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Part II — Algebraic Topology
Based on lectures by H
...
They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine
...
The fundamental group of a space, homomorphisms induced by maps of spaces,
change of base point, invariance under homotopy equivalence
...
Path-lifting and homotopy-lifting properties, and
their application to the calculation of fundamental groups
...
*Construction
of the universal covering of a path-connected, locally simply connected space*
...

[5]
The Seifert-Van Kampen theorem
Free groups, generators and relations for groups, free products with amalgamation
...
Applications to the
calculation of fundamental groups
...
[3]
Homology
Simplicial homology, the homology groups of a simplex and its boundary
...
*Proof of functoriality for continuous maps, and of
homotopy invariance*
...

The Mayer-Vietoris theorem
...
Rational homology groups; the
Euler-Poincar´e characteristic and the Lefschetz fixed-point theorem
...
1 Some recollections and conventions
...
2 Cell complexes
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0 Motivation
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1 Homotopy
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2 Paths
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3 The fundamental group
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1 Covering space
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2 The fundamental group of the
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its applications

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4 Some group theory
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5 Seifert-van Kampen theorem
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4 The fundamental group of all surfaces
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1 Simplicial complexes
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2 Simplicial approximation
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1 Simplicial homology
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2 Some homological algebra
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3 Homology calculations
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4 Mayer-Vietoris sequence
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5 Continuous maps and homotopy invariance
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6 Homology of spheres and applications
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7 Homology of surfaces
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8 Rational homology, Euler and Lefschetz numbers

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0

Introduction

II Algebraic Topology

Introduction

In topology, a typical problem is that we have two spaces X and Y , and we want
to know if X ∼
= Y , i
...
if X and Y are homeomorphic
...
But what if they
are not? How can we prove that two spaces are not homeomorphic?
For example, are Rm and Rn homeomorphic (for m 6= n)? Intuitively, they
should not be, since they have different dimensions, and in fact they are not
...
It turns out we are much better
at algebra than topology
...
For example, we will be able to reduce the problem of whether Rm
and Rn are homeomorphic (for m =
6 n) to the question of whether Z and {e}
are isomorphic, which is very easy
...

Let Dn be the n dimensional unit disk, and S n−1 be the n−1 dimensional unit
sphere
...
Alternatively, this
says that we cannot continuously map the disk onto the boundary sphere such
that the boundary sphere is fixed by the map
...
This is something we can prove in 5 seconds
...

In algebraic topology, we will be developing a lot of machinery to do this sort
of translation
...
It will take some hard work,
and will be rather tedious and boring at the beginning
...

In case you are completely uninterested in topology, and don’t care if Rm and
n
R are homeomorphic, further applications of algebraic topology include solving
equations
...
If you are not
interested in these either, you may as well drop this course
...
1

II Algebraic Topology

Definitions
Some recollections and conventions

We will start with some preliminary definitions and conventions
...
In this course, the word map will always refer to continuous
maps
...

We are going to build a lot of continuous maps in algebraic topology
...
The gluing lemma tells us that
this works
...
If f : X → Y is a function of topological spaces,
X = C ∪ K, C and K are both closed, then f is continuous if and only if the
restrictions f |C and f |K are continuous
...
Suppose f is continuous
...
So f |C is continuous
...

If f |C and f |K are continuous, then for any closed A ⊆ Y , we have
−1
f −1 (A) = f |−1
C (A) ∪ f |K (A),

which is closed
...

This lemma is also true with “closed” replaced with “open”
...

We will also need the following technical lemma about metric spaces
...
Let (X, d) be a compact metric space
...
Then there is some δ such that for each x ∈ X, there is some α ∈ A
such that Bδ (x) ⊆ Uα
...

Proof
...
Then for each n ∈ N, there is some xn ∈ X such that
B1/n (xn ) is not contained in any Uα
...
Suppose this subsequence converges to y
...
Since Uα
is open, there is some r > 0 such that Br (y) ⊆ Uα
...
But then
B1/n (xn ) ⊆ Br (y) ⊆ Uα
...


1
...
Even if we require them
to be compact, Hausdorff etc, we can often still produce really ugly topological
spaces with weird, unexpected behaviour
...

To build cell complexes, we are not just gluing maps, but spaces
...
For a space X, and a map f : S n−1 → X, the
space obtained by attaching an n-cell to X along f is
X ∪f Dn = (X q Dn )/∼,
where the equivalence relation ∼ is the equivalence relation generated by x ∼ f (x)
for all x ∈ S n−1 ⊆ Dn (and q is the disjoint union)
...
So we are just sticking a disk onto X by attaching the boundary of
the disk onto a sphere within X
...
A (finite) cell complex is a space X obtained by
(i) Start with a discrete finite set X (0)
...

X
= X
q
α

For example, given the X (0) above, we can attach some loops and lines to
obtain the following X (1)

We can add surfaces to obtain the following X (2)

(iii) Stop at some X = X (k)
...

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Definitions

II Algebraic Topology

To define non-finite cell complexes, we just have to remove the words “finite” in
the definition and remove the final stopping condition
...
So instead let’s look at a
non-cell complex
...
The following is not a cell complex: we take R2 , and add a circle with
radius 12 and center (0, 21 )
...
We obtain something like

This is known as the Hawaiian Earring
...
However, in the definition of a cell
complex, the cells are supposed to be completely unrelated and disjoint, apart
from intersecting at the origin
...

