Part II Algebraic Topology
Based on lectures by H. Wilton
Notes taken by Dexter Chua
Michaelmas 2015
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Part IB Analysis II is essential, and Metric and Topological Spaces is highly desirable
The fundamental group
Homotopy of continuous functions and homotopy equivalence between topological
spaces. The fundamental group of a space, homomorphisms induced by maps of spaces,
change of base point, invariance under homotopy equivalence. [3]
Covering spaces
Covering spaces and covering maps. Path-lifting and homotopy-lifting properties, and
their application to the calculation of fundamental groups. The fundamental group
of the circle; topological proof of the fundamental theorem of algebra. *Construction
of the universal covering of a path-connected, locally simply connected space*. The
correspondence between connected coverings of
X
and conjugacy classes of subgroups
of the fundamental group of X. [5]
The Seifert-Van Kampen theorem
Free groups, generators and relations for groups, free products with amalgamation.
Statement *and proof* of the Seifert-Van Kampen theorem. Applications to the
calculation of fundamental groups. [4]
Simplicial complexes
Finite simplicial complexes and subdivisions; the simplicial approximation theorem. [3]
Homology
Simplicial homology, the homology groups of a simplex and its boundary. Functorial
properties for simplicial maps. *Proof of functoriality for continuous maps, and of
homotopy invariance*. [4]
Homology calculations
The homology groups of
S
n
, applications including Brouwer’s fixed-point theorem.
The Mayer-Vietoris theorem. *Sketch of the classification of closed combinatorical
surfaces*; determination of their homology groups. Rational homology groups; the
Euler-Poincar´e characteristic and the Lefschetz fixed-point theorem. [5]
Contents
0 Introduction
1 Definitions
1.1 Some recollections and conventions
1.2 Cell complexes
2 Homotopy and the fundamental group
2.0 Motivation
2.1 Homotopy
2.2 Paths
2.3 The fundamental group
3 Covering spaces
3.1 Covering space
3.2 The fundamental group of the circle and its applications
3.3 Universal covers
3.4 The Galois correspondence
4 Some group theory
4.1 Free groups and presentations
4.2 Another view of free groups
4.3 Free products with amalgamation
5 Seifert-van Kampen theorem
5.1 Seifert-van Kampen theorem
5.2 The effect on π
1
of attaching cells
5.3 A refinement of the Seifert-van Kampen theorem
5.4 The fundamental group of all surfaces
6 Simplicial complexes
6.1 Simplicial complexes
6.2 Simplicial approximation
7 Simplicial homology
7.1 Simplicial homology
7.2 Some homological algebra
7.3 Homology calculations
7.4 Mayer-Vietoris sequence
7.5 Continuous maps and homotopy invariance
7.6 Homology of spheres and applications
7.7 Homology of surfaces
7.8 Rational homology, Euler and Lefschetz numbers
0 Introduction
In topology, a typical problem is that we have two spaces
X
and
Y
, and we want
to know if
X
=
Y
, i.e. if
X
and
Y
are homeomorphic. If they are homeomorphic,
we can easily show this by writing down a homeomorphism. But what if they
are not? How can we prove that two spaces are not homeomorphic?
For example, are
R
m
and
R
n
homeomorphic (for
m 6
=
n
)? Intuitively, they
should not be, since they have different dimensions, and in fact they are not.
But how can we actually prove this?
The idea of algebraic topology is to translate these non-existence problems in
topology to non-existence problems in algebra. It turns out we are much better
at algebra than topology. It is much easier to show that two groups are not
isomorphic. For example, we will be able to reduce the problem of whether
R
m
and
R
n
are homeomorphic (for
m 6
=
n
) to the question of whether
Z
and
{e}
are isomorphic, which is very easy.
While the statement that
R
m
6
=
R
n
for
n 6
=
m
is intuitively obvious, algebraic
topology can be used to prove some less obvious results.
Let
D
n
be the
n
dimensional unit disk, and
S
n1
be the
n
1 dimensional unit
sphere. We will be able to show that there is no continuous map
F
:
D
n
S
n1
such that the composition
S
n1
D
n
S
n1
F
is the identity, where the first arrow is the inclusion map. Alternatively, this
says that we cannot continuously map the disk onto the boundary sphere such
that the boundary sphere is fixed by the map.
Using algebraic topology, we can translate this statement into an algebraic
statement: there is no homomorphism F : {0} Z such that
Z {0} Z
F
is the identity. This is something we can prove in 5 seconds.
By translating a non-existence problem of a continuous map to a non-existence
problem of a homomorphism, we have made our life much easier.
In algebraic topology, we will be developing a lot of machinery to do this sort
of translation. However, this machinery is not easy. It will take some hard work,
and will be rather tedious and boring at the beginning. So keep in mind that
the point of all that hard work is to prove all these interesting theorems.
In case you are completely uninterested in topology, and don’t care if
R
m
and
R
n
are homeomorphic, further applications of algebraic topology include solving
equations. For example, we will be able to prove the fundamental theorem of
algebra (but we won’t), as well as Brouwer’s fixed point theorem (which says
that every continuous function from
D
2
D
2
has a fixed point). If you are not
interested in these either, you may as well drop this course.
1 Definitions
1.1 Some recollections and conventions
We will start with some preliminary definitions and conventions.
Definition (Map). In this course, the word map will always refer to continuous
maps. We are doing topology, and never care about non-continuous functions.
We are going to build a lot of continuous maps in algebraic topology. To do
so, we will often need to glue maps together. The gluing lemma tells us that
this works.
Lemma (Gluing lemma). 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.
Proof. Suppose f is continuous. Then for any closed A Y , we have
f|
1
C
(A) = f
1
(A) C,
which is closed. So f|
C
is continuous. Similarly, f |
K
is continuous.
If f |
C
and f |
K
are continuous, then for any closed A Y , we have
f
1
(A) = f|
1
C
(A) f|
1
K
(A),
which is closed. So f is continuous.
This lemma is also true with “closed” replaced with “open”. The proof is
the same proof with “closed” replaced with “open”.
We will also need the following technical lemma about metric spaces.
Lemma. Let (
X, d
) be a compact metric space. Let
U
=
{U
α
}
αA
be an open
cover of
X
. Then there is some
δ
such that for each
x X
, there is some
α A
such that B
δ
(x) U
α
. We call δ a Lebesgue number of this cover.
Proof.
Suppose not. Then for each
n N
, there is some
x
n
X
such that
B
1/n
(
x
n
) is not contained in any
U
α
. Since
X
is compact, the sequence (
x
n
)
has a convergent subsequence. Suppose this subsequence converges to y.
Since
U
is an open cover, there is some
α A
such that
y U
α
. Since
U
α
is open, there is some
r >
0 such that
B
r
(
y
)
U
α
. But then we can find a
sufficiently large n such that
1
n
<
r
2
and d(x
n
, y) <
r
2
. But then
B
1/n
(x
n
) B
r
(y) U
α
.
Contradiction.
1.2 Cell complexes
In topology, we can construct some really horrible spaces. Even if we require them
to be compact, Hausdorff etc, we can often still produce really ugly topological
spaces with weird, unexpected behaviour. In algebraic topology, we will often
restrict our attention to some nice topological spaces, known as cell complexes.
To build cell complexes, we are not just gluing maps, but spaces.
Definition (Cell attachment). For a space
X
, and a map
f
:
S
n1
X
, the
space obtained by attaching an n-cell to X along f is
X
f
D
n
= (X q D
n
)/,
where the equivalence relation
is the equivalence relation generated by
x f
(
x
)
for all x S
n1
D
n
(and q is the disjoint union).
Intuitively, a map
f
:
S
n1
X
just picks out a subset of
X
that looks like
the sphere. So we are just sticking a disk onto
X
by attaching the boundary of
the disk onto a sphere within X.
Definition (Cell complex). A (finite) cell complex is a space X obtained by
(i) Start with a discrete finite set X
(0)
.
(ii)
Given
X
(n1)
, form
X
(n)
by taking a finite set of maps
{f
α
:
S
n1
X
(n1)
} and attaching n-cells along the f
α
:
X
(n)
=
X
(n1)
q
a
α
D
n
α
!
/{x f
α
(x)}.
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)
. The minimum such k is the dimension of X.
To define non-finite cell complexes, we just have to remove the words “finite” in
the definition and remove the final stopping condition.
We have just had an example of a cell complex above, and it is easy to
produce many more examples, such as the
n
-sphere. So instead let’s look at a
non-cell complex.
Example. The following is not a cell complex: we take
R
2
, and add a circle with
radius
1
2
and center (0
,
1
2
). Then we add another circle with radius
1
4
and center
(0
,
1
4
), then a circle with radius
1
8
and center (0
,
1
8
) etc. We obtain something like
This is known as the Hawaiian Earring.
Why is this not an (infinite) cell complex? We did obtain it by attaching
lots of 1-cells to the single point (0
,
0). However, in the definition of a cell
complex, the cells are supposed to be completely unrelated and disjoint, apart
from intersecting at the origin. However, here the circles clump together at the
origin.
In particular, if we take the following sequence (0
,
1)
,
(0
,
1
2
)
,
(0
,
1
4
)
, · · ·
, it
converges to (0
,
0). If this were a cell complex, then this shouldn’t happen
because the cells are unrelated, and picking a point from each cell should not
produce a convergent sequence (if you are not convinced, if we actually did
produce by attaching cells, then note that during the attaching process, we
needn’t have attached them this way. We could have made it such that the
n
th
cell has radius
n
. Then clearly picking the topmost point of each cell will not
produce a convergent sequence).
We will see that the Hawaiian Earring will be a counterexample to a lot of
our theorems here.
2 Homotopy and the fundamental group
This will be our first trick to translate topological spaces into groups.
2.0 Motivation
Recall that we wanted to prove that
R
n
6
=
R
m
for
n 6
=
m
. Let’s first do the
simple case, where m = 1, n = 2. We want to show that R 6
=
R
2
.
This is not hard. We know that
R
is a line, while
R
2
is a plane. Let’s try to
remove a point from each of them. If we remove a point from
R
, the space stops
being path connected. However, removing a point does not have this effect on
R
2
. Since being path connected is a topological property, we have now showed
that R and R
2
are not homeomorphic.
Unfortunately, this does not extend very far. We cannot use this to show
that R
2
and R
3
are not homeomorphic. What else can we do?
Notice that when we remove a point from
R
2
, sure it is still connected, but
something has changed.
Consider a circle containing the origin in
R
2
\ {
0
}
. If the origin were there,
we can keep shrinking the circle down until it becomes a point. However, we
cannot do this if the origin is missing.
The strategy now is to exploit the fact that
R
2
\ {
0
}
has circles which cannot be
deformed to points.
2.1 Homotopy
We have just talked about the notion of “deforming” circles to a point. We can
think of a circle in
X
as a map
S
1
X
, and we want to “deform” this map to
a point. This process of deformation is known as homotopy. Here we are going
to use the interval [0, 1] R a lot, and we will just call it I.
Notation.
I = [0, 1] R.
Definition (Homotopy). Let
f, g
:
X Y
be maps. 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).
We think of the interval
I
as time. For each time
t
,
H
(
· , t
) defines a map
X Y
.
So we want to start from f, move with time, and eventually reach g.
If such an
H
exists, we say
f
is homotopic to
g
, and write
f ' g
. If we want
to make it explicit that the homotopy is H, we write f '
H
g.
As mentioned at the beginning, by calling
H
a map, we are requiring it to
be continuous.
Sometimes, we don’t want a general homotopy. 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. In general, we have
Definition (Homotopy
rel
A). 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).
This notion will find itself useful later, but we don’t have to pay too much
attention to this yet.
Our notation suggests that homotopy is an equivalence relation. Indeed, we
have
Proposition. For spaces
X, Y
, and
A X
, the “homotopic
rel A
relation is
an equivalence relation. In particular, when
A
=
, homotopy is an equivalence
relation.
Proof.
(i) Reflexivity: f ' f since H(x, t) = f(x) is a homotopy.
(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.
(iii)
Transitivity: Suppose
f, g, h
:
X Y
and
f '
H
g rel A
,
g '
H
0
h rel A
.
We want to show that
f ' h rel A
. The idea is to “glue” the two maps
together.
We know how to continuously deform
f
to
g
, and from
g
to
h
. So we just
do these one after another. We define H
00
: X × I Y by
H
00
(x, t) =
(
H(x, 2t) 0 t
1
2
H
0
(x, 2t 1)
1
2
t 1
This is well-defined since
H
(
x,
1) =
g
(
x
) =
H
0
(
x,
0). This is also continuous
by the gluing lemma. It is easy to check that
H
00
is a homotopy
rel A
.
We now have a notion of equivalence of maps two maps are equivalent if
they are homotopic. We can extend the notion of homotopy to spaces as well.
Recall that when we defined homeomorphism, we required that there be some
f, g such that f g = id, g f = id. Here, we replace equality by homotopy.
Definition (Homotopy equivalence). A map
f
:
X Y
is a homotopy equiva-
lence if there exists a
g
:
Y X
such that
f g ' id
Y
and
g f ' id
X
. We call
g a homotopy inverse for f.
If a homotopy equivalence
f
:
X Y
exists, we say that
X
and
Y
are
homotopy equivalent and write X ' Y .
We are soon going to prove that this is indeed an equivalence relation on
spaces, but we first look at some examples of homotopy equivalent spaces. Clearly,
homeomorphic spaces are homotopy equivalent. However, we will see that we
can do much more “violent” things to a space and still be homotopy equivalent.
Example. Let
X
=
S
1
,
Y
=
R
2
\ {
0
}
. We have a natural inclusion map
i
:
X Y
. To obtain a map
Y X
, we can project each point onto the circle.
r(y)
y
In particular, we define r : Y X by
r(y) =
y
kyk
.
We immediately have
r i
=
id
X
. We will now see that
i r ' id
Y
. The
composition
i r
first projects each object into
S
1
, and then includes it back
into
R
2
\ {
0
}
. So this is just the projection map. We can define a homotopy
H : Y × I Y by
H(y, t) =
y
t + (1 t)kyk
.
This is continuous, and H( · , 0) = i r, H( · , 1) = id
Y
.
As we have said, homotopy equivalence can do really “violent” things to a
space. We started with a 2-dimensional
R
2
\ {
0
}
space, and squashed it into a
one-dimensional sphere.
Hence we see that homotopy equivalence doesn’t care about dimensions.
Dimensions seem to be a rather fundamental thing in geometry, and we are
discarding it here. So what is left? What does homotopy equivalence preserve?
While
S
1
and
R
2
\{
0
}
seem rather different, they have something in common
they both have a “hole”. We will later see that this is what homotopy equivalence
preserves.
Example. Let
Y
=
R
n
,
X
=
{
0
}
=
. Let
X Y
be the inclusion map, and
r
:
Y X
be the unique map that sends everything to
{
0
}
. Again, we have
r i = id
X
. We can also obtain a homotopy from i r to id
Y
by
H(y, t) = ty.
Again, from the point of view of homotopy theory,
R
n
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. However, by giving up these structures,
it is easier to focus on what we really want to care about holes. For example,
it is often much easier to argue about the one-point space
than the whole of
R
2
! 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.
In general things homotopy equivalent to a single point are known as con-
tractible spaces.
Notation. denotes the one-point space {0}.
Definition (Contractible space). If X ' , then X is contractible.
We now show that homotopy equivalence of spaces is an equivalence relation.
To do this, we first need a lemma.
Lemma. Consider the spaces and arrows
X Y Z
f
0
f
1
g
0
g
1
If f
0
'
H
f
1
and g
0
'
H
0
g
1
, then g
0
f
0
' g
1
f
1
.
Proof.
We will show that
g
0
f
0
' g
0
f
1
' g
1
f
1
. Then we are done since
homotopy between maps is an equivalence relation. So we need to write down
two homotopies.
(i) Consider the following composition:
X × I Y Z
H
g
0
It is easy to check that this is the first homotopy we need to show
g
0
f
0
'
g
0
f
1
.
(ii) The following composition is a homotopy from g
0
f
1
to g
1
f
1
:
X × I Y × I Z
f
1
×id
I
H
0
Proposition. Homotopy equivalence of spaces is an equivalence relation.
Proof.
Symmetry and reflexivity are trivial. To show transitivity, let
f
:
X Y
and
h
:
Y Z
be homotopy equivalences, and
g
:
Y X
and
k
:
Z Y
be their homotopy inverses. We will show that
h f
:
X Z
is a homotopy
equivalence with homotopy inverse g k. We have
(h f) (g k) = h (f g) k ' h id
Y
k = h k ' id
Z
.
Similarly,
(g k) (h f ) = g (k h) f ' g id
Y
f = g f ' id
X
.
So done.
Definition (Retraction). Let
A X
be a subspace. A retraction
r
:
X A
is a map such that
r i
=
id
A
, where
i
:
A X
is the inclusion. If such an
r
exists, we call A a retract of X.
This map sends everything in
X
to
A
without moving things in
A
. Roughly
speaking, if such a retraction exists, then A is no more complicated than X.
Definition (Deformation retraction). The retraction
r
is a deformation retrac-
tion if
i r ' id
X
. A deformation retraction is strong if we require this homotopy
to be a homotopy rel A.
Roughly, this says that A is as complicated as X.
Example. Take
X
any space, and
A
=
{x} X
. Then the constant map
r
:
X A
is a retraction. If
X
is contractible, then
A
is a deformation retract
of X.
2.2 Paths
Definition (Path). A path in a space
X
is a map
γ
:
I X
. If
γ
(0) =
x
0
and
γ(1) = x
1
, we say γ is a path from x
0
to x
1
, and write γ : x
0
x
1
.
If γ(0) = γ(1), then γ is called a loop (based at x
0
).
x
0
x
1
Note that this map does not have to be injective. It can be self-intersecting
or do all sorts of weird stuff.
Recall that the basic idea of algebraic topology is to assign to each space
X
an (algebraic) object, which is (hopefully) easier to deal with. In homotopy
theory, we do so using the idea of paths.
To do so, we need to be able to perform operations on paths.
Definition (Concatenation of paths). If we have two paths
γ
1
from
x
0
to
x
1
;
and γ
2
from x
1
to x
2
, we define the concatenation to be
(γ
1
· γ
2
)(t) =
(
γ
1
(2t) 0 t
1
2
γ
2
(2t 1)
1
2
t 1.
This is continuous by the gluing lemma.
Note that we concatenate left to right, but function composition goes from
right to left.
x
0
x
1
x
2
Definition (Inverse of path). The inverse of a path γ : I X is defined by
γ
1
(t) = γ(1 t).
This is exactly the same path but going in the opposite direction.
What else do we need? We need an identity.
Definition (Constant path). The constant path at a point
x X
is given by
c
x
(t) = x.
We haven’t actually got a good algebraic system. We have
γ
and
γ
1
, but
when we compose them, we get a path from
x
1
to
x
2
and back, and not the
identity. Also, we are not able to combine arbitrary paths in a space.
Before we make these into proper algebraic operations, we will talk about
something slightly different. We can view this as a first attempt at associating
things to topological spaces.
Definition (Path components). We can define a relation on
X
:
x
1
x
2
if
there exists a path from
x
1
to
x
2
. By the concatenation, inverse and constant
paths,
is an equivalence relation. The equivalence classes [
x
] are called path
components. We denote the quotient X/ by π
0
(X).
In the above space, we have three path components.
This isn’t really a very useful definition, since most spaces we care about are
path-connected, i.e. only have one path component. However, this is a first step
at associating something to spaces. We can view this as a “toy model” for the
more useful definitions we will later have.
One important property of this
π
0
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.
Proposition. For any map f : X Y , there is a well-defined function
π
0
(f) : π
0
(X) π
0
(Y ),
defined by
π
0
(f)([x]) = [f(x)].
Furthermore,
(i) If f ' g, then π
0
(f) = π
0
(g).
(ii) For any maps A B C
h k
, we have π
0
(k h) = π
0
(k) π
0
(h).
(iii) π
0
(id
X
) = id
π
0
(X)
Proof.
To show this is well-defined, suppose [
x
] = [
y
]. Then let
γ
:
I X
be a
path from x to y. Then f γ is a path from f (x) to f (y). So [f(x)] = [f (y)].
(i)
If
f ' g
, let
H
:
X × I Y
be a homotopy from
f
to
g
. Let
x X
. Then
H
(
x, ·
) is a path from
f
(
x
) to
g
(
x
). So [
f
(
x
)] = [
g
(
x
)], i.e.
π
0
(
f
)([
x
]) =
π
0
(g)([x]). So π
0
(f) = π
0
(g).
(ii) π
0
(k h)([x]) = π
0
(k) π
0
(h)([x]) = [k(h(x))].
(iii) π
0
(id
X
)([x]) = [id
X
(x)] = [x]. So π
0
(id
X
) = id
π
0
(X)
.
Corollary. If f : X Y is a homotopy equivalence, then π
0
(f) is a bijection.
Example. The two point space
X
=
{−
1
,
1
}
is not contractible, because
|π
0
(X)| = 2, but |π
0
()| = 1.
This is a rather silly example, since we can easily prove it directly. However,
this is an example to show how we can use this machinery to prove topological
results.
Now let’s return to our operations on paths, and try to make them algebraic.
Definition (Homotopy of paths). Paths
γ, γ
0
:
I X
are homotopic as paths if
they are homotopic rel
{
0
,
1
} I
, i.e. the end points are fixed. We write
γ ' γ
0
.
x
0
x
1
Note that we would necessarily want to fix the two end points. Otherwise, if we
allow end points to move, we can shrink any path into a constant path, and our
definition of homotopy would be rather silly.
This homotopy works well with our previous operations on paths.
Proposition. Let
γ
1
, γ
2
:
I X
be paths,
γ
1
(1) =
γ
2
(0). Then if
γ
1
' γ
0
1
and
γ
2
' γ
0
2
, then γ
1
· γ
2
' γ
0
1
· γ
0
2
.
x
0
x
1
x
2
γ
1
γ
2
γ
0
1
γ
0
2
Proof. Suppose that γ
1
'
H
1
γ
0
1
and γ
2
'
H
2
γ
0
2
. Then we have the diagram
γ
1
γ
2
γ
0
1
γ
0
2
x
0
x
1
x
2
H
1
H
2
We can thus construct a homotopy by
H(s, t) =
(
H
1
(s, 2t) 0 t
1
2
H
2
(s, 2t 1)
1
2
t 1
.
