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such that h(
x
,0) = f(
x
) and h(
x
,1) = g(
x
) for all
x
X
. In that case, we shall also say
that
f is homotopic to g
and write f
g.
If we define f
t
:
X
Æ
Y
by f
t
(
x
) = h(
x
,t), then we can see that the existence of h is
equivalent with a one-parameter family of maps connecting f and g and we can think
of h as deforming f into g. See Figure 5.12(a).
5.7.1. Example.
Consider the maps f, g :
D
2
Æ
D
2
given by f(
p
) =
0
and g(
p
) =
p
.
The map h :
D
2
¥ [0,1] Æ
D
2
defined by h(
p
,t) = t
p
is a homotopy between them. In
other words, the identity map of
D
2
is homotopic to a constant map.
5.7.2. Theorem.
is an equivalence relation on the con-
tinuous maps from one topological space to another.
The homotopy relation
Proof.
We must show that the relation is reflexive, symmetric, and transitive.
Reflexivity:
If f :
X
Æ
Y
is a continuous map, then h :
X
¥ [0,1] Æ
Y
defined by h(
x
,t)
= f(
x
) is a homotopy between f and f.
Symmetry:
Let f, g :
X
Æ
Y
be continuous maps and assume that h :
X
¥ [0,1] Æ
Y
is
a homotopy between f and g. Define k :
X
¥ [0,1] Æ
Y
by k(x,t) = h(x,1-t).
Then k is a homotopy between g and f.
Transitivity:
Let f, g, h :
X
Æ
Y
be continuous maps and assume that a, b :
X
¥ [0,1]
Æ
Y
are homotopies between f and g and g and h, respectively. Define
g :
X
¥ [0,1] Æ
Y
by
()
=
(
)
Œ
[
]
g
xt
,
a
b
x t
,
2
,
t
0 1 2
,
(
)
Œ
[
]
=
xt
,
21
-
,
t
121
,
.
Then g is a homotopy between f and h.
The theorem is proved.
h(X¥0)
Y
p ¥ [0,1]
p ¥ 0
h(X¥t)
Y
y
0
h
X¥0
h
y
t
= h(p,t)
h(X¥1)
0
p ¥ t
y
1
X¥[0,1]
t
p ¥ 1
X¥1
1
(b)
(a)
Figure 5.12.
Homotopies between maps.