In particular, if we take the following sequence (0, 1), (0, 21 ), (0, 14 ), · · · , it
converges to (0, 0)
...
We could have made it such that the nth
cell has radius n
...

We will see that the Hawaiian Earring will be a counterexample to a lot of
our theorems here
...


2
...
Let’s first do the
simple case, where m = 1, n = 2
...

This is not hard
...
Let’s try to
remove a point from each of them
...
However, removing a point does not have this effect on
R2
...

Unfortunately, this does not extend very far
...
What else can we do?
Notice that when we remove a point from R2 , sure it is still connected, but
something has changed
...
If the origin were there,
we can keep shrinking the circle down until it becomes a point
...


The strategy now is to exploit the fact that R2 \ {0} has circles which cannot be
deformed to points
...
1

Homotopy

We have just talked about the notion of “deforming” circles to a point
...
This process of deformation is known as homotopy
...

Notation
...

Definition (Homotopy)
...
A homotopy from f to g
is a map
H :X ×I →Y
such that
H(x, 0) = f (x),

H(x, 1) = g(x)
...
For each time t, H( · , t) defines a map X → Y
...

If such an H exists, we say f is homotopic to g, and write f ' g
...

As mentioned at the beginning, by calling H a map, we are requiring it to
be continuous
...
We might want to make sure
that when we are deforming from a path f to g, the end points of the path don’t
move
...
We say f is homotopic to g rel A, written f '
g rel A, if for all a ∈ A ⊆ X, we have
H(a, t) = f (a) = g(a)
...

Our notation suggests that homotopy is an equivalence relation
...
For spaces X, Y , and A ⊆ X, the “homotopic rel A” relation is
an equivalence relation
...

Proof
...

(ii) Symmetry: if H(x, t) is a homotopy from f to g, then H(x, 1 − t) is a
homotopy from g to f
...

We want to show that f ' h rel A
...

We know how to continuously deform f to g, and from g to h
...
We define H 00 : X × I → Y by
(
H(x, 2t)
0 ≤ t ≤ 12
00
H (x, t) =
H 0 (x, 2t − 1) 12 ≤ t ≤ 1
This is well-defined since H(x, 1) = g(x) = H 0 (x, 0)
...
It is easy to check that H 00 is a homotopy rel A
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We can extend the notion of homotopy to spaces as well
...
Here, we replace equality by homotopy
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A map f : X → Y is a homotopy equivalence if there exists a g : Y → X such that f ◦ g ' idY and g ◦ f ' idX
...

If a homotopy equivalence f : X → Y exists, we say that X and Y are
homotopy equivalent and write X ' Y
...
Clearly,
homeomorphic spaces are homotopy equivalent
...

Example
...
We have a natural inclusion map
i : X ,→ Y
...

y
r(y)

In particular, we define r : Y → X by
r(y) =

y

...
We will now see that i ◦ r ' idY
...
So this is just the projection map
...

t + (1 − t)kyk
This is continuous, and H( · , 0) = i ◦ r, H( · , 1) = idY
...
We started with a 2-dimensional R2 \ {0} space, and squashed it into a
one-dimensional sphere
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Dimensions seem to be a rather fundamental thing in geometry, and we are
discarding it here
...
We will later see that this is what homotopy equivalence
preserves
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Let Y = Rn , X = {0} = ∗
...
Again, we have
r ◦ i = idX
...

Again, from the point of view of homotopy theory, Rn is just the same as a
point! You might think that this is something crazy to do — we have just given
up a lot of structure of topological spaces
...
For example,
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Homotopy and the fundamental group

II Algebraic Topology

it is often much easier to argue about the one-point space ∗ than the whole of
R2 ! By studying properties that are preserved by homotopy equivalence, and
not just homeomorphism, we can simplify our problems by reducing complicated
spaces to simpler ones via homotopy equivalence
...

Notation
...

Definition (Contractible space)
...

We now show that homotopy equivalence of spaces is an equivalence relation
...

Lemma
...

Proof
...
Then we are done since
homotopy between maps is an equivalence relation
...

(i) Consider the following composition:
X ×I

H

Y

g0

Z

It is easy to check that this is the first homotopy we need to show g0 ◦ f0 '
g0 ◦ f1
...
Homotopy equivalence of spaces is an equivalence relation
...
Symmetry and reflexivity are trivial
...
We will show that h ◦ f : X → Z is a homotopy
equivalence with homotopy inverse g ◦ k
...

Similarly,
(g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f ' g ◦ idY ◦ f = g ◦ f ' idX
...

Definition (Retraction)
...
A retraction r : X → A
is a map such that r ◦ i = idA , where i : A ,→ X is the inclusion
...

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Homotopy and the fundamental group

II Algebraic Topology

This map sends everything in X to A without moving things in A
...

Definition (Deformation retraction)
...
A deformation retraction is strong if we require this homotopy
to be a homotopy rel A
...

Example
...
Then the constant map
r : X → A is a retraction
...


2
...
A path in a space X is a map γ : I → X
...

If γ(0) = γ(1), then γ is called a loop (based at x0 )
...
It can be self-intersecting
or do all sorts of weird stuff
...
In homotopy
theory, we do so using the idea of paths
...

