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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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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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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

3

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

3

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

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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 have an intermediate
simply connected space, let X
˜ x
group H such that π1 (X,
˜0 ) = 1 ≤ H ≤ π1 (X, x0 )
...
Now we
action of π1 (X, x0 )
...
We define ∼H on X
the orbit relation for the action of H, i
...
x
˜ ∼H y˜ if there is some h ∈ H such
˜ be the quotient space X/∼
¯ H
...
We then let X
We can now do the messy algebra to show that this is the covering space we
want
...
e
...

Now we want to prove injectivity
...

˜ x
Suppose we have path-connected spaces (Y, y0 ), (X, x0 ) and (X,
˜0 ), with
˜ x
˜0 ) → (X, x0 ) a covering map
...

˜ x
˜0 ) → (X, x0 ) be a covering map of pathLemma (Lifting criterion)
...
If f : (Y, y0 ) → (X, x0 ) is a continuous map, then there is a (unique)
˜ x
lift f˜ : (Y, y0 ) → (X,
˜0 ) such that the diagram below commutes (i
...
p ◦ f˜ = f ):
˜ x
(X,
˜0 )
f˜

(Y, y0 )

f

p

(X, x0 )

if and only if the following condition holds:
˜ x
f∗ π1 (Y, y0 ) ≤ p∗ π1 (X,
˜0 )
...
So this lemma is
really about existence
...
g
...
So paths and
homotopies can always be lifted
...
One direction is easy: if f˜ exists, then f = p ◦ f˜
...
So we
know that im f∗ ⊆ im p∗
...

In the other direction, uniqueness follows from the uniqueness of lifts
...
We define f˜ as follows:
Given a y ∈ Y , there is some path αy : y0
y
...
By path lifting, this path lifts uniquely to β˜y in X
...
Note that if f exists, then this must be what f sends y to
...


35

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Covering spaces

II Algebraic Topology

Suppose we picked a different path αy0 : y0
y
...

˜ x
˜0 ) says f ◦ γ is the image of a
Our condition that f∗ π1 (Y, y0 ) ≤ p∗ π1 (X,
0
˜
˜
˜
˜ and hence have the same
loop in X
...
So this shows that f˜ is well-defined
...
First, observe that any open set U ⊆ X
can be written as a union of V˜ such that p|V˜ : V˜ → p(V˜ ) is a homeomorphism
...

Let y ∈ f˜−1 (V˜ ), and let x = f (y)
...
We claim that W ⊆ f˜−1 (V˜ )
...
Then f sends
˜ is given by p|−1 (f (γ)),
this to a path from x to f (z)
...
So it follows that f˜(z) = p|V−1
(f
(z))
∈ V˜
...
What this statement properly means is made precise in the following
proposition:
˜1, x
˜2, X
˜ 2 ) be path-connected based spaced,
˜1 ), (X
Proposition
...
Then we have
˜1, x
˜ 2 , x˜2 )
p1∗ π1 (X
˜1 ) = p2∗ π1 (X
if and only if there is some homeomorphism h such that the following diagram
commutes:
h
˜1, x
˜2, x
(X
˜1 )
(X
˜2 )
p1

p2

(X, x0 )
i
...
p1 = p2 ◦ h
...
We are saying that we can find a nice homeomorphism
that works well with the covering map p
...
If such a homeomorphism exists, then clearly the subgroups are equal
...
By symmetry, we can get h−1 = p˜2
...

36

3

Covering spaces

II Algebraic Topology

Since idX˜ 1 is also a lift, by the uniqueness of lifts, we know p˜2 ◦ p˜1 is the identity
map
...

˜1, x
(X
˜1 )
p˜2

˜1, x
(X
˜1 )

p˜1

˜2, x
(X
˜2 )

p2

p1

(X, x0 )

p1

Now what we would like to do is to forget about the basepoints
...
This doesn’t actually change the
group itself, but if we are talking about subgroups, conjugation can send a
subgroup into a different subgroup
...

Proposition
...


37

4

Some group theory

4

II Algebraic Topology

Some group theory

Algebraic topology is about translating topology into group theory
...
Well, maybe you do, but not the right
group theory
...
1

Free groups and presentations

Recall that in IA Groups, we defined, say, the dihedral group to be
D2n = hr, s | rn = s2 = e, srs = r−1 i
...
This is, in some sense, the
“freest” group we can have
...

