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M
453
S
A
3
S
26, 2004
Exercise 1 from Sections 1416, page 91
First consider
A
as a subspace of
Y
. In this topology, a subset
B
of
A
is open if and only if
it is of the form
B
=
A
∩
V
, where
V
is an open set in
Y
. Now
Y
is a subspace of
X
, so
V
is open precisely when it is of the form
V
=
Y
∩
U
for some
U
open in
X
. Putting these
together, we get
B
=
A
∩
(
Y
∩
U
)
=
A
∩
U
, since
A
⊂
Y
. Hence we reach the conclusion that
B
is open if and only if
B
=
A
∩
U
for some open set
U
of
X
, which is the very definition
for
B
to be open in the subspace topology
A
inherits from
X
.
s
Exercise 4 from Sections 1416, page 92
We’ll check that
π
1
:
X
×
Y
→
X
is an open map; the argument for
π
2
is exactly the same.
Let
W
be an open set in
X
×
Y
. Using the definition of the product topology, we can write
W
as a union of basis elements:
W
=
u
α
∈
J
U
α
×
V
α
,
where
J
is some indexing set,
U
α
is open in
X
, and
V
α
is open in
Y
. Unions behave well
under a function, namely
f
(
U
α
A
α
)
=
U
α
f
(
A
α
), and
π
1
(
U
α
×
V
α
)
=
U
α
, so
π
1
(
W
)
=
π
1
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 Spring '05
 SMITH
 Topology

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