3Four major tools of (co)homology

III Algebraic Topology



3.5 Applications
We now use all these tools to do lots of computations. The first thing we compute
will be the homology of spheres.
Theorem. We have
H
i
(S
1
) =
(
Z i = 0, 1
0 otherwise
.
Proof. We can split S
1
up as
A
B
qp
We want to apply Mayer-Vietoris. We have
A
=
B
=
R ' , A B
=
R
a
R ' {p}
a
{q}.
We obtain
0 0
· · · H
1
(A B) H
1
(A) H
1
(B) H
1
(S
1
)
H
0
(A B) H
0
(A) H
0
(B) H
0
(S
1
) 0
Z Z Z Z Z
i
A
i
B
Notice that the map into
H
1
(
S
1
) is zero. So the kernel of
is trivial, i.e.
is an
injection. So
H
1
(
S
1
) is isomorphic to the image of
, which is, by exactness, the
kernel of i
A
i
B
. So we want to know what this map does.
We know that
H
0
(
A B
)
=
Z Z
is generated by
p
and
q
, and the inclusion
map sends each of
p
and
q
to the unique connected components of
A
and
B
. So
the homology classes are both sent to (1
,
1)
H
0
(
A
)
H
0
(
B
)
=
Z Z
. We then
see that the kernel of
i
A
i
B
is generated by (
p q
), and is thus isomorphic
to Z. So H
1
(S
1
)
=
Z.
By looking higher up the exact sequence, we see that all other homology
groups vanish.
We can do the sphere in general in the same way.
Theorem. For any n 1, we have
H
i
(S
n
) =
(
Z i = 0, n
0 otherwise
.
Proof. We again cut up S
n
as
A = S
n
\ {N}
=
R
n
' ,
B = S
n
\ {S}
=
R
n
' ,
where N and S are the north and south poles. Moreover, we have
A B
=
R × S
n1
' S
n1
So we can “induct up” using the Mayer-Vietoris sequence:
· · · H
i
(S
n1
) H
i
() H
i
() H
i
(S
n
)
H
i1
(S
n1
) H
i1
() H
i1
() H
i1
(S
n
) · · ·
Now suppose
n
2, as we already did
S
1
already. If
i >
1, then
H
i
(
) = 0 =
H
i1
(). So the Mayer-Vietoris map
H
i
(S
n
) H
i1
(S
n1
)
is an isomorphism.
All that remains is to look at
i
= 0
,
1. The
i
= 0 case is trivial. For
i
= 1,
we look at
0
· · · H
1
() H
1
() H
1
(S
n
)
H
0
(S
n1
) H
0
() H
0
() H
0
(S
n
) 0
Z Z Z Z
f
To conclude that
H
1
(
S
n
) is trivial, it suffices to show that the map
f
is injective.
By picking the canonical generators, it is given by 1
7→
(1
,
1). So we are done.
Corollary.
If
n 6
=
m
, then
S
n1
6' S
m1
, since they have different homology
groups.
Corollary. If n 6= m, then R
n
6
=
R
m
.
Now suppose we have a map
f
:
S
n
S
n
. As always, it induces a map
f
:
H
n
(
S
n
)
H
n
(
S
n
). Since
H
n
(
S
n
)
=
Z
, we know the map
f
is given by
multiplication by some integer. We call this the degree of the map f.
Definition
(Degree of a map)
.
Let
f
:
S
n
S
n
be a map. The degree
deg
(
f
)
is the unique integer such that under the identification
H
n
(
S
n
)
=
Z
, the map
f
is given by multiplication by deg(f).
In particular, for n = 1, the degree is the winding number.
Note that there are two ways we can identify
H
n
(
S
n
) and
Z
, which differ by
a sign. For the degree to be well-defined, and not just defined up to a sign, we
must ensure that we identify both H
n
(S
n
) in the same way.
Using the degree, we can show that certain maps are not homotopic to each
other. We first note the following elementary properties:
Proposition.
(i) deg(id
S
n
) = 1.
(ii) If f is not surjective, then deg(f) = 0.
(iii) We have deg(f g) = (deg f)(deg g).
(iv) Homotopic maps have equal degrees.
Proof.
(i) Obvious.
(ii) If f is not surjective, then f can be factored as
S
n
S
n
\ {p} S
n
f
,
where
p
is some point not in the image of
f
. But
S
n
\ {p}
is contractible.
So f
factors as
f
: H
n
(S
n
) H
n
() = 0 H
n
(S
n
) .
So f
is the zero homomorphism, and is thus multiplication by 0.
(iii) This follows from the functoriality of H
n
.
(iv) Obvious as well.