To solve all our previous problems about operations on paths not behaving
well, we can look at paths up to homotopy.
Proposition. Let γ
0
: x
0
x
1
, γ
1
: x
1
x
2
, γ
2
: x
2
x
3
be paths. Then
(i) (γ
0
· γ
1
) · γ
2
' γ
0
· (γ
1
· γ
2
)
(ii) γ
0
· c
x
1
' γ
0
' c
x
0
· γ
0
.
(iii) γ
0
· γ
1
0
' c
x
0
and γ
1
0
· γ
0
' c
x
1
.
Proof.
(i) Consider the following diagram:
x
0
x
3
γ
0
γ
1
γ
2
γ
0
γ
1
γ
2
x
1
x
2
(ii) Consider the following diagram:
γ
0
c
x
1
γ
0
x
0
x
1
x
1
(iii) Consider the following diagram:
γ
0
γ
1
0
c
x
0
x
0
x
0
Turning these into proper proofs is left as an exercise for the reader.
2.3 The fundamental group
The idea is to take spaces and turn them into groups. We want to try to do
this using paths. We’ve seen that if we want to do this, we should not work
directly with paths, but paths up to homotopy. According to our proposition,
this operation satisfies associativity, inverses and identity. The last issue we have
to resolve is that we can’t actually put two paths together unless they have the
same start and end points.
The idea is to fix one of the points
x
0
in our space, and only think about
loops that start and end at x
0
. So we can always join two paths together.
This tells us that we aren’t going to just think about spaces, but spaces with
basepoints. We’ve now got everything we need.
Definition (Fundamental group). Let
X
be a space and
x
0
X
. The fun-
damental group of
X
(based at
x
0
), denoted
π
1
(
X, x
0
), is the set of homotopy
classes of loops in
X
based at
x
0
(i.e.
γ
(0) =
γ
(1) =
x
0
). The group operations
are defined as follows:
We define an operation by [
γ
0
][
γ
1
] = [
γ
0
· γ
1
]; inverses by [
γ
]
1
= [
γ
1
]; and
the identity as the constant path e = [c
x
0
].
Often, when we write the homotopy classes of paths [
γ
], we just get lazy and
write γ.
Theorem. The fundamental group is a group.
Proof. Immediate from our previous lemmas.
Often in mathematics, after defining a term, we give lots of examples of
it. Unfortunately, it is rather difficult to prove that a space has a non-trivial
fundamental group, until we have developed some relevant machinery. Hence we
will have to wait for a while before we have some concrete examples. Instead,
we will look at some properties of the fundamental group first.
Definition (Based space). A based space is a pair (
X, x
0
) of a space
X
and a
point x
0
X, the basepoint. A map of based spaces
f : (X, x
0
) (Y, y
0
)
is a continuous map
f
:
X Y
such that
f
(
x
0
) =
y
0
. A based homotopy is a
homotopy rel {x
0
}.
Recall that for
π
0
, to every map
f
:
X Y
, we can associate a function
π
0
(f) : π
0
(X) π
0
(Y ). We can do the same for π
1
.
Proposition. To a based map
f : (X, x
0
) (Y, y
0
),
there is an associated function
f
= π
1
(f) : π
1
(X, x
0
) π
1
(Y, y
0
),
defined by [γ] 7→ [f γ]. Moreover, it satisfies
(i) π
1
(f) is a homomorphism of groups.
(ii) If f ' f
0
, then π
1
(f) = π
1
(f
0
).
(iii)
For any maps
(A, a) (B, b) (C, c)
h k
, we have
π
1
(
k h
) =
π
1
(k) π
1
(h).
(iv) π
1
(id
X
) = id
π
1
(X,x
0
)
Proof. Exercise.
In category-theoretic language, we say that π
1
is a functor.
So far so good. However, to define the fundamental group, we had to make a
compromise and pick a basepoint. But we just care about the space. 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.
The first observation is that
π
1
(
X, x
0
) only “sees” the path component of
x
0
,
since all loops based at
x
0
can only live inside the path component of
x
0
. Hence,
the first importance of picking a basepoint is picking a path component.
For all practical purposes, we just assume that
X
is path connected, since if
we weren’t, the fundamental group just describes a particular path component
of the original space.
Now we want to compare fundamental groups with different basepoints.
Suppose we have two basepoints
x
0
and
x
1
. Suppose we have a loop
γ
at
x
0
.
How can we turn this into a loop based at
x
1
? This is easy. We first pick a path
u
:
x
0
x
1
. Then we can produce a new loop at
x
1
by going along
u
1
to
x
0
,
take the path γ, and then return to x
0
by u, i.e. consider u
1
· γ · u.
x
0
x
1
u
Proposition. A path u : x
0
x
1
induces a group isomorphism
u
#
: π
1
(X, x
0
) π
1
(X, x
1
)
by
[γ] 7→ [u
1
· γ · u].
This satisfies
(i) If u ' u
0
, then u
#
= u
0
#
.
(ii) (c
x
0
)
#
= id
π
1
(X,x
0
)
(iii) If v : x
1
x
2
. Then (u · v)
#
= v
#
u
#
.
(iv) If f : X Y with f(x
0
) = y
0
, f (x
1
) = y
1
, then
(f u)
#
f
= f
u
#
: π
1
(X, x
0
) π
1
(Y, y
1
).
A nicer way of writing this is
π
1
(X, x
0
) π
1
(Y, y
0
)
π
1
(X, x
1
) π
1
(Y, y
1
)
f
u
#
(fu)
#
f
The property says that the composition is the same no matter which way
we go from
π
1
(
X, x
0
) to
π
1
(
Y, y
1
). We say that the square is a commutative
diagram. These diagrams will appear all of the time in this course.
(v)
If
x
1
=
x
0
, then
u
#
is an automorphism of
π
1
(
X, x
0
) given by conjugation
by u.
It is important (yet difficult) to get the order of concatenation and composition
right. Path concatenation is from left to right, while function composition is
from right to left.
Proof.
Yet another exercise. Note that (
u
1
)
#
= (
u
#
)
1
, which is why we have
an isomorphism.
The main takeaway is that if
x
0
and
x
1
are in the same path component,
then
π
1
(X, x
0
)
=
π
1
(X, x
1
).
So the basepoint isn’t really too important. However, we have to be careful.
While the two groups are isomorphic, the actual isomorphism depends on which
path
u
:
x
0
x
1
we pick. So there is no natural isomorphism between the two
groups. In particular, we cannot say “let
α π
1
(
X, x
0
). Now let
α
0
be the
corresponding element in π
1
(X, x
1
)”.
We can also see that if (
X, x
0
) and (
Y, y
0
) are based homotopy equivalent,
then
π
1
(
X, x
0
)
=
π
1
(
Y, y
0
). Indeed, if they are homotopy equivalent, then there
are some f : X Y , g : Y X such that
f g ' id
Y
, g f ' id
X
.
So
f
g
= id
π
1
(Y,y
0
)
, g
f
= id
π
1
(X,x
0
)
,
and f
and g
are the isomorphisms we need.
However, can we do this with non-based homotopies? Suppose we have
a space
X
, and a space
Y
, and functions
f, g
:
X Y
, with a homotopy
H : X × I Y from f to g. If this were a based homotopy, we know
f
= g
: π
1
(X, x
0
) π
1
(Y, f(x
0
)).
Now we don’t insist that this homotopy fixes
x
0
, and we could have
f
(
x
0
)
6
=
g
(
x
0
). How can we relate
f
:
π
1
(
X, x
0
)
π
1
(
Y, f
(
x
0
)) and
g
:
π
1
(
X, x
0
)
π
1
(Y, g(x
0
))?
First of all, we need to relate the groups
π
1
(
Y, f
(
x
0
)) and
π
1
(
Y, g
(
x
0
)). To
do so, we need to find a path from
f
(
x
0
) to
g
(
x
0
). To produce this, we can use
the homotopy
H
. We let
u
:
f
(
x
0
)
g
(
x
0
) with
u
=
H
(
x
0
, ·
). Now we have
three maps f
, g
and u
#
. Fortunately, these do fit together well.
x
0
X Y
f(x
0
)
g(x
0
)
f
g
Lemma. The following diagram commutes:
π
1
(Y, f(x
0
))
π
1
(X, x
0
)
π
1
(Y, g(x
0
))
u
#
f
g
In algebra, we say
g
= u
#
f
.
Proof. Suppose we have a loop γ : I X based at x
0
.
We need to check that
g
([γ]) = u
#
f
([γ]).
In other words, we want to show that
g γ ' u
1
· (f γ) · u.
To prove this result, we want to build a homotopy.
Consider the composition:
F : I × I X × I Y.
γ×id
I
H
Our plan is to exhibit two homotopic paths
+
and
in I × I such that
F
+
= g γ, F
= u
1
· (f γ) · u.
This is in general a good strategy
X
is a complicated and horrible space we
don’t understand. So to construct a homotopy, we make ourselves work in a
much nicer space I × I.
Our
+
and
are defined in a rather simple way.
I × I
+
F
Y
u
1
f
u
g
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).
Note that
+
and
are homotopic as paths. If this is not obvious, we can
manually check the homotopy
L(s, t) = t
+
(s) + (1 t)
(s).
This works because I × I is convex. Hence F
+
'
F L
F
as paths.
Now we check that the compositions
F
±
are indeed what we want. We
have
F
+
(s) = H(γ(s), 1) = g γ(s).
Similarly, we can show that
F
(s) = u
1
· (f γ) · u(s).
So done.
It is worth looking at this proof hard and truly understand what is going
on, since this is a really good example of how we can construct interesting
homotopies.
With this lemma, we can show that fundamental groups really respect
homotopies.
Theorem. If
f
:
X Y
is a homotopy equivalence, and
x
0
X
, then the
induced map
f
: π
1
(X, x
0
) π
1
(Y, f(x
0
)).
is an isomorphism.
While this seems rather obvious, it is actually non-trivial if we want to do
it from scratch. While we are given a homotopy equivalence, we are given no
guarantee that the homotopy respects our basepoints. So this proof involves
some real work.
Proof.
Let
g
:
Y X
be a homotopy inverse. So
f g '
H
id
Y
and
gf '
H
0
id
X
.
x
0
X Y
f(x
0
)
g f(x
0
)
u
0
f
g
We have no guarantee that
g f
(
x
0
) =
x
0
, but we know that our homotopy
H
0
gives us u
0
= H
0
(x
0
, · ) : x
0
g f(x
0
).
Applying our previous lemma with id
X
for f and g f for g”, we get
u
0
#
(id
X
)
= (g f)
Using the properties of the
operation, we get that
g
f
= u
0
#
.
However, we know that
u
0
#
is an isomorphism. So
f
is injective and
g
is
surjective.
Doing it the other way round with
f g
instead of
g f
, we know that
g
is
injective and f
is surjective. So both of them are isomorphisms.
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.
We now use the fundamental group to define several simple notions.
Definition (Simply connected space). A space
X
is simply connected if it is
path connected and π
1
(X, x
0
)
=
1 for some (any) choice of x
0
X.
Example. Clearly, a point
is simply connected since there is only one path on
(the constant path). Hence, any contractible space is simply connected since
it is homotopic to . For example, R
n
is simply connected for all n.
There is a useful characterization of simply connected spaces:
Lemma. A path-connected space
X
is simply connected if and only if for any
x
0
, x
1
X, there exists a unique homotopy class of paths x
0
x
1
.
Proof.
Suppose
X
is simply connected, and let
u, v
:
x
0
x
1
be paths. Now
note that
u · v
1
is a loop based at
x
0
, it is homotopic to the constant path, and
v
1
· v is trivially homotopic to the constant path. So we have
u ' u · v
1
· v ' v.
On the other hand, suppose there is a unique homotopy class of paths
x
0
x
1
for all
x
0
, x
1
X
. Then in particular there is a unique homotopy class of loops
based at x
0
. So π
1
(X, x
0
) is trivial.
3 Covering spaces
We can ask ourselves a question what are groups? We can write down a
definition in terms of operations and axioms, but this is not what groups are.
Groups were created to represent symmetries of objects. In particular, a group
should be acting on something. If not, something wrong is probably going on.
We have created the fundamental group. 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. So far we
have not exhibited a non-trivial fundamental group. 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.
These things we act on are covering spaces.
3.1 Covering space
Intuitively, a covering space of
X
is a pair (
˜
X, p
:
˜
X X
), such that if we take
any
x
0
X
, there is some neighbourhood
U
of
x
0
such that the pre-image of
the neighbourhood is “many copies” of U.
U
x
0
˜
X
p
X
Definition (Covering space). A covering space of
X
is a pair (
˜
X, p
:
˜
X X
),
such that each x X has a neighbourhood U which is evenly covered.
Whether we require
p
to be surjective is a matter of taste. However, it will
not matter if X is path-connected, which is the case we really care about.
Definition (Evenly covered). U X is evenly covered by p :
˜
X X if
p
1
(U)
=
a
αΛ
V
α
,
where p|
V
α
: V
α
U is a homeomorphism, and each of the V
α
˜
X is open.
Example. Homeomorphisms are covering maps. Duh.
Example. Consider
p
:
R S
1
C
defined by
t 7→ e
2πit
. This is a covering
space.
1
p
1
(1)
R
p
S
1
Here we have p
1
(1) = Z.
As we said, we are going to use covering spaces to determine non-trivial
fundamental groups. Before we do that, we can guess what the fundamental
group of
S
1
is. If we have a loop on
S
1
, it could be nothing, it can loop around
the circle once, or loop around many times. 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. So it is not unreasonable to guess
that
π
1
(
S
1
,
1)
=
Z
. However, we also know that
p
1
(1) =
Z
. In fact, for any
point
z S
1
,
p
1
(
z
) is just
Z
many copies” of
z
. We will show that this is not
a coincidence.
Example. Consider
p
n
:
S
1
S
1
(for any
n Z \ {
0
}
) defined by
z 7→ z
n
. We
can consider this as “winding” the circle
n
times, or as the following covering
map:
S
1
p
n
S
1
where we join the two red dots together.
This time the pre-image of 1 would be
n
copies of 1, instead of
Z
copies of 1.
Example. Consider
X
=
RP
2
, which is the real projective plane. This is defined
by
S
2
/
, where we identify every
x x
, i.e. every pair of antipodal points.
We can also think of this as the space of lines in
R
3
. This is actually really
difficult to draw (and in fact impossible to fully embed in
R
3
), so we will not
attempt. There is, however, a more convenient way of thinking about
RP
2
. We
just use the definition we imagine an
S
2
, but this time, instead of a point
being a “point” on the sphere, it is a pair of antipodal points.
U
U
We define
p
:
S
2
RP
2
to be the quotient map. Then this is a covering map,
since the pre-image of a small neighbourhood of any
x
0
is just two copies of the
neighbourhood.
As we mentioned, a covering space of
X
is (locally) like many “copies” of
X
. Hence, given any function
f
:
Y X
, a natural question is whether we can
“lift” it to a function f : Y
˜
X, by mapping to one of the copies of X in
˜
X.
This is not always possible, but when it is, we call this a lift.
Definition (Lifting). Let
f
:
Y X
be a map, and
p
:
˜
X X
a covering
space. A lift of
f
is a map
˜
f
:
Y
˜
X
such that
f
=
p
˜
f
, i.e. the following
diagram commutes:
˜
X
Y X
p
˜
f
f
We can visualize this in the case where
Y
is the unit interval
I
and the map
is just a path in X.
˜
X
X
p
Y
f
˜
f
f(Y )
˜
f(Y )
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. This is made precise by the following lemma:
Lemma. Let
p
:
˜
X X
be a covering map,
f
:
Y X
be a map, and
˜
f
1
,
˜
f
2
be both lifts of f. Then
S = {y Y :
˜
f
1
(y) =
˜
f
2
(y)}
is both open and closed. In particular, if
Y
is connected,
˜
f
1
and
˜
f
2
agree either
everywhere or nowhere.
This is sort of a “uniqueness statement” for a lift. If we know a point in the
lift, then we know the whole path. This is since once we’ve decided our starting
point, i.e. which “copy” of
X
we work in, the rest of
˜
f
has to follow what
f
does.
Proof.
First we show it is open. Let
y
be such that
˜
f
1
(
y
) =
˜
f
2
(
y
). Then
there is an evenly covered open neighbourhood
U X
of
f
(
y
). Let
˜
U
be
such that
˜
f
1
(
y
)
˜
U
,
p
(
˜
U
) =
U
and
p|
˜
U
:
˜
U U
is a homeomorphism. Let
V =
˜
f
1
1
(
˜
U)
˜
f
1
2
(
˜
U). We will show that
˜
f
1
=
˜
f
2
on V .
Indeed, by construction
p|
˜
U
˜
f
1
|
V
= p|
˜
U
˜
f
2
|
V
.
Since p|
˜
U
is a homeomorphism, it follows that
˜
f
1
|
V
=
˜
f
2
|
V
.
Now we show
S
is closed. Suppose not. Then there is some
y
¯
S \ S
. So
˜
f
1
(
y
)
6
=
˜
f
2
(
y
). Let
U
be an evenly covered neighbourhood of
f
(
y
). Let
p
1
(
U
) =
`
U
α
.
Let
˜
f
1
(
y
)
U
β
and
˜
f
2
(
y
)
U
γ
, where
β 6
=
γ
. Then
V
=
˜
f
1
1
(
U
β
)
˜
f
1
2
(
U
γ
) is
an open neighbourhood of
y
, and hence intersects
S
by definition of closure. So
there is some
x V
such that
˜
f
1
(
x
) =
˜
f
2
(
x
). But
˜
f
1
(
x
)
U
β
and
˜
f
2
(
x
)
U
γ
,
and hence
U
β
and
U
γ
have a non-trivial intersection. This is a contradiction. So
S is closed.
We just had a uniqueness statement. 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. However, it turns out homotopies
can always be lifted.
Lemma (Homotopy lifting lemma). Let
p
:
˜
X X
be a covering space,
H
:
Y × I X
be a homotopy from
f
0
to
f
1
. Let
˜
f
0
be a lift of
f
0
. Then there
exists a unique homotopy
˜
H : Y × I
˜
X such that
(i)
˜
H( · , 0) =
˜
f
0
; and
(ii)
˜
H is a lift of H, i.e. p
˜
H = H.
This lemma might be difficult to comprehend at first. We can look at the
special case where
Y
=
. Then a homotopy is just a path. So the lemma
specializes to
Lemma (Path lifting lemma). Let
p
:
˜
X X
be a covering space,
γ
:
I X
a
path, and
˜x
0
˜
X
such that
p
(
˜x
0
) =
x
0
=
γ
(0). Then there exists a unique path
˜γ : I
˜
X such that
(i) ˜γ(0) = ˜x
0
; and
(ii) ˜γ is a lift of γ, i.e. p ˜γ = γ.
This is exactly the picture we were drawing before. We just have to start
at a point
˜x
0
, and then everything is determined because locally, everything
upstairs in
˜
X
is just like
X
. Note that we have already proved uniqueness. So
we just need to prove existence.
In theory, it makes sense to prove homotopy lifting, and path lifting comes
immediately as a corollary. However, the proof of homotopy lifting is big and
scary. 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.
Proof. Let
S = {s I : ˜γ exists on [0, s] I}.
Observe that
(i) 0 S.
(ii) S
is open. If
s S
and
˜γ
(
s
)
V
β
p
1
(
U
), we can define
˜γ
on some
small neighbourhood of s by
˜γ(t) = (p|
V
β
)
1
γ(t)
(iii) S
is closed. If
s 6∈ S
, then pick an evenly covered neighbourhood
U
of
γ
(
s
).
Suppose
γ
((
s ε, s
))
U
. So
s
ε
2
6∈ S
. So (
s
ε
2
,
1]
S
=
. So
S
is
closed.
Since
S
is both open and closed, and is non-empty, we have
S
=
I
. So
˜γ
exists.
How can we promote this to a proof of the homotopy lifting lemma? At
every point
y Y
, we know what to do, since we have path lifting. So
˜
H
(
y, ·
)
is defined. So the thing we have to do is to show that this is continuous. Steps
of the proof are
(i)
Use compactness of
I
to argue that the proof of path lifting works on small
neighbourhoods in Y .
(ii)
For each
y
, we pick an open neighbourhood
U
of
y
, and define a good path
lifting on U × I.
(iii)
By uniqueness of lifts, these path liftings agree when they overlap. So we
have one big continuous lifting.
With the homotopy lifting lemma in our toolkit, we can start to use it to
do stuff. So far, we have covering spaces and fundamental groups. 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.
At least one payoff of this work is that we are going to exhibit some non-trivial
fundamental groups.
We have just showed that we are allowed to lift homotopies. However, what
we are really interested in is homotopy as paths. The homotopy lifting lemma
does not tell us that the lifted homotopy preserves basepoints. This is what we
are going to show.
Corollary. Suppose
γ, γ
0
:
I X
are paths
x
0
x
1
and
˜γ, ˜γ
0
:
I
˜
X
are lifts
of γ and γ
0
respectively, both starting at ˜x
0
p
1
(x
0
).
If
γ ' γ
0
as paths, then
˜γ
and
˜γ
0
are homotopic as paths. In particular,
˜γ(1) = ˜γ
0
(1).
Note that if we cover the words “as paths” and just talk about homotopies,
then this is just the homotopy lifting lemma. So we can view this as a stronger
form of the homotopy lifting lemma.
Proof.
The homotopy lifting lemma gives us an
˜
H
, a lift of
H
with
˜
H
(
· ,
0) =
˜γ
.
γ
γ
0
c
x
0
c
x
1
H
lift
˜γ
˜
γ
0
c
˜x
0
c
˜x
1
˜
H
In this diagram, we by assumption know the bottom of the
˜
H
square is
˜γ
. To
show that this is a path homotopy from
˜γ
to
˜γ
0
, we need to show that the other
edges are c
˜x
0
, c
˜x
1
and ˜γ
0
respectively.
