n-1.

X^n/X^n-1 is a wedge sum of n-spheres, one for each n-cell of X

Answer 97

A

l.e.s. of pair … pg 170

Answer 98

A

l.e.s of pair. Need to distinguish finite dim complex from infinite dim complex pg 171

Answer 99

A

The idea is to weave together the homology exact sequences for (X^n, X^n-1) and (X^n-1, (X^n-2)

pg 172

Answer 100

A

pg 173 chase around diagram

Answer 101

A

Very nice for computations. Get a basis for cellular n-chain group by choosing an orientation for each n-cell and taking free group on these n-cells.

The boundary map goes from n-cell to (n-1)-cells. Boundary of n-cell is S^n-1 this is attached into X^n-1. To understand value of boundary map on a specific (n-1)-cell e, collapse everything in X^(n-1) except e, get S^n-1. Then boundary is just degree of this map

pgs 174 - 178

Answer 102

A

pgs 178 - 181

Different for n even vs. n odd

Answer 103

A

Let c_n be the number of n-cells in a CW complex. The Euler characterstic is the alternating sum a0 - a1 + a2 - a3 …

What is Euler characteristic of closed surfaces?

Recall familiar case for polyhedron v - e + f

Maybe say a little about Gauss-Bonnet theorem - integrating Gaussian curvature of Riemannian manifold over entire manifold always equals 2 pi * Euler characteristic

pg 181 - 182

Answer 104

A

Suppose X is any space, A,B < X. Suppose X = int(A) U int(B). Then there exists a l.e.s. –> H_n(A intersect B) —> H_n(A) + H_n(B) —> H_n(X) —> H_n-1(A intersect B) —> …

Natural with respect to continuous maps of spaces.

Proof. Has close connection to excision thm. Use a nice s.e.s. of chain complexes

pg 183 -184

Answer 105

A

pg 185

A cycle c in X can be written as a sum of a chain in A and a chain in B, c = a + b. 0 = Boundary c (since cycle) = boundary a + boundary b. So boundary a = -boundary b is in A intersect B. The connecting hom sends c = a+b —> boundary a = -boundary b

Answer 106

A

As long as A intersect B != 0 this works

pg 186

Answer 107

A

Use 2 hemispheres and induction

pg 186

Answer 108

A

Let h:S^k –> S^n be any embedding. Then reduced homology of S^n - h(S^k) = Z if i = n-k-1, 0 otherwise

Special cases:

Jordan curve theorem: h: S^1 –> S^2
Jordan-Brower separation theorem: h: S^n-1 –> S^n

Both of these just say the complement of a codimension 1 sphere has 2 components. Each with the homology of a point.

Now in the case of S^1 —> S^2, all embeddings are topologically equivalent - in particular both components of S^2 - S^1 are contractible

However, if dim n > 2, then S^n-1 may separate S^n into two components, one of which is not contractible - Alexander’s Horned Sphere. — this can’t be seen using homology.

Proof. We first show that if h: D^k –> S^n is an embedding, then S^n - h(D^k) is acyclic in reduced homology. Removing a disc makes homology trivial - induction on k

pgs 187 - 193

Answer 109

A

Think about…

Answer 110

A

Thm. If U < R^n is open and h:U –> R^n is a continuous injection, then h(U) is open in R^n

So every cts injection of U into R^n is an open mapping

pg 193 - 194

Answer 111

A

We can use any abelian group for coefficients. Formally everything works out the same. Good choice of coefficients may yield richer supply of algebraic invariants

On the other hand H_n(X, G) is algebraically completely determined by knowledge of the homology groups of X with integer coefficients - Universal Coefficients Theorem

Answer 112

A

Pros

highly computable - linear algebra
rich combinatorial/algebraic structure

Cons

intuition less clear - what is a ‘cycle’ geometrically?
incomplete invariant - not enough to pin down homotopy type

Cobordism - replace simplicies with manifolds
Homotopy groups - complete invariant

Whitehead Thm: If f:X –> Y induces an iso on pi_n for all n, then f is a homotopy equivalence.

Hurewicz Thm - If X is path-connect, then H_1(X, Z) is the abelianization of p_1(X)

ss pg 1 - 4

Answer 113

A

A group with trivial abelinization. Any non-abelian simple group e.g. A_5 since no nontrivial normal subgroups - loose info in abelianization

Answer 114

A

A homotopy f_t : I –> X s.t. f_t(0) = x0 and f_t(1) = x1 for all t - the endpoints are glued down. i.e. a homotopy rel to boundary I = {0,1}

[f] = path homotopy class of f

pg 6

Answer 115

A

If f(1) = g(0) then glue together

f . g (s) = f(2s) if s <= 1/2
g(2s-1) if s>= 1/2

Answer 116

A

As a set = all path-homotopy classes of loops in X based at x0.

