(FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL
< HOPF DEGREE THEOREM >
GEUNHO GIM
1. Without loss of generality, we can assume that x = z = 0. Let A = df0 . By regularity
at 0, A is a linear isomorphism. Write f (x) = Ax + u(x) where u(x) = o(|x|). Choose
a small such that |Ax| > |u(x)| on ∂B(0, ). By using the homotopy
Ax + tu(x