Now
˜
H
(
· ,
1) is a lift of
H
(
· ,
1) =
γ
0
, starting at
˜x
0
. Since lifts are unique,
we must have
˜
H
(
· ,
1) =
˜γ
0
. So this is indeed a homotopy between
˜γ
and
˜γ
0
.
Now we need to check that this is a homotopy of paths.
We know that
˜
H
(0
, ·
) is a lift of
H
(0
, ·
) =
c
x
0
. We are aware of one lift of
c
x
0
, namely
c
˜x
0
. By uniqueness of lifts, we must have
˜
H
(0
, ·
) =
c
˜x
0
. Similarly,
˜
H(1, · ) = c
˜x
1
. So this is a homotopy of paths.
So far, our picture of covering spaces is like this:
x
0
x
1
Except. . . is it? Is it possible that we have four copies of
x
0
but just three copies
of
x
1
? This is obviously possible if
X
is not path connected the component
containing
x
0
and the one containing
x
1
are completely unrelated. But what if
X is path connected?
Corollary. If
X
is a path connected space,
x
0
, x
1
X
, then there is a bijection
p
1
(x
0
) p
1
(x
1
).
Proof.
Let
γ
:
x
0
x
1
be a path. We want to use this to construct a bijection
between each preimage of
x
0
and each preimage of
x
1
. The obvious thing to do
is to use lifts of the path γ.
x
0
x
1
γ
Define a map
f
γ
:
p
1
(
x
0
)
p
1
(
x
1
) that sends
˜x
0
to the end point of the
unique lift of γ at ˜x
0
.
The inverse map is obtained by replacing
γ
with
γ
1
, i.e.
f
γ
1
. To show this
is an inverse, suppose we have some lift
˜γ
:
˜x
0
˜x
1
, so that
f
γ
(
˜x
0
) =
˜x
1
. Now
notice that
˜γ
1
is a lift of
γ
1
starting at
˜x
1
and ending at
˜x
0
. So
f
γ
1
(
˜x
1
) =
˜x
0
.
So f
γ
1
is an inverse to f
γ
, and hence f
γ
is bijective.
Definition (
n
-sheeted). A covering space
p
:
˜
X X
of a path-connected space
X is n-sheeted if |p
1
(x)| = n for any (and hence all) x X.
Each covering space has a number associated to it, namely the number of
sheets. Is there any number we can assign to fundamental groups? Well, the
index of a subgroup might be a good candidate. We’ll later see if this is the case.
One important property of covering spaces is the following:
Lemma. If p :
˜
X X is a covering map and ˜x
0
˜
X, then
p
: π
1
(
˜
X, ˜x
0
) π
1
(X, x
0
)
is injective.
Proof.
To show that a group homomorphism
p
is injective, we have to show
that if p
(x) is trivial, then x must be trivial.
Consider a based loop
˜γ
in
˜
X
. We let
γ
=
p ˜γ
. If
γ
is trivial, i.e.
γ ' c
x
0
as
paths, the homotopy lifting lemma then gives us a homotopy upstairs between
˜γ
and c
˜x
0
. So ˜γ is trivial.
As we have originally said, our objective is to make our fundamental group
act on something. We are almost there already.
Let’s look again at the proof that there is a bijection between
p
1
(
x
0
) and
p
1
(
x
1
). What happens if
γ
is a loop? For any
˜x
0
p
1
(
x
0
), we can look at
the end point of the lift. This end point may or may not be our original
˜x
0
. So
each loop γ “moves” our ˜x
0
to another ˜x
0
0
.
However, we are not really interested in paths themselves. We are interested
in equivalence classes of paths under homotopy of paths. However, this is fine.
If
γ
is homotopic to
γ
0
, then this homotopy can be lifted to get a homotopy
between
˜γ
and
˜γ
0
. In particular, these have the same end points. So each (based)
homotopy class gives a well-defined endpoint.
˜
X
X
p
x
0
γ
˜γ
˜x
0
˜x
0
0
Now this gives an action of
π
1
(
X, x
0
) on
p
1
(
x
0
)! Note, however, that this will
not be the sort of actions we are familiar with. We usually work with left-actions,
where the group acts on the left, but now we will have right-actions, which
may confuse you a lot. To see this, we have to consider what happens when we
perform two operations one after another, which you shall check yourself. We
write this action as ˜x
0
· [γ].
When we have an action, we are interested in two things the orbits, and
the stabilizers. This is what the next lemma tells us about.
Lemma. Suppose X is path connected and x
0
X.
(i)
The action of
π
1
(
X, x
0
) on
p
1
(
x
0
) is transitive if and only if
˜
X
is path
connected. Alternatively, we can say that the orbits of the action correspond
to the path components.
(ii) The stabilizer of ˜x
0
p
1
(x
0
) is p
(π
1
(
˜
X, ˜x
0
)) π
1
(X, x
0
).
(iii) If
˜
X is path connected, then there is a bijection
p
(π
1
(
˜
X, ˜x
0
))\π
1
(X, x
0
) p
1
(x
0
).
Note that
p
(
π
1
(
˜
X, ˜x
0
))
\π
1
(
X, x
0
) is not a quotient, but simply the set of
cosets. We write it the “wrong way round” because we have right cosets
instead of left cosets.
Note that this is great! If we can find a covering space
p
and a point
x
0
such that
p
1
(
x
0
) is non-trivial, then we immediately know that
π
1
(
X, x
0
) is
non-trivial!
Proof.
(i)
If
˜x
0
, ˜x
0
0
p
1
(
x
0
), then since
˜
X
is path connected, we know that there
is some
˜γ
:
˜x
0
˜x
0
0
. Then we can project this to
γ
=
p ˜γ
. Then
γ
is
a path from
x
0
x
0
, i.e. a loop. Then by the definition of the action,
˜x
0
· [γ] = ˜γ(1) = ˜x
0
0
.
(ii)
Suppose [
γ
]
stab
(
˜x
0
). Then
˜γ
is a loop based at
˜x
0
. So
˜γ
defines
[˜γ] π
1
(
˜
X, ˜x
0
) and γ = p ˜γ.
(iii) This follows directly from the orbit-stabilizer theorem.
We now want to use this to determine that the fundamental group of a space
is non-trivial. We can be more ambitious, and try to actually find
π
1
(
X, x
0
).
In the best possible scenario, we would have
π
1
(
˜
X, ˜x
0
) trivial. Then we have a
bijection between
π
1
(
X, x
0
) and
p
1
(
x
0
). In other words, we want our covering
space
˜
X to be simply connected.
Definition (Universal cover). A covering map
p
:
˜
X X
is a universal cover
if
˜
X is simply connected.
We will look into universal covers in depth later and see what they really are.
Corollary. If
p
:
˜
X X
is a universal cover, then there is a bijection
:
π
1
(X, x
0
) p
1
(x
0
).
Note that the orbit-stabilizer theorem does not provide a canonical bijection
between
p
1
(
x
0
) and
p
π
1
(
˜
X, ˜x
0
)
\π
1
(
X, x
0
). To obtain a bijection, we need to
pick a starting point
˜x
0
p
1
(
x
0
). So the above bijection
depends on a choice
of ˜x
0
.
3.2
The fundamental group of the circle and its applica-
tions
Finally, we can exhibit a non-trivial fundamental group. We are going to consider
the space S
1
and a universal covering R.
1
p
1
(1)
R
p
S
1
Then our previous corollary gives
Corollary. There is a bijection π
1
(S
1
, 1) p
1
(1) = Z.
What’s next? We just know that
π
1
(
S
1
,
1) is countably infinite, but can we
work out the group structure?
We can, in fact, prove a stronger statement:
Theorem. The map : π
1
(S
1
, 1) p
1
(1) = Z is a group isomorphism.
Proof.
We know it is a bijection. So we need to check it is a group homomorphism.
The idea is to write down representatives for what we think the elements should
be.
˜u
2
2
1
0
1
2
R
p
S
1
u
2
Let
˜u
n
:
I R
be defined by
t 7→ nt
, and let
u
n
=
p ˜u
n
. Since
R
is simply
connected, there is a unique homotopy class between any two points. So for any
[
γ
]
π
1
(
S
1
,
1), if
˜γ
is the lift to
R
at 0 and
˜γ
(1) =
n
, then
˜γ ' ˜u
n
as paths. So
[γ] = [u
n
].
To show that this has the right group operation, we can easily see that
^u
m
· u
n
= ˜u
m+n
, since we are just moving by n + m in both cases. Therefore
([u
m
][u
n
]) = ([u
m
· u
m
]) = m + n = ([u
m+n
]).
So is a group isomorphism.
What have we done? In general, we might be given a horrible, crazy loop
in
S
1
. It would be rather difficult to work with it directly in
S
1
. So we pull it
up to the universal covering
R
. Since
R
is nice and simply connected, we can
easily produce a homotopy that “straightens out” the path. We then project
this homotopy down to S
1
, to get a homotopy from γ to u
n
.
It is indeed possible to produce a homotopy directly inside
S
1
from each loop
to some
u
n
, but that would be tedious work that involves messing with a lot of
algebra and weird, convoluted formulas.
With the fundamental group of the circle, we do many things. An immediate
application is that we can properly define the “winding number” of a closed
curve. Since
C \ {
0
}
is homotopy equivalent to
S
1
, its fundamental group is
Z
as well. Any closed curve
S
1
C \ {
0
}
thus induces a group homomorphism
Z Z
. Any such group homomorphism must be of the form
t 7→ nt
, and the
winding number is given by
n
. If we stare at it long enough, it is clear that this
is exactly the number of times the curve winds around the origin.
Also, we have the following classic application:
Theorem (Brouwer’s fixed point theorem). Let
D
2
=
{
(
x, y
)
R
2
:
x
2
+
y
2
1
}
be the unit disk. If
f
:
D
2
D
2
is continuous, then there is some
x D
2
such
that f (x) = x.
Proof. Suppose not. So x 6= f(x) for all x D
2
.
x
f(x)
g(x)
We define
g
:
D
2
S
1
as in the picture above. Then we know that
g
is
continuous and
g
is a retraction from
D
2
onto
S
1
. In other words, the following
composition is the identity:
S
1
D
2
S
1
ι
id
S
1
g
Then this induces a homomorphism of groups whose composition is the identity:
Z {0} Z
ι
id
Z
g
But this is clearly nonsense! So we must have had a fixed point.
But we have a problem. What about
D
3
? Can we prove a similar theorem?
Here the fundamental group is of little use, since we can show that the funda-
mental group of
S
n
for
n
2 is trivial. Later in the course, we will be able to
prove this theorem for higher dimensions, when we have developed more tools
to deal with stuff.
3.3 Universal covers
We have defined universal covers mysteriously as covers that are simply connected.
We have just shown that
p
:
R S
1
is a universal cover. In general, what do
universal covers look like?
Let’s consider a slightly more complicated example. 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
. How can we visualize this?
First of all, how can we visualize a torus? Often, we just picture it as the
surface of a doughnut. Alternatively, we can see it as a quotient of the square,
where we identify the following edges:
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 space just extends indefinitely for ever,
somewhat like
R
2
. 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.
If you live inside the square, the universe actually looks like this:
As we said, this looks somewhat likes
R
2
, but we know that this is not
R
2
,
since we can see some symmetry in this space. Whenever we move one unit
horizontally or vertically, we get back to “the same place”. In fact, we can move
horizontally by
n
units and vertically by
m
units, for any
n, m Z
, and still get
back to the same place. This space has a huge translation symmetry. What is
this symmetry? It is exactly Z × Z.
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.
Hopefully, you are convinced that universal covers are nice. We would like to
say that universal covers always exist. However, this is not always true.
Firstly, we should think what would having a universal cover imply?
Suppose
X
has a universal cover
˜
X
. Pick any point
x
0
X
, and pick an
evenly covered neighbourhood
U
in
X
. This lifts to some
˜
U
˜
X
. If we draw a
teeny-tiny loop
γ
around
x
0
inside
U
, we can lift this
γ
to
˜γ
in
˜
U
. But we know
that
˜
X
is simply connected. So
˜γ
is homotopic to the constant path. Hence
γ
is also homotopic to the constant path. So all loops (contained in
U
) at
x
0
are
homotopic to the constant path.
It seems like for every
x
0
X
, there is some neighbourhood of
x
0
that is
simply connected. Except that’s not what we just showed above. The homotopy
from
˜γ
to the constant path is a homotopy in
˜
X
, and can pass through anything
in
˜
X
, not just
˜
U
. Hence the homotopy induced in
X
is also a homotopy in
X
,
not a homotopy in
U
. So
U
itself need not be simply connected. What we have
is a slightly weaker notion.
Definition (Locally simply connected).
X
is locally simply connected if for all
x
0
X, there is some neighbourhood U of x
0
such that U is simply connected.
As we mentioned, what we actually want is a weaker condition.
Definition (Semi-locally simply connected).
X
is semi-locally simply connected
if for all
x
0
X
, there is some neighbourhood
U
of
x
0
such that any loop
γ
based at x
0
is homotopic to c
x
0
as paths in X.
We have just argued that if a universal cover
p
:
˜
X X
exists, then
X
is
semi-locally simply connected. This is really not interesting, since we don’t care
if a space is semi-locally simply connected. What is important is that this is the
other direction. We still need one more additional condition:
Definition (Locally path connected). A space
X
is locally path connected if for
any point
x
and any neighbourhood
V
of
x
, there is some open path connected
U V such that x U.
It is important to note that a path connected space need not be locally path
connected. It is an exercise for the reader to come up with a counterexample.
Theorem. If
X
is path connected, locally path connected and semi-locally
simply connected, then X has a universal covering.
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
. If we use this definition,
then we can prove this result easily using Zorn’s lemma. However, that proof
is not too helpful since it does not tell us where the universal covering comes
from. Instead, we will provide a constructive proof (sketch) that will hopefully
be more indicative of what universal coverings are like.
Proof.
(idea) We pick a basepoint
x
0
X
for ourselves. Suppose we have a
universal covering
˜
X
. Then this lifts to some
˜x
0
in
˜
X
. If we have any other point
˜x
˜
X
, since
˜
X
should be path connected, there is a path
˜α
:
˜x
0
˜x
. If we
have another path, then since
˜
X
is simply connected, the paths are homotopic.
Hence, we can identify each point in
˜
X with a path from ˜x
0
, i.e.
{points of
˜
X} {paths ˜α from ˜x
0
˜
X}/'.
This is not too helpful though, since we are defining
˜
X
in terms of things in
˜
X
.
However, by path lifting, we know that paths
˜α
from
˜x
0
in
˜
X
biject with paths
α
from
x
0
in
X
. Also, by homotopy lifting, homotopies of paths in
X
can be
lifted to homotopies of paths in
˜
X. So we have
{points of
˜
X} {paths α from x
0
X}/'.
So we can produce our
˜
X by picking a basepoint x
0
X, and defining
˜
X = {paths α : I X such that α(0) = x
0
}/'.
The covering map p :
˜
X X is given by [α] 7→ α(1).
One then has to work hard to define the topology, and then show this is
simply connected.
3.4 The Galois correspondence
Recall that at the beginning, we wanted to establish a correspondence between
covering spaces and fundamental groups. We have already established the result
that covering maps are injective on
π
1
. Therefore, given a (based) covering
space
p
: (
˜
X, ˜x
0
)
(
X, x
0
), we can give a subgroup
p
π
1
(
˜
X, ˜x
0
)
π
1
(
X, x
0
). 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.
We can have the following table of correspondences:
Covering spaces Fundamental group
(Based) covering spaces Subgroups of π
1
Number of sheets Index
Universal covers Trivial subgroup
We now want to look at some of these correspondences.
Recall that we have shown that
π
1
(
X, x
0
) acts on
p
1
(
x
0
). However, this is
not too interesting an action, since
p
1
(
x
0
) is a discrete group with no structure.
Having groups acting on a cube is fun because the cube has some structure. So
we want something more “rich” for π
1
(X, x
0
) to act on.
We note that we can make
π
1
(
X, x
0
) “act on” the universal cover. How?
Recall that in the torus example, each item in the fundamental group corresponds
to translating the whole universal covering by some amount. In general, a point
on
˜
X
can be thought of as a path
α
on
X
starting from
x
0
. Then it is easy to
make a loop
γ
:
x
0
x
0
act on this: use the concatenation
γ ·α
: [
γ
]
·
[
α
] = [
γ ·α
].
˜x
0
˜x
0
0
˜γ
α
˜
X
X
p
x
0
γ
α
We will use this idea and return to the initial issue of making subgroups corre-
spond to covering spaces. We want to show that this is surjective every sub-
group arises from some cover. We want to say “For any subgroup
H π
1
(
X, x
0
),
there is a based covering map
p
: (
˜
X, ˜x
0
)
(
X, x
0
) such that
p
π
1
(
˜
X, ˜x
0
) =
H
”.
Except, this cannot possibly be true, since by taking the trivial subgroup, this
would imply that there is a universal covering for every space. So we need some
additional assumptions.
Proposition. Let
X
be a path connected, locally path connected and semi-
locally simply connected space. For any subgroup
H π
1
(
X, x
0
), there is a
based covering map p : (
˜
X, ˜x
0
) (X, x
0
) such that p
π
1
(
˜
X, ˜x
0
) = H.
Proof.
Since
X
is a path connected, locally path connected and semi-locally
simply connected space, let
¯
X
be a universal covering. We have an intermediate
group
H
such that
π
1
(
˜
X, ˜x
0
) = 1
H π
1
(
X, x
0
). How can we obtain a
corresponding covering space?
Note that if we have
¯
X
and we want to recover
X
, we can quotient
¯
X
by the
action of
π
1
(
X, x
0
). Since
π
1
(
X, x
0
) acts on
¯
X
, so does
H π
1
(
X, x
0
). Now we
can define our covering space by taking quotients. We define
H
on
¯
X
to be
the orbit relation for the action of
H
, i.e.
˜x
H
˜y
if there is some
h H
such
that ˜y = h˜x. We then let
˜
X be the quotient space
¯
X/
H
.
We can now do the messy algebra to show that this is the covering space we
want.
We have just showed that every subgroup comes from some covering space,
i.e. the map from the set of covering spaces to the subgroups of
π
1
is surjective.
Now we want to prove injectivity. To do so, we need a generalization of the
homotopy lifting lemma.
Suppose we have path-connected spaces (
Y, y
0
), (
X, x
0
) and (
˜
X, ˜x
0
), with
f
: (
Y, y
0
)
(
X, x
0
) a continuous map,
p
: (
˜
X, ˜x
0
)
(
X, x
0
) a covering map.
When does a lift of
f
to
˜
f
: (
Y, y
0
)
(
˜
X, ˜x
0
) exist? The answer is given by the
lifting criterion.
Lemma (Lifting criterion). Let
p
: (
˜
X, ˜x
0
)
(
X, x
0
) be a covering map of path-
connected based spaces, and (
Y, y
0
) a path-connected, locally path connected
based space. If
f
: (
Y, y
0
)
(
X, x
0
) is a continuous map, then there is a (unique)
lift
˜
f
: (
Y, y
0
)
(
˜
X, ˜x
0
) such that the diagram below commutes (i.e.
p
˜
f
=
f
):
(
˜
X, ˜x
0
)
(Y, y
0
) (X, x
0
)
p
f
˜
f
if and only if the following condition holds:
f
π
1
(Y, y
0
) p
π
1
(
˜
X, ˜x
0
).
Note that uniqueness comes from the uniqueness of lifts. So this lemma is
really about existence.
Also, note that the condition holds trivially when
Y
is simply connected, e.g.
when it is an interval (path lifting) or a square (homotopy lifting). So paths and
homotopies can always be lifted.
Proof.
One direction is easy: if
˜
f
exists, then
f
=
p
˜
f
. So
f
=
p
˜
f
. So we
know that im f
im p
. So done.
In the other direction, uniqueness follows from the uniqueness of lifts. So we
only need to prove existence. We define
˜
f as follows:
Given a
y Y
, there is some path
α
y
:
y
0
y
. Then
f
maps this to
β
y
:
x
0
f
(
y
) in
X
. By path lifting, this path lifts uniquely to
˜
β
y
in
˜
X
. Then
we set
˜
f
(
y
) =
˜
β
y
(1). Note that if
˜
f
exists, then this must be what
˜
f
sends
y
to.
What we need to show is that this is well-defined.
Suppose we picked a different path
α
0
y
:
y
0
y
. Then this
α
0
y
would have
differed from α
y
by a loop γ in Y .
Our condition that
f
π
1
(
Y, y
0
)
p
π
1
(
˜
X, ˜x
0
) says
f γ
is the image of a
loop in
˜
X
. So
˜
β
y
and
˜
β
0
y
also differ by a loop in
˜
X
, and hence have the same
end point. So this shows that
˜
f is well-defined.
Finally, we show that
˜
f
is continuous. 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.
Thus, it suffices to show that if
p|
˜
V
:
˜
V p
(
˜
V
) =
V
is a homeomorphism, then
˜
f
1
(
˜
V ) is open.
Let
y
˜
f
1
(
˜
V
), and let
x
=
f
(
y
). Since
f
1
(
V
) is open and
Y
is locally
path-connected, we can pick an open
W f
1
(
V
) such that
y W
and
W
is
path connected. We claim that W
˜
f
1
(
˜
V ).
Indeed, if
z W
, then we can pick a path
γ
from
y
to
z
. Then
f
sends
this to a path from
x
to
f
(
z
). The lift of this path to
˜
X
is given by
p|
1
˜
V
(
f
(
γ
)),
whose end point is p|
1
˜
V
(f(z))
˜
V . So it follows that
˜
f(z) = p|
1
˜
V
(f(z))
˜
V .
