Exercises III
Joshua Egger
MATH191
Exercise 3.3:
Prove by hand that the homology of the sphere
H
k
(
S
n
) is zero whenever
n
6
=
k,
0.
More can be said:
Claim:
H
k
(
S
n
) =
(
Z
2
n
= 0
, n
0
n >
0
Proof.
We start with
S
0
. Since
S
0
is the disjoint union of two points, it is clear that
H
k
(
S
0
) =
(
Z
2
⊕
Z
2
k
= 0
0
k >
0
Now consider the
n
sphere
S
n
and define the sets
U
:=
S
n
\ {
the “north pole”
}
and
V
:=
S
n
\
{
the “south pole”
}
. Then
U
and
V
are contractible,
U
∪
V
=
S
n
and the intersection
U
∩
V
∼
=
S
n

1
(note: homotopy invariance of homology implies that
H
•
(
U
∩
V
)
∼
=
H
•
(
S
n

1
)), so we can apply the
Mayer Vietoris sequence. Let’s start with
S
1
. We know that
H
1
(
S
1
)
∼
=
Z
2
due to contractibility of
S
1
,
so we need to look at the piece
· · ·
H
n
(
U
)
⊕
H
n
(
V
)
→
H
n
(
S
1
)
→
H
n

1
(
U
∪
V
)
→ · · ·
But
H
n
(
U
)
⊕
H
n
(
V
) =
H
n

1
(
U
∪
V
) = 0 so this is also true for
H
n
(
S
1
) where
n
≥
2. Now we induct,
take
n
≥
2 and suppose the result holds for
S
k
where
k
≤
m

1. To find
H
1
(
S
n
) we look at the piece
· · · →
H
1
(
U
)
⊕
H
1
(
V
)
→
H
1
(
S
n
)
→
H
0
(
S
n

1
)
→
H
0
(
U
)
⊕
H
0
(
V
)
→
0
Since
S
n

1
, U
, and
V
are all contractible, the map from
H
0
(
S
n

1
)
→
H
0
(
U
)
⊕
H
0
(
V
) is isomorphic to
the injection.
Z
2
→
Z
2
⊕
Z
2
. This implies that
H
1
(
S
n
)
∼
= 0
Finally, we look at the segment
· · · →
H
k
(
U
)
⊕
H
k
(
V
)
→
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 Spring '08
 Staff
 Math