Define group operation [f][g] = [f . g]

Check: well-defined and group axioms

The group axiom checks come down to fact that reparameterization does not impact homotopy type

pgs 7 - 9

Answer 117

A

Let r_t be a def retract onto x0, then for any loop f based at x0, r_t after f is a path homotopy from f to c, [f] = 1

pg 9

Answer 118

A

Say x0 and x1 are joined by a path h, then B_h[f] = [h . f . h bar] is an isomorphism of groups

prove: well-defined, homomorphism, bijection - think conjugation of groups = isomorphism

pg 10 - 11

Answer 119

A

X simply connected if path-connected and pi_1(X) = 1

Answer 120

A

pg 12 -13

Answer 121

A

A covering space of X is a space X hat together with a map p: X hat –> X (the covering projection) s.t. X has an open cover {U_alpha} by evenly covered set U_alpha

Evenly covered - U_alpha has preimage a disjoint union of open sets in X hat each mapped homeomorphically to U_alpha via p

Examples

exp: C –> C - {0}
x –> z^2 in C
R - S^1

pg 13 - 16

Answer 122

A

An action of a group G on a space X is a homomorphism G –> Homeo(X)

A covering space action is an action of G on X s.t. each x in X has a nbd U whose G-translates are pairwise disjoint
i.e. points in orbit are separated - spaced out. g1(U) intersect g2(U) != 0 => g1 = g2 or equivalently U intersect g(U) != 0 =>g = identity

The orbit space is the quotient X/G where X/~ where x ~ g(x) for all x, g with quotient topology. That is points of X/G are orbits Gx = {gx | g in G}

Z action on R –> S^1
Z^2 action on R^2 –> T^2

pg 16-20

Answer 123

A

So X is a covering space of the orbit space X/G

The disjoint translates {g(U) | g in G} are identified to a single open set in X/G. Preimage is then clearly disjoint set of open neighborhoods homeomorphic to U. Every point has a nbd like this by def of covering space action

Answer 124

A

yes g(x) = x for any g,x implies g = 1. No fixed points

Answer 125

A

pi_1(X/G) = G

First we define phi : G –> pi_1(X/G). Choose some x0 in X to act as basepoint. Now since X simply connected, there exists a unique homotopy class of paths gamma from x0 to g(x0) for each g in G. Then p gamma is a loop in X/G based at p(x0). Define phi(g) = [p gamma]. Now show phi is group isomorphism.

pg 21 -24

Answer 126

A

A map f hat : Y –> X hat s.t. f = p f hat. A map upstairs that projects nicely downstairs

Path Lifting: {x} x I –> X
Homotopy Lifting: I x I –> X
pg 22

These are special cases of the Homotopy Lifting Theorem (pg25)

Given a space Y, a covering space p: X hat –> X, a map F : Y x I –> X, and a lift F hat : Y x {0} –> X hat, then F hat has a unique extension to a map F hat : Y x I –> X hat lifting F.

i.e. if we can lift one end of the map F, Y x {0} end, then the rest of the map lifts uniquely

Answer 127

A

pg 32 -33

Answer 128

A

Given group G and generating set S, define a graph (directed, labeled)
vertices: elements of G
edges : (g, gs) for all g in G, s in S

i.e. use generating set S to move around the graph - reach different elements of G

A circuit in a graph is a closed edge path with no repeated vertex

A tree is a connected graph with no circuit

Examples

Z
Z^2
Free group –> tree

Answer 129

A

idea is to pick some root x0, then look at level 1, level 2, etc and def retract edges back to previous level. No loops so this all works nicely…pg 36-37

Answer 130

A

The key is that F(S) acts on its Cayley graph, a tree via covering space action - quotient is the wedge sum of circles. Since trees are simply connect pi_1(T / F(S) ) = F(S)
pg 37
Van Kampen pgs 67-68

Answer 131

A

pi_1(X) x pi_1(Y)

38

Answer 132

A

Covariant functor from pointed Top spaces, pointed cts maps to Groups, homomorphism

What is induced hom?

pgs 39 - 40

Answer 133

A

The big problem we are addressing here is that fundamental group always has issues with basepoint.

pg 40 -42

Answer 134

A

Choose a maximal tree T in graph, if necessary appealing to Zorn’s lemma. Then pi_1 is free group with basis one element for each edge not contained in T.

Just retract maximal tree to get a wedge of circles

pg 43-44

Answer 135

A

If X is any path-connect space, then H_1(X, Z) is the abelianization of pi_1(X)

The Hurewicz map h: pi_1(X, xo) –> H_1(X) takes the path homotopy class [f] to the homology class [f] (notice any loop may be considered a singular 1-simplex - f(0) = f(1) so this is actually a cycle boundary f = 0)

h is a homomorphism of groups, if X path-connected, h induces an isomorphism [pi_1(X, x0)]ab –> H_1(X)ab = H_1(X)

pgs 45 - 51

Answer 136

A

If {G_a} is a family of groups, the free product *_a G_a is given as set of words g1…gm with each g_i in some G_a.