Definition (Concatenation of paths)
...

This is continuous by the gluing lemma
...


x0

x2
x1

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II Algebraic Topology

Definition (Inverse of path)
...

This is exactly the same path but going in the opposite direction
...

Definition (Constant path)
...

We haven’t actually got a good algebraic system
...
Also, we are not able to combine arbitrary paths in a space
...
We can view this as a first attempt at associating
things to topological spaces
...
We can define a relation on X: x1 ∼ x2 if there
exists a path from x1 to x2
...
The equivalence classes [x] are called path components
...


In the above space, we have three path components
...
e
...
However, this is a first step
at associating something to spaces
...

One important property of this π0 is that not only does it associate a set to
each topological space, but also associates a function between the corresponding
sets to each continuous map
...
For any map f : X → Y , there is a well-defined function
π0 (f ) : π0 (X) → π0 (Y ),
defined by
π0 (f )([x]) = [f (x)]
...

(ii) For any maps A

h

B

k

C , we have π0 (k ◦ h) = π0 (k) ◦ π0 (h)
...
To show this is well-defined, suppose [x] = [y]
...
Then f ◦ γ is a path from f (x) to f (y)
...

(i) If f ' g, let H : X × I → Y be a homotopy from f to g
...
Then
H(x, · ) is a path from f (x) to g(x)
...
e
...
So π0 (f ) = π0 (g)
...

(iii) π0 (idX )([x]) = [idX (x)] = [x]
...

Corollary
...

Example
...

This is a rather silly example, since we can easily prove it directly
...

Now let’s return to our operations on paths, and try to make them algebraic
...
Paths γ, γ 0 : I → X are homotopic as paths if
they are homotopic rel {0, 1} ⊆ I, i
...
the end points are fixed
...


x0

x1

Note that we would necessarily want to fix the two end points
...

This homotopy works well with our previous operations on paths
...
Let γ1 , γ2 : I → X be paths, γ1 (1) = γ2 (0)
...


γ10

γ20

x0

x2

x1
γ1

γ2

Proof
...
Then we have the diagram
γ10

x0

H1

γ20

x1

γ1

H2
γ2

13

x2

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Homotopy and the fundamental group

II Algebraic Topology

We can thus construct a homotopy by
(
H1 (s, 2t)
0 ≤ t ≤ 12
H(s, t) =

...

Proposition
...
Then

(i) (γ0 · γ1 ) · γ2 ' γ0 · (γ1 · γ2 )
(ii) γ0 · cx1 ' γ0 ' cx0 · γ0
...

Proof
...

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Homotopy and the fundamental group

2
...
We want to try to do
this using paths
...
According to our proposition,
this operation satisfies associativity, inverses and identity
...

The idea is to fix one of the points x0 in our space, and only think about
loops that start and end at x0
...

This tells us that we aren’t going to just think about spaces, but spaces with
basepoints
...

Definition (Fundamental group)
...
The fundamental group of X (based at x0 ), denoted π1 (X, x0 ), is the set of homotopy
classes of loops in X based at x0 (i
...
γ(0) = γ(1) = x0 )
...

Often, when we write the homotopy classes of paths [γ], we just get lazy and
write γ
...
The fundamental group is a group
...
Immediate from our previous lemmas
...
Unfortunately, it is rather difficult to prove that a space has a non-trivial
fundamental group, until we have developed some relevant machinery
...
Instead,
we will look at some properties of the fundamental group first
...
A based space is a pair (X, x0 ) of a space X and a
point x0 ∈ X, the basepoint
...
A based homotopy is a
homotopy rel {x0 }
...
We can do the same for π1
...
To a based map
f : (X, x0 ) → (Y, y0 ),
there is an associated function
f∗ = π1 (f ) : π1 (X, x0 ) → π1 (Y, y0 ),
defined by [γ] 7→ [f ◦ γ]
...

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Homotopy and the fundamental group

II Algebraic Topology

(ii) If f ' f 0 , then π1 (f ) = π1 (f 0 )
...


(B, b)

k

(C, c) , we have π1 (k ◦ h) =

(iv) π1 (idX ) = idπ1 (X,x0 )
Proof
...

In category-theoretic language, we say that π1 is a functor
...
However, to define the fundamental group, we had to make a
compromise and pick a basepoint
...
We don’t
want a basepoint! Hence, we should look carefully at what happens when we
change the basepoint, and see if we can live without it
...
Hence,
the first importance of picking a basepoint is picking a path component
...

Now we want to compare fundamental groups with different basepoints
...
Suppose we have a loop γ at x0
...
We first pick a path
u : x0
x1
...
e
...


x0

x1
u

Proposition
...

This satisfies
(i) If u ' u0 , then u# = u0#
...
Then (u · v)# = v# ◦ u#
...


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Homotopy and the fundamental group

II Algebraic Topology

A nicer way of writing this is
f∗

π1 (X, x0 )
u#

π1 (Y, y0 )
(f ◦u)#

f∗

π1 (X, x1 )

π1 (Y, y1 )

The property says that the composition is the same no matter which way
we go from π1 (X, x0 ) to π1 (Y, y1 )
...
These diagrams will appear all of the time in this course
...

