3Covering spaces
II Algebraic Topology
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.