As a corollary, we obtain the renowned Brouwer’s fixed point theorem, a
highly non-constructive fixed point existence theorem proved by a constructivist.
Corollary
(Brouwer’s fixed point theorem)
.
Any map
f
:
D
n
D
n
has a fixed
point.
Proof.
Suppose
f
has no fixed point. Define
r
:
D
n
S
n1
=
D
n
by taking
the intersection of the ray from f(x) through x with D
n
. This is continuous.
x
f(x)
r(x)
Now if x D
n
, then r(x) = x. So we have a map
S
n1
= D
n
D
n
D
n
= S
n1
i r
,
and the composition is the identity. This is a contradiction contracting
D
n
to a point, this gives a homotopy from the identity map
S
n1
S
n1
to the
constant map at a point. This is impossible, as the two maps have different
degrees.
A more manual argument to show this would be to apply
H
n1
to the chain
of maps above to obtain a contradiction.
We know there is a map of degree 1, namely the identity, and a map of degree
0, namely the constant map. How about other degrees?
Proposition.
A reflection
r
:
S
n
S
n
about a hyperplane has degree
1. As
before, we cover S
n
by
A = S
n
\ {N}
=
R
n
' ,
B = S
n
\ {S}
=
R
n
' ,
where we suppose the north and south poles lie in the hyperplane of reflection.
Then both A and B are invariant under the reflection. Consider the diagram
H
n
(S
n
) H
n1
(A B) H
n1
(S
n1
)
H
n
(S
n
) H
n1
(A B) H
n1
(S
n1
)
MV
r
r
r
MV
where the S
n1
on the right most column is given by contracting A B to the
equator. Note that
r
restricts to a reflection on the equator. By tracing through
the isomorphisms, we see that
deg
(
r
) =
deg
(
r|
equator
). So by induction, we only
have to consider the case when n = 1. Then we have maps
0 H
1
(S
1
) H
0
(A B) H
0
(A) H
0
(B)
0 H
1
(S
1
) H
0
(A B) H
0
(A) H
0
(B)
MV
r
r
r
r
MV
Now the middle vertical map sends
p 7→ q
and
q 7→ p
. Since
H
1
(
S
1
) is given by
the kernel of
H
0
(
A B
)
H
0
(
A
)
H
0
(
B
), and is generated by
p q
, we see
that this sends the generator to its negation. So this is given by multiplication
by 1. So the degree is 1.
Corollary. The antipodal map a : S
n
S
n
given by
a(x
1
, · · · , x
n+1
) = (x
1
, · · · , x
n+1
)
has degree (1)
n+1
because it is a composition of (n + 1) reflections.
Corollary
(Hairy ball theorem)
. S
n
has a nowhere 0 vector field iff
n
is odd.
More precisely, viewing
S
n
R
n+1
, a vector field on
S
n
is a map
v
:
S
n
R
n+1
such that hv(x), xi = 0, i.e. v(x) is perpendicular to x.
Proof. If n is odd, say n = 2k 1, then
v(x
1
, y
1
, x
2
, y
2
, · · · , x
k
, y
k
) = (y
1
, x
1
, y
2
, x
2
, · · · , y
k
, x
k
)
works.
Conversely, if
v
:
S
n
R
n+1
\ {
0
}
is a vector field, we let
w
=
v
|v|
:
S
n
S
n
. We can construct a homotopy from
w
to the antipodal map by “linear
interpolation”, but in a way so that it stays on the sphere. We let
H : [0, π] × S
n
S
n
(t, x) 7→ cos(t)x + sin(t)w(x)
Since
w
(
x
) and
x
are perpendicular, it follows that this always has norm 1, so
indeed it stays on the sphere.
Now we have
H(0, x) = x, H(π, x) = x.
So this gives a homotopy from
id
to
a
, which is a contradiction since they have
different degrees.
So far, we’ve only been talking about spheres all the time. We now move on
to something different.
Example. Let K be the Klein bottle. We cut them up by
A
B
Then both A and B are cylinders, and their intersection is two cylinders:
A B
s
t
A B
We have a long exact sequence
0 H
2
(K) H
1
(A B) H
1
(A) H
1
(B)
H
1
(K) H
0
(A B) H
0
(A) H
0
(B)
H
0
(K) 0
(i
A
,i
B
)
j
A
j
B
(i
A
,i
B
)
j
A
j
B
We plug in the numbers to get
0 H
2
(K) Z Z Z Z
H
1
(K) Z Z Z Z Z 0
(i
A
,i
B
)
j
A
j
B
(i
A
,i
B
)
j
A
j
B
Now let’s look at the first (
i
A
, i
B
) map. We suppose each
H
1
is generated by
the left-to-right loops. Then we have
i
A
(s) = 1
i
A
(t) = 1
i
B
(s) = 1
i
B
(t) = 1
So the matrix of (i
A
, i
B
) is
1 1
1 1
.