Definition (Alphabet and words)
...
We assume that
S ∩ S −1 = ∅
...
e
...

Example
...
Then words could be the empty word ∅, or a, or
aba−1 b−1 , or aa−1 aaaaabbbb, etc
...

When we see things like aa−1 , we would want to cancel them
...

Definition (Elementary reduction)
...

Since each reduction shortens the word, and the word is finite in length, we
cannot keep reducing for ever
...

Definition (Reduced word)
...

Example
...

Note that there is an inclusion map S → S ∗ that sends the symbol sα to the
word sα
...
The free group on the set S, written F (S), is the set
of reduced words on S ∗ together with some operations:
(i) Multiplication is given by concatenation followed by elementary reduction
to get a reduced word
...

−1
−1 −1
(iii) The inverse of x1 · · · xn is x−1
= sα
...

Note that we have not showed that multiplication is well-defined — we might
reduce the same word in different ways and reach two different reduced words
...
This is a cleaner
way to define the free group without messing with alphabets and words, but is
(for most people) less intuitive
...
We will state this definition as a lemma
...
If G is a group and φ : S → G is a set map, then there exists a unique
homomorphism f : F (S) → G such that the following diagram commutes:
F (S)
f

S

φ

G

where the arrow not labeled is the natural inclusion map that sends sα (as a
symbol from the alphabet) to sα (as a word)
...
Clearly if f exists, then f must send each sα to φ(sα ) and s−1

...
So if f exists, it must be unique
...

This is well-defined if we define F (S) to be the set of all reduced words, since
each reduced word has a unique representation (since it is defined to be the
representation itself)
...
Then
xy = x1 · · · xn y1 · · · ym
...

So f is a homomorphism
...
We can show that F (S) is the
unique group satisfying the conditions of this lemma (up to isomorphism), by
taking G = F (S) and using the uniqueness properties
...
Let S be a set, and let R ⊆ F (S) be any
subset
...
e
...
This can be given explicitly by
( n
)
Y
−1
hhRii =
gi ri gi : n ∈ N, ri ∈ R, gi ∈ F (S)
...

This is just the usual notation we have for groups
...

Again, we can define this with a universal property
...
If G is a group and φ : S → G is a set map such that f (r) = 1 for all
±1
±1
±1
r ∈ R (i
...
if r = s±1
φ(s2 )±1 · · · φ(sm )±1 = 1),
1 s2 · · · sm , then φ(r) = φ(s1 )
then there exists a unique homomorphism f : hS | Ri → G such that the
following triangle commutes:
hS | Ri
f

S

φ

G

Proof is similar to the previous one
...

Example (The stupid canonical presentation)
...
We can view
the group as a set, and hence obtain a free group F (G)
...
Then by the universal property of the free group, there is a
surjection f : F (G) → G
...
Then hG | Ri is a presentation
for G, since the first isomorphism theorem says
G∼
= F (G)/ ker f
...
For example, even the simplest non-trivial
group Z/2 will be written as a quotient of a free group with two generators
...

Example
...

Example
...


4
...
We are now going to do this using topology
...
For the following illustration, we will just assume
S = {a, b}, but what we will do works for any set S
...
This is a cell complex, with one 0-cell and |S|
1-cells
...
We will call the 0-cells and 1-cells vertices and
edges, and call the whole thing a graph
...
e
...
Moreover, this graph is connected
and simply connected, i
...
it’s a tree
...
So it must have 4
edges attached to it
...
So X
˜ looks like this:
out
...
It is easy to show this is really a covering map
...
Notice that
˜ starting at x
every word w ∈ S ∗ denotes a unique “edge path” in X
˜0 , where an
edge path is a sequence of oriented edges e˜1 , · · · , e˜n such that the “origin” of
e˜i+1 is equal to the “terminus” of e˜i
...


x
˜0

We can note a few things:
˜ is connected
...


(ii) If an edge-path γ˜ fails to be locally injective, we can simplify it
...
So it fails to be locally injective if two
consecutive edges in the path are the same edge with opposite orientations:
e˜i

e˜i+1

e˜i−1

e˜i+2

We can just remove the redundant lines and get
e˜i−1

e˜i+2

This reminds us of two things — homotopy of paths and elementary
reduction of words
...

