between the higher homotopy groups of a topological space and its universal
covering space.
Definition 4.12. Let X and
(
X be connected topological spaces. The pair (
(
X, p),
or simply
(
X , is called the covering space of X if there exists a continuous map
p :
(
X → X such that
(1) p is surjective (onto)
(2) for each x ∈ X , there exists a connected open set U ⊂ X containing
x, such that p
−1
(U) is a disjoint union of open sets in
(
X, each of which is
mapped homeomorphically onto U by p.
In particular, if
(
X is simply connected, (
(
X , p) is called the universal
covering space of X.[Remarks: Certain groups are known to be topological
spaces. They are called topological groups. For example SO(n) and SU(n) are
topological groups. If X and
(
X in definition 4.12 happen to be topological groups
and p :
(
X → X to be a group homomorphism, the (universal) covering space is
called the (universal) covering group.]
For example,
is the universal covering space of S
1
, see section 4.3. Since
S
1
is identified with U(1), is a universal covering group of U(1) if is regarded
as an additive group. The map p :
→ U(1) may be p : x → e
i2π x
. Clearly p
is surjective and if U ={e
i2π x
|x ∈ (x
0
− 0.1, x
0
+ 0.1)},then
p
−1
(U) =
+
n∈
(x
0
− 0.1 + n, x
0
+ 0.1 +n)
which is a disjoint union of open sets of
. Itiseasytoshowthat p is also a
homomorphism with respect to addition in
and multiplication in U(1). Hence,
(
, p) is the universal covering group of U(1) = S
1
.
Theorem 4.10. Let (
(
X, p) be the universal covering space of a connected
topological space X .Ifx
0
∈ X and ˜x
0
∈
(
X are base points such that p( ˜x
0
) = x
0
,
the induced homomorphism
p
∗
: π
n
(
(
X , (x
0
) → π
n
(X, x
0
) (4.44)
is an isomorphism for n ≥ 2. [War n in g: This theorem cannot be applied if n = 1;
π
1
( ) ={e} while π
1
(S
1
) = .]
The proof is given in Croom (1978). For example, we have π
n
( ) ={e}
since
is contractible. Then we find
π
n
(S
1
)
∼
=
π
n
(U(1)) ={e} n ≥ 2. (4.45)
Example 4.12. Let S
n
={x ∈
n+1
||x|
2
= 1}. The real projective space P
n
is
obtained from S
n
by identifying the pair of antipodal points (x , −x ). It is easy to