Math 132  Topology II: Smooth Manifolds. Spring 2017.
Week 6 Summary
Reading
§
1.5, 2.1, 2.2
GP
References
GP

Differential Topology
, Guillemin & Pollack
1
Lecture 15, February 27
1.1
Transversality
Question:
Let
f
∶
X
→
Y
be smooth,
Z
⊆
Y
a submanifold. When is
f

1
(
Z
)
a submani
fold?
Example 1.1.
1. If
Z
=
{
z
}
⊆
Y
, with
z
∈
Y
a regular value of
f
∶
X
→
Y
.
Then,
f

1
(
Z
)
is a submanifold by the Preimage Theorem.
2. If
Z
⊆
Y
is open then
f

1
(
Z
)
⊆
X
is open in
X
, hence a submanifold.
Remark 1.2.
The examples given above are the extreme cases
dim
Z
∶
0
?
?
dim
Y
Z
⊆
Y
{
z
}
reg. value
?
?
Z
open
Definition 1.3.
Let
f
∶
X
→
Y
be smooth,
Z
⊆
Y
manifolds. We say that
f
is transversal
to
Z
, written
f
Z
, if, for every
x
∈
f

1
(
Z
)
,
im
(
df
)
x
+
T
f
(
x
)
Z
=
T
f
(
x
)
Y.
If
j
∶
X
→
Y
is the inclusion of the submanifold
X
⊆
Y
, then we say that
X
is transver
sal to
Z
, written if
X
Z
, if
j
is transversal to
Z
.
Example 1.4.
1. Let
f
∶
R
→
R
2
,t
(
t,t
2
)
,
Z
=
{(
x,
0
)
x
∈
R
}
.
Then,
Z
is not
transversal to
f
: we have
f

1
(
Z
)
=
{
0
}
. Then,
T
(
0
,
0
)
Z
=
span
(
e
1
)
,
im
(
df
)
0
=
colJac
0
f
=
span
(
e
1
)
.
Hence,
T
(
0
,
0
)
Z
+
im
(
df
)
0
=
span
(
e
1
)
≠
R
2
.
2. Let
f
be as above,
Z
=
{(
x,x
)
x
∈
R
}
. Then,
f

1
(
Z
)
=
{
0
,
1
}
.
(
t
=
0
)
∶
T
(
0
,
0
)
Z
=
span
(
e
1
+
e
2
)
,
im
(
df
)
0
=
colJac
0
f
=
span
(
e
1
)
,
⇒
T
(
0
,
0
)
Z
+
im
(
df
)
0
=
span
(
e
1
+
e
2
,e
1
)
=
R
2
.
(
t
=
1
)
∶
T
(
1
,
1
)
Z
=
span
(
e
1
+
e
2
)
,
im
(
df
)
1
=
colJac
1
f
=
span
(
e
1
+
2
e
2
)
,
⇒
T
(
1
,
1
)
Z
+
im
(
df
)
1
=
R
2
.
Hence,
f
Z
.
1