Now we prove that every subgroup of
π
1
comes from exactly one covering
space. What this statement properly means is made precise in the following
proposition:
Proposition. Let (
X, x
0
), (
˜
X
1
, ˜x
1
), (
˜
X
2
, ˜x
2
) be path-connected based spaced,
and p
i
: (
˜
X
i
, ˜x
i
) (X, x
0
) be covering maps. Then we have
p
1
π
1
(
˜
X
1
, ˜x
1
) = p
2
π
1
(
˜
X
2
, ˜x
2
)
if and only if there is some homeomorphism
h
such that the following diagram
commutes:
(
˜
X
1
, ˜x
1
) (
˜
X
2
, ˜x
2
)
(X, x
0
)
h
p
1
p
2
i.e. p
1
= p
2
h.
Note that this is a stronger statement than just saying the two covering
spaces are homeomorphic. We are saying that we can find a nice homeomorphism
that works well with the covering map p.
Proof.
If such a homeomorphism exists, then clearly the subgroups are equal. If
the subgroups are equal, we rotate our diagram a bit:
(
˜
X
2
, ˜x
2
)
(
˜
X
1
, ˜x
1
) (X, x
0
)
p
2
p
1
h=˜p
1
Then
h
=
˜p
1
exists by the lifting criterion. By symmetry, we can get
h
1
=
˜p
2
.
To show
˜p
2
is indeed the inverse of
˜p
1
, note that
˜p
2
˜p
1
is a lift of
p
2
˜p
1
=
p
1
.
Since
id
˜
X
1
is also a lift, by the uniqueness of lifts, we know
˜p
2
˜p
1
is the identity
map. Similarly, ˜p
1
˜p
2
is also the identity.
(
˜
X
1
, ˜x
1
)
(
˜
X
1
, ˜x
1
) (
˜
X
2
, ˜x
2
) (X, x
0
)
p
1
p
1
˜p
1
˜p
2
p
2
Now what we would like to do is to forget about the basepoints. What
happens when we change base points? Recall that the effect of changing base
points is that we will conjugate our group. This doesn’t actually change the
group itself, but if we are talking about subgroups, conjugation can send a
subgroup into a different subgroup. Hence, if we do not specify the basepoint,
we don’t get a subgroup of π
1
, but a conjugacy class of subgroups.
Proposition. Unbased covering spaces correspond to conjugacy classes of
subgroups.
4 Some group theory
Algebraic topology is about translating topology into group theory. Unfortu-
nately, you don’t know group theory. Well, maybe you do, but not the right
group theory. So let’s learn group theory!
4.1 Free groups and presentations
Recall that in IA Groups, we defined, say, the dihedral group to be
D
2n
= hr, s | r
n
= s
2
= e, srs = r
1
i.
What does this expression actually mean? Can we formally assign a meaning to
this expression?
We start with a simple case the free group. This is, in some sense, the
“freest” group we can have. It is defined in terms of an alphabet and words.
Definition (Alphabet and words). We let
S
=
{s
α
:
α
Λ
}
be our alphabet,
and we have an extra set of symbols
S
1
=
{s
1
α
:
α
Λ
}
. We assume that
S S
1
= . What do we do with alphabets? We write words with them!
We define
S
to be the set of words over
S S
1
, i.e. it contains
n
-tuples
x
1
· · · x
n
for any 0 n < , where each x
i
S S
1
.
Example. Let
S
=
{a, b}
. Then words could be the empty word
, or
a
, or
aba
1
b
1
, or
aa
1
aaaaabbbb
, etc. We are usually lazy and write
aa
1
aaaaabbbb
as aa
1
a
5
b
4
.
When we see things like
aa
1
, we would want to cancel them. This is called
elementary reduction.
Definition (Elementary reduction). An elementary reduction takes a word
us
α
s
1
α
v and gives uv, or turns us
1
α
s
α
v into uv.
Since each reduction shortens the word, and the word is finite in length, we
cannot keep reducing for ever. Eventually, we reach a reduced state.
Definition (Reduced word). A word is reduced if it does not admit an elementary
reduction.
Example. , a, aba
1
b
1
are reduced words, while aa
1
aaaaabbbb is not.
Note that there is an inclusion map
S S
that sends the symbol
s
α
to the
word s
α
.
Definition (Free group). 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. For example, (
aba
1
b
1
)
·
(
bab
) =
aba
1
b
1
bab
=
ab
2
(ii) The identity is the empty word .
(iii) The inverse of x
1
· · · x
n
is x
1
n
· · · x
1
1
, where, of course, (s
1
α
)
1
= s
α
.
The elements of S are called the generators of F (S).
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.
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 is a cleaner
way to define the free group without messing with alphabets and words, but is
(for most people) less intuitive. This definition also does not make it clear that
the free group
F
(
S
) of any set
S
exists. We will state this definition as a lemma.
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)
S G
f
φ
where the arrow not labeled is the natural inclusion map that sends
s
α
(as a
symbol from the alphabet) to s
α
(as a word).
Proof.
Clearly if
f
exists, then
f
must send each
s
α
to
φ
(
s
α
) and
s
1
α
to
φ
(
s
α
)
1
.
Then the values of f on all other elements must be determined by
f(x
1
· · · x
n
) = f(x
1
) · · · f(x
n
)
since
f
is a homomorphism. So if
f
exists, it must be unique. So it suffices to
show that this f is a well-defined homomorphism.
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).
To show this is a homomorphism, suppose
x = x
1
· · · x
n
a
1
· · · a
k
, y = a
1
k
· · · a
1
1
y
1
· · · y
m
,
where y
1
6= x
1
n
. Then
xy = x
1
· · · x
n
y
1
· · · y
m
.
Then we can compute
f(x)f(y) =
φ(x
1
) · · · φ(x
n
)φ(a
1
) · · · φ(a
k
)

φ(a
k
)
1
· · · φ(a
1
)
1
φ(y
1
) · · · φ(y
m
)
= φ(x
1
) · · · φ(x
n
) · · · φ(y
1
) · · · φ(y
m
)
= f(xy).
So f is a homomorphism.
We call this a “universal property” of
F
(
S
). 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.
Definition (Presentation of a group). Let
S
be a set, and let
R F
(
S
) be any
subset. We denote by
hhRii
the normal closure of
R
, i.e. the smallest normal
subgroup of F (S) containing R. This can be given explicitly by
hhRii =
(
n
Y
i=1
g
i
r
i
g
1
i
: n N, r
i
R, g
i
F (S)
)
.
Then we write
hS | Ri = F (S)/hhRii.
This is just the usual notation we have for groups. For example, we can write
D
2n
= hr, s | r
n
, s
2
, srsri.
Again, we can define this with a universal property.
Lemma. If
G
is a group and
φ
:
S G
is a set map such that
f
(
r
) = 1 for all
r R
(i.e. if
r
=
s
±1
1
s
±1
2
· · · s
±1
m
, then
φ
(
r
) =
φ
(
s
1
)
±1
φ
(
s
2
)
±1
· · · φ
(
s
m
)
±1
= 1),
then there exists a unique homomorphism
f
:
hS | Ri G
such that the
following triangle commutes:
hS | Ri
S G
f
φ
Proof is similar to the previous one.
In some sense, this says that all the group
hS | Ri
does is it satisfies the
relations in R, and nothing else.
Example (The
stupid
canonical presentation). Let
G
be a group. We can view
the group as a set, and hence obtain a free group
F
(
G
). There is also an obvious
surjection
G G
. Then by the universal property of the free group, there is a
surjection
f
:
F
(
G
)
G
. Let
R
=
ker f
=
hhRii
. Then
hG | Ri
is a presentation
for G, since the first isomorphism theorem says
G
=
F (G)/ ker f.
This is a really stupid example. For example, even the simplest non-trivial
group
Z/
2 will be written as a quotient of a free group with two generators.
However, this tells us that every group has a presentation.
Example. ha, b | bi
=
hai
=
Z.
Example. With a bit of work, we can show that ha, b | ab
3
, ba
2
i = Z/5.
4.2 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. We are now going to do this using topology.
Again let
S
be a set. For the following illustration, we will just assume
S = {a, b}, but what we will do works for any set S. We define X by
a
b
x
0
We call this a “rose with 2 petals”. This is a cell complex, with one 0-cell and
|S|
1-cells. For each
s S
, we have one 1-cell,
e
s
, and we fix a path
γ
s
: [0
,
1]
e
s
that goes around the 1-cell once. We will call the 0-cells and 1-cells vertices and
edges, and call the whole thing a graph.
What’s the universal cover of
X
? Since we are just lifting a 1-complex, the
result should be a 1-complex, i.e. a graph. Moreover, this graph is connected
and simply connected, i.e. it’s a tree. We also know that every vertex in the
universal cover is a copy of the vertex in our original graph. So it must have 4
edges attached to it. 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
out. This should be the case in
˜
X as well. So
˜
X looks like this:
˜x
0 aa
b
b
a
b
b
The projection map is then obvious we send all the vertices in
˜
X
to
x
0
X
,
and then the edges according to the labels they have, in a way that respects the
direction of the arrow. It is easy to show this is really a covering map.
We are now going to show that this tree “is” the free group. Notice that
every word
w S
denotes a unique “edge path” in
˜
X
starting at
˜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
.
For example, the following path corresponds to w = abb
1
b
1
ba
1
b
1
.
˜x
0
We can note a few things:
(i)
˜
X is connected. So for all ˜x p
1
(x
0
), there is an edge-path ˜γ : ˜x
0
˜x.
(ii)
If an edge-path
˜γ
fails to be locally injective, we can simplify it. 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. So it fails to be locally injective if two
consecutive edges in the path are the same edge with opposite orientations:
˜e
i1
˜e
i
˜e
i+1
˜e
i+2
We can just remove the redundant lines and get
˜e
i1
˜e
i+2
This reminds us of two things homotopy of paths and elementary
reduction of words.
(iii)
Each point
˜x p
1
(
x
0
) is joined to
˜x
0
by a unique locally injective
edge-path.
(iv)
For any
w S
, from (ii), we know that
˜γ
is locally injective if and only if
w is reduced.
We can thus conclude that there are bijections
F (S) p
1
(x
0
) π
1
(X, x
0
)
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] π
1
(X, x
0
) such that ˜x
0
· [γ] = ˜x.
So there is a bijection between
F
(
S
) and
π
1
(
X, x
0
). It is easy to see that the
operations on
F
(
S
) and
π
1
(
X, x
0
) are the same, since they are just concatenating
words or paths. So this bijection identifies the two group structures. So this
induces an isomorphism F (S)
=
π
1
(X, x
0
).
4.3 Free products with amalgamation
We managed to compute the fundamental group of the circle. But we want
to find the fundamental group of more things. Recall that at the beginning,
we defined cell complexes, and said these are the things we want to work with.
Cell complexes are formed by gluing things together. So we want to know what
happens when we glue things together.
Suppose a space
X
is constructed from two spaces
A, B
by gluing (i.e.
X
=
A B
). We would like to describe
π
1
(
X
) in terms of
π
1
(
A
) and
π
1
(
B
). To
understand this, we need to understand how to “glue” groups together.
Definition (Free product). Suppose we have groups
G
1
=
hS
1
| R
1
i, G
2
=
hS
2
|
R
2
i, where we assume S
1
S
2
= . The free product G
1
G
2
is defined to be
G
1
G
2
= hS
1
S
2
| R
1
R
2
i.
This is not a really satisfactory definition. A group can have many different
presentations, and it is not clear this is well-defined. However, it is clear that
this group exists. Note that there are natural homomorphisms
j
i
:
G
i
G
1
G
2
that send generators to generators. Then we can have the following universal
property of the free product:
Lemma.
G
1
G
2
is the group such that for any group
K
and homomorphisms
φ
i
:
G
i
K
, there exists a unique homomorphism
f
:
G
1
G
2
K
such that
the following diagram commutes:
G
2
G
1
G
1
G
2
K
φ
2
j
2
φ
1
j
1
f
Proof.
It is immediate from the universal property of the definition of presenta-
tions.
Corollary. The free product is well-defined.
Proof.
The conclusion of the universal property can be seen to characterize
G
1
G
2
up to isomorphism.
Again, we have a definition in terms of a concrete construction of the group,
without making it clear this is well-defined; then we have a universal property
that makes it clear this is well-defined, but not clear that the object actually
exists. Combining the two would give everything we want.
However, this free product is not exactly what we want, since there is little
interaction between
G
1
and
G
2
. In terms of gluing spaces, this corresponds to
gluing
A
and
B
when
A B
is trivial (i.e. simply connected). What we really
need is the free product with amalgamation, as you might have guessed from
the title.
Definition (Free product with amalgamation). Suppose we have groups
G
1
,
G
2
and H, with the following homomorphisms:
H G
2
G
1
i
2
i
1
The free product with amalgamation is defined to be
G
1
H
G
2
= G
1
G
2
/hh{(j
2
i
2
(h))
1
(j
1
i
1
)(h) : h H}ii.
Here we are attempting to identify things “in H as the same, but we need
to use the maps
j
k
and
i
k
to map the things from
H
to
G
1
G
2
. So we want to
say “for any
h
,
j
1
i
1
(
h
) =
j
2
i
2
(
h
)”, or (
j
2
i
2
(
h
))
1
(
j
1
i
1
)(
h
) =
e
. So we
quotient by the (normal closure) of these things.
We have the universal property
Lemma.
G
1
H
G
2
is the group such that for any group
K
and homomorphisms
φ
i
:
G
i
K
, there exists a unique homomorphism
G
1
H
G
2
K
such that the
following diagram commutes:
H G
2
G
1
G
1
H
G
2
K
i
2
i
1
φ
2
j
2
φ
1
j
1
f
This is the language we will need to compute fundamental groups. These
definitions will (hopefully) become more concrete as we see more examples.
5 Seifert-van Kampen theorem
5.1 Seifert-van Kampen theorem
The Seifert-van Kampen theorem is the theorem that tells us what happens
when we glue spaces together.
Here we let X = A B, where A, B, A B are path-connected.
x
0
A B
We pick a basepoint
x
0
A B
for convenience. Since we like diagrams, we can
write this as a commutative diagram:
A B B
A X
where all arrows are inclusion (i.e. injective) maps. We can consider what
happens when we take the fundamental groups of each space. Then we have the
induced homomorphisms
π
1
(A B, x
0
) π
1
(B, x
0
)
π
1
(A, x
0
) π
1
(X, x
0
)
We might guess that π
1
(X, x
0
) is just the free product with amalgamation
π
1
(X, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
The Seifert-van Kampen theorem says that, under mild hypotheses, this guess is
correct.
Theorem (Seifert-van Kampen theorem). Let
A, B
be open subspaces of
X
such
that
X
=
A B
, and
A, B, A B
are path-connected. Then for any
x
0
A B
,
we have
π
1
(X, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
Note that by the universal property of the free product with amalgamation,
we by definition know that there is a unique map
π
1
(
A, x
0
)
π
1
(AB,x
0
)
π
1
(
B, x
0
)
π
1
(X, x
0
). The theorem asserts that this map is an isomorphism.
Proof is omitted because time is short.
Example. Consider a higher-dimensional sphere
S
n
=
{
v
R
n+1
:
|
v
|
= 1
}
for
n 2. We want to find π
1
(S
n
).
The idea is to write
S
n
as a union of two open sets. We let
n
= e
1
S
n
R
n+1
be the North pole, and
s
=
e
1
be the South pole. We let
A
=
S
n
\ {n}
,
and
B
=
S
n
\ {s}
. By stereographic projection, we know that
A, B
=
R
n
. The
hard part is to understand the intersection.
To do so, we can draw a cylinder
S
n1
×
(
1
,
1), and project our
A B
onto the cylinder. We can similarly project the cylinder onto
A B
. So
A B
=
S
n1
× (1, 1) ' S
n1
, since (1, 1) is contractible.
n
s
We can now apply the Seifert-van Kampen theorem. Note that this works only
if S
n1
is path-connected, i.e. n 2. Then this tells us that
π
1
(S
n
)
=
π
1
(R
n
)
π
1
(S
n1
)
π
1
(R
n
)
=
1
π
1
(S
n1
)
1
It is easy to see this is the trivial group. We can see this directly from the
universal property of the amalgamated free product, or note that it is the quotient
of 1 1, which is 1.
So for n 2, π
1
(S
n
)
=
1.
We have found yet another simply connected space. However, this is unlike
our previous examples. Our previous spaces were simply connected because they
were contractible. However, we will later see that
S
n
is not contractible. So this
is genuinely a new, interesting example.
Why did we go though all this work to prove that
π
1
(
S
n
)
=
1? It feels like
we can just prove this directly pick a point that is not in the curve as the
North pole, project stereographically to
R
n
, and contract it there. However,
the problem is that space-filling curves exist. We cannot guarantee that we can
pick a point not on the curve! It is indeed possible to prove directly that given
any curve on
S
n
(with
n
1), we can deform it slightly so that it is no longer
surjective. Then the above proof strategy works. However, using Seifert-van
Kampen is much neater.
Example (
RP
n
). Recall that
RP
n
=
S
n
/ id}
, and the quotient map
S
n
RP
n
is a covering map. Now that we have proved that
S
n
is simply connected,
we know that S
n
is a universal cover of RP
n
.
For any x
0
RP
n
, we have a bijection
π
1
(RP
n
, x
0
) p
1
(x
0
).
Since
p
1
(
x
0
) has two elements by definition, we know that
|π
1
(
RP
n
, x
0
)
|
= 2.
So π
1
(RP
n
, x
0
) = Z/2.
You will prove a generalization of this in example sheet 2.
Example (Wedge of two circles). We are going to consider the operation of
wedging. Suppose we have two topological spaces, and we want to join them
together. The natural way to join them is to take the disjoint union. What if
we have based spaces? If we have (
X, x
0
) and (
Y, y
0
), we cannot just take the
disjoint union, since we will lose our base point. What we do is take the wedge
sum, where we take the disjoint union and then identify the base points:
X Y
x
0
y
0
X Y
x
0
y
0
Suppose we take the wedge sum of two circles
S
1
S
1
. We would like to pick
A, B
to be each of the circles, but we cannot, since
A
and
B
have to be open.
So we take slightly more, and get the following:
x
0
A B
Each of A and B now look like this:
We see that both
A
and
B
retract to the circle. So
π
1
(
A
)
=
π
1
(
B
)
=
Z
, while
A B is a cross, which retracts to a point. So π
1
(A B) = 1.
Hence by the Seifert-van Kampen theorem, we get
π
1
(S
1
S
1
, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
)
=
Z
1
Z
=
Z Z
=
F
2
,
where
F
2
is just
F
(
S
) for
|S|
= 2. We can see that
Z Z
=
F
2
by showing that
they satisfy the same universal property.
Note that we had already figured this out when we studied the free group,
where we realized F
2
is the fundamental group of this thing.
More generally, as long as
x
0
, y
0
in
X
and
Y
are “reasonable”,
π
1
(
X Y
)
=
π
1
(X) π
1
(Y ).
Next, we would exhibit some nice examples of the covering spaces of
S
1
S
1
,
i.e. the “rose with 2 petals”.
Recall that π
1
(S
1
S
1
, x
0
)
=
F
2
=
ha, bi.
Example. Consider the map
φ
:
F
2
Z/
3 which sends
a 7→
1
, b 7→
1. Note
that 1 is not the identity, since this is an abelian group, and the identity is 0.
This exists since the universal property tells us we just have to say where the
generators go, and the map exists (and is unique).
Now
ker φ
is a subgroup of
F
2
. So there is a based covering space of
S
1
S
1
corresponding to it, say,
˜
X. Let’s work out what it is.
First, we want to know how many sheets it has, i.e. how many copies of
x
0
we have. There are three, since we know that the number of sheets is the index
of the subgroup, and the index of
ker φ
is
|Z/
3
|
= 3 by the first isomorphism
theorem.
˜x
0
Let’s try to lift the loop
a
at
˜x
0
. Since
a 6∈ ker φ
=
π
1
(
˜
X, ˜x
0
),
a
does not lift to
a loop. So it goes to another vertex.
˜x
0
a
Similarly, a
2
6∈ ker φ = π
1
(
˜
X, ˜x
0
). So we get the following
˜x
0
a
a
Since a
3
ker φ, we get a loop labelled a.
˜x
0
a
a
a
Note that
ab
1
ker φ
. So
ab
1
gives a loop. So
b
goes in the same direction as
a:
˜x
0
a
a
a
b
b
b
This is our covering space.
This is a fun game to play at home:
(i) Pick a group G (finite groups are recommended).
(ii) Then pick α, β G and let φ : F
2
G send a 7→ α, b 7→ β.
(iii) Compute the covering space corresponding to φ.
5.2 The effect on π
1
of attaching cells
We have started the course by talking about cell complexes, but largely ignored
them afterwards. Finally, we are getting back to them.
The process of attaching cells is as follows: we start with a space
X
, get a
function
S
n1
X
, and attach
D
n
to
X
by gluing along the image of
f
to get
X
f
D
n
:
0
y
0
x
0
Since we are attaching stuff, we can use the Seifert-van Kampen theorem to
analyse this.
Theorem. If
n
3, then
π
1
(
X
f
D
n
)
=
π
1
(
X
). More precisely, the map
π
1
(
X, x
0
)
π
1
(
X
f
D
n
, x
0
) induced by inclusion is an isomorphism, where
x
0
is a point on the image of f .
This tells us that attaching a high-dimensional disk does not do anything to
the fundamental group.
Proof.
Again, the difficulty of applying Seifert-van Kampen theorem is that we
need to work with open sets.
Let 0
D
n
be any point in the interior of
D
n
. We let
A
=
X
f
(
D
n
\ {
0
}
).
Note that
D
n
\{
0
}
deformation retracts to the boundary
S
n1
. So
A
deformation
retracts to X. Let B =
˚
D, the interior of D
n
. Then
A B =
˚
D
n
\ 0
=
S
n1
× (1, 1)
We cannot use
y
0
as our basepoint, since this point is not in
A B
. Instead,
pick an arbitrary
y
1
A B
. Since
D
n
is path connected, we have a path
γ
:
y
1
y
0
, and we can use this to recover the fundamental groups based at
y
0
.
Now Seifert-van Kampen theorem says
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
)
π
1
(AB,y
1
)
π
1
(B, y
1
).