This has determinant 2, so it is injective and has trivial kernel. So we must have
H
2
(K) = 0. We also have
H
1
(A) H
1
(B)
im(i
A
, i
B
)
=
ha, bi
h−a + b, a + bi
=
hai
h2ai
=
Z/2Z.
Now the second (i
A
, i
B
) is given by
1 1
1 1
,
whose kernel is isomorphic to Z. So we have
0 Z/2Z H
1
(K) Z 0 .
So we have H
1
(K)
=
Z Z/2Z.
Finally, we look at some applications of relative homology.
Lemma.
Let
M
be a
d
-dimensional manifold (i.e. a Hausdorff, second-countable
space locally homeomorphic to R
d
). Then
H
n
(M, M \ {x})
=
(
Z n = d
0 otherwise
.
This is known as the local homology.
This gives us a homological way to define the dimension of a space.
Proof.
Let
U
be an open neighbourhood isomorphic to
R
d
that maps
x 7→
0.
We let Z = M \ U . Then
Z = Z M \ {x} =
˚
(M \ {x}).
So we can apply excision, and get
H
n
(U, U \ {x}) = H
n
(M \ Z, (M \ {x}) \ Z)
=
H
n
(M, M \ {x}).
So it suffices to do this in the case
M
=
R
d
and
x
= 0. The long exact sequence
for relative homology gives
H
n
(R
d
) H
n
(R
d
, R
d
\ {0}) H
n1
(R
d
\ {0}) H
n1
(R
d
) .
Since H
n
(R
d
) = H
n1
(R
d
) = 0 for n 2 large enough, it follows that
H
n
(R
d
, R
d
\ {0})
=
H
n1
(R
d
\ {0})
=
H
n1
(S
d1
),
and the result follows from our previous computation of the homology of
S
d1
.
We will have to check the lower-degree terms manually, but we shall not.
These relative homology groups are exactly the homology group of spheres.
So we might be able to define a degree-like object for maps between manifolds.
However, there is the subtlety that the identification with
Z
is not necessarily
canonical, and this involves the problem of orientation. For now, we will just
work with S
n
.
Definition
(Local degree)
.
Let
f
:
S
d
S
d
be a map, and
x S
d
. Then
f
induces a map
f
: H
d
(S
d
, S
d
\ {x}) H
d
(S
d
, S
d
\ {f(x)}).
We identify
H
d
(
S
d
, S
d
\{x}
)
=
H
d
(
S
d
)
=
Z
for any
x
via the inclusion
H
d
(
S
d
)
H
d
(
S
d
, S
d
\ {x}
), which is an isomorphism by the long exact sequence. Then
f
is given by multiplication by a constant
deg
(
f
)
x
, called the local degree of
f
at
x.
What makes the local degree useful is that we can in fact compute the degree
of a map via local degrees.
Theorem. Let f : S
d
S
d
be a map. Suppose there is a y S
d
such that
f
1
(y) = {x
1
, · · · , x
k
}
is finite. Then
deg(f) =
k
X
i=1
deg(f)
x
i
.
Proof.
Note that by excision, instead of computing the local degree at
x
i
via
H
d
(
S
d
, S
d
\ {x}
), we can pick a neighbourhood
U
i
of
x
i
and a neighbourhood
V
of x
i
such that f(U
i
) V , and then look at the map
f
: H
d
(U
i
, U
i
\ {x
i
}) H
d
(V, V \ {y})
instead. Moreover, since
S
d
is Hausdorff, we can pick the
U
i
such that they are
disjoint. Consider the huge commutative diagram:
H
d
(S
d
) H
d
(S
d
)
H
d
(S
d
, S
d
\ {x
1
, · · · , x
k
}) H
d
(S
d
, S
d
\ {y})
H
d
(
`
U
i
,
`
(U
i
\ x
i
))
L
k
i=1
H
d
(U
i
, U
i
\ x
i
) H
d
(V, V \ {y})
f
f
excision
L
f
Consider the generator 1 of
H
d
(
S
d
). By definition, its image in
H
d
(
S
d
) is
deg
(
f
).
Also, its image in
L
H
d
(
U
i
, U
i
\ {x
i
}
) is (1
, · · · ,
1). The bottom horizontal map
sends this to
P
deg(f)
x
i
. So by the isomorphisms, it follows that
deg(f) =
k
X
i=1
deg(f)
x
.