A word is reduced if

No g_i is identity
g_i and g_i+1 come from different groups

Operation: concatenate and simplify

Examples:

Free groups - free product of Z’s
Z_2 * Z_2

pg 52-53

Answer 137

A

Given any group H and homomorphisms phi_a : G_a –> H, there exists a unique homomorphism phi : *_a G_a –> H induced by phi_a’s s.t. g1 … gm —> phi_a1(g1)…phi_am(gm)

pg 53

Answer 138

A

The group of isometries of R generated by two reflections a(x) = -x and b(x) = 1-x

isomorphic to Z_2 * Z_2

pg 54

fundamental group of wedge sum of RP^2 pg 69

Answer 139

A

(a) IF X is the union of path-connected open sets {A_a} each containing x0 and s.t. each A_a intersect A_b is path-connected, THEN every loop in X based at x0 is homotopic to a product of loops each from some A_a.

INTERPRETATION: Each inclusion A_a –> X induces pi_1(A_a) –> pi_1(X). These induce a map phi from the free product of pi_1(A_a) to pi_1(X). (a) states that this map is surjective. The loops in pi_1(A_a)’s generate pi_1(X).

(b) phi is often not injective. Ker(phi) consists of “relations” between words in the A_a’s. The kernel of phi is the normal subgroup N generated by all elements i_ab(w) i_ba(w)^-1 for all w in pi_1(A_a intersect A_b) for all a,b
pg 56-57, pg 62-66

Answer 140

A

Van Kampen on S^n using open cover S^n - {N} and S^n - {S} North and South poles. pg 58-59, 67
Using CW structure point and n-cell - generator, no relators - pg 70

Answer 141

A

First observe that since S^n is compact, the only groups that can have covering space actions on S^n are finite groups. Use degree of maps S^n –> S^n. deg: G –> {+ - 1}
pg 59-61

Answer 142

A

For CW complex we have a natural presentation for pi_1 where:

1-cells = generators
2-cells = relators
n-cells (n>2) have no effect at all!

Thm. (a) If Y is formed from X by attaching 2-cells via attaching paths phi_a : S^1 –> X, let N = normal subgroup generated by the loops {phi_a}. Then pi_1(X) –> pi_1(Y) [map induced by inclusion] is surjective with kernel = N i.e. pi_1(Y) = pi_1(X) / N

(b) If Y is formed from X by attaching n-cells for some n > 2, then pi_1(X) –> pi_1(Y) [induced by inclusion] is an isomorphism
(c) A cell complex Y with 2 skeleton X satisfies pi_1(X) –> pi_1(Y) [induced by inclusion] is an isomorphism.

pgs 70, 73 - 76

Answer 143

A

M_g has 2g generators = 1 cells and 1 relator:
F(a1, b1, … , ag,bg) / < < [a1, b1] … [ag, bg] > >
N_g has g generators and 1 relator:
F(a1, … , ag) / < < a1^2 … ag^2 > >

Answer 144

A

For any group G with presentation G = < s_a | r_b > there is a 2-complex X with pi_1(X) = G.

1 vertex
1-cell for each generator
2-cell for each relator

pg 77

Answer 145

A

Every knot is an embedding of S^1 so the knots themselves are homeomorphic, what is different is how S^1 is embedded into S^3. Thus, we study the complement of the knot.

We use S^3 to compactify R^3. Normally just think of R^3 and work locally.

Can realize S^3 as the union of two solid tori - glued together along boundary torus.

Recall by Alexander’s Thm, homology cannot distinguish different knots. pi_1 can actually be useful here. Have nice family of torus knots K_m,n where (m,n) = 1

Computation of this is an involved use of Van Kampen

Get pi_1(X_m,n) = < a , b | a^m = b^n > … to see that we get distinct groups, mod out by center < a^m = b^n>, get < a , b | a^m = b^n = 1 > = Z_m * Z_n.

pgs 78 - 86

Answer 146

A

A based cover p: (X hat, xo hat) —> (X, xo) corresponds to a subgroup of pi_1(X, x0)

The map p_* : pi_1(X hat) –> pi_1(X) is always injective. p_*(pi_1(X hat)) is a subgroup of pi_1(X)
The index of the subgroup p_*(pi_1(X hat)) in pi_1(X) equals the number of sheets of the cover
Change of basepoint gives conjugate subgroups
If the symmetries of X hat act transitively on the vertices, then p_*( pi_1(X hat) ) is a normal subgroup N of pi_1(X)
Every subgroup of pi_1(X) corresponds to some based cover X hat –> X “Galois correspondence” i.e. the lattice of subgroups is in 1-to-1 correspondence with the lattice of based covers. The trivial subgroup 1 corresponds to the simply connected cover - called the universal cover

Answer 147

A

Consider only orientable - If genus g > 0, then this is homeomorphic to R^2.