Since
B
is just a disk, and
A B
is simply connected (
n
3 implies
S
n1
is
simply connected), their fundamental groups are trivial. So we get
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
).
We can now use γ to change base points from y
1
to y
0
. So
π
1
(X
f
D
n
, y
0
)
=
π
1
(A, y
0
)
=
π
1
(X, y
0
).
The more interesting case is when we have smaller dimensions.
Theorem. If
n
= 2, then the natural map
π
1
(
X, x
0
)
π
1
(
X
f
D
n
, x
0
) is
surjective, and the kernel is
hh
[
f
]
ii
. Note that this statement makes sense, since
S
n1
is a circle, and f : S
n1
X is a loop in X.
This is what we would expect, since if we attach a disk onto the loop given
by f, this loop just dies.
Proof. As before, we get
π
1
(X
f
D
n
, y
1
)
=
π
1
(A, y
1
)
π
1
(AB,y
1
)
π
1
(B, y
1
).
Again,
B
is contractible, and
π
1
(
B, y
1
)
=
1. However,
π
1
(
A B, y
1
)
=
Z
. Since
π
1
(A B, y
1
) is just (homotopic to) the loop induced by f, it follows that
π
1
(A, y
1
)
π
1
(AB,y
1
)
1 = (π
1
(A, y
1
) 1)/hhπ
1
(A B, y
1
)ii
=
π
1
(X, x
0
)/hhfii.
In summary, we have
π
1
(X
f
D
n
) =
(
π
1
(X) n 3
π
1
(X)/hhfii n = 2
This is a useful result, since this is how we build up cell complexes. If we want to
compute the fundamental groups, we can just work up to the two-cells, and know
that the higher-dimensional cells do not have any effect. Moreover, whenever
X
is a cell complex, we should be able to write down the presentation of π
1
(X).
Example. Let
X
be the 2-torus. Possibly, our favorite picture of the torus is
(not a doughnut):
This is already a description of the torus as a cell complex!
We start with our zero complex X
(0)
:
We then add our 1-cells to get X
(1)
:
a
b
We now glue our square to the cell complex to get X = X
(2)
:
matching up the colors and directions of arrows.
So we have our torus as a cell complex. What is its fundamental group?
There are many ways we can do this computation, but this time we want to do
it as a cell complex.
We start with
X
(0)
. This is a single point. So its fundamental group is
π
1
(X
(0)
) = 1.
When we add our two 1-cells, we get π
1
(X
(1)
) = F
2
=
ha, bi.
Finally, to get
π
1
(
X
), we have to quotient out by the boundary of our square,
which is just aba
1
b
1
. So we have
π
1
(X
(2)
) = F
2
/hhaba
1
b
1
ii = ha, b | aba
1
b
1
i
=
Z
2
.
We have the last congruence since we have two generators, and then we make
them commute by quotienting the commutator out.
This procedure can be reversed given a presentation of a group, we can just
add the right edges and squares to produce a cell complex with that presentation.
Corollary. For any (finite) group presentation
hS | Ri
, there exists a (finite)
cell complex (of dimension 2) X such that π
1
(X)
=
hS | Ri.
There really isn’t anything that requires that finiteness in this proof, but
finiteness makes us feel more comfortable.
Proof.
Let
S
=
{a
1
, · · · , a
m
}
and
R
=
{r
1
, · · · , r
n
}
. We start with a single point,
and get our
X
(1)
by adding a loop about the point for each
a
i
S
. We then get
our 2-cells
e
2
j
for
j
= 1
, · · · , n
, and attaching them to
X
(1)
by
f
i
:
S
1
X
(1)
given by a based loop representing r
i
F (S).
Since all groups have presentations, this tells us that all groups are funda-
mental groups of some spaces (at least those with finite presentations).
5.3 A refinement of the Seifert-van Kampen theorem
We are going to make a refinement of the theorem so that we don’t have to
worry about that openness problem. We first start with a definition.
Definition (Neighbourhood deformation retract). A subset
A X
is a neigh-
bourhood deformation retract if there is an open set
A U X
such that
A
is
a strong deformation retract of
U
, i.e. there exists a retraction
r
:
U A
and
r ' id
U
rel A.
This is something that is true most of the time, in sufficiently sane spaces.
Example. If
Y
is a subcomplex of a cell complex, then
Y
is a neighbourhood
deformation retract.
Theorem. Let
X
be a space,
A, B X
closed subspaces. Suppose that
A
,
B
and
A B
are path connected, and
A B
is a neighbourhood deformation
retract of A and B. Then for any x
0
A B.
π
1
(X, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
x
0
A B
This is just like Seifert-van Kampen theorem, but usually easier to apply, since
we no longer have to “fatten up” our A and B to make them open.
Proof.
Pick open neighbourhoods
A B U A
and
A B V B
that strongly deformation retract to
A B
. Let
U
be such that
U
retracts
to
A B
. Since
U
retracts to
A
, it follows that
U
is path connected since
path-connectedness is preserved by homotopies.
Let
A
0
=
AV
and
B
0
=
BU
. Since
A
0
= (
X \B
)
V
, and
B
0
= (
X \A
)
U
,
it follows that A
0
and B
0
are open.
Since
U
and
V
retract to
A B
, we know
A
0
' A
and
B
0
' B
. Also,
A
0
B
0
= (
AV
)
(
B U
) =
U V ' A B
. In particular, it is path connected.
So by Seifert van-Kampen, we get
π
1
(A B) = π
1
(A
0
, x
0
)
π
1
(A
0
B
0
,x
0
)
π
1
(B
0
, x
0
) = π
1
(A, x
0
)
π
1
(AB,x
0
)
π
1
(B, x
0
).
This is basically what we’ve done all the time when we enlarge our
A
and
B
to become open.
Example. Let
X
=
S
1
S
1
, the rose with two petals. Let
A, B
=
S
1
be the
circles.
x
0
A B
Then since
{x
0
}
=
A B
is a neighbourhood deformation retract of
A
and
B
,
we know that
π
1
X
=
π
1
S
1
π
1
S
1
.
5.4 The fundamental group of all surfaces
We have found that the torus has fundamental group
Z
2
, but we already knew
this, since the torus is just
S
1
× S
1
, and the fundamental group of a product is
the product of the fundamental groups, as you have shown in the example sheet.
So we want to look at something more interesting. We look at all surfaces.
We start by defining what a surface is. It is surprisingly difficult to get
mathematicians to agree on how we can define a surface. Here we adopt the
following definition:
Definition (Surface). A surface is a Hausdorff topological space such that every
point has a neighbourhood U that is homeomorphic to R
2
.
Some people like
C
more that
R
, and sometimes they put
C
2
instead of
R
2
in
the definition, which is confusing, since that would have two complex dimensions
and hence four real dimensions. Needless to say, the actual surfaces will also
be different and have different homotopy groups. We will just focus on surfaces
with two real dimensions.
To find the fundamental group of all surfaces, we rely on the following
theorem that tells us what surfaces there are.
Theorem (Classification of compact surfaces). If
X
is a compact surface, then
X is homeomorphic to a space in one of the following two families:
(i)
The orientable surface of genus
g
, Σ
g
includes the following (please excuse
my drawing skills):
A more formal definition of this family is the following: we start with the
2-sphere, and remove a few discs from it to get
S
2
\
g
i=1
D
2
. Then we take
g tori with an open disc removed, and attach them to the circles.
(ii)
The non-orientable surface of genus
n
,
E
n
=
{RP
2
, K, · · · }
(where
K
is
the Klein bottle). This has a similar construction as above: we start with
the sphere S
2
, make a few holes, and then glue obius strips to them.
It would be nice to be able to compute fundamental groups of these guys.
To do so, we need to write them as polygons with identification.
Example. To obtain a surface of genus two, written Σ
2
, we start with what we
had for a torus:
a
b
a
b
If we just wanted a torus, we are done (after closing the loop), but now we want
a surface with genus 2, so we add another torus:
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
To visualize how this works, imagine cutting this apart along the dashed line.
This would give two tori with a hole, where the boundary of the holes are just the
dashed line. Then gluing back the dashed lines would give back our orientable
surface with genus 2.
In general, to produce Σ
g
, we produce a polygon with 4g sides. Then we get
π
1
Σ
g
= ha
1
, b
1
, · · · , a
g
, b
g
| a
1
b
1
a
1
1
b
1
1
· · · a
g
b
g
a
1
g
b
1
g
i.
We do we care? The classification theorem tells us that each surface is homeomor-
phic to some of these orientable and non-orientable surfaces, but it doesn’t tell us
there is no overlap. It might be that Σ
6
=
Σ
241
, via some weird homeomorphism
that destroys some holes.
However, this result lets us know that all these orientable surfaces are
genuinely different. While it is difficult to stare at this fundamental group
and say that
π
1
Σ
g
6
=
π
1
Σ
g
0
for
g 6
=
g
0
, we can perform a little trick. We can
take the abelianization of the group
π
1
Σ
g
, where we further quotient by all
commutators. Then the abelianized fundamental group of Σ
g
will simply be
Z
2g
. These are clearly distinct for different values of
g
. So all these surfaces are
distinct. Moreover, they are not even homotopy equivalent.
The fundamental groups of the non-orientable surfaces is left as an exercise
for the reader.
6 Simplicial complexes
So far, we have taken a space
X
, and assigned some things to it. The first was
easy
π
0
(
X
). It was easy to calculate and understand. We then spent a lot of
time talking about
π
1
(
X
). What are they good for? A question we motivated
ourselves with was to prove
R
m
=
R
n
implies
n
=
m
.
π
0
was pretty good for the
case when
n
= 1. If
R
m
=
R
, then we would have
R
m
\ {
0
}
=
R \ {
0
} ' S
0
. We
know that
|π
0
(
S
0
)
|
= 2, while
|π
0
(
R
m
\ {
0
}
)
|
= 1 for
m 6
= 1. This is just a fancy
way of saying that R \ {0} is disconnected while R
m
\ {0} is not for m 6= 1.
We can just add 1 to
n
, and add 1 to our subscript. If
R
m
=
R
2
, then we
have
R
m
\{
0
}
=
R
2
\{
0
} ' S
1
. We know that
π
1
(
S
1
)
=
Z
, while
π
1
(
R
m
\{
0
}
)
=
π
1
(S
m1
)
=
1 unless m = 2.
The obvious thing to do is to create some
π
n
(
X
). But as you noticed,
π
1
took us quite a long time to define, and was really hard to compute. As we
would expect, this only gets harder as we get to higher dimensions. This is
indeed possible, but will only be done in Part III courses.
The problem is that π
n
works with groups, and groups are hard. There are
too many groups out there. We want to do some easier algebra, and a good
choice is linear algebra. Linear algebra is easy. In the rest of the course, we
will have things like
H
0
(
X
) and
H
1
(
X
), instead of
π
n
, which are more closely
related to linear algebra.
Another way to motivate the abandoning of groups is as follows: recall last
time we saw that if
X
is a finite cell, reasonably explicitly defined, then we can
write down a presentation
hS | Ri
for
π
1
(
X
). This sounds easy, except that we
don’t understand presentations. In fact there is a theorem that says there is no
algorithm that decides if a particular group presentation is actually the trivial
group.
So even though we can compute the fundamental group in terms of presenta-
tions, this is not necessarily helpful. On the other hand, linear algebra is easy.
Computers can do it. So we will move on to working with linear algebra instead.
This is where homology theory comes in. It takes a while for us to define it,
but after we finish developing the machinery, things are easy.
6.1 Simplicial complexes
There are many ways of formulating homology theory, and these are all equivalent
at least for sufficiently sane spaces (such as cell complexes). In this course, we
will be using simplicial homology, which is relatively more intuitive and can be
computed directly. The drawback is that we will have to restrict to a particular
kind of space, known as simplicial complexes. This is not a very serious restriction
per se, since many spaces like spheres are indeed simplicial complexes. However,
the definition of simplicial homology is based on exactly how we view our space
as a simplicial complex, and it will take us quite a lot of work to show that the
simplicial homology is indeed a property of the space itself, and not how we
represent it as a simplicial complex.
We now start by defining simplicial complexes, and developing some general
theory of simplicial complexes that will become useful later on.
Definition (Affine independence). A finite set of points
{a
1
, · · · , a
n
} R
m
is
affinely independent iff
n
X
i=1
t
i
a
i
= 0 with
n
X
i=1
t
i
= 0 t
i
= 0 for all i.
Example. When n = 3, the following points are affinely independent:
The following are not:
The proper way of understanding this definition is via the following lemma:
Lemma.
a
0
, · · · , a
n
R
m
are affinely independent if and only if
a
1
a
0
, · · · , a
n
a
0
are linearly independent.
Alternatively,
n
+ 1 affinely independent points span an
n
-dimensional thing.
Proof. Suppose a
0
, · · · , a
n
are affinely independent. Suppose
n
X
i=1
λ
i
(a
i
a
0
) = 0.
Then we can rewrite this as
n
X
i=1
λ
i
!
a
0
+ λ
1
a
1
+ · · · + λ
n
a
n
= 0.
Now the sum of the coefficients is 0. So affine independence implies that all
coefficients are 0. So a
1
a
0
, · · · , a
n
a
0
are linearly independent.
On the other hand, suppose
a
1
a
0
, · · · , a
n
a
0
are linearly independent.
Now suppose
n
X
i=0
t
i
a
i
= 0,
n
X
i=0
t
i
= 0.
Then we can write
t
0
=
n
X
i=1
t
i
.
Then the first equation reads
0 =
n
X
i=1
t
i
!
a
0
+ t
1
a
1
+ · · · + t
n
a
n
=
n
X
i=1
t
i
(a
i
a
0
).
So linear independence implies all t
i
= 0.
The relevance is that these can be used to define simplices (which are simple,
as opposed to complexes).
Definition (
n
-simplex). An
n
-simplex is the convex hull of (
n
+ 1) affinely
independent points a
0
, · · · , a
n
R
m
, i.e. the set
σ = ha
0
, · · · , a
n
i =
(
n
X
i=0
t
i
a
i
:
n
X
i=0
t
i
= 1, t
i
0
)
.
The points
a
0
, · · · , a
n
are the vertices, and are said to span
σ
. The (
n
+ 1)-tuple
(t
0
, · · · , t
n
) is called the barycentric coordinates for the point
P
t
i
a
i
.
Example. When n = 0, then our 0-simplex is just a point:
When n = 1, then we get a line:
When n = 2, we get a triangle:
When n = 3, we get a tetrahedron:
The key motivation of this is that simplices are determined by their vertices.
Unlike arbitrary subspaces of
R
n
, they can be specified by a finite amount of
data. We can also easily extract the faces of the simplices.
Definition (Face, boundary and interior). A face of a simplex is a subset (or
subsimplex) spanned by a subset of the vertices. The boundary is the union of
the proper faces, and the interior is the complement of the boundary.
The boundary of
σ
is usually denoted by
σ
, while the interior is denoted by
˚σ, and we write τ σ when τ is a face of σ.
In particular, the interior of a vertex is the vertex itself. Note that these
notions of interior and boundary are distinct from the topological notions of
interior and boundary.
Example. The standard
n
-simplex is spanned by the basis vectors
{
e
0
, · · · ,
e
n
}
in R
n+1
. For example, when n = 2, we get the following:
We will now glue simplices together to build complexes, or simplicial com-
plexes.
Definition. A (geometric) simplicial complex is a finite set
K
of simplices in
R
n
such that
(i) If σ K and τ is a face of σ, then τ K.
(ii) If σ, τ K, then σ τ is either empty or a face of both σ and τ .
Definition (Vertices). The vertices of
K
are the zero simplices of
K
, denoted
V
K
.
Example. This is a simplicial complex:
These are not:
Technically, a simplicial complex is defined to be a set of simplices, which
are just collections of points. It is not a subspace of
R
n
. Hence we have the
following definition:
Definition (Polyhedron). The polyhedron defined by
K
is the union of the
simplices in K, and denoted by |K|.
We make this distinction because distinct simplicial complexes may have the
same polyhedron, such as the following:
Just as in cell complexes, we can define things like dimensions.
Definition (Dimension and skeleton). The dimension of
K
is the highest
dimension of a simplex of
K
. The
d
-skeleton
K
(d)
of
K
is the union of the
n-simplices in K for n d.
Note that since these are finite and live inside
R
n
, we know that
|K|
is always
compact and Hausdorff.
Usually, when we are given a space, say
S
n
, it is not defined to be a simplicial
complex. We can “make” it a simplicial complex by a triangulation.
Definition (Triangulation). A triangulation of a space
X
is a homeomorphism
h : |K| X, where K is some simplicial complex.
Example. Let
σ
be the standard
n
-simplex. The boundary
σ
is homeomorphic
to
S
n1
(e.g. the boundary of a (solid) triangle is the boundary of the triangle,
which is also a circle). This is called the simplicial (n 1)-sphere.
We can also triangulate our S
n
in a different way:
Example. In
R
n+1
, consider the simplices
e
0
, · · · , ±
e
n
i
for each possible
combination of signs. So we have 2
n+1
simplices in total. Then their union
defines a simplicial complex K, and
|K|
=
S
n
.
The nice thing about this triangulation is that the simplicial complex is
invariant under the antipodal map. So not only can we think of this as a
triangulation of the sphere, but a triangulation of RP
n
as well.
As always, we don’t just look at objects themselves, but also maps between
them.
Definition (Simplicial map). A simplicial map
f
:
K L
is a function
f
:
V
K
V
L
such that if
ha
0
, · · · , a
n
i
is a simplex in
K
, then
{f
(
a
0
)
, · · · , f
(
a
n
)
}
spans a simplex of L.
The nice thing about simplicial maps is that we only have to specify where
the vertices go, and there are only finitely many vertices. So we can completely
specify a simplicial map by writing down a finite amount of information.
It is important to note that we say
{f
(
a
0
)
, · · · , f
(
a
n
)
}
as a set span a simplex
of L. In particular, they are allowed to have repeats.
Example. Suppose we have the standard 2-simplex K as follows:
a
0
a
1
a
2
The following does not define a simplicial map because
ha
1
, a
2
i
is a simplex in
K, but {f(a
1
), f(a
2
)} does not span a simplex:
f(a
0
), f(a
1
) f(a
2
)
On the other hand, the following is a simplicial map, because now
{f
(
a
1
)
, f
(
a
2
)
}
spans a simplex, and note that
{f
(
a
0
)
, f
(
a
1
)
, f
(
a
2
)
}
also spans a 1-simplex
because we are treating the collection of three vertices as a set, and not a
simplex.
f(a
0
), f(a
1
) f(a
2
)
Finally, we can also do the following map:
f(a
0
), f(a
1
) f(a
2
)
The following lemma is obvious, but we will need it later on.
Lemma. If
K
is a simplicial complex, then every point
x |K|
lies in the
interior of a unique simplex.
As we said, simplicial maps are nice, but they are not exactly what we want.
We want to have maps between spaces.
Lemma. A simplicial map
f
:
K L
induces a continuous map
|f|
:
|K| |L|
,
and furthermore, we have
|f g| = |f| |g|.
There is an obvious way to define this map. We know how to map vertices,
and then just extend everything linearly.
Proof. For any point in a simplex σ = ha
0
, · · · , a
n
i, we define
|f|
n
X
i=0
t
i
a
i
!
=
n
X
i=0
t
i
f(a
i
).
The result is in
L
because
{f
(
a
i
)
}
spans a simplex. It is not difficult to see this
is well-defined when the point lies on the boundary of a simplex. This is clearly
continuous on σ, and is hence continuous on |K| by the gluing lemma.
The final property is obviously true by definition.
6.2 Simplicial approximation
This is all very well, but we are really interested in continuous maps. So given a
continuous map
f
, we would like to find a related simplicial map
g
. In particular,
we want to find a simplicial map
g
that “approximates”
f
in some sense. The
definition we will write down is slightly awkward, but it turns out this is the
most useful definition.
Definition (Open star and link). Let
x |K|
. The open star of
x
is the union
of all the interiors of the simplices that contain x, i.e.
St
K
(x) =
[
xσK
˚σ.
The link of
x
, written
Lk
K
(
x
), is the union of all those simplices that do not
contain x, but are faces of a simplex that does contain x.
Definition (Simplicial approximation). Let
f
:
|K| |L|
be a continuous map
between the polyhedra. A function
g
:
V
K
V
L
is a simplicial approximation
to f if for all v V
K
,
f(St
K
(v)) St
L
(g(v)). ()
The following lemma tells us why this is a good definition:
Lemma. If
f
:
|K| |L|
is a map between polyhedra, and
g
:
V
K
V
L
is a simplicial approximation to
f
, then
g
is a simplicial map, and
|g| ' f
.
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 .
Proof.
First we want to check
g
is really a simplicial map if it satisfies (
). Let
σ
=
ha
0
, · · · , a
n
i
be a simplex in
K
. We want to show that
{g
(
a
0
)
, · · · , g
(
a
n
)
}
spans a simplex in L.
Pick an arbitrary
x ˚σ
. Since
σ
contains each
a
i
, we know that
x St
K
(
a
i
)
for all i. Hence we know that
f(x)
n
\
i=0
f(St
K
(a
i
))
n
\
i=0
St
L
(g(a
i
)).
Hence we know that there is one simplex, say,
τ
that contains all
g
(
a
i
) whose
interior contains
f
(
x
). Since each
g
(
a
i
) is a vertex in
L
, each
g
(
a
i
) must be a
vertex of τ . So they span a face of τ, as required.
We now want to prove that
|g| ' f
. We let
H
:
|K| × I |L| R
m
be
defined by
(x, t) 7→ t|g|(x) + (1 t)f(x).
This is clearly continuous. So we need to check that
im H |L|
. But we
know that both
|g|
(
x
) and
f
(
x
) live in
τ
and
τ
is convex. It thus follows that
H(x × I) τ |L|.