It is important (yet difficult) to get the order of concatenation and composition
right
...

Proof
...
Note that (u−1 )# = (u# )−1 , which is why we have
an isomorphism
...

So the basepoint isn’t really too important
...

While the two groups are isomorphic, the actual isomorphism depends on which
path u : x0
x1 we pick
...
In particular, we cannot say “let α ∈ π1 (X, x0 )
...

We can also see that if (X, x0 ) and (Y, y0 ) are based homotopy equivalent,
then π1 (X, x0 ) ∼
= π1 (Y, y0 )
...


So
f∗ ◦ g∗ = idπ1 (Y,y0 ) ,

g∗ ◦ f∗ = idπ1 (X,x0 ) ,

and f∗ and g∗ are the isomorphisms we need
...
If this were a based homotopy, we know
f∗ = g∗ : π1 (X, x0 ) → π1 (Y, f (x0 ))
...
How can we relate f∗ : π1 (X, x0 ) → π1 (Y, f (x0 )) and g∗ : π1 (X, x0 ) →
π1 (Y, g(x0 ))?
First of all, we need to relate the groups π1 (Y, f (x0 )) and π1 (Y, g(x0 ))
...
To produce this, we can use
the homotopy H
...
Now we have
three maps f∗ , g∗ and u#
...

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Homotopy and the fundamental group

X

II Algebraic Topology

Y

f

f (x0 )
x0
g(x0 )
g
Lemma
...

Proof
...

We need to check that
g∗ ([γ]) = u# ◦ f∗ ([γ])
...

To prove this result, we want to build a homotopy
...


Our plan is to exhibit two homotopic paths `+ and `− in I × I such that
F ◦ `+ = g ◦ γ,

F ◦ `− = u−1 · (f ◦ γ) · u
...
So to construct a homotopy, we make ourselves work in a
much nicer space I × I
...


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II Algebraic Topology

I ×I
`+

Y
g

F

`−

u−1

u

f

More precisely, `+ is the path s 7→ (s, 1), and `− is the concatenation of the
paths s 7→ (0, 1 − s), s 7→ (s, 0) and s 7→ (1, s)
...
If this is not obvious, we can
manually check the homotopy
L(s, t) = t`+ (s) + (1 − t)`− (s)
...
Hence F ◦ `+ 'F ◦L F ◦ `− as paths
...
We
have
F ◦ `+ (s) = H(γ(s), 1) = g ◦ γ(s)
...

So done
...

With this lemma, we can show that fundamental groups really respect
homotopies
...
If f : X → Y is a homotopy equivalence, and x0 ∈ X, then the
induced map
f∗ : π1 (X, x0 ) → π1 (Y, f (x0 ))
...

While this seems rather obvious, it is actually non-trivial if we want to do
it from scratch
...
So this proof involves
some real work
...
Let g : Y → X be a homotopy inverse
...


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Homotopy and the fundamental group

X

II Algebraic Topology

Y

f

x0
f (x0 )
0

u

g ◦ f (x0 )
g
We have no guarantee that g ◦ f (x0 ) = x0 , but we know that our homotopy H 0
gives us u0 = H 0 (x0 , · ) : x0
g ◦ f (x0 )
...


However, we know that u0# is an isomorphism
...

Doing it the other way round with f ◦ g instead of g ◦ f , we know that g∗ is
injective and f∗ is surjective
...

With this theorem, we can finally be sure that the fundamental group
is a property of the space, without regards to the basepoint (assuming path
connectedness), and is preserved by arbitrary homotopies
...

Definition (Simply connected space)
...

Example
...
Hence, any contractible space is simply connected since
it is homotopic to ∗
...

There is a useful characterization of simply connected spaces:
Lemma
...

Proof
...
Now
note that u · v −1 is a loop based at x0 , it is homotopic to the constant path, and
v −1 · v is trivially homotopic to the constant path
...

On the other hand, suppose there is a unique homotopy class of paths x0
x1
for all x0 , x1 ∈ X
...
So π1 (X, x0 ) is trivial
...

Groups were created to represent symmetries of objects
...
If not, something wrong is probably going on
...
So what do they act on? Can we
find something on which these fundamental groups act on?
An answer to this would also be helpful in more practical terms
...
This would be easy if we
can make the group act on something — if the group acts non-trivially on our
thing, then clearly the group cannot be trivial
...


3
...

˜
X

p
U
x0

X

˜ p:X
˜ → X),
Definition (Covering space)
...

Whether we require p to be surjective is a matter of taste
...

˜ → X if
Definition (Evenly covered)
...

where p|Vα : Vα → U is a homeomorphism, and each of the Vα ⊆ X
Example
...
Duh
...
Consider p : R → S 1 ⊆ C defined by t 7→ e2πit
...


21

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Covering spaces

II Algebraic Topology

R
p−1 (1)

p

1

S1

Here we have p−1 (1) = Z
...
Before we do that, we can guess what the fundamental
group of S 1 is
...
So we can characterize each loop by
the number of times it loops around the circle, and it should not be difficult to
convince ourselves that two loops that loop around the same number of times
can be continuously deformed to one another
...
However, we also know that p−1 (1) = Z
...
We will show that this is not
a coincidence
...
Consider pn : S 1 → S 1 (for any n ∈ Z \ {0}) defined by z 7→ z n
...