Not all conformally equivalent - relate to unformization thm which says open disc, complex plane, or Riemann sphere are the options.

Also relate to hyperbolic geometry which gives S^2, R^2, hyperbolic plane

pg 93

Answer 148

A

An isomorphism of covering spaces is a homeomorphism f: X1 hat –> X2 hat s.t. p1 = p2f - draw diagram

pg 95

Answer 149

A

Simple application of homotopy lifting property

pg 96

Answer 150

A

Consists of all homotopy classes of loops in X at x0 whose lift to X hat starting at x0 hat is a loop.

Some loops in X lift to a path - unravel. Others stay a loop. All we can do in moving up is unravel loops

pg 96-97

Answer 151

A

Recall: # sheets = #p^-1(x) if X, X hat path connected, then this number is constant.

Index = number of right cosets

pg 97 -99

Answer 152

A

conjugate subgroups

pg 100

Answer 153

A

EXISTENCE: Assume Y is path-connected and locally path-connected. f lifts <=> im(f_) < im(p_)

UNIQUENESS: Two lifts that agree at one point are equal at every point.

Note: Our proofs relied on: path lifting, connected, locally path-connectedness of Y.

pg 101 - 106

Answer 154

A

Free group has covering space action on tree T = Cayley(F,S) which is simply connected. Let H be a subgroup of F. H acts on T with quotient T/H a graph. H = pi_1(T/H) is a free group - already saw fundamental group of any graph is free

“Any cover of a graph is a graph” its fundamental group is free

pg 110

Answer 155

A

This is an “automorphism” of covers - permute sheets of cover - “shuffle the deck” a map f: X’ –> X’ s.t. p = pf

“Label preserving symmetry”

The set of all deck transformations is a group acting on X’ by a covering space action

pg 110-111

Answer 156

A

Prop: Two path-connected covers p, p’ of X are base-point preserving isomorphic <=> im(p_) = im(p’_)

=> Easy
<= Follows from lifting property

This shows that the Galois correspondence is well-define and one-to-one. The set of base-point preserving isomorphism classes of covers of X injects into the set of subgroups of pi_1(X, x0). It remains to show this is surjective - i.e. there is a cover corresponding to every subgroup of pi_1(X, x0).

If we ignore base-points, covers p, p’ isomorphic <=> im(p_) and im(p’_) are conjugate in pi_1(X, x0).

pg 106 -108

Prop: If X is the orbit space of a covering space action of G on a simply connected space X tilda, then every subgroup of pi_1(X, x0) has a corresponding cover - i.e. surjective

Recall: G = pi_1(X, x0). If H < G, then we have a covering space action of H on X tilda. The map X tilda –> X tilda / G factors as X tilda –> X tilda / H —> X tilda / G

pg 109

Prop. Let X be path-connect and locally path-connected. Assume X has a simply connected cover X’ –> X. Then every subgroup of pi_1(X, x0) is realized as im(p_*) for some cover p: X hat –> X.

Pf. Consider the deck transformation group G(X’) -acts transitively by lifting criterion since s.c.

pg 112- 113

Answer 157

A

A universal cover is just simply connected cover of X

X is semilocally simply connected if each point x in X has a nbd U s.t. i_* : pi_1(U,x) –> pi_1(X, x) is the trivial map i.e. all loops in U are nullhomotopic in X

Answer 158

A

Let X be path connect and locally path connected. Then X has a simply connected cover <=> X is semilocally simply connected

=> Every point has a evenly covered nbd U…
<= Notice if we have a simply connected cover X’ –> X, there is a 1-1 correspondence:
{points of X’} {homotopy classes of paths in X starting at x0}

Thus, we define a universal covering space for X has a set { [gamma] | gamma is a path in X starting at x0} the set of all homotopy classes of paths starting at x0. Define p :x’ –> X by [gamma] –> gamma(1)

Now define topology and check s.c.

pgs 115 - 119

Answer 159

A

A covering space p: X’ –> X is regular or normal if for each x in X and all x1’, x2’ in p^-1(x), there exists a deck transformation taking x1’ –> x2’

Answer 160

A

Normal <=> H is a normal subgroup

Group of deck transformations G(X’) = N(H) / H the normalizer of H in pi_1(X,x0) - notice if H normal then G(X’) = pi_1(X, x0) / H.

Answer 161

A

Notice by excision, H_n(M, M - {x} ) - the local homology at x - is isomorphic to Z for every point x of a manifold M. (locally looks like R^n).

A LOCAL ORIENTATION of M at x is a choice of one of the possible generators for H_n(M, M - {x} ) = +1 or -1. Two ways to wrap around the hole.