To prove the last part, it suffices to show that every simplicial approximation
to a simplicial map must be the map itself. Then the homotopy is
rel M
by
the construction above. This is easily seen to be true if
g
is a simplicial
approximation to
f
, then
f
(
v
)
f
(
St
K
(
v
))
St
L
(
g
(
v
)). Since
f
(
v
) is a vertex
and
g
(
v
) is the only vertex in
St
L
(
g
(
v
)), we must have
f
(
v
) =
g
(
v
). So done.
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. Is this possible? Let’s take a
nice example. Let’s consider the following K:
How many homotopy classes of continuous maps are there
K K
? Countably
many, one for each winding number. However, there can only be at most 3
3
= 27
simplicial maps. The problem is that we don’t have enough vertices to realize all
those interesting maps. The idea is to refine our simplicial complexes. 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. After we subdivide it once, we can
realize more homotopy classes of maps. 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.
Definition (Barycenter). The barycenter of σ = ha
0
, · · · , a
n
i is
ˆσ =
n
X
i=0
1
n + 1
a
i
.
Definition (Barycentric subdivision). The (first) barycentric subdivision
K
0
of
K is the simplicial complex:
K
0
= {hˆσ
0
, · · · , ˆσ
n
i : σ
i
K and σ
0
< σ
1
< · · · < σ
n
}.
If you stare at this long enough, you will realize this is exactly what we have
drawn above.
The
r
th barycentric subdivision
K
(r)
is defined inductively as the barycentric
subdivision of the r 1th barycentric subdivision, i.e.
K
(r)
= (K
(r1)
)
0
.
Proposition. |K| = |K
0
| and K
0
really is a simplicial complex.
Proof. Too boring to be included in lectures.
We now have a slight problem. Even though
|K
0
|
and
|K|
are equal, the
identity map from |K
0
| to |K| is not a simplicial map.
To solve this problem, we can choose any function
K V
K
by
σ 7→ v
σ
with
v
σ
σ
, i.e. a function that sends any simplex to any of its vertices. Then we can
define
g
:
K
0
K
by sending
ˆσ 7→ v
σ
. Then this is a simplicial map, and indeed
a simplicial approximation to the identity map
|K
0
| |K|
. We will revisit this
idea later when we discuss homotopy invariance.
The key theorem is that as long as we are willing to perform barycentric
subdivisions, then we can always find a simplicial approximation.
Theorem (Simplicial approximation theorem). Let
K
and
L
be simplicial
complexes, and
f
:
|K| |L|
a continuous map. Then there exists an
r
and
a simplicial map
g
:
K
(r)
L
such that
g
is a simplicial approximation of
f
.
Furthermore, if
f
is already simplicial on
M K
, then we can choose
g
such
that |g||
M
= f|
M
.
The first thing we have to figure out is how far we are going to subdivide.
To do this, we want to quantify how “fine” our subdivisions are.
Definition (Mesh). Let K be a simplicial complex. The mesh of K is
µ(K) = max{kv
0
v
1
k : hv
0
, v
1
i K}.
We have the following lemma that tells us how large our mesh is:
Lemma. Let dim K = n, then
µ(K
(r)
)
n
n + 1
r
µ(K).
The key point is that as
r
, the mesh goes to zero. So indeed we can
make our barycentric subdivisions finer and finer. The proof is purely technical
and omitted.
Now we can prove the simplicial approximation theorem.
Proof of simplicial approximation theorem.
Suppose we are given the map
f
:
|K| |L|
. We have a natural cover of
|L|
, namely the open stars of all vertices.
We can use f to pull these back to |K| to obtain a cover of |K|:
{f
1
(St
L
(w)) : w V
L
}.
The idea is to barycentrically subdivide our
K
such that each open star of
K
is
contained in one of these things.
By the Lebesgue number lemma, there exists some
δ
, the Lebesgue number
of the cover, such that for each x |K|, B
δ
(x) is contained in some element of
the cover. By the previous lemma, there is an r such that µ(K
(r)
) < δ.
Now since the mesh
µ
(
K
(r)
) is the smallest distance between any two vertices,
the radius of every open star
St
K
(r)
(
x
) is at most
µ
(
K
(r)
). Hence it follows that
St
K
(r)
(
x
)
B
δ
(
x
) for all vertices
x V
K
(r)
. Therefore, for all
x V
K
(r)
, there
is some w V
L
such that
St
K
(r)
(x) B
δ
(x) f
1
(St
L
(w)).
Therefore defining g(x) = w, we get
f(St
K
(r)
(x)) St
L
(g(x)).
So g is a simplicial approximation of f.
The last part follows from the observation that if
f
is a simplicial map, then
it maps vertices to vertices. So we can pick g(v) = f(v).
7 Simplicial homology
7.1 Simplicial homology
For now, we will forget about simplicial approximations and related fluff, but
just note that it is fine to assume everything can be considered to be simplicial.
Instead, we are going to use this framework to move on and define some new
invariants of simplicial complexes
K
, known as
H
n
(
K
). These are analogous
to
π
0
, π
1
, · · ·
, but only use linear algebra in the definitions, and are thus much
simpler. The drawback, however, is that the definitions are slightly less intuitive
at first sight.
Despite saying “linear algebra”, we won’t be working with vector spaces most
of the time. Instead, we will be using abelian groups, which really should be
thought of as
Z
-modules. At the end, we will come up with an analogous theory
using
Q
-modules, i.e.
Q
-vector spaces, and most of the theory will carry over.
Using
Q
makes some of our work easier, but as a result we would have lost some
information. In the most general case, we can replace
Z
with any abelian group,
but we will not be considering any of these in this course.
Recall that we defined an
n
-simplex as the collection of
n
+ 1 vertices that
span a simplex. As a simplex, when we permute the vertices, we still get the
same simplex. What we want to do is to remember the orientation of the simplex.
In particular, we want think of the simplices (a, b) and (b, a) as different:
a
b b
a
Hence we define oriented simplices.
Definition (Oriented
n
-simplex). An oriented
n
-simplex in a simplicial complex
K
is an (
n
+ 1)-tuple (
a
0
, · · · , a
n
) of vertices
a
i
V
k
such that
ha
0
, · · · , a
n
i K
,
where we think of two (
n
+ 1)-tuples (
a
0
, · · · , a
n
) and (
a
π(0)
, · · · , a
π(n)
) as the
same oriented simplex if π S
n
is an even permutation.
We often denote an oriented simplex as
σ
, and then
¯σ
denotes the same
simplex with the opposite orientation.
Example. As oriented 2-simplices, (
v
0
, v
1
, v
2
) and (
v
1
, v
2
, v
0
) are equal, but
they are different from (
v
2
, v
1
, v
0
). We can imagine the two different orientations
of the simplices as follows:
v
0
v
2
v
1
v
0
v
1
v
2
One and two dimensions are the dimensions where we can easily visualize
the orientation. This is substantially harder in higher dimensions, and often we
just work with the definition instead.
Definition (Chain group
C
n
(
K
)). Let
K
be a simplicial complex. For each
n 0, we define C
n
(K) as follows:
Let
{σ
1
, · · · , σ
`
}
be the set of
n
-simplices of
K
. For each
i
, choose an
orientation on
σ
i
. That is, choose an order for the vertices (up to an even
permutation). This choice is not important, but we need to make it. Now when
we say σ
i
, we mean the oriented simplex with this particular orientation.
Now let
C
n
(
K
) be the free abelian group with basis
{σ
1
, · · · , σ
`
}
, i.e.
C
n
(
K
)
=
Z
`
. So an element in C
n
(K) might look like
σ
3
7σ
1
+ 52σ
64
28σ
1000000
.
In other words, an element of C
n
(K) is just a formal sum of n-simplices.
For convenience, we define
C
`
(
K
) = 0 for
<
0. This will save us from
making exceptions for n = 0 cases later.
For each oriented simplex, we identify σ
i
with ¯σ
i
, at least when n 1.
In this definition, we have to choose a particular orientation for each of our
simplices. If you don’t like making arbitrary choices, we could instead define
C
n
(K) as some quotient, but it is slightly more complicated.
Note that if there are no
n
-simplices (e.g. when
n
=
1), we can still
meaningfully talk about C
n
(K), but it’s just 0.
Example. We can think of elements in the chain group
C
1
(
X
) as “paths” in
X
.
For example, we might have the following simplex:
v
0
v
1
v
2
σ
1
σ
2
σ
3
σ
6
Then the path
v
0
v
1
v
2
v
0
v
1
around the left triangle is represented
by the member
σ
1
σ
2
+ σ
3
+ σ
1
= 2σ
1
σ
2
+ σ
3
.
Of course, with this setup, we can do more random things, like adding 57 copies
of
σ
6
to it, and this is also allowed. So we could think of these as disjoint union
of paths instead.
When defining fundamental groups, we had homotopies that allowed us to
“move around”. Somehow, we would like to say that two of these paths are
“equivalent” under certain conditions. To do so, we need to define the boundary
homomorphisms.
Definition (Boundary homomorphisms). We define boundary homomorphisms
d
n
: C
n
(K) C
n1
(K)
by
(a
0
, · · · , a
n
) 7→
n
X
i=0
(1)
i
(a
0
, · · · , ˆa
i
, · · · , a
n
),
where (
a
0
, · · · , ˆa
i
, · · · , a
n
) = (
a
0
, · · · , a
i1
, a
i+1
, · · · , a
n
) is the simplex with
a
i
removed.
This means we remove each vertex in turn and add them up. This is clear if
we draw some pictures in low dimensions:
v
0
v
1
v
0
v
1
If we take a triangle, we get
v
0
v
2
v
1
v
0
v
2
v
1
An important property of the boundary map is that the boundary of a boundary
is empty:
Lemma. d
n1
d
n
= 0.
In other words, im d
n+1
ker d
n
.
Proof.
This just involves expanding the definition and working through the
mess.
With this in mind, we will define the homology groups as follows:
Definition (Simplicial homology group
H
n
(
K
)). The
n
th simplicial homology
group H
n
(K) is defined as
H
n
(K) =
ker d
n
im d
n+1
.
This is a nice, clean definition, but what does this mean geometrically?
Somehow,
H
k
(
K
) describes all the
k
-dimensional holes” in
|K|
. Since we
are going to draw pictures, we are going to start with the easy case of
k
= 1.
Our
H
1
(
K
) is made from the kernel of
d
1
and the image of
d
2
. First, we give
these things names.
Definition (Chains, cycles and boundaries). The elements of
C
k
(
K
) are called
k
-chains of
K
, those of
ker d
k
are called
k
-cycles of
K
, and those of
im d
k+1
are
called k-boundaries of K.
Suppose we have some
c ker d
k
. In other words,
dc
= 0. If we interpret
c
as a “path”, if it has no boundary, then it represents some sort of loop, i.e. a
cycle. For example, if we have the following cycle:
e
0
e
2
e
1
We have
c = (e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
).
We can then compute the boundary as
dc = (e
1
e
0
) + (e
2
e
1
) + (e
0
e
2
) = 0.
So this c is indeed a cycle.
Now if
c im d
2
, then
c
=
db
for some 2-chain
b
, i.e.
c
is the boundary of
some two-dimensional thing. This is why we call this a 1-boundary. For example,
suppose we have our cycle as above, but is itself a boundary of a 2-chain.
v
0
v
2
v
1
We see that a cycle that has a boundary has been “filled in”. Hence the “holes”
are, roughly, the cycles that haven’t 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”.
Example. Let
K
be the standard simplicial 1-sphere, i.e. we have the following
in R
3
.
e
0
e
1
e
2
Our simplices are thus
K = {he
0
i, he
1
i, he
2
i, he
0
, e
1
i, he
1
, e
2
i, he
2
, e
0
i}.
Our chain groups are
C
0
(K) = h(e
0
), (e
1
), (e
2
)i
=
Z
3
C
1
(K) = h(e
0
, e
1
), (e
1
, e
2
), (e
2
, e
0
)i
=
Z
3
.
All other chain groups are zero. 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. However, you are probably clueful
enough to distinguish the two uses.
Hence, the only non-zero boundary map is
d
1
: C
1
(K) C
0
(K).
We can write down its matrix with respect to the given basis.
1 0 1
1 1 0
0 1 1
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. We have
H
0
(K) =
ker(d
0
: C
0
(K) C
1
(K))
im(d
1
: C
1
(K) C
0
(K))
=
C
0
(K)
im d
1
=
Z
3
im d
1
.
After doing some row operations with our matrix, we see that the image of
d
1
is
a two-dimensional subspace generated by the image of two of the edges. Hence
we have
H
0
(K) = Z.
What does this
H
0
(
K
) represent? We initially said that
H
k
(
K
) should represent
the
k
-dimensional holes, but when
k
= 0, this is simpler. As for
π
0
,
H
0
just
represents the path components of
K
. We interpret this to mean
K
has one
path component. In general, if
K
has
r
path components, then we expect
H
0
(
K
)
to be Z
r
.
Similarly, we have
H
1
(K) =
ker d
1
im d
2
=
ker d
1
.
It is easy to see that in fact we have
ker d
1
= h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
=
Z.
So we also have
H
1
(K)
=
Z.
We see that this
H
1
(
K
) is generated by precisely the single loop in the triangle.
The fact that
H
1
(
K
) is non-trivial means that we do indeed have a hole in the
middle of the circle.
Example. Let L be the standard 2-simplex (and all its faces) in R
3
.
e
0
e
1
e
2
Now our chain groups are
C
0
(L) = C
0
(K)
=
Z
3
=
h(e
0
), (e
1
), (e
2
)i
C
1
(L) = C
1
(K)
=
Z
3
=
h(e
0
, e
1
), (e
1
, e
2
), (e
2
, e
0
)i
C
2
(L)
=
Z = h(e
0
, e
1
, e
2
)i.
Since
d
1
is the same as before, the only new interesting boundary map is
d
2
. We
compute
d
2
((e
0
, e
1
, e
2
)) = (e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
).
We know that
H
0
(
L
) depends only on
d
0
and
d
1
, which are the same as for
K
.
So
H
0
(L)
=
Z.
Again, the interpretation of this is that
L
is path-connected. The first homology
group is
H
1
(L) =
ker d
1
im d
2
=
h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
=
0.
This precisely illustrates the fact that the “hole” is now filled in L.
Finally, we have
H
2
(L) =
ker d
2
im d
3
= ker d
2
=
0.
This is zero since there aren’t any two-dimensional holes in L.
We have hopefully gained some intuition on what these homology groups
mean. We are now going to spend a lot of time developing formalism.
7.2 Some homological algebra
We will develop some formalism to help us compute homology groups
H
k
(
K
) in
lots of examples. Firstly, we axiomatize the setup we had above.
Definition (Chain complex and differentials). A chain complex
C
·
is a sequence
of abelian groups
C
0
, C
1
, C
2
, · · ·
equipped with maps
d
n
:
C
n
C
n1
such that
d
n1
d
n
= 0 for all n. We call these maps the differentials of C
·
.
0 C
0
C
1
C
2
· · ·
d
0
d
1
d
2
d
3
Whenever we have some of these things, we can define homology groups in
exactly the same way as we defined them for simplicial complexes.
Definition (Cycles and boundaries). The space of n-cycles is
Z
n
(C) = ker d
n
.
The space of n-boundaries is
B
n
(C) = im d
n+1
.
Definition (Homology group). The
n
-th homology group of
C
·
is defined to be
H
n
(C) =
ker d
n
im d
n+1
=
Z
n
(C)
B
n
(C)
.
In mathematics, when we have objects, we don’t just talk about the objects
themselves, but also functions between them. Suppose we have two chain
complexes. For the sake of simplicity, we write the chain maps of both as d
n
.
In general, we want to have maps between these two sequences. This would
correspond to having a map
f
i
:
C
i
D
i
for each
i
, satisfying some commuta-
tivity relations.
Definition (Chain map). A chain map
f
·
:
C
·
D
·
is a sequence of homo-
morphisms f
n
: C
n
D
n
such that
f
n1
d
n
= d
n
f
n
for all n. In other words, the following diagram commutes:
0 C
0
C
1
C
2
· · ·
0 D
0
D
1
D
2
· · ·
d
0
f
0
d
1
f
1
d
2
f
2
d
3
d
0
d
1
d
2
d
3
We want to have homotopies. So far, chain complexes seem to be rather rigid,
but homotopies themselves are rather floppy. How can we define homotopies for
chain complexes? It turns out we can have a completely algebraic definition for
chain homotopies.
Definition (Chain homotopy). A chain homotopy between chain maps
f
·
, g
·
:
C
·
D
·
is a sequence of homomorphisms h
n
: C
n
D
n+1
such that
g
n
f
n
= d
n+1
h
n
+ h
n1
d
n
.
We write f
·
' g
·
if there is a chain homotopy between f
·
and g
·
.
The arrows can be put in the following diagram, which is not commutative:
C
n1
C
n
D
n
D
n+1
h
n1
d
n
f
n
g
n
h
n
d
n+1
The intuition behind this definition is as follows: suppose
C
·
=
C
·
(
K
) and
D
·
=
C
·
(
L
) for
K, L
simplicial complexes, and
f
·
and
g
·
are “induced” by
simplicial maps f, g : K L (if f maps an n-simplex σ to a lower-dimensional
simplex, then
f
σ
= 0). How can we detect if
f
and
g
are homotopic via the
homotopy groups?
Suppose
H
:
|K| × I |L|
is a homotopy from
f
to
g
. We further suppose
that
H
actually comes from a simplicial map
K × I L
(we’ll skim over the
technical issue of how we can make
K × I
a simplicial complex. Instead, you are
supposed to figure this out yourself in example sheet 3).
Let σ be an n-simplex of K, and here is a picture of H(σ × I):
Let
h
n
(σ) = H(σ × I).
We think of this as an (
n
+ 1)-chain. What is its boundary? We’ve got the
vertical sides plus the top and bottom. The bottom should just be
f
(
σ
), and
the top is just
g
(
σ
), since
H
is a homotopy from
f
to
g
. How about the sides?
They are what we get when we pull the boundary
σ
up with the homotopy,
i.e.
H
(
σ × I
) =
h
n1
d
n
(
σ
). Now note that
f
(
σ
) and
g
(
σ
) have opposite
orientations, so we get the result
d
n+1
h
n
(σ) = h
n1
d
n
(σ) + g
n
(σ) f
n
(σ).
Rearranging and dropping the σs, we get
d
n+1
h
n
h
n1
d
n
= g
n
f
n
.
This looks remarkably like our definition for chain homotopies of maps, with the
signs a bit wrong. So in reality, we will have to fiddle with the sign of
h
n
a bit
to get it right, but you get the idea.
Lemma. A chain map f
·
: C
·
D
·
induces a homomorphism:
f
: H
n
(C) H
n
(D)
[c] 7→ [f(c)]
Furthermore, if f
·
and g
·
are chain homotopic, then f
= g
.
Proof.
Since the homology groups are defined as the cycles quotiented by the
boundaries, to show that
f
defines a homomorphism, we need to show
f
sends
cycles to cycles and boundaries to boundaries. This is an easy check. If
d
n
(
σ
) = 0,
then
d
n
(f
n
(σ)) = f
n
(d
n
(σ)) = f
n
(0) = 0.
So f
n
(σ) Z
n
(D).
Similarly, if σ is a boundary, say σ = d
n
(τ), then
f
n
(σ) = f
n
(d
n
(τ)) = d
n
(f
n
(τ)).
So f
n
(σ) is a boundary. It thus follows that f
is well-defined.
Now suppose
h
n
is a chain homotopy between
f
and
g
. For any
c Z
n
(
C
),
we have
g
n
(c) f
n
(c) = d
n+1
h
n
(c) + h
n1
d
n
(c).
Since c Z
n
(C), we know that d
n
(c) = 0. So
g
n
(c) f
n
(c) = d
n+1
h
n
(c) B
n
(D).
Hence
g
n
(
c
) and
f
n
(
c
) differ by a boundary. So [
g
n
(
c
)]
[
f
n
(
c
)] = 0 in
H
n
(
D
),
i.e. f
(c) = g
(c).
The following statements are easy to check:
Proposition.
(i) Being chain-homotopic is an equivalence relation of chain maps.
(ii) If a
·
: A
·
C
·
is a chain map and f
·
' g
·
, then f
·
a
·
' g
·
a
·
.
(iii) If f : C
·
D
·
and g : D
·
A
·
are chain maps, then
g
f
= (g
·
f
·
)
.
(iv) (id
C
·
)
= id
H
(C)
.
The last two statements can be summarized in fancy language by saying that
H
n
is a functor.
Definition (Chain homotopy equivalence). Chain complexes
C
·
and
D
·
are
chain homotopy equivalent if there exist
f
·
:
C
·
D
·
and
g
·
:
D
·
C
·
such
that
f
·
g
·
' id
D
·
, g
·
f
·
' id
C
·
.
We should think of this in exactly the same way as we think of homotopy
equivalences of spaces. The chain complexes themselves are not necessarily the
same, but the induced homology groups will be.
Lemma. Let
f
·
:
C
·
D
·
be a chain homotopy equivalence, then
f
:
H
n
(
C
)
H
n
(D) is an isomorphism for all n.
Proof.
Let
g
·
be the homotopy inverse. Since
f
·
g
·
' id
D
·
, we know
f
g
=
id
H
(D)
. Similarly,
g
f
=
id
H
(C)
. So we get isomorphisms between
H
n
(
C
)
and H
n
(D).
7.3 Homology calculations
We’ll get back to topology and compute some homologies.
Here, K is always a simplicial complex, and C
·
= C
·
(K).
Lemma. Let
f
:
K L
be a simplicial map. Then
f
induces a chain map
f
·
: C
·
(K) C
·
(L). Hence it also induces f
: H
n
(K) H
n
(L).
Proof.
This is fairly obvious, except that simplicial maps are allowed to “squash”
simplices, so
f
might send an
n
-simplex to an (
n
1)-simplex, which is not in
D
n
(L). We solve this problem by just killing these troublesome simplices.
Let
σ
be an oriented
n
-simplex in
K
, corresponding to a basis element of
C
n
(K). Then we define
f
n
(σ) =
(
f(σ) f(σ) is an n-simplex
0 f(σ) is a k-simplex for k < n
.