(iv) For any w ∈ S ∗ , from (ii), we know that γ˜ is locally injective if and only if
w is reduced
...

So there is a bijection between F (S) and π1 (X, x0 )
...
So this bijection identifies the two group structures
...


4
...
But we want
to find the fundamental group of more things
...

Cell complexes are formed by gluing things together
...

Suppose a space X is constructed from two spaces A, B by gluing (i
...

X = A ∪ B)
...
To
understand this, we need to understand how to “glue” groups together
...
Suppose we have groups G1 = hS1 | R1 i, G2 = hS2 |
R2 i, where we assume S1 ∩ S2 = ∅
...

This is not a really satisfactory definition
...
However, it is clear that
this group exists
...
So we can completely
specify a simplicial map by writing down a finite amount of information
...
In particular, they are allowed to have repeats
...
Suppose we have the standard 2-simplex K as follows:
a2

a0

a1

The following does not define a simplicial map because ha1 , a2 i is a simplex in
K, but {f (a1 ), f (a2 )} does not span a simplex:
f (a0 ), f (a1 )

f (a2 )

On the other hand, the following is a simplicial map, because now {f (a1 ), f (a2 )}
spans a simplex, and note that {f (a0 ), f (a1 ), f (a2 )} also spans a 1-simplex
because we are treating the collection of three vertices as a set, and not a
simplex
...

Lemma
...

As we said, simplicial maps are nice, but they are not exactly what we want
...

Lemma
...

There is an obvious way to define this map
...

Proof
...

i=0

i=0

The result is in L because {f (ai )} spans a simplex
...
This is clearly
continuous on σ, and is hence continuous on |K| by the gluing lemma
...

60

6

Simplicial complexes

6
...
So given a
continuous map f , we would like to find a related simplicial map g
...
The
definition we will write down is slightly awkward, but it turns out this is the
most useful definition
...
Let x ∈ |K|
...
e
...

x∈σ∈K

The link of x, written LkK (x), is the union of all those simplices that do not
contain x, but are faces of a simplex that does contain x
...
Let f : |K| → |L| be a continuous map
between the polyhedra
...

(∗)
The following lemma tells us why this is a good definition:
Lemma
...

Furthermore, if f is already simplicial on some subcomplex M ⊆ K, then we get
g|M = f |M , and the homotopy can be made rel M
...
First we want to check g is really a simplicial map if it satisfies (∗)
...
We want to show that {g(a0 ), · · · , g(an )}
spans a simplex in L
...
Since σ contains each ai , we know that x ∈ StK (ai )
for all i
...


i=0

Hence we know that there is one simplex, say, τ that contains all g(ai ) whose
interior contains f (x)
...
So they span a face of τ , as required
...
We let H : |K| × I → |L| ⊆ Rm be
defined by
(x, t) 7→ t|g|(x) + (1 − t)f (x)
...
So we need to check that im H ⊆ |L|
...
It thus follows that
H(x × I) ⊆ τ ⊆ |L|
...
Then the homotopy is rel M by
the construction above
...
Since f (v) is a vertex
and g(v) is the only vertex in StL (g(v)), we must have f (v) = g(v)
...

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What’s the best thing we might hope for at this point? It would be great if
every map were homotopic to a simplicial map
...
Let’s consider the following K:

How many homotopy classes of continuous maps are there K → K? Countably
many, one for each winding number
...
The problem is that we don’t have enough vertices to realize all
those interesting maps
...
Suppose
we have the following simplex:

What we do is to add a point in the center of each simplex, and join them up:

This is known as the barycentric subdivision
...
We will show that for any map, as long
as we are willing to barycentrically subdivide the simplex many times, we can
find a simplicial approximation to it
...
The barycenter of σ = ha0 , · · · , an i is
σ
ˆ=

n
X
i=0

1
ai
...
The (first) barycentric subdivision K 0 of
K is the simplicial complex:
K 0 = {hˆ
σ0 , · · · , σ
ˆn i : σi ∈ K and σ0 < σ1 < · · · < σn }
...