An ORIENTATION of M is a function x –> mu_x assigning a local orientation to each point of M which is LOCALLY CONSISTENT - Each x has an open neighborhood B s.t. all of the chosen mu_y’s for y in B are images of the same generator of H_n(M, M-B) = H_n( R^n, R^n - B) = Z under inclusion: (R^n, R^n - B) –> (R^n, R^n - {y}). i.e. within B, all generators “wind” in same direction

M is ORIENTABLE if an orientation exists.

pg 128-130

Answer 162

A

If M is a CLOSED, ORIENTABLE manifold of dimension n, then H_k(M, Z) = H^n-k(M, Z).

If nonorientable, can use Z_2 coefficients and get same result.

n-dimensional manifold = Hausdorff, second countable, locally homeo to R^n

Closed = compact without boundary

A FUNDAMENTAL CLASS is a generator of H_n(M, Z) whose image in H_n(M, M-{x}) is the chosen orientation

pg 136

Answer 163

A

Consider connected M.

Every manifold M has an orientable 2-sheeted covering space.

M is orientable <=> double cover has two components.
M nonorientable <=> double cover is connected

It follows that if M is simply connected, M is orientable - can’t have connected double cover.

Further, if pi_1(M) does not contain an index 2 subgroup, then M is orientable - no 2-sheeted covering

Examples

S^n –> RP^n if n is even
T^2 –> Klien bottle

pg 130-131

Answer 164

A

This provides the intuition for Poincare Duality.

Draw picture of dual cell structures on S^2 as a cube…

0-cells 2-cells
1-cells 1-cells
2-cells 0-cells

Pairing of cells and dual cells gives an identification C_0 = C_2, C_1 = C_1, C_2 = C*_0 of cellular chain groups

Use Z_2 coefficients to avoid orientation issues

Boundary map for C_i becomes coboundary map for C*_2-i

The argument works for any closed n-manifold with dual cell structures.

Answer 165

A

aka Eilenburg-MacLane Space - any path connected space X with contractible universal cover X tilda and pi_1(X) = G.

Recall - universal cover normally only simply connected, extra strength of contractible gives much more powerful structure

Examples

S^1 is a K(Z, 1) space
Any closed surface except S^2 and RP^2 is a K(G,1) space - note if M_g has g >2, then M_g covers M_2, so they both have same universal cover
RP^2 not K(G,1) space

pgs 138-139

Answer 166

A

Recall S^inf is union of all S^n = {(x1,x2,x3,…) in R^inf | only finitely many terms nonzero and sum xi^2 = 1}

Key is the shift map
pg 140

Answer 167

A

Using the fact that S^inf is contractible, we view S^inf as unit sphere in C^inf which gives an easy action of Z_n on S^inf by scalar mult by nth root of unity. The quotien S^inf / Z_n has fundamental group Z_n and contractible universal cover S^inf so is a K(Z_n, 1)-space

Answer 168

A

This says the homotopy type of a K(G,1) space depends only on the group G - so can be seen as a topological model of the group

The key to the proof is the following proposition:

Prop. Let X be a connected CW complex and Y be a K(G,1) space. Then every homomorphism of groups pi_1(X, x0) –> pi_1(Y, y0) is induced by a map of spaces (X, x0) –> (Y, y0) which is unique up to homotopy preserving both basepoints

Pf of prop.
Let H = pi_1(X, x0) and G = pi_1(Y, y0)
Key: A map X –> Y inducing phi: H –> G is equivalent to a phi-equivariant map X tilda –> Y tilda.

Then since Y tilda contractible, we really have no obstruction to building any map we like
pg 142 - 145

Answer 169

A

These provide simplest examples of the following phenomena:

Manifolds which agree on pi_1 and all homology groups but are not homotopy equivalent - L(5, 1) and L(5, 2)
Manifolds which agree on all homotopy groups but are not homeomorphic - L(7, 1) and L(7, 2)

We require gcd(m,l) = 1. Then view S^3 as { (z, w) in C^2 : |z|^2+|w|^2 = 1}. Z_m = <a> acts on S^3 via (z, w) –> (az, z^l w) where a is primative mth root of unity. This is a covering space action with quotient a closed 3-manifold L(m, l) = S^3 / Z_m.</a>

If m=2 we get RP^3

pi_1(L(m,l) ) = Z_m

H_0 = Z
H_1 = Z_m
H_2 = 0
H_3 = Z

KEY: The fundamental group and homology groups can’t see the value of l. Our algebraic invariants can’t distinguish these spaces.

These spaces are homeomorphic <=> m’ = m and l’ = l^+-1 mod m

Homotopy equivalent <=> m = m’ and l’ = +-k^2 l mod m for some k in Z

These are not K(G, 1) spaces. In fact, Z_m does not admit a finite dimensional K(G, 1) space. It then follows that if X is any finite dimensional K(G, 1) CW complex, then G is torsion free.

pg 146 - 153</a>

Answer 170

A

pg 154 - 159

Answer 171

A

By conjugation

Answer 172

A

I*_120 = the preimage of the icosahedral group I_60 < SO(3, R) under the hom SU(2) –> SO(3, R)

Note I_60 is isomorphic to A_5 - simple of order 60
and we can prove I*_120 is perfect - has trivial abelianization

Answer 173

A

This is the first example of a HOMOLOGY SPHERE - a closed n-manifold with the same integral homology as S^n - which is not homeomorphic to S^n.