This time the pre-image of 1 would be n copies of 1, instead of Z copies of 1
...
Consider X = RP2 , which is the real projective plane
...
e
...

We can also think of this as the space of lines in R3
...
There is, however, a more convenient way of thinking about RP2
...


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Covering spaces

II Algebraic Topology

U

U
We define p : S 2 → RP2 to be the quotient map
...

As we mentioned, a covering space of X is (locally) like many “copies” of
X
...

˜
“lift” it to a function f : Y → X,
This is not always possible, but when it is, we call this a lift
...
Let f : Y → X be a map, and p : X
˜ such that f = p ◦ f˜, i
...
the following
space
...

f˜(Y )
f˜

˜
X
p

f (Y )

Y

X

f

It feels that if we know which “copy” of X we lifted our map to, then we
already know everything about f˜, since we are just moving our f from X to
that particular copy of X
...
Let p : X
be both lifts of f
...
In particular, if Y is connected, f˜1 and f˜2 agree either
everywhere or nowhere
...
If we know a point in the
lift, then we know the whole path
...
e
...

Proof
...
Let y be such that f˜1 (y) = f˜2 (y)
...
Let U
˜ , p(U
˜ ) = U and p| ˜ : U
˜ → U is a homeomorphism
...
We will show that f˜1 = f˜2 on V
...

Since p|U˜ is a homeomorphism, it follows that
f˜1 |V = f˜2 |V
...
Suppose not
...
So f˜1`
(y) 6=
f˜2 (y)
...
Let p−1 (U ) = Uα
...
Then V = f˜1−1 (Uβ ) ∩ f˜2−1 (Uγ ) is
an open neighbourhood of y, and hence intersects S by definition of closure
...
But f˜1 (x) ∈ Uβ and f˜2 (x) ∈ Uγ ,
and hence Uβ and Uγ have a non-trivial intersection
...
So
S is closed
...
How about existence? Given a map,
is there guarantee that we can lift it to something? Moreover, if I have fixed a
“copy” of X I like, can I also lift my map to that copy? We will later come up
with a general criterion for when lifts exist
...

˜ → X be a covering space,
Lemma (Homotopy lifting lemma)
...
Let f˜0 be a lift of f0
...
e
...

(ii) H
This lemma might be difficult to comprehend at first
...
Then a homotopy is just a path
...
Let p : X
˜ such that p(˜
path, and x
˜0 ∈ X
x0 ) = x0 = γ(0)
...
e
...

This is exactly the picture we were drawing before
...
Note that we have already proved uniqueness
...

24

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Covering spaces

II Algebraic Topology

In theory, it makes sense to prove homotopy lifting, and path lifting comes
immediately as a corollary
...
So instead, we will prove path lifting, which is something we can more
easily visualize and understand, and then use that to prove homotopy lifting
...
Let
S = {s ∈ I : γ˜ exists on [0, s] ⊆ I}
...

(ii) S is open
...
If s 6∈ S, then pick an evenly covered neighbourhood U of γ(s)
...
So s − 2ε 6∈ S
...
So S is
closed
...
So γ˜
exists
...
So H(y,
·)
is defined
...
Steps
of the proof are
(i) Use compactness of I to argue that the proof of path lifting works on small
neighbourhoods in Y
...

(iii) By uniqueness of lifts, these path liftings agree when they overlap
...

With the homotopy lifting lemma in our toolkit, we can start to use it to
do stuff
...
We are now
going to build a bridge between these two, and show how covering spaces can be
used to reflect some structures of the fundamental group
...

We have just showed that we are allowed to lift homotopies
...
The homotopy lifting lemma
does not tell us that the lifted homotopy preserves basepoints
...

˜ are lifts
Corollary
...

If γ ' γ 0 as paths, then γ˜ and γ˜ 0 are homotopic as paths
...


25

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Covering spaces

II Algebraic Topology

Note that if we cover the words “as paths” and just talk about homotopies,
then this is just the homotopy lifting lemma
...

˜ a lift of H with H(
˜ · , 0) = γ˜
...
The homotopy lifting lemma gives us an H,
γ˜0

γ0

cx 0

H

lift

cx 1

cx˜0

γ

˜
H

cx˜1

γ˜

˜ square is γ˜
...

˜ · , 1) is a lift of H( · , 1) = γ 0 , starting at x
Now H(
˜0
...
So this is indeed a homotopy between γ˜ and γ˜ 0
...

˜
We know that H(0,
· ) is a lift of H(0, · ) = cx0
...
By uniqueness of lifts, we must have H(0,
· ) = cx˜0
...
So this is a homotopy of paths
...
is it? Is it possible that we have four copies of x0 but just three copies
of x1 ? This is obviously possible if X is not path connected — the component
containing x0 and the one containing x1 are completely unrelated
...
If X is a path connected space, x0 , x1 ∈ X, then there is a bijection
p−1 (x0 ) → p−1 (x1 )
...
We want to use this to construct a bijection
Proof
...
The obvious thing to do
is to use lifts of the path γ
...

The inverse map is obtained by replacing γ with γ −1 , i
...
fγ −1
...
Now
notice that γ˜ −1 is a lift of γ −1 starting at x
˜1 and ending at x
˜0
...

