M
453
S
A
11
D
3, 2004
Exercise 2 from Section 51, page 330
This is a special case of problem 3 below.
±
Exercise 3 from Section 51, page 330
(a) The formula
F
(
x
,
t
)
=
(1

t
)
x
is a homotopy between the identity map on either
I
or
R
to the constant map at 0.
(b) Suppose
X
is contractible and let
F
:
X
×
I
→
X
be a nullhomotopy with
F
(
x
,
0)
=
x
and
F
(
x
,
1)
=
x
0
for all
x
. If
x
,
y
∈
X
,
F
(
x
,
t
) and
F
(
y
,
t
) are paths from
x
and
y
to
x
0
respectively. Then
F
(
x
,
t
)
*
F
(
y
,
t
) is a path from
x
to
y
.
(c)
Y
is contractible, so
i
Y
'
e
y
0
where
e
y
0
is the constant map at
y
0
. Thus given any map
f
:
X
→
Y
, we have
i
Y
◦
f
'
e
y
0
◦
f
. Since
i
Y
◦
f
=
f
, we see that any map is
homotopic to the constant map that takes all of
X
to
y
0
.
(d) Let
i
X
'
e
x
0
and [
f
]
,
[
g
]
∈
[
X
,
Y
]. Then
f
=
f
◦
i
X
'
f
◦
e
x
0
=
e
f
(
x
0
)
and similarly
g
'
e
g
(
x
0
)
. Hence we need only show that
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 Spring '05
 SMITH
 Math, Topology, Topological space, Bk, path lifting lemma, constant map

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