Sigma^3 = S^3 / I*_120

Where I*_120 acts by left multiplication.

See homology is same as S^3 using abelinization, universal coeficients, and Poincare

See not S^3 since pi_1 is I*_120

Answer 174

A

Hurwicz theorem gives H_1 = pi_1 abelianized

universal coefficients + Poincare duality fill in H^0 –> H_3 and H^1 —> H_2
H_0 = Z since connected

Answer 175

A

The homotopy class of maps f: (I^n, boundary I^n) –> (X, x0) where the homotopies must map boundary I^n to x0 for all t.

Equivalent to mapping (S^n , s0) –> (X, x0) but easier to compute with.

n= 0 : pi_0(X, x0) I^0 is a point, boundary of I^0 is empty set. So we map ({x}, empty set) –> (X, x0) - gives set of path components of X - no natural group structure

n=1 : covered heavily a group - often non-abelian

n > 1 : an abelian group

pg 166

Answer 176

A

First define + on set by copying pi_1 concatenation definition in just the first coordinate of f. Interpret also using spheres… pg 168

Group structure is easy, commutativity proof relies on 2 + dimensions to move around maps in
pg 167

Answer 177

A

For any gamma path in X from x0 to x1, we can define gamma f ; (I^n, boundary I^n) —> (X, x0) by following gamma on each radial line segment.

Then if [f] in pi_1(X, x0) –> [gamma f] is an automorphjism of pi_n(X, x0) - gives action of pi_1 on pi_n

Answer 178

A

A covering space projection p : (X tilda, x0 tilda) —> (X, x0) induces isomorphisms p_* on pi_n for all n > 1.

Hatcher 351

Answer 179

A

pi_n of a product of spaces is isomorphic to product of pi_n for each individual space

Hatcher 352

Answer 180

A

X is aspherical if X is path-connected and pi_n(X) = 0 for all n >= 2

X is n-connected if pi_n(X) = 0 for all k <= n

Examples

K(G, 1) -spaces are aspherical
0-connected = path connected
1-connected = simply connected

pg 169

Answer 181

A

I^n = unit cube [0,1]^n
I^n-1 = the set {(s1, ... , sn) | sn = 0} in I^n
J^n-1 = closure of (boundary I^n) - I^n-1 = all faces of boundary I^n except I^n-1

Now p_n(X, A, x0) = homotopy classes of maps f : (I^n, boundary I^n, J^n-1) —> (X, A, x0) with homotopies of the same type (X, A, x0)

Answer 182

A

The trivial elements of relative homotopy are the balls that can be pushed entirely into A without moving boundary (think 2-disks). A disk in A can be homotoped to x0 easily.

=> We are given a homotopy f_t : (D^n, S^n-1, s0) —> (X, A, x0) from f to a constant map c(y) = x0. Inside the domain of f_t, D^n x I, there is a family of discs starting at D^n x {0} and ending at (D^n x {1} ) union (S^n-1 x I) all sharing same boundary S^n-1 x {0}. This gives us a homotopy D^n x I which pushes f into A keeping f fixed on S^n-1

<= Now suppose f is homotopic rel S^n-1 to g with image in A. Clearly [f] = [g] in pi_n(X, A, xo). We just need to show [g] = 0 in pi_n. g: (D^n, S^n-1, s0) –> (X, A, x0) is a map with image in A. D^n def retracts to s0 (straight-line homotopy) so we can just compose g with the d.r. to get a homotopy shrinking D^n to a point without leaving A.

pg 171 - 172

Answer 183

A

Very simple compared to homology exact sequence of pair.

First, note pi_n(X, {x0}, x0) = pi_n(X, x0) (X, x0)
(X, x0) = (X, {x0}, x0) –> (X, A, x0)

Boundary map induced by restriction of
(D^n, S^n-1, s0) –> (X, A , x0) to (S^n, s0) –> (A, x0)

Can very easily see what is happening on boundary - roughly matches the intuition for homology exact sequence

Thm. For any pointed pair of spaces (X, A, x0) where x0 in A < X, the following sequence is exact:

… –> pi_n(A, x0) –> pi_n(X, x0) –> pi_n(X, A, x0) –> pi_n-1(A, x0) –> … –> pi_1(X, x0) –> pi_1(X, A, x0) –> pi_0(A, x0) –> pi_0(X, x0)

Notice above pi_1(X, x0) is the last group in this l.e.s.

Need to check boundary^2 = 0 and exactness

pgs 173 - 176

Answer 184

A

Thm. Every map f: X –> Y of CW complexes is homotopic to a cellular map [i.e. maps X^(n) to Y^(n) for all n].

If f is already cellular on a subcomplex A < X, then exists homotopy rel A.