The rth barycentric subdivision K (r) is defined inductively as the barycentric
subdivision of the r − 1th barycentric subdivision, i
...

K (r) = (K (r−1) )0
...
|K| = |K|0 and K 0 really is a simplicial complex
...
Too boring to be included in lectures
...
Even though |K 0 | and |K| are equal, the
identity map from |K 0 | to |K| is not a simplicial map
...
e
...
Then we
can define g : K 0 → K by sending σ
ˆ 7→ vσ
...

The key theorem is that as long as we are willing to perform barycentric
subdivisions, then we can always find a simplicial approximation
...
Le K and L be simplicial complexes, and f : |K| → |L| a continuous map
...

Furthermore, if f is already simplicial on M ⊆ K, then we can choose g such
that |g||M = f |M
...

To do this, we want to quantify how “fine” our subdivisions are
...
Let K be a simplicial complex
...

We have the following lemma that tells us how large our mesh is:
Lemma
...


The key point is that as r → ∞, the mesh goes to zero
...
The proof is purely technical
and omitted
...

Proof of simplicial approximation theorem
...
We have a natural cover of |L|, namely the open stars of all vertices
...

The idea is to barycentrically subdivide our K such that each open star of K is
contained in one of these things
...
By the previous lemma, there is an r such that µ(K (r) ) < δ
...
Hence it follows that
StK (r) (x) ⊆ Bδ (x) for all vertices x ∈ VK (r)
...

Therefore defining g(x) = w, we get
f (StK (r) (x)) ⊆ StL (g(x))
...

The last part follows from the observation that if f is a simplicial map, then
it maps vertices to vertices
...


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Simplicial homology

7
7
...

Instead, we are going to use this framework to move on and define some new
invariants of simplicial complexes K, known as Hn (K)
...
The drawback, however, is that the definitions are slightly less intuitive
at first sight
...
Instead, we will be using abelian groups, which really should be
thought of as Z-modules
...
e
...

Using Q makes some of our work easier, but as a result we would have lost some
information
...

Recall that we defined an n-simplex as the collection of n + 1 vertices that
span a simplex
...
What we want to do is to remember the orientation of the simplex
...

Definition (Oriented n-simplex)
...

We often denote an oriented simplex as σ, and then σ
¯ denotes the same
simplex with the opposite orientation
...
As oriented 2-simplices, (v0 , v1 , v2 ) and (v1 , v2 , v0 ) are equal, but
they are different from (v2 , v1 , v0 )
...
This is substantially harder in higher dimensions, and often we
just work with the definition instead
...
Let K be a simplicial complex
...
For each i, choose an
orientation on σi
...
This choice is not important, but we need to make it
...

Now let Cn (K) be the free abelian group with basis {σ1 , · · · , σ` }, i
...
Cn (K) ∼
=
Z`
...

In other words, an element of Cn (K) is just a formal sum of n-simplices
...
This will save us from
making exceptions for n = 0 cases later
...

In this definition, we have to choose a particular orientation for each of our
simplices
...

Note that if there are no n-simplices (e
...
when n = −1), we can still
meaningfully talk about Cn (K), but it’s just 0
...
We can think of elements in the chain group C1 (X) as “paths” in
X
...

Of course, with this setup, we can do more random things, like adding 57 copies
of σ6 to it, and this is also allowed
...

When defining fundamental groups, we had homotopies that allowed us to
“move around”
...
To do so, we need to define the boundary
homomorphisms
...
We define boundary homomorphisms
dn : Cn (K) → Cn−1 (K)
by
(a0 , · · · , an ) 7→

n
X

(−1)i (a0 , · · · , a
ˆi , · · · , an ),

i=0

where (a0 , · · · , a
ˆi , · · · , an ) = (a0 , · · · , ai−1 , ai+1 , · · · , an ) is the simplex with ai
removed
...
This is clear if
we draw some pictures in low dimensions:
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II Algebraic Topology

v0

−v0

v1

v1

If we take a triangle, we get
v1

v1

v0

v2

v0

v2

An important property of the boundary map is that the boundary of a boundary
is empty:
Lemma
...

