Graphics Reference
In-Depth Information
The easy case is where one can find an open neighborhood
U
of
x
0
over which the
covering space is trivial and that contains the curve g(t). The set p
-1
(
U
) will consist of
disjoint open sets in
Y
that are homeomorphic copies of
U
. Let
V
be the one that con-
tains
y
0
and let p
V
= pΩ
V
. Then ˜ = p
V
-1
°
g is the unique curve we seek. For the general
case, we separate the proof into two parts.
We prove uniqueness first. Let ˜
1
and ˜
2
be two liftings of g that start at
y
0
. Con-
sider the sets
=Œ
[]
()
=
{
˜
˜
()
}
=Œ
[]
()
π
{
˜
˜
()
}
A
t
01
,
g
t
g
t
and
B
t
01
,
g
t
g
t
.
1
2
1
2
These are obviously disjoint sets whose union is [0,1]. Using continuity, it is easy to
show that both of these sets are open in [0,1]. Since 0 Œ
A
,
A
is nonempty. But [0,1]
is connected and so
B
must be the empty set and we have proved uniqueness.
To prove the existence of a ˜, consider the set
=Œ
[]
{
[]
}
C
t
01
,
there
is a lifting of over
g
0
,
t
.
Figure 7.26 should help the reader follow the rest of the argument. Since the cover-
ing space is trivial over an open neighborhood of
x
0
and we know how to lift paths
over such a neighborhoods, the set
C
will contain a small neighborhood of 0 and
hence, if c is the supremum of
C
, then 0 < c £ 1. If c = 1, we are done. Assume that
c < 1. Choose a neighborhood
U
of g(c) over which the covering space is trivial. Choose
e>0, such that [c - 2e,c + 2e] Ã [0,1] and g([c - 2e,c + 2e]) Ã
U
. By the definition of
c, there is a lifting
˜
:
[
]
Æ
Y
g
0c-
,
e
of
[
]
Æ
X
g
:
0c-
,
e
.
Because the curve gΩ[c - 2e,c + 2e] lies in
U
it can be lifted to
Y
, that is, the lifting ˜
can be extended to a lifting of gΩ[0,c + 2e]. This contradicts the fact that c was the
supremum of the set
C
and so c < 1 is impossible.
lifting of g | [c - e, c + 2e]
~
y
0
= g (0)
g (c - e)
Y
p
U
Figure 7.26.
Proving the existence of path
liftings.
y
0
= g (0)
X
g (c - e)
g (c)
g (c + 2e)