Notice this is part of a long line of results like this where we show a continuous map between spaces in some category can be homotoped to the desired map from that category. i.e. cts between smooth manifolds homotopic to smooth etc.

The trickiest part here is dealing with the possibility of space filling curves/surfaces. We need to show any lower dimensional cell which fills up a higher dimensional cell is homotopic to one that does not fill up the higher dimensional cell - slightly more general proof pgs 180 - 183. Basic idea is that we can pull a map of a lower dimensional space taught to piecewise linear.

With this handled, Cellular approximation is proved on pgs 183 - 184. The key is that since we can homotope a space filling map to miss a point, we can further push it all the way down onto boundary of cell…

Answer 185

A

This follows immediately from cellular approximation. We use a cell structure on S^n : one 0-cell, one n-cell.
Then the k-skeleton is a point

Answer 186

A

The ORDER of an open cover U of a space X is the largest k s.t. U has k+1 sets with nonempty intersection

If X is a metric space, MESH(U) = sup (diam u) for u in U

A REFINEMENT of U is an open cover V s.t. for all v in V, there exists a u in U s.t. v < u.

Answer 187

A

The minimum order of an open cover provides some measure of dimension - this was how Lesbgue thought of it. Notice for any epsilon > 0 I^n has an open cover of order <= n and mesh < epsilon. Any compact K in R^n with open cover U has a finite refinement of order <=n

So the order really is capturing dimension in R^n - combinatorial complexity of how different sets can overlap each other

pg 179

Answer 188

A

Recall, a map f:S^n –> S^n induces f_* : H_n(S^n) –> H_n(S^n) DEGREE of f = d means f_* is multiplication by d.

Thm. Two maps S^n –> S^n (n > 0) are homotopic <=> they have the same degree.

=> Easy. Homotopic maps induce the same map on homology.

<= Induction on n. For n =1, we already know pi_1(S^1) = Z (degree) by covering space theory.

Prop 1. Let x1, … , xm, y1, …, ym in S^n be distinct points (n>1). Then there exists a homeomorphism of S^n homotopic to 1 that maps x_i to y_i for all i.

Prop 2. For n > 1 every map S^n –> S^n is homotopic to one that preserves hemispheres i.e. the closed upper [lower] hemisphere maps to itself (forces equator to go to equator)

Proof of Thm. Recall from fall semester that a map f: S^n –> S^n has a suspension Sf : S^n+1 –> S^n+1 s.t. deg(f) = deg(Sf). We claim each map f: S^n –> S^n is homotopic to the iterated suspension of the map S^1 to S^1 taking z –> z^d for some d in Z. Prove by induction on n. Base n=1 is done.

pg 185 - 191

Answer 189

A

Simple application of Brouwer-Hopf Degree thm.

Answer 190

A

A FIBER BUNDLE is a map p: E –> B s.t. “all fibers p^-1(b) are homeomorphic top the same space F.” Locally looks like a product. Each point of B has a nbd U s.t. p^-1(U) = U x F. Have commutative diagram… pg 192

TOTAL SPACE = E the big fibered space
BASE SPACE = B the space we project onto
FIBER = F the preimage of a point of B
LOCAL TRIVIALIZATIONS = the maps h: p^-1(U) –> U x F

F –> E –> B

Answer 191

A

Covering spaces: A bundle with a discrete fiber is the same as a covering space with constant number of sheets
Mobius band: I = fiber, Mobius = total, S^1 = base
Klein bottle: S^1 = fiber, Klien = total, S^1 = base
Projective spaces: S^0 –> S^n –> RP^n
S^1 –> S^2n+1 –> CP^n
S^3 –> S^4n+3 –> HP^n

pgs 193 - 195

Answer 192

A

There are only 4 finite dimensional real division algebras: R, C, H, O of dimensions 1, 2, 4, 8 over R. We get a Hopf fibration for R, C, H, O but can’t go further. A theorem of Adams states that 4 Hopf fibrations are the only fiber bundles F –> E –> B with F, E, B all spheres.

These just come from sphere coverings of the one dimensional projective spaces which just compactify R, C, H, O to S^1, S^2, S^4, S^8 respectively by adding point at infinity.

Real :S^0 –> S^1 –> S^1 = RP^n
Complex: S^1 –> S^3 –> S^2 = CP^1 the projection from S^3 to S^2 just takes (z,w) –> z/w the slope of the line
Quaternionic: S^3 –> S^7 –> S^4 = HP^1
Octonian: S^7 –> S^15 –> S^8 = OP^1

pgs 196 - 199

Answer 193

A

… –> pi_n(F) –> pi_n(E) –> pi_n(B) –> pi_n-1(F) –> … –> pi_0(B).

