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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
() =
(
)
Π[
]
g
xt
,
a
b
x t
,
2
,
t
0 1 2
,
(
)
Π[
]
=
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.
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