Definition 4.9. If a point a ∈ X is a deformation retract of X , X is said to be
contractible.
Let c
a
: X →{a} be a constant map. If X is contractible, there exists a
homotopy H : X × I → X such that H (x , 0) = c
a
(x ) = a and H (x, 1) =
id
X
(x ) = x for any x ∈ X and, moreover, H (a, t) = a for any t ∈ I .The
homotopy H is called the contraction.
Example 4.2. X =
n
is contractible to the origin 0. In fact, if we define
H :
n
× I → by H (x , t) = tx,wehave(i)H (x , 0) = 0andH (x , 1) = x
for any x ∈ X and (ii) H (0, 1) = 0foranyt ∈ I . Now it is clear that any convex
subset of
n
is contractible.
Exercise 4.2. Let D
2
={(x, y) ∈
2
|x
2
+ y
2
≤ 1}. Show that the unit circle
S
1
is a deformation retract of D
2
−{0}. Show also that the unit sphere S
n
is a
deformation retract of D
n+1
−{0},whereD
n+1
={x ∈
n+1
||x |≤1}.
Theorem 4.4. The fundamental group of a contractible space X is trivial,
π
1
(X, x
0
)
∼
=
{e}. In particular, the fundamental group of
n
is trivial,
π
1
(
n
, x
0
)
∼
=
{e}.
Proof. A contractible space has the same fundamental group as a point {p} and a
point has a trivial fundamental group.
If an arcwise connected space X has a trivial fundamental group, X is said
to be simply connected, see section 2.3.
4.3 Examples of fundamental groups
There does not exist a routine procedure to compute the fundamental groups,
in general. However, in certain cases, they are obtained by relatively simple
considerations. Here we look at the fundamental groups of the circle S
1
and
related spaces.
Let us express S
1
as {z ∈ ||z|=1}. Define a map p : → S
1
by
p : x → exp(ix). Under p, the point 0 ∈
is mapped to 1 ∈ S
1
,whichis
taken to be the base point. We imagine that
wraps around S
1
under p,see
figure 4.9. If x, y ∈
satisfies x − y = 2πm(m ∈ ), they are mapped to the
same point in S
1
. Then we write x ∼ y. This is an equivalence relation and the
equivalence class [x ]={y|x − y = 2πm for some m ∈
} is identified with
a point exp(ix ) ∈ S
1
. It then follows that S
1
∼
=
/2π .Let
˜
f : → be
a continuous map such that
˜
f (0) = 0and
˜
f (x + 2π) ∼
˜
f (x ). It is obvious
that
˜
f (x + 2π) =
˜
f (x ) + 2nπ for any x ∈
,wheren is a fixed integer. If
x ∼ y (x − y = 2πm),wehave
(
f (x) −
(
f (y) =
(
f (y + 2πm) −
(
f (y)
=
(
f (y) +2π mn −
(
f (y) = 2πmn