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(
) =-
(
)
rx x
,
,...,
x
x x
,
,...,
x
12
n
+
1
12
n
+
1
has degree -1.
Proof. It is not hard to prove the theorem by simple computations using an appro-
priate triangulation (K,j) of S n and the simplicial map of K corresponding to r, but
the details are rather messy and we do not repeat them here. See [AgoM76]. Intu-
itively, think of S n as a cell complex with two n-cells that consist of two hemispheres
(the parts with the x-coordinates nonnegative or nonpositive). Then a generator of
H n ( S n ) can be represented by a cycle that consists of the sum of these two cells ori-
ented appropriately. The reflection will then map each of these oriented n-cells into
the other, but with the opposite orientation.
Let n ≥ 1. The antipodal map f : S n
Æ S n , f( p ) =- p , has degree
7.5.1.3. Theorem.
(-1) n+1 .
Consider the reflections r i : S n
Æ S n , 1 £ i £ n+1, defined by
Proof.
(
) =
(
)
rx x
,
,...,
x
x
,...,
x
,
-
x x
,
,...,
x
.
i
12
n
+
1
1
i
-
1
i
i
+
1
n
+
1
Clearly, f = r 1 ° r 2 ° ... ° r n+1 . Therefore, the theorem follows easily from Theorems
7.5.1.1(4) and 7.5.1.2.
Actually, the case where n is odd can be proved directly without appealing to
Theorem 7.5.1.2. In terms of coordinates,
(
) =-
(
)
fx x
,
,...,
x
x
,
-
x
,...,
-
x
.
12
n
+
1
1
2
n
+
1
If n = 1, then f is just a rotation through 180 degrees and is homotopic to the
identity. The map
1
¥ [ Æ
1
h:
S
01
,
S
defined by
(
) = (
) --
(
) + (
)
(
)
hx x t
,
,
cos
p
t
x
,
x
sin
p
t x
,
-
x
12
1
2
2
1
is one such homotopy. In the case of an arbitrary odd n, we have an even number of
coordinates and we can again define a homotopy between f and the identity by using
a map like h for each pair of coordinates x 2i-1 and x 2i , i = 1,2, ...(n + 1)/2.
Using the properties of the degree of a map, we can easily deduce some
well-known theorems. See also Section 8.5.
S n is not a retract of D n+1 .
7.5.1.4. Theorem.
Proof. Suppose that we have a retraction r : D n+1 Æ S n . Let i : S n Æ D n+1 be the natural
inclusion map. The maps
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