More precisely, if σ = (a
0
, · · · , a
n
), then
f
n
(σ) =
(
(f(a
0
), · · · , f (a
n
)) f(a
0
), · · · , f (a
n
) spans an n-simplex
0 otherwise
.
We then extend f
n
linearly to obtain f
n
: C
n
(K) C
n
(L).
It is immediate from this that this satisfies the chain map condition, i.e.
f
·
commutes with the boundary operators.
Definition (Cone). A simplicial complex is a cone if, for some v
0
V
k
,
|K| = St
K
(v
0
) Lk
K
(v
0
).
v
0
We see that a cone ought to be contractible we can just squash it to the point
v
0
. This is what the next lemma tells us.
Lemma. If
K
is a cone with cone point
v
0
, then inclusion
i
:
{v
0
} |K|
induces
a chain homotopy equivalence i
·
: C
n
({v
0
}) C
n
(K). Therefore
H
n
(K) =
(
Z n = 0
0 n > 0
The homotopy inverse to
i
·
is given by
r
·
, where
r
:
k 7→ v
0
is the only map.
It is clear that
r
·
i
·
=
id
, and we need to show that
i
·
r
·
' id
. This chain
homotopy can be defined by
h
n
:
C
n
(
K
)
C
n+1
(
K
), where
h
n
associates to
any simplex σ in Lk
K
(v
0
) the simplex spanned by σ and v
0
. Details are left to
the reader.
Corollary. If
n
is the standard
n
-simplex, and
L
consists of
n
and all its
faces, then
H
k
(L) =
(
Z k = 0
0 k > 0
Proof. K is a cone (on any vertex).
What we would really like is an example of non-trivial homology groups.
Moreover, we want them in higher dimensions, and not just the examples we got
for fundamental groups. An obvious candidate is the standard n-sphere.
Corollary. Let
K
be the standard (
n
1)-sphere (i.e. the proper faces of
L
from above). Then for n 2, we have
H
k
(K) =
Z k = 0
0 0 < k < n 1
Z k = n 1
.
Proof. We write down the chain groups for K and L.
0 C
0
(L) C
1
(L) · · · C
n1
(L) C
n
(L)
0 C
0
(K) C
1
(K) · · · C
n1
(K) C
n
(K) = 0
=
d
L
1
=
d
L
n1
=
d
L
n
d
K
1
d
K
n1
For
k < n
1, we have
C
k
(
K
) =
C
k
(
L
) and
C
k+1
(
K
) =
C
k+1
(
L
). Also, the
boundary maps are equal. So
H
k
(K) = H
k
(L) = 0.
We now need to compute
H
n1
(K) = ker d
K
n1
= ker d
L
n1
= im d
L
n
.
We get the last equality since
ker d
L
n1
im d
L
n
= H
n1
(L) = 0.
We also know that
C
n
(
L
) is generated by just one simplex (
e
0
, · · · , e
n
). So
C
n
(
L
)
=
Z
. Also
d
L
n
is injective since it does not kill the generator (
e
0
, · · · , e
n
).
So
H
n1
(K)
=
im d
L
n
=
Z.
This is very exciting, because at last, we have a suggestion for a non-trivial
invariant of
S
n1
for
n >
2. We say this is just a “suggestion”, since the simplicial
homology is defined for simplicial complexes, and we are not guaranteed that if
we put a different simplicial complex on
S
n1
, we will get the same homology
groups. So the major technical obstacle we still need to overcome is to see that
H
k
are invariants of the polyhedron
|K|
, not just
K
, and similarly for maps.
But this will take some time.
We will quickly say something we’ve alluded to already:
Lemma (Interpretation of
H
0
).
H
0
(
K
)
=
Z
d
, where
d
is the number of path
components of K.
Proof.
Let
K
be our simplicial complex and
v, w V
k
. We note that by definition,
v, w
represent the same homology class in
H
0
(
K
) if and only if there is some
c
such that
d
1
c
=
w v
. The requirement that
d
1
c
=
w v
is equivalent to saying
c
is a path from
v
to
w
. So [
v
] = [
w
] if and only if
v
and
w
are in the same path
component of K.
Most of the time, we only care about path-connected spaces, so
H
0
(
K
)
=
Z
.
7.4 Mayer-Vietoris sequence
The Mayer-Vietoris theorem is exactly like the Seifert-van Kampen theorem
for fundamental groups, which tells us what happens when we glue two spaces
together.
Suppose we have a space K = M N.
M N
We will learn how to compute the homology of the union
M N
in terms of
those of M, N and 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.
The situation is somewhat similar here. We will need to learn some algebra in
order to state the Mayer-Vietoris theorem. The objects we need are known as
exact sequences.
Definition (Exact sequence). A pair of homomorphisms of abelian groups
A B C
f g
is exact (at B) if
im f = ker g.
A collection of homomorphisms
· · · A
i
A
i+1
A
i+2
· · ·
f
i1
f
i
f
i+1
f
i+2
is exact at A
i
if
ker f
i
= im f
i1
.
We say it is exact if it is exact at every A
i
.
Recall that we have seen something similar before. When we defined the
chain complexes, we had
d
2
= 0, i.e.
im d ker d
. Here we are requiring exact
equivalence, which is something even better.
Algebraically, we can think of an exact sequence as chain complexes with
trivial homology groups. Alternatively, we see the homology groups as measuring
the failure of a sequence to be exact.
There is a particular type of exact sequences that is important.
Definition (Short exact sequence). A short exact sequence is an exact sequence
of the form
0 A B C 0
f g
What does this mean?
The kernel of
f
is equal to the image of the zero map, i.e.
{
0
}
. So
f
is
injective.
The image of
g
is the kernel of the zero map, which is everything. So
g
is
surjective.
im f = ker g.
Since we like chain complexes, we can produce short exact sequences of chain
complexes.
Definition (Short exact sequence of chain complexes). A short exact sequence
of chain complexes is a pair of chain maps i
·
and j
·
0 A
·
B
·
C
·
0
i
·
j
·
such that for each k,
0 A
k
B
k
C
k
0
i
k
j
k
is exact.
Note that by requiring the maps to be chain maps, we imply that
i
k
and
j
k
commute with the boundary maps of the chain complexes.
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). If we have a short exact sequence of complexes
0 A
·
B
·
C
·
0
i
·
j
·
then a miracle happens to their homology groups. In particular, there is a long
exact sequence (i.e. an exact sequence that is not short)
· · · H
n
(A) H
n
(B) H
n
(C)
H
n1
(A) H
n1
(B) H
n1
(C) · · ·
i
j
i
j
where
i
and
j
are induced by
i
·
and
j
·
, and
is a map we will define in the
proof.
Having an exact sequence is good, since if we know most of the terms in an
exact sequence, we can figure out the remaining ones. 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. Yet, we still need to know some of
them, and since they are defined in the proof, you need to remember the proof.
Note, however, if we replace
Z
in the definition of chain groups by a field
(e.g.
Q
), then all the groups become vector spaces. Then everything boils down
to the rank-nullity theorem. 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. If, at any point, homology
groups confuse you, then you can try to work with homology groups over
Q
and
get a feel for what homology groups are like, since this is easier.
We are first going to apply this result to obtain the Mayer-Vietoris theorem.
Theorem (Mayer-Vietoris theorem). Let
K, L, M, N
be simplicial complexes
with K = M N and L = M N. We have the following inclusion maps:
L M
N K.
i
j
k
`
Then there exists some natural homomorphism
:
H
n
(
K
)
H
n1
(
L
) that
gives the following long exact sequence:
· · · H
n
(L) H
n
(M) H
n
(N) H
n
(K)
H
n1
(L) H
n1
(M) H
n1
(N) H
n1
(K) · · ·
· · · H
0
(M) H
0
(N) H
0
(K) 0
i
+j
k
`
i
+j
k
`
k
`
Here
A B
is the direct sum of the two (abelian) groups, which may also be
known as the Cartesian product.
Note that unlike the Seifert-van Kampen theorem, this does not require the
intersection
L
=
M N
to be (path) connected. This restriction was needed
for Seifert-van Kampen since the fundamental group is unable to see things
outside the path component of the basepoint, and hence it does not like non-path
connected spaces well. However, homology groups don’t have these problems.
Proof.
All we have to do is to produce a short exact sequence of complexes. We
have
0 C
n
(L) C
n
(M) C
n
(N) C
n
(K) 0
i
n
+j
n
k
n
`
n
Here
i
n
+
j
n
:
C
n
(
L
)
C
n
(
M
)
C
n
(
N
) is the map
x 7→
(
x, x
), while
k
n
n
:
C
n
(
M
)
C
n
(
N
)
C
n
(
K
) is the map (
a, b
)
7→ a b
(after applying the
appropriate inclusion maps).
It is easy to see that this is a short exact sequence of chain complexes. The
image of
i
n
+
j
n
is the set of all elements of the form (
x, x
), and the kernel of
k
n
n
is also these. It is also easy to see that
i
n
+
j
n
is injective and
k
n
n
is
surjective.
At first sight, the Mayer-Vietoris theorem might look a bit scary to use, since
it involves all homology groups of all orders at once. However, this is actually
often a good thing, since we can often use this to deduce the higher homology
groups from the lower homology groups.
Yet, to properly apply the Mayer-Vietoris sequence, we need to understand
the map
. To do so, we need to prove the snake lemma.
Theorem (Snake lemma). If we have a short exact sequence of complexes
0 A
·
B
·
C
·
0
i
·
j
·
then there is a long exact sequence
· · · H
n
(A) H
n
(B) H
n
(C)
H
n1
(A) H
n1
(B) H
n1
(C) · · ·
i
j
i
j
where
i
and
j
are induced by
i
·
and
j
·
, and
is a map we will define in the
proof.
The method of proving this is sometimes known as “diagram chasing”, where
we just “chase” around commutative diagrams to find the elements we need.
The idea of the proof is as follows in the short exact sequence, we can think
of
A
as a subgroup of
B
, and
C
as the quotient
B/A
, by the first isomorphism
theorem. So any element of
C
can be represented by an element of
B
. We
apply the boundary map to this representative, and then exactness shows that
this must come from some element of
A
. We then check carefully that this is
well-defined, i.e. does not depend on the representatives chosen.
Proof.
The proof of this is in general not hard. It just involves a lot of checking
of the details, such as making sure the homomorphisms are well-defined, are
actually homomorphisms, are exact at all the places etc. The only important
and non-trivial part is just the construction of the map
.
First we look at the following commutative diagram:
0 A
n
B
n
C
n
0
0 A
n1
B
n1
C
n1
0
i
n
d
n
j
n
d
n
d
n
i
n1
j
n1
To construct
:
H
n
(
C
)
H
n1
(
A
), let [
x
]
H
n
(
C
) be a class represented
by
x Z
n
(
C
). We need to find a cycle
z A
n1
. By exactness, we know the
map
j
n
:
B
n
C
n
is surjective. So there is a
y B
n
such that
j
n
(
y
) =
x
.
Since our target is
A
n1
, we want to move down to the next level. So consider
d
n
(
y
)
B
n1
. We would be done if
d
n
(
y
) is in the image of
i
n1
. By exactness,
this is equivalent to saying
d
n
(
y
) is in the kernel of
j
n1
. Since the diagram is
commutative, we know
j
n1
d
n
(y) = d
n
j
n
(y) = d
n
(x) = 0,
using the fact that
x
is a cycle. So
d
n
(
y
)
ker j
n1
=
im i
n1
. Moreover, by
exactness again,
i
n1
is injective. So there is a unique
z A
n1
such that
i
n1
(z) = d
n
(y). We have now produced our z.
We are not done. We have
[
x
] = [
z
] as our candidate definition, but we
need to check many things:
(i) We need to make sure
is indeed a homomorphism.
(ii) We need d
n1
(z) = 0 so that [z] H
n1
(A);
(iii)
We need to check [
z
] is well-defined, i.e. it does not depend on our choice
of y and x for the homology class [x].
(iv) We need to check the exactness of the resulting sequence.
We now check them one by one:
(i)
Since everything involved in defining
are homomorphisms, it follows
that
is also a homomorphism.
(ii) We check d
n1
(z) = 0. To do so, we need to add an additional layer.
0 A
n
B
n
C
n
0
0 A
n1
B
n1
C
n1
0
0 A
n2
B
n2
C
n2
0
i
n
d
n
j
n
d
n
d
n
i
n1
d
n1
j
n1
d
n1
d
n1
i
n2
j
n2
We want to check that
d
n1
(
z
) = 0. We will use the commutativity of the
diagram. In particular, we know
i
n2
d
n1
(z) = d
n1
i
n1
(z) = d
n1
d
n
(y) = 0.
By exactness at
A
n2
, we know
i
n2
is injective. So we must have
d
n1
(z) = 0.
(iii) (a)
First, in the proof, suppose we picked a different
y
0
such that
j
n
(
y
0
) =
j
n
(
y
) =
x
. Then
j
n
(
y
0
y
) = 0. So
y
0
y ker j
n
=
im i
n
. Let
a A
n
be such that i
n
(a) = y
0
y. Then
d
n
(y
0
) = d
n
(y
0
y) + d
n
(y)
= d
n
i
n
(a) + d
n
(y)
= i
n1
d
n
(a) + d
n
(y).
Hence when we pull back
d
n
(
y
0
) and
d
n
(
y
) to
A
n1
, the results differ
by the boundary
d
n
(
a
), and hence produce the same homology class.
(b)
Suppose [
x
0
] = [
x
]. We want to show that
[
x
] =
[
x
0
]. This time,
we add a layer above.
0 A
n+1
B
n+1
C
n+1
0
0 A
n
B
n
C
n
0
0 A
n1
B
n1
C
n1
0
i
n+1
d
n+1
j
n+1
d
n+1
d
n+1
i
n
d
n
j
n
d
n
d
n
i
n1
j
n1
By definition, since [x
0
] = [x], there is some c C
n+1
such that
x
0
= x + d
n+1
(c).
By surjectivity of
j
n+1
, we can write
c
=
j
n+1
(
b
) for some
b B
n+1
.
By commutativity of the squares, we know
x
0
= x + j
n
d
n+1
(b).
The next step of the proof is to find some
y
such that
j
n
(
y
) =
x
.
Then
j
n
(y + d
n+1
(b)) = x
0
.
So the corresponding
y
0
is
y
0
=
y
+
d
n+1
(
b
). So
d
n
(
y
) =
d
n
(
y
0
), and
hence
[x] =
[x
0
].
(iv)
This is yet another standard diagram chasing argument. When reading
this, it is helpful to look at a diagram and see how the elements are chased
along. It is even more beneficial to attempt to prove this yourself.
(a) im i
ker j
: This follows from the assumption that i
n
j
n
= 0.
(b) ker j
im i
: Let [
b
]
H
n
(
B
). Suppose
j
([
b
]) = 0. Then there is
some
c C
n+1
such that
j
n
(
b
) =
d
n+1
(
c
). By surjectivity of
j
n+1
,
there is some
b
0
B
n+1
such that
j
n+1
(
b
0
) =
c
. By commutativity,
we know j
n
(b) = j
n
d
n+1
(b
0
), i.e.
j
n
(b d
n+1
(b
0
)) = 0.
By exactness of the sequence, we know there is some
a A
n
such
that
i
n
(a) = b d
n+1
(b
0
).
Moreover,
i
n1
d
n
(a) = d
n
i
n
(a) = d
n
(b d
n+1
(b
0
)) = 0,
using the fact that
b
is a cycle. Since
i
n1
is injective, it follows that
d
n
(a) = 0. So [a] H
n
(A). Then
i
([a]) = [b] [d
n+1
(b
0
)] = [b].
So [b] im i
.
(c) im j
ker
: Let [
b
]
H
n
(
B
). To compute
(
j
([
b
])), we first
pull back
j
n
(
b
) to
b B
n
. Then we compute
d
n
(
b
) and then pull it
back to
A
n+1
. However, we know
d
n
(
b
) = 0 since
b
is a cycle. So
(j
([b])) = 0, i.e.
j
= 0.
(d) ker
im j
: Let [
c
]
H
n
(
C
) and suppose
([
c
]) = 0. Let
b B
n
be such that
j
n
(
b
) =
c
, and
a A
n1
such that
i
n1
(
a
) =
d
n
(
b
).
By assumption,
([
c
]) = [
a
] = 0. So we know
a
is a boundary,
say
a
=
d
n
(
a
0
) for some
a
0
A
n
. Then by commutativity we know
d
n
(b) = d
n
i
n
(a
0
). In other words,
d
n
(b i
n
(a
0
)) = 0.
So [b i
n
(a
0
)] H
n
(B). Moreover,
j
([b i
n
(a
0
)]) = [j
n
(b) j
n
i
n
(a
0
)] = [c].
So [c] im j
.
(e) im
ker i
: Let [
c
]
H
n
(
C
). Let
b B
n
be such that
j
n
(
b
) =
c
,
and a A
n1
be such that i
n
(a) = d
n
(b). Then
([c]) = [a]. Then
i
([a]) = [i
n
(a)] = [d
n
(b)] = 0.
So i
= 0.
(f) ker i
im
: Let [
a
]
H
n
(
A
) and suppose
i
([
a
]) = 0. So we can
find some
b B
n+1
such that
i
n
(
a
) =
d
n+1
(
b
). Let
c
=
j
n+1
(
b
). Then
d
n+1
(c) = d
n+1
j
n+1
(b) = j
n
d
n+1
(b) = j
n
i
n
(a) = 0.
So [
c
]
H
n
(
C
). Then [
a
] =
([
c
]) by definition of
. So [
a
]
im
.
7.5 Continuous maps and homotopy invariance
This is the most technical part of the homology section of the course. The goal is
to see that the homology groups
H
(
K
) depend only on the polyhedron, and not
the simplicial structure on it. Moreover, we will show that they are homotopy
invariants of the space, and that homotopic maps
f ' g
:
|K| |L|
induce
equal maps
H
(
K
)
H
(
L
). Note that this is a lot to prove. At this point, we
don’t even know arbitrary continuous maps induce any map on the homology.
We only know simplicial maps do.
We start with a funny definition.
Definition (Contiguous maps). Simplicial maps
f, g
:
K L
are contiguous if
for each
σ K
, the simplices
f
(
σ
) and
g
(
σ
) (i.e. the simplices spanned by the
image of the vertices of σ) are faces of some some simplex τ L.
σ
τ
g
f
The significance of this definition comes in two parts: simplicial approxima-
tions of the same map are contiguous, and contiguous maps induce the same
maps on homology.
Lemma. If
f, g
:
K L
are simplicial approximations to the same map
F
,
then f and g are contiguous.
Proof.
Let
σ K
, and pick some
s ˚σ
. Then
F
(
s
)
˚τ
for some
τ L
.
Then the definition of simplicial approximation implies that for any simplicial
approximation f to F , f(σ) spans a face of τ.
Lemma. If f, g : K L are continguous simplicial maps, then
f
= g
: H
n
(K) H
n
(L)
for all n.
Geometrically, the homotopy is obvious. If
f
(
σ
) and
g
(
σ
) are both faces of
τ
,
then we just pick the homotopy as
τ
. The algebraic definition is less inspiring,
and it takes some staring to see it is indeed what we think it should be.
Proof. We will write down a chain homotopy between f and g:
h
n
((a
0
, · · · , a
n
)) =
n
X
i=0
(1)
i
[f(a
0
), · · · , f (a
i
), g(a
i
), · · · , g(a
n
)],
where the square brackets means the corresponding oriented simplex if there are
no repeats, 0 otherwise.
We can now check by direct computation that this is indeed a chain homotopy.
Now we know that if
f, g
:
K L
are both simplicial approximations to the
same
F
, then they induce equal maps on homology. However, we do not yet know
that continuous maps induce well-defined maps on homologies, since to produce
simplicial approximations, we needed to perform barycentric subdivision, and
we need to make sure this does not change the homology.
K
K
0
Given a barycentric subdivision, we need to choose a simplicial approximation
to the identity map
a
:
K
0
K
. It turns out this is easy to do, and we can do
it almost arbitrarily.
Lemma. Each vertex
ˆσ K
0
is a barycenter of some
σ K
. Then we choose
a
(
ˆσ
) to be an arbitrary vertex of
σ
. This defines a function
a
:
V
K
0
V
K
.
This a is a simplicial approximation to the identity. Moreover, every simplicial
approximation to the identity is of this form.
Proof. Omitted.
Next, we need to show this gives an isomorphism on the homologies.
Proposition. Let
K
0
be the barycentric subdivision of
K
, and
a
:
K
0
K
a simplicial approximation to the identity map. Then the induced map
a
:
H
n
(K
0
) H
n
(K) is an isomorphism for all n.
Proof.
We first deal with
K
being a simplex
σ
and its faces. Now that
K
is just
a cone (with any vertex as the cone vertex), and
K
0
is also a cone (with the
barycenter as the cone vertex). Therefore
H
n
(K)
=
H
n
(K
0
)
=
(
Z n = 0
0 n > 0
So only
n
= 0 is (a little) interesting, but it is easy to check that
a
is an
isomorphism in this case, since it just maps a vertex to a vertex, and all vertices
in each simplex are in the same homology class.
To finish the proof, note that
K
is built up by gluing up simplices, and
K
0
is built by gluing up subdivided simplices. So to understand their homology
groups, we use the Mayer-Vietoris sequence.
Given a complicated simplicial complex
K
, let
σ
be a maximal dimensional
simplex of
K
. We let
L
=
K \ {σ}
(note that
L
includes the boundary of
σ
).
We let S = {σ and all its faces} K and T = L S.
