Graphics Reference
In-Depth Information
say that F is an
orientation-reversing bundle map
if
˜
defines an orientation-reversing
vector space isomorphism on each fiber of x
1
.
The pullback of an oriented vector bundle has a natural orientation because fibers
get mapped isomorphically onto fibers and so we can simply pull back the orienta-
tion with the inverse of the isomorphism. More precisely,
Definition.
Let (x,s) be an oriented n-plane bundle and let f :
B
1
Æ
B
(x) be a map.
The
induced orientation
f*s on f*x is defined as follows: Let (
˜
,f) : f*xÆxbe the canon-
ical map with fiber maps
˜
F
. If
b
1
Œ
B
1
, then
˜
-
1
( ( )
=
()
(
(
()
)
)
f
*
s
b
f
b
s
f
b
.
1
F
1
1
*
8.9.9. Theorem.
Any vector bundle x=(
E
,p,
B
) over a simply connected space
B
is
orientable.
Proof.
See Figure 8.30. Fix a point
b
0
in
B
. Choose an orientation s
0
in p
-1
(
b
0
). Let
b
Œ
B
. We shall define an orientaton s for x so that s(
b
0
) =s
0
. Since
B
is path-
connected, there is a path g :
I
Æ
B
so that g(0) =
b
0
and g(1) =
b
. By Corollary 8.9.5,
g*x is a trivial bundle over
I
, which means that g*x admits a unique orientation h
g
so
that h
g
(0) is mapped to s
0
by the canonical map from g*x to x. See Exercise 8.9.5 for
the uniqueness part. Let s(
b
) be the orientation of p
-1
(
b
) to which h
g
(1) is mapped by
the canonical map from g*x to x. We need to show that
(1) s is a well-defined map, and
(2) s is a continuously varying choice of orientations in each fiber.
We shall only prove (1) and leave (2) as an exercise for the reader. Suppose that there
is another path l :
I
Æ
B
so that l(0) =
b
0
and l(1) =
b
. Again, choose the unique ori-
entation h
l
for l*x so that h
l
(0) is mapped to s
0
by the canonical map from l*x to x.
We must show that h
l
(1) maps to s(
b
). Since
B
is simply connected, there is a
homotopy
I ¥ I ¥ R
p
-1
(b)
E
p
-1
(b
0
)
E(h*x)
(t,1) ¥ R
p
-1
(l(t))
p
g
g (t)
I ¥ 0
h
(t,0)
1 ¥ I
b
I ¥ I
B
b
0
l(t)
0 ¥ I
I ¥ 1
(t,1)
l
Figure 8.30.
Proving Theorem 8.9.9.
