Math 132 - Topology II: Smooth Manifolds. Spring 2017.
Diagnostic Quiz. February 15, 1.30-2pm
Name:
SOLUTION
1. Let
X
=
{(
x,y,z
)
z
=
x
2
+
y
2
}
⊆
R
3
. Consider the smooth map
f
∶
X
→
R
2
,
(
x,y,z
)
(
y,z
)
.
(a) Compute
T
(
a,b,c
)
X
, where
(
a,b,c
)
∈
X
.
(b) Define what it means for
f
to be an immersion.
(c) Determine the points
(
a,b,c
)
∈
X
where
f
is an immersion.
Solution:
(a) We choose a parameterisation
φ
∶
R
2
→
X ,
(
u,v
)
(
u,v,u
2
+
v
2
)
.
φ
is smooth and has inverse
ψ
∶
X
→
R
2
,
(
x,y,z
)
(
x,y
)
. As
ψ
extends to the
smooth function Ψ
∶
R
3
→
R
2
,
(
x,y,z
)
(
x,y
)
we get that
ψ
is smooth.
Then,
T
(
a,b,c
)
X
=
im
(
dφ
)
(
a,b
)
=
col Jac
(
a,b
)
φ
=
col
1
0
0
1
2
a
2
b
(b)
f
is an immersion if, for every
x
∈
X
,
(
df
)
x
is injective.
(c) Let
(
a,b,c
)
∈
X
. Let
v
=
λ
1
1
0
2
a
+
λ
2
0
1
2
b
∈
T
(
a,b,c
)
X
. The map
f
extends to
smooth
F
∶
R
3
→
R
2
,
(
x,y,z
)
(
y,z
)
. Thus,
(
df
)
(
a,b,c
)
(
v
)
=
(
Jac
(
a,b,c
)
F
)
v
=
0
1
0
0
0
1
λ
1
1
0
2
a
+
λ
2
0
1
2
b
=
λ
2
2
(
λ
1
a
+
λ
2
b
)
f
is an immersion at those
(
a,b,c
)
where ker
(
df
)
(
a,b,c
)
=
{
0
}
:
- If
a
=
0 then we see that span
1
0
2
a
⊆
ker
(
df
)
(
a,b,c
)
;
- Assume
a
≠
0, then ker
(
df
)
(
a,b,c
)
=
{
0
}
.