We can do similarly for
K
0
, and let
L
0
, S
0
, T
0
be the corresponding barycentric
subdivisions. We have
K
=
L S
and
K
0
=
L
0
S
0
(and
L
0
S
0
=
T
0
). By the
previous lemma, we see our construction of
a
gives
a
(
L
0
)
L
,
a
(
S
0
)
S
and
a(T
0
) T . So these induce maps of the corresponding homology groups
H
n
(T
0
) H
n
(S
0
) H
n
(L
0
) H
n
(K
0
) H
n1
(T
0
) H
n1
(S
0
) H
n1
(L
0
)
H
n
(T ) H
n
(S) H
n
(L) H
n
(K) H
n1
(T ) H
n1
(S) H
n1
(L)
a
a
a
a
a
a
a
By induction, we can assume all but the middle maps are isomorphisms. By
the five lemma, this implies the middle map is an isomorphism, where the five
lemma is as follows:
Lemma (Five lemma). Consider the following commutative diagram:
A
1
B
1
C
1
D
1
E
1
A
2
B
2
C
2
D
2
E
2
a
b
c
d
e
If the top and bottom rows are exact, and
a, b, d, e
are isomorphisms, then
c
is
also an isomorphism.
Proof. Exercise in example sheet.
We now have everything we need to move from simplical maps to continuous
maps. Putting everything we have proved so far together, we obtain the following
result:
Proposition. To each continuous map
f
:
|K| |L|
, there is an associated
map f
: H
n
(K) H
n
(L) (for all n) given by
f
= s
ν
1
K,r
,
where
s
:
K
(r)
L
is a simplicial approximation to
f
, and
ν
K,r
:
H
n
(
K
(r)
)
H
n
(
K
) is the isomorphism given by composing maps
H
n
(
K
(i)
)
H
n
(
K
(i1)
)
induced by simplical approximations to the identity.
Furthermore:
(i) f
does not depend on the choice of r or s.
(ii) If g : |M | |K| is another continuous map, then
(f g)
= f
g
.
Proof. Omitted.
Corollary. If
f
:
|K| |L|
is a homeomorphism, then
f
:
H
n
(
K
)
H
n
(
L
) is
an isomorphism for all n.
Proof. Immediate from (ii) of previous proposition.
This is good. We know now that homology groups is a property of the space
itself, not simplicial complexes. For example, we have computed the homology
groups of a particular triangulation of the
n
-sphere, and we know it is true for
all triangulations of the n-sphere.
We’re not exactly there yet. We have just seen that homology groups are
invariant under homeomorphisms. What we would like to know is that they are
invariant under homotopy. In other words, we want to know homotopy equivalent
maps induce equal maps on the homology groups.
The strategy is:
(i) Show that “small” homotopies don’t change the maps on H
n
(ii) Note that all homotopies can be decomposed into “small” homotopies.
Lemma. Let
L
be a simplicial complex (with
|L| R
n
). Then there is an
ε
=
ε
(
L
)
>
0 such that if
f, g
:
|K| |L|
satisfy
kf
(
x
)
g
(
x
)
k < ε
, then
f
= g
: H
n
(K) H
n
(L) for all n.
The idea of the proof is that if
kf
(
x
)
g
(
x
)
k
is small enough, we can
barycentrically subdivide
K
such that we get a simplicial approximation to both
f and g.
Proof.
By the Lebesgue number lemma, there is an
ε >
0 such that each ball of
radius 2ε in |L| lies in some star St
L
(w).
Now apply the Lebesgue number lemma again to
{f
1
(
B
ε
(
y
))
}
y∈|L|
, an open
cover of |K|, and get δ > 0 such that for all x |K|, we have
f(B
δ
(x)) B
ε
(y) B
2ε
(y) St
L
(w)
for some y |L| and St
L
(w). Now since g and f differ by at most ε, we know
g(B
δ
(x)) B
2ε
(y) St
L
(w).
Now subdivide r times so that µ(K
(r)
) <
1
2
δ. So for all v V
K
(r)
, we know
St
K
(r)
(v) B
δ
(v).
This gets mapped by both
f
and
g
to
St
L
(
w
) for the same
w V
L
. We define
s : V
K
(r)
V
L
sending v 7→ w.
Theorem. Let f ' g : |K| |L|. Then f
= g
.
Proof.
Let
H
:
|K| × I |L|
. Since
|K| × I
is compact, we know
H
is uniformly
continuous. Pick
ε
=
ε
(
L
) as in the previous lemma. Then there is some
δ
such
that |s t| < δ implies |H(x, s) H(x, t)| < ε for all x |K|.
Now choose 0 =
t
0
< t
1
< · · · < t
n
= 1 such that
t
i
t
i1
< δ
for all
i
.
Define
f
i
:
|K| |L|
by
f
i
(
x
) =
H
(
x, t
i
). Then we know
kf
i
f
i1
k < ε
for all
i. Hence (f
i
)
= (f
i1
)
. Therefore (f
0
)
= (f
n
)
, i.e. f
= g
.
This is good, since we know we can not only deal with spaces that are
homeomorphic to complexes, but spaces that are homotopic to complexes. This
is important, since all complexes are compact, but we would like to talk about
non-compact spaces such as open balls as well.
Definition (
h
-triangulation and homology groups). An
h
-triangulation of a
space
X
is a simplicial complex
K
and a homotopy equivalence
h
:
|K| X
.
We define H
n
(X) = H
n
(K) for all n.
Lemma. H
n
(X) is well-defined, i.e. it does not depend on the choice of K.
Proof. Clear from previous theorem.
We have spent a lot of time and effort developing all the technical results
and machinery of homology groups. We will now use them to do stuff.
Digression historical motivation
At first, we motivated the study of algebraic topology by saying we wanted to
show that
R
n
and
R
m
are not homeomorphic. However, historically, this is not
why people studied algebraic topology, since there are other ways to prove this
is true.
Initially, people were interested in studying knots. These can be thought of
as embeddings
S S
3
. We really should just think of
S
3
as
R
3
{∞}
, where
the point at infinity is just there for convenience.
The most interesting knot we know is the unknot U :
A less interesting but still interesting knot is the trefoil T .
One question people at the time asked was whether the trefoil knot is just the
unknot in disguise. It obviously isn’t, but can we prove it? In general, given two
knots, is there any way we can distinguish if they are the same?
The idea is to study the fundamental groups of the knots. It would obviously
be silly to compute the fundamental groups of
U
and
T
themselves, since they
are both isomorphic to
S
1
and the groups are just
Z
. Instead, we look at the
fundamental groups of the complements. It is not difficult to see that
π
1
(S
3
\ U)
=
Z,
and with some suitable machinery, we find
π
1
(S
3
\ T )
=
ha, b | a
3
b
2
i.
Staring at it hard enough, we can construct a surjection
π
1
(
S
3
\ T
)
S
3
. This
tells us
π
1
(
S
3
\ T
) is non-abelian, and is certainly not
Z
. So we know
U
and
T
are genuinely different knots.
7.6 Homology of spheres and applications
Lemma. The sphere S
n1
is triangulable, and
H
k
(S
n1
)
=
(
Z k = 0, n 1
0 otherwise
Proof.
We already did this computation for the standard (
n
1)-sphere
n
,
where
n
is the standard n-simplex.
We immediately have a few applications
Proposition. R
n
6
=
R
m
for m 6= n.
Proof. See example sheet 4.
Theorem (Brouwer’s fixed point theorem (in all dimensions)). There is no
retraction
D
n
onto
D
n
=
S
n1
. So every continuous map
f
:
D
n
D
n
has a
fixed point.
Proof.
The proof is exactly the same as the two-dimensional case, with homology
groups instead of the fundamental group.
We first show the second part from the first. Suppose
f
:
D
n
D
n
has no
fixed point. Then the following g : D
n
D
n
is a continuous retraction.
x
f(x)
g(x)
So we now show no such continuous retraction can exist. Suppose
r
:
D
n
D
n
is a retraction, i.e. r i ' id : D
n
D
n
.
S
n1
D
n
S
n1
i r
We now take the (n 1)th homology groups to obtain
H
n1
(S
n1
) H
n1
(D
n
) H
n1
(S
n1
).
i
r
Since
r i
is homotopic to the identity, this composition must also be the identity,
but this is clearly nonsense, since
H
n1
(
S
n1
)
=
Z
while
H
n1
(
D
n
)
=
0. So
such a continuous retraction cannot exist.
Note it is important that we can work with continuous maps directly, and
not just their simplicial approximations. Otherwise, here we can only show
that every simplicial approximation of
f
has a fixed point, but this does not
automatically entail that f itself has a fixed point.
For the next application, recall from the first example sheet that if
n
is odd,
then the antipodal map
a
:
S
n
S
n
is homotopic to the identity. What if
n
is
even?
The idea is to consider the effect on the homology group:
a
: H
n
(S
n
) H
n
(S
n
).
By our previous calculation, we know
a
is a map
a
:
Z Z
. If
a
is homotopic
to the identity, then
a
should be homotopic to the identity map. We will now
compute a
and show that it is multiplication by 1 when n is even.
To do this, we want to use a triangulation that is compatible with the
antipodal map. The standard triangulation clearly doesn’t work. Instead, we
use the following triangulation h : |K| S
n
:
The vertices of K are given by
V
K
= e
0
, ±e
1
, · · · , ±e
n
}.
This triangulation works nicely with the antipodal map, since this maps a vertex
to a vertex. To understand the homology group better, we need the following
lemma:
Lemma. In the triangulation of
S
n
given by vertices
V
K
=
e
0
, ±
e
1
, · · · , ±
e
n
}
,
the element
x =
X
ε∈{±1}
n+1
ε
0
· · · ε
n
(ε
0
e
0
, · · · , ε
n
e
n
)
is a cycle and generates H
n
(S
n
).
Proof.
By direct computation, we see that
dx
= 0. So
x
is a cycle. To show it
generates
H
n
(
S
n
), we note that everything in
H
n
(
S
n
)
=
Z
is a multiple of the
generator, and since
x
has coefficients
±
1, it cannot be a multiple of anything
else (apart from x). So x is indeed a generator.
Now we can prove our original proposition.
Proposition. If n is even, then the antipodal map a 6' id.
Proof.
We can directly compute that
a
x
= (
1)
n+1
x
. If
n
is even, then
a
=
1,
but id
= 1. So a 6' id.
7.7 Homology of surfaces
We want to study compact surfaces and their homology groups. To work with
the simplicial homology, we need to assume they are triangulable. We will not
prove this fact, and just assume it to be true (it really is).
Recall we have classified compact surfaces, and have found the following
orientable surfaces Σ
g
.
We also have non-compact versions of these, known as
F
g
, where we take the
above ones and cut out a hole:
We want to compute the homology groups of these Σ
g
. One way to do so is
to come up with specific triangulations of the space, and then compute the
homology groups directly. However, there is a better way to do it. Given a Σ
g
,
we can slice it apart along a circle and write it as
Σ
g
= F
g1
F
1
.
Then, we need to compute
H
(
F
g
). Fortunately, this is not too hard, since it
turns out
F
g
is homotopic to a relatively simple space. Recall that we produced
Σ
g
by starting with a 4g-gon and gluing edges:
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
Now what is
F
g
?
F
g
is Σ
g
with a hole cut out of it. We’ll cut the hole out at
the center, and be left with
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
We can now expand the hole to the boundary, and get a deformation retraction
from F
g
to its boundary:
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
Gluing the edges together, we obtain the rose with 2
g
petals, which we shall call
X
2g
:
Using the Mayer-Vietoris sequence (exercise), we see that
H
n
(F
g
)
=
H
n
(X
2g
)
=
Z n = 0
Z
2g
n = 1
0 n > 1
We now compute the homology groups of Σ
g
. The Mayer-Vietoris sequence gives
the following
0 H
2
(S
1
) H
2
(F
g1
) H
2
(F
1
) H
2
g
)
H
1
(S
1
) H
1
(F
g1
) H
1
(F
1
) H
1
g
)
H
0
(S
1
) H
0
(F
g1
) H
0
(F
1
) H
0
g
) 0
We can put in the terms we already know, and get
0 H
2
g
) Z Z
2g
H
1
g
) Z Z
2
Z 0
By exactness, we know
H
2
g
) =
ker{Z Z
2g
}
. We now note that this map
is indeed the zero map, by direct computation this map sends
S
1
to the
boundary of the hole of
F
g1
and
F
1
. If we look at the picture, after the
deformation retraction, this loop passes through each one-cell twice, once with
each orientation. So these cancel, and give 0. So H
2
g
) = Z.
To compute
H
1
g
), for convenience, we break it up into a short exact
sequence, noting that the function Z Z
2g
is zero:
0 Z
2g
H
1
g
) ker(Z Z
2
) 0.
We now claim that
Z Z
2
is injective this is obvious, since it sends 1
7→
(1
,
1).
So the kernel is zero. So
H
1
g
)
=
Z
2g
.
This is a typical application of the Mayer-Vietoris sequence. We write down
the long exact sequence, and put in the terms we already know. This does not
immediately give us the answer we want we will need to understand one or
two of the maps, and then we can figure out all the groups we want.
7.8 Rational homology, Euler and Lefschetz numbers
So far, we have been working with chains with integral coefficients. It turns out
we can use rational coefficients instead. In the past,
C
n
(
K
) was an abelian group,
or a
Z
-module. If we use rational coefficients, since
Q
is a field, this becomes a
vector space, and we can use a lot of nice theorems about vector spaces, such as
the rank-nullity theorem. Moreover, we can reasonably talk about things like
the dimensions of these homology groups.
Definition (Rational homology group). For a simplicial complex
K
, we can de-
fine the rational
n
-chain group
C
n
(
K, Q
) in the same way as
C
n
(
K
) =
C
n
(
K, Z
).
That is,
C
n
(
K, Q
) is the vector space over
Q
with basis the
n
-simplices of
K
(with a choice of orientation).
We can define d
n
, Z
n
, B
n
as before, and the rational nth homology group is
H
n
(K; Q)
=
Z
n
(K; Q)
B
n
(K; Q)
.
Now our homology group is a vector space, and it is much easier to work with.
However, the consequence is that we will lose some information. Fortunately,
the way in which we lose information is very well-understood and rather simple.
Lemma. If H
n
(K)
=
Z
k
F for F a finite group, then H
n
(K; Q)
=
Q
k
.
Proof. Exercise.
Hence, when passing to rational homology groups, we lose all information
about the torsion part. In some cases this information loss is not very significant,
but in certain cases, it can be. For example, in RP
2
, we have
H
n
(RP
2
)
=
Z n = 0
Z/2 n = 1
0 n > 1
If we pass on to rational coefficients, then we have lost everything in
H
1
(
RP
2
),
and RP
2
looks just like a point:
H
n
(RP
2
; Q)
=
H
n
(; Q).
Example. We have
H
n
(S
n
; Q)
=
(
Q k = 0, n
0 otherwise
.
We also have
H
n
g
, Q)
=
Q k = 0, 2
Q
2g
k = 1
0 otherwise
.
In this case, we have not lost any information because there was no torsion part
of the homology groups.
However, for the non-orientable surfaces, one can show that
H
k
(E
n
) =
Z k = 0
Z
n1
× Z/2 k = 1
0 otherwise
,
so
H
k
(E
n
; Q) =
Q k = 0
Q
n1
k = 1
0 otherwise
.
This time, this is different from the integral coefficient case, where we have
an extra Z
2
term in H
1
.
The advantage of this is that the homology “groups” are in fact vector spaces,
and we can talk about things like dimensions. Also, maps on homology groups
are simply linear maps, i.e. matrices, and we can study their properties with
proper linear algebra.
Recall that the Euler characteristic of a polyhedron is defined as “faces
edges + vertices”. This works for two-dimensional surfaces, and we would like to
extend this to higher dimensions. There is an obvious way to do so, by counting
higher-dimensional surfaces and putting the right signs. However, if we define it
this way, it is not clear that this is a property of the space itself, and not just a
property of the triangulation. Hence, we define it as follows:
Definition (Euler characteristic). The Euler characteristic of a triangulable
space X is
χ(X) =
X
i0
(1)
i
dim
Q
H
i
(X; Q).
This clearly depends only on the homotopy type of
X
, and not the triangula-
tion. We will later show this is equivalent to what we used to have.
More generally, we can define the Lefschetz number.
Definition (Lefschetz number). Given any map
f
:
X X
, we define the
Lefschetz number of f as
L(f) =
X
i0
(1)
i
tr(f
: H
i
(X; Q) H
i
(X; Q)).
Why is this a generalization of the Euler characteristic? Just note that the
trace of the identity map is the number of dimensions. So we have
χ(X) = L(id).
Example. We have
χ(S
n
) =
(
2 n even
0 n odd
We also have
χ
g
) = 2 2g, χ(E
n
) = 2 n.
Example. If
α
:
S
n
S
n
is the antipodal map, we saw that
α
:
H
n
(
S
n
)
H
n
(S
n
) is multiplication by (1)
n+1
. So
L(α) = 1 + (1)
n
(1)
n+1
= 1 1 = 0.
We see that even though the antipodal map has different behaviour for
different dimensions, the Lefschetz number ends up being zero all the time. We
will soon see why this is the case.
Why do we want to care about the Lefschetz number? The important thing
is that we will use this to prove a really powerful generalization of Brouwer’s
fixed point theorem that allows us to talk about things that are not balls.
Before that, we want to understand the Lefschetz number first. To define the
Lefschetz number, we take the trace of
f
, and this is a map of the homology
groups. However, we would like to understand the Lefschetz number in terms of
the chain groups, since these are easier to comprehend. Recall that the homology
groups are defined as quotients of the chain groups, so we would like to know
what happens to the trace when we take quotients.
Lemma. Let
V
be a finite-dimensional vector space and
W V
a subspace.
Let
A
:
V V
be a linear map such that
A
(
W
)
W
. Let
B
=
A|
W
:
W W
and C : V/W V/W the induced map on the quotient. Then
tr(A) = tr(B) + tr(C).
Proof. In the right basis,
A =
B A
0
0 C
.
What this allows us to do is to not look at the induced maps on homology,
but just the maps on chain complexes. This makes our life much easier when it
comes to computation.
Corollary. Let f
·
: C
·
(K; Q) C
·
(K; Q) be a chain map. Then
X
i0
(1)
i
tr(f
i
: C
i
(K) C
i
(K)) =
X
i0
(1)
i
tr(f
: H
i
(K) H
i
(K)),
with homology groups understood to be over Q.
This is a great corollary. The thing on the right is the conceptually right
thing to have homology groups are nice and are properties of the space itself,
not the triangulation. However, to actually do computations, we want to work
with the chain groups and actually calculate with chain groups.
Proof. There is an exact sequence
0 B
i
(K; Q) Z
i
(K; Q) H
i
(K; Q) 0
This is since
H
i
(
K, Q
) is defined as the quotient of
Z
i
over
B
i
. We also have
the exact sequence
0 Z
i
(K; Q) C
i
(K; Q) B
i1
(K; Q) 0
d
i
This is true by definition of
B
i1
and
Z
i
. Let
f
H
i
, f
B
i
, f
Z
i
, f
C
i
be the various
maps induced by f on the corresponding groups. Then we have
L(|f|) =
X
i0
(1)
i
tr(f
H
i
)
=
X
i0
(1)
i
(tr(f
Z
i
) tr(f
B
i
))
=
X
i0
(1)
i
(tr(f
C
i
) tr(f
B
i1
) tr(f
B
i
)).
Because of the alternating signs in dimension, each
f
B
i
appears twice in the sum
with opposite signs. So all f
B
i
cancel out, and we are left with
L(|f|) =
X
i0
(1)
i
tr(f
C
i
).
Since
tr
(
id |
C
i
(K;Q)
) =
dim
Q
C
i
(
K
;
Q
), which is just the number of
i
-simplices,
this tells us the Euler characteristic we just defined is the usual Euler character-
istic, i.e.
χ(X) =
X
i0
(1)
i
number of i-simplices.
Finally, we get to the important theorem of the section.
Theorem (Lefschetz fixed point theorem). Let
f
:
X X
be a continuous
map from a triangulable space to itself. If L(f) 6= 0, then f has a fixed point.
Proof.
We prove the contrapositive. Suppose
f
has no fixed point. We will show
that L(f ) = 0. Let
δ = inf{|x f(x)| : x X}
thinking of
X
as a subset of
R
n
. 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).
Choose a triangulation L : |K| X such that µ(K) <
δ
2
. We now let
g : K
(r)
K
be a simplicial approximation to
f
. Since we picked our triangulation to be so
fine, for x σ K, we have
|f(x) g(x)| <
δ
2
since the mesh is already less than
δ
2
. Also, we know
|f(x) x| δ.
So we have
|g(x) x| >
δ
2
.
So we must have
g
(
x
)
6∈ σ
. The conclusion is that for any
σ K
, we must have
g(σ) σ = .
Now we compute L(f) = L(|g|). The only complication here is that g is a map
from
K
(r)
to
K
, and the domains and codomains are different. So we need to
compose it with
s
i
:
C
i
(
K
;
Q
)
C
i
(
K
(r)
;
Q
) induced by inverses of simplicial
approximations to the identity map. Then we have
L(|g|) =
X
i0
(1)
i
tr(g
: H
i
(X; Q) H
i
(X; Q))
=
X
i0
(1)
i
tr(g
i
s
i
: C
i
(K; Q) C
i
(K; Q))
Now note that
s
i
takes simplices of
σ
to sums of subsimplices of
σ
. So
g
i
s
i
takes every simplex off itself. So each diagonal terms of the matrix of
g
i
s
i
is
0! Hence the trace is
L(|g|) = 0.
Example. If
X
is any contractible polyhedron (e.g. a disk), then
L
(
f
) = 1 for
any
f
:
X X
, which is obvious once we contract the space to a point. So
f
has a fixed point.
Example. Suppose
G
is a path-connected topological group, i.e.
G
is a group
and a topological space, and inverse and multiplication are continuous maps.
If g 6= 1, then the map
r
g
: G G
γ 7→ γg
has no fixed point. This implies
L
(
r
g
) = 0. However, since the space is path
connected,
r
g
' id
, where the homotopy is obtained by multiplying elements
along the path from e to r
g
. So
χ(G) = L(id
G
) = L(r
g
) = 0.
So if G 6= 1, then in fact χ(G) = 0.
This is quite fun. We have worked with many surfaces. Which can be
topological groups? The torus is, since it is just
S
1
× S
1
, and
S
1
is a topological
group. The Klein bottle? Maybe. However, other surfaces cannot since they
don’t have Euler characteristic 0.