In other words, im dn+1 ⊆ ker dn
...
This just involves expanding the definition and working through the
mess
...
The nth simplicial homology
group Hn (K) is defined as
Hn (K) =

ker dn

...
Since we
are going to draw pictures, we are going to start with the easy case of k = 1
...
First, we give
these things names
...
The elements of Ck (K) are called
k-chains of K, those of ker dk are called k-cycles of K, and those of im dk+1 are
called k-boundaries of K
...
In other words, dc = 0
...
e
...
For example, if we have the following cycle:
e1

e0

e2

We have
c = (e0 , e1 ) + (e1 , e2 ) + (e2 , e0 )
...

67

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So this c is indeed a cycle
...
e
...
This is why we call this a 1-boundary
...

v1

v0

v2

We see that a cycle that has a boundary has been “filled in”
...
Hence we define the homology
group as the cycles quotiented by the boundaries, and we interpret its elements
as k-dimensional “holes”
...
Let K be the standard simplicial 1-sphere, i
...
we have the following
in R3
...

Our chain groups are
C0 (K) = h(e0 ), (e1 ), (e2 )i ∼
= Z3
C1 (K) = h(e0 , e1 ), (e1 , e2 ), (e2 , e0 )i ∼
= Z3
...
Note that our notation is slightly confusing
here, since the brackets h · i can mean the simplex spanned by the vertices, or
the group generated by certain elements
...

Hence, the only non-zero boundary map is
d1 : C1 (K) → C0 (K)
...



−1 0
1
 1 −1 0 
0
1 −1
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II Algebraic Topology

We have now everything we need to know about the homology groups, and we
just need to do some linear algebra to figure out the image and kernel, and thus
the homology groups
...

=
=
im(d1 : C1 (K) → C0 (K))
im d1
im d1

After doing some row operations with our matrix, we see that the image of d1 is
a two-dimensional subspace generated by the image of two of the edges
...

What does this H0 (K) represent? We initially said that Hk (K) should represent
the k-dimensional holes, but when k = 0, this is simpler
...
We interpret this to mean K has one
path component
...

Similarly, we have
ker d1 ∼
H1 (K) =
= ker d1
...

So we also have

H1 (K) ∼
= Z
...

The fact that H1 (K) is non-trivial means that we do indeed have a hole in the
middle of the circle
...
Let L be the standard 2-simplex (and all its faces) in R3
...


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Since d1 is the same as before, the only new interesting boundary map is d2
...

We know that H0 (L) depends only on d0 and d1 , which are the same as for K
...

Again, the interpretation of this is that L is path-connected
...

im d2
h(e0 , e1 ) + (e1 , e2 ) + (e2 , e0 )i
This precisely illustrates the fact that the “hole” is now filled in L
...

im d3
This is zero since there aren’t any two-dimensional holes in L
...
We are now going to spend a lot of time developing formalism
...
2

Some homological algebra

We will develop some formalism to help us compute homology groups Hk (K) in
lots of examples
...

Definition (Chain complex and differentials)
...
We call these maps the differentials of C·
...

Definition (Cycles and boundaries)
...

The space of n-boundaries is
Bn (C) = im dn+1
...
The n-th homology group of C· is defined to be
Hn (C) =

ker dn
Zn (C)
=

...
Suppose we have two chain
complexes
...

In general, we want to have maps between these two sequences
...

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Definition (Chain map)
...
In other words, the following diagram commutes:
0

d0

C0

d1

f0

0

d0

d2

C1
f1

D0

d1

C2

d3

···

f2

D1

d2

D2

d3

···

We want to have homotopies
...
So [v] = [w] if and only if v and w are in the same path
component of K
...


7
...

Suppose we have a space K = M ∪ N
...

Recall that to state the Seifert-van Kampen theorem, we needed to learn
some new group-theoretic notions, such as free products with amalgamation
...
We will need to learn some algebra in
order to state the Mayer-Vietoris theorem
...

Definition (Exact sequence)
...

A collection of homomorphisms
fi−1

···

Ai

fi

Ai+1

fi+1

Ai+2

fi+2

···

is exact at Ai if
ker fi = im fi−1
...

Recall that we have seen something similar before
...
e
...
Here we are requiring exact
equivalence, which is something even better
...
Alternatively, we see the homology groups as measuring
the failure of a sequence to be exact
...

