7Simplicial homology
II Algebraic Topology
7.6 Homology of spheres and applications
Lemma. The sphere S
n−1
is triangulable, and
H
k
(S
n−1
)
∼
=
(
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
n−1
. 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
n−1
D
n
S
n−1
i r
We now take the (n − 1)th homology groups to obtain
H
n−1
(S
n−1
) H
n−1
(D
n
) H
n−1
(S
n−1
).
i
∗
r
∗
Since
r ◦i
is homotopic to the identity, this composition must also be the identity,
but this is clearly nonsense, since
H
n−1
(
S
n−1
)
∼
=
Z
while
H
n−1
(
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.