Examples
1. If F = discrete and B = path connected, then we are looking at covering spaces. Since pi_n(F) = pi_n-1(F) for n>1, we recover the isomorphism of pi_n(E) with pi_n(B)

If F = path-connected and B = aspherical, then we get a s.e.s. of pi_1
0 –> pi_1(F) –> pi_1(E) –> pi_1(B) –> 0
The Hopf fibration pi_n(S^3) –> pi_n(S^2) is an isomorphism if n > 2. In particular pi_3(S^2) = Z! The first nontrivial example of a higher homotopy group pi_n+k(S^n) - very different than homology - still an open problem to understand.
HOMOGENEOUS SPACES - let G be a Lie group acting by diffeomorphisms on a manifold M. If the action is transitive, M is a homogeneous space. M = G / H for H = Stab(x0) orbit/stabalizer thm
Then H –> G –> M = G/H is a fiber bundle
O(n) acts on S^n-1 transitively with H = O(n-1) for all n >0 we get a bundle O(n-1) –> O(n) –> S^n-1 so pi_i(O(n-1) ) = pi_i( O(n) ) for i < n-2. In particular, if we hold i constant and take n–> inf, we see that pi_i stabilizes for large n. Does not depend on n.

pgs 200 - 202

Answer 194

A

As with covering space theory, fiber bundle theory relies heavily on our ability to lift homotopies from base space to total space.

We say a map p: E –> B has the HOMOTOPY LIFTING PROPERTY w.r.t. a space X if any homotopy X x I –> B with a lift X x {0} –> E extends to a lift X x I –> E

Note: covering spaces have HLP w.r.t any X.

The most important/useful case is when X is a disk. We say p: E –> B is a (Serre) FIBRATION if it has HLP w.r.t. X = D^n for all n.

pg 203

Answer 195

A

Example on pg 204 shows B x F –> B is a fibration so locally a fiber bundle is a fibration. We just need to show this local property can be extended to a global fibration.

pg208-209

Answer 196

A

Let (X, A) be a CW-pair, (Y,B) any topological pair B != 0. Assume for each n s.t. X-A has cells of dim n pi_n(Y, B, y0) = 0 for all y0 in B. Then every map f : (X, A) –> (Y, B) is homotopic rel A to a map X –> B.

Be sure to discuss corollary that (Y, X) CW pair and X –> Y induces isos on pi_n for all n, then X is a def retract of Y.

pgs 209-210

Answer 197

A

A map f: X –> Y of connected CW complexes that induces isomorphisms on pi_n for all n is a homotopy equivalence.

PF. By cellular approximation f is homotopic to a cellular map g: X –>Y. Consider mapping cylinder of g: X–>Y.

Since g cellular, Mg is a CW complex and (Mg, X) is a CW pair
Mg def retracts to Y
By corollary of compression lemma, Mg d.r. onto X.

Hence X and Y are homotopy equivalent.

NOTE: It is not enough to show pi_n(X) = pi_n(Y) for all n. There mist be a map f:X –> Y inducing isomorphisms f_*

An important corollary:

A CW complex is contractible <=> pi_n(X) = 0 for all n.
Just apply Whitehead to the constant map X –> point.

pg 211 -212

Answer 198

A

Yes! Whitehead requires the existence of a map f: X –> Y inducing the isomorphisms of homotopy groups.

3 -dim Lens spaces L(5,1) vs. L(5,2)
S^2 and S^3 x CP^inf

Answer 199

A

A weak homotopy equivalence f: X –> Y is a map inducing isomorphisms on pi_n for all n.

For CW complexes, a weak homotopy equivalence is a homotopy equivalence by Whitehead.

Answer 200

A

By universal coefficients, inducing isos on H_n => inducing isos on H^n. Replacing Y with the mapping cylnder Mf, it suffices to consider the special case X < Y and f: X –> Y is the inclusion.

By l.e.s. in pi_n for (X,Y) inducing iso on pi_n <=> pi_n(Y, X) = 0

By l.e.s. in H_n for (X,Y) inducing iso on H_n <=> H_n(Y,X) = 0.

So we just need to show p_n(Y,X) = 0 for all n => H_n(Y,X) = 0 for all n.

pg 215-216

Answer 201

A

Given a topological space X, a CW approximation to X is a CW complex Z and a weak homotopy equivalence f:Z –>X.

THM. Every topological space has a CW approximation.

For studying pi_n, H_n, H^n of any space, we can simply pass to a CW approximation and work within the category of CW complexes and cellular maps. Much easier to deal with.

pgs 217 -

Answer 202

A

Yes. Start with a presentation 2-complex then attach 3 cells to kill all elements of pi_2(X^(2) ), 4 cells to kill … so X has pi_n(X) = 0 for all n > 1. Then X tilda has pi_n zero for all n so X tilda is contractible by Whitehead

We already saw that any two K(G,1) complexes for the same space are homotopy equivalent. Thus we can study groups using topology!

pg 219 -222

Brainscape's Knowledge GenomeTM

Algebraic Topology Flashcards

Brainscape's Knowledge Genome^TM