So fγ −1 is an inverse to fγ , and hence fγ is bijective
...
A covering space p : X
−1
X is n-sheeted if |p (x)| = n for any (and hence all) x ∈ X
...
Is there any number we can assign to fundamental groups? Well, the
index of a subgroup might be a good candidate
...

One important property of covering spaces is the following:
˜ → X is a covering map and x
˜ then
Lemma
...

Proof
...

˜ We let γ = p ◦ γ˜
...
e
...

0
paths, the homotopy lifting lemma then gives us a homotopy upstairs between γ˜
and cx˜0
...

As we have originally said, our objective is to make our fundamental group
act on something
...

Let’s look again at the proof that there is a bijection between p−1 (x0 ) and
−1
p (x1 )
...
This end point may or may not be our original x
˜0
...

However, we are not really interested in paths themselves
...
However, this is fine
...
In particular, these have the same end points
...


27

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Covering spaces

II Algebraic Topology

˜
X

x
˜0
γ˜
x
˜00

p
γ

x0

X

Now this gives an action of π1 (X, x0 ) on p−1 (x0 )! Note, however, that this will
not be the sort of actions we are familiar with
...
To see this, we have to consider what happens when we
perform two operations one after another, which you shall check yourself
...

When we have an action, we are interested in two things — the orbits, and
the stabilizers
...

Lemma
...

˜ is path
(i) The action of π1 (X, x0 ) on p−1 (x0 ) is transitive if and only if X
connected
...

˜ x
(ii) The stabilizer of x
˜0 ∈ p−1 (x0 ) is p∗ (π1 (X,
˜0 )) ⊆ π1 (X, x0 )
...

˜ x
Note that p∗ π1 (X,
˜0 )\π1 (X, x0 ) is not a quotient, but simply the set of
cosets
...

Note that this is great! If we can find a covering space p and a point x0
such that p−1 (x0 ) is non-trivial, then we immediately know that π1 (X, x0 ) is
non-trivial!
Proof
...
Then we can project this to γ = p ◦ γ˜
...
e
...
Then by the definition of the action,
x
˜0 · [γ] = γ˜ (1) = x
˜00
...
Then γ˜ is a loop based at x˜0
...

(iii) This follows directly from the orbit-stabilizer theorem
...
We can be more ambitious, and try to actually find π1 (X, x0 )
...
Then we have a
bijection between π1 (X, x0 ) and p−1 (x0 )
...

space X
˜ → X is a universal cover
Definition (Universal cover)
...

if X
We will look into universal covers in depth later and see what they really are
...
If p : X
π1 (X, x0 ) → p−1 (x0 )
...
To obtain a bijection, we need to
pick a starting point x
˜0 ∈ p−1 (x0 )
...


3
...
We are going to consider
the space S 1 and a universal covering R
...
There is a bijection π1 (S 1 , 1) → p−1 (1) = Z
...
The map ` : π1 (S 1 , 1) → p−1 (1) = Z is a group isomorphism
...
We know it is a bijection
...

The idea is to write down representatives for what we think the elements should
be
...
Since R is simply
connected, there is a unique homotopy class between any two points
...
So
[γ] = [un ]
...
Therefore
m · un = u
`([um ][un ]) = `([um · um ]) = m + n = `([um+n ])
...

What have we done? In general, we might be given a horrible, crazy loop
in S 1
...
So we pull it
up to the universal covering R
...
We then project
this homotopy down to S 1 , to get a homotopy from γ to un
...

With the fundamental group of the circle, we do many things
...
Since C \ {0} is homotopy equivalent to S 1 , its fundamental group is Z
as well
...
Any such group homomorphism must be of the form t 7→ nt, and the
winding number is given by n
...

Also, we have the following classic application:
Theorem (Brouwer’s fixed point theorem)
...
If f : D2 → D2 is continuous, then there is some x ∈ D2 such
that f (x) = x
...
Suppose not
...


30

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Covering spaces

II Algebraic Topology

f (x)
x
g(x)
We define g : D2 → S 1 as in the picture above
...
In other words, the following
composition is the identity:
S1

ι

D2

g

S1

idS 1

Then this induces a homomorphism of groups whose composition is the identity:
Z

ι∗

{0}

g∗

Z

idZ

But this is clearly nonsense! So we must have had a fixed point
...
What about D3 ? Can we prove a similar theorem?
Here the fundamental group is of little use, since we can show that the fundamental group of S n for n ≥ 2 is trivial
...


3
...

We have just shown that p : R → S 1 is a universal cover
...
What would be a
universal cover of the torus S 1 × S 1 ? An obvious guess would be p × p : R × R →
S 1 × S 1
...
Alternatively, we can see it as a quotient of the square,
where we identify the following edges:

31

3

Covering spaces

II Algebraic Topology

Then what does it feel like to live in the torus? If you live in a torus and look
around, you don’t see a boundary
...
The difference is that in the torus, you aren’t actually seeing
free space out there, but just seeing copies of the same space over and over again
...
Whenever we move one unit
horizontally or vertically, we get back to “the same place”
...
This space has a huge translation symmetry
...

We see that if we live inside the torus S 1 × S 1 , it feels like we are actually
living in the universal covering space R × R, except that we have an additional
symmetry given by the fundamental group Z × Z
...
We would like to
say that universal covers always exist
...