Definition (Short exact sequence)
...
e
...
So f is
injective
...
So g is
surjective
...

Since we like chain complexes, we can produce short exact sequences of chain
complexes
...
A short exact sequence
of chain complexes is a pair of chain maps i· and j·
0

A·

i·

B·
76

j·

C·

0

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Simplicial homology

II Algebraic Topology

such that for each k,
0

Ak

ik

Bk

jk

Ck

0

is exact
...

The reason why we care about short exact sequences of chain complexes
(despite them having such a long name) is the following result:
Theorem (Snake lemma)
...
In particular, there is a long
exact sequence (i
...
an exact sequence that is not short)
···

Hn (A)

i∗

j∗

Hn (B)

Hn (C)

∂∗

Hn−1 (A)

i∗

j∗

Hn−1 (B)

Hn−1 (C)

···

where i∗ and j∗ are induced by i· and j· , and ∂∗ is a map we will define in the
proof
...
To do this, we also need
to understand the maps i∗ , j∗ and ∂∗ , but we don’t need to understand all, since
we can deduce some of them with exactness
...

Note, however, if we replace Z in the definition of chain groups by a field (e
...

Q), then all the groups become vector spaces
...
Of course, this does not get us the right answer in exams,
since we want to have homology groups over Z, and not Q, but this helps us to
understand the exact sequences somewhat
...

We are first going to apply this result to obtain the Mayer-Vietoris theorem
...
Let K, L, M, N be simpl
Definition (Lefschetz number)
...

i≥0

Why is this a generalization of the Euler characteristic? Just note that the
trace of the identity map is the number of dimensions
...

Example
...


7

Simplicial homology

II Algebraic Topology

Example
...
So
L(α) = 1 + (−1)n (−1)n+1 = 1 − 1 = 0
...
We
will soon see why this is the case
...

Before that, we want to understand the Lefschetz number first
...
However, we would like to understand the Lefschetz number in terms of
the chain groups, since these are easier to comprehend
...

Lemma
...

Let A : V → V be a linear map such that A(W ) ⊆ W
...
Then
tr(A) = tr(B) + tr(C)
...
In the right basis,

A=


A0

...
This makes our life much easier when it
comes to computation
...
Let f· : C· (K; Q) → C· (K; Q) be a chain map
...

This is a great corollary
...
However, to actually do computations, we want to work
with the chain groups and actually calculate with chain groups
...
There is an exact sequence
0

Bi (K; Q)

Zi (K; Q)

Hi (K; Q)

0

This is since Hi (K, Q) is defined as the quotient of Zi over Bi
...
Let fiH , fiB , fiZ , fiC be the various
maps induced by f on the corresponding groups
...


i≥0

Because of the alternating signs in dimension, each fiB appears twice in the sum
with opposite signs
...

i≥0

Since tr(id |Ci (K;Q) ) = dimQ Ci (K; Q), which is just the number of i-simplices,
this tells us the Euler characteristic we just defined is the usual Euler characteristic, i
...

X
χ(X) =
(−1)i number of i-simplices
...

Theorem (Lefschetz fixed point theorem)
...
If L(f ) 6= 0, then f has a fixed point
...
We prove the contrapositive
...
We will show
that L(f ) = 0
...
We know this is non-zero, since f has no fixed
point, and X is compact (and hence the infimum point is achieved by some x)
...
We now let
g : K (r) → K
be a simplicial approximation to f
...
Also, we know
|f (x) − x| ≥ δ
...

2
So we must have g(x) 6∈ σ
...

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Now we compute L(f ) = L(|g|)
...
So we need to
compose it with si : Ci (K; Q) → Ci (K (r) ; Q) induced by inverses of simplicial
approximations to the identity map
...
So gi ◦ si
takes every simplex off itself
...

Example
...
g
...
So f
has a fixed point
...
Suppose G is a path-connected topological group, i
...
X is a group
and a topological space, and inverse and multiplication are continuous maps
...
This implies L(rg ) = 0
...
So
χ(G) = L(idG ) = L(rg ) = 0
...

This is quite fun
...
Which can be
topological groups? The torus is, since it is just S 1 × S 1 , and S 1 is a topological
group
...
However, other surfaces cannot since they
don’t have Euler characteristic 0

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