Firstly, we should think — what would having a universal cover imply?
˜ Pick any point x0 ∈ X, and pick an
Suppose X has a universal cover X
...

˜ If we draw a
evenly covered neighbourhood U in X
...
But we know
˜ is simply connected
...
Hence γ
that X
is also homotopic to the constant path
...

It seems like for every x0 ∈ X, there is some neighbourhood of x0 that is
simply connected
...
The homotopy
˜ and can pass through anything
from γ˜ to the constant path is a homotopy in X,

32

3

Covering spaces

II Algebraic Topology

˜ not just U
˜
...
So U itself need not be simply connected
...

Definition (Locally simply connected)
...

As we mentioned, what we actually want is a weaker condition
...
X is semi-locally simply connected
if for all x0 ∈ X, there is some neighbourhood U of x0 such that any loop γ
based at x0 is homotopic to cx0 as paths in X
...
This is really not interesting, since we don’t care
if a space is semi-locally simply connected
...
We still need one more additional condition:
Definition (Locally path connected)
...

It is important to note that a path connected space need not be locally path
connected
...

Theorem
...

Note that we can alternatively define a universal covering as a covering space
of X that is also a covering space of all other covers of X
...
However, that proof
is not too helpful since it does not tell us where the universal covering comes
from
...

Proof
...
Suppose we have a
˜ Then this lifts to some x
˜ If we have any other point
universal covering X
...

˜ since X
˜ should be path connected, there is a path α
x
˜ ∈ X,
˜:x
˜0
x
˜
...

have another path, then since X
˜ with a path from x
Hence, we can identify each point in X
˜0 , i
...

˜ ←→ {paths α
˜
{points of X}
˜ from x
˜0 ∈ X}/'
...

˜
This is not too helpful though, since we are defining X
˜
However, by path lifting, we know that paths α
˜ from x
˜0 in X biject with paths
α from x0 in X
...

˜ ←→ {paths α from x0 ∈ X}/'
...

X
˜ → X is given by [α] 7→ α(1)
...

33

3

Covering spaces

3
...
We have already established the result
that covering maps are injective on π1
...
It
turns out that as long as we define carefully what we mean for based covering
spaces to be “the same”, this is a one-to-one correspondence — each subgroup
corresponds to a covering space
...

Recall that we have shown that π1 (X, x0 ) acts on p−1 (x0 )
...

Having groups acting on a cube is fun because the cube has some structure
...

We note that we can make π1 (X, x0 ) “act on” the universal cover
...
In general, a point
˜ can be thought of as a path α on X starting from x0
...

α

˜
X

x
˜00

x
˜0 γ˜

p
γ

α
x0

X

We will use this idea and return to the initial issue of making subgroups correspond to covering spaces
...
We want to say “For any subgroup H ≤ π1 (X, x0 ),
˜ x
˜ x
there is a based covering map p : (X,
˜0 ) → (X, x0 ) such that p∗ π1 (X,
˜0 ) = H”
...
So we need some
additional assumptions
...
Let X be a path connected, locally path connected and semilocally simply connected space
...


34

3

Covering spaces

II Algebraic Topology

Proof
...

We will show that this is indeed well-defined later, using topology!
Some people like to define the free group in a different way
...
This definition also does not make it clear that
the free group F (S) of any set S exists
...

Lemma
...

−1
Proof
...

α to φ(sα )
Then the values of f on all other elements must be determined by

f (x1 · · · xn ) = f (x1 ) · · · f (xn )
since f is a homomorphism
...
So it suffices to
show that this f is a well-defined homomorphism
...

To show this is a homomorphism, suppose
x = x1 · · · xn a1 · · · ak ,

−1
y = a−1
k · · · a1 y1 · · · ym ,

where y1 6= x−1
n
...

Then we can compute




f (x)f (y) = φ(x1 ) · · · φ(xn )φ(a1 ) · · · φ(ak ) φ(ak )−1 · · · φ(a1 )−1 φ(y1 ) · · · φ(ym )
= φ(x1 ) · · · φ(xn ) · · · φ(y1 ) · · · φ(ym )
= f (xy)
...

We call this a “universal property” of F (S)
...


39

4

Some group theory

II Algebraic Topology

Definition (Presentation of a group)
...
We denote by hhRii the normal closure of R, i
...
the smallest normal
subgroup of F (S) containing R
...

i=1

Then we write
hS | Ri = F (S)/hhRii
...
For example, we can write
D2n = hr, s | rn , s2 , srsri
...

Lemma
...
e
...

In some sense, this says that all the group hS | Ri does is it satisfies the
relations in R, and nothing else
...
Let G be a group
...
There is also an obvious
surjection G → G
...
Let R = ker f = hhRii
...

This is a really stupid example
...

However, this tells us that every group has a presentation
...
ha, b | bi ∼
= hai ∼
= Z
...
With a bit of work, we can show that ha, b | ab−3 , ba−2 i = Z/5
...
2

Another view of free groups

Recall that we have not yet properly defined free groups, since we did not show
that multiplication is well-defined
...

Again let S be a set
...
We define X by
40

4

Some group theory

II Algebraic Topology

x0

a

b

We call this a “rose with 2 petals”
...
For each s ∈ S, we have one 1-cell, es , and we fix a path γs : [0, 1] → es
that goes around the 1-cell once
...

What’s the universal cover of X? Since we are just lifting a 1-complex, the
result should be a 1-complex, i
...
a graph
...
e
...
We also know that every vertex in the
universal cover is a copy of the vertex in our original graph
...
So it has to look like something this:

x
˜0

In X, we know that at each vertex, there should be an edge labeled a going in;
an edge labeled a going out; an edge labeled b going in; an edge labeled b going
˜ as well
...
This should be the case in X

b
x
˜0 a

a

b

a

b
b

˜ to x0 ∈ X,
The projection map is then obvious — we send all the vertices in X
and then the edges according to the labels they have, in a way that respects the
direction of the arrow
...

41

4

Some group theory

II Algebraic Topology

We are now going to show that this tree “is” the free group
...

For example, the following path corresponds to w = abb−1 b−1 ba−1 b−1
...
So for all x
(i) X
˜ ∈ p−1 (x0 ), there is an edge-path γ˜ : x
˜0

x
˜
...
How can
an edge fail to be locally injective? It is fine if we just walk along a path,
since we are just tracing out a line
...

(iii) Each point x
˜ ∈ p−1 (x0 ) is joined to x
˜0 by a unique locally injective
edge-path
...

We can thus conclude that there are bijections
F (S)

p−1 (x0 )

42

π1 (X, x0 )

4

Some group theory

II Algebraic Topology

that send x
˜ to the word w ∈ F (S) such that γ˜w is a locally injective edge-path
to x
˜0
x
˜; and x
˜ to [γ] ∈ π1 (X, x0 ) such that x
˜0 · [γ] = x
˜
...
It is easy to see that the
operations on F (S) and π1 (X, x0 ) are the same, since they are just concatenating
words or paths
...
So this
induces an isomorphism F (S) ∼
= π1 (X, x0 )
...
3

Free products with amalgamation

We managed to compute the fundamental group of the circle
...
Recall that at the beginning,
we defined cell complexes, and said these are the things we want to work with
...
So we want to know what
happens when we glue things together
...
e
...
We would like to describe π1 (X) in terms of π1 (A) and π1 (B)
...

Definition (Free product)
...
The free product G1 ∗ G2 is defined to be
G1 ∗ G2 = hS1 ∪ S2 | R1 ∪ R2 i
...
A group can have many different
presentations, and it is not clear this is well-defined
...
Note that there are natural homomorphisms ji : Gi → G1 ∗ G2
that send genera
(ii) We need dn−1 (z) = 0 so that [z] ∈ Hn−1 (A);
(iii) We need to check [z] is well-defined, i
...
it does not depend on our choice
of y and x for the homology class [x]
...

We now check them one by one:
(i) Since everything involved in defining ∂∗ are homomorphisms, it follows
that ∂∗ is also a homomorphism
...
To do so, we need to add an additional layer
...
We will use the commutativity of the
diagram
...

By exactness at An−2 , we know in−2 is injective
...

(iii) (a) First, in the proof, suppose we picked a different y 0 such that jn (y 0 ) =
jn (y) = x
...
So y 0 − y ∈ ker jn = im in
...
Then
dn (y 0 ) = dn (y 0 − y) + dn (y)
= dn ◦ in (a) + dn (y)
= in−1 ◦ dn (a) + dn (y)
...

(b) Suppose [x0 ] = [x]
...
This time,
we add a layer above
...

By surjectivity of jn+1 , we can write c = jn+1 (b) for some b ∈ Bn+1
...

The next step of the proof is to find some y such that jn (y) = x
...

So the corresponding y 0 is y 0 = y + dn+1 (b)
...

80

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Simplicial homology

II Algebraic Topology

(iv) This is yet another standard diagram chasing argument
...
It is even more beneficial to attempt to prove this yourself
...

(b) ker j∗ ⊆ im i∗ : Let [b] ∈ Hn (B)
...
Then there is
some c ∈ Cn+1 such that jn (b) = dn+1 (c)
...
By commutativity,
we know jn (b) = jn ◦ dn+1 (b0 ), i
...

jn (b − dn+1 (b0 )) = 0
...

Moreover,
in−1 ◦ dn (a) = dn ◦ in (a) = dn (b − dn+1 (b0 )) = 0,
using the fact that b is a cycle
...
So [a] ∈ Hn (A)
...

So [b] ∈ im i∗
...
To compute ∂∗ (j∗ ([b])), we first
pull back jn (b) to b ∈ Bn
...
However, we know dn (b) = 0 since b is a cycle
...
e
...

(d) ker ∂∗ ⊆ im j∗ : Let [c] ∈ Hn (C) and suppose ∂∗ ([c]) = 0
...

By assumption, ∂∗ ([c]) = [a] = 0
...
Then by commutativity we know
dn (b) = dn ◦ in (a0 )
...

So [b − in (a0 )] ∈ Hn (B)
...

So [c] ∈ im j∗
...
Let b ∈ Bn be such that jn (b) = c,
and a ∈ An−1 be such that in (a) = dn (b)
...
Then
i∗ ([a]) = [in (a)] = [dn (b)] = 0
...

(f) ker i∗ ⊆ im ∂∗ : Let [a] ∈ Hn (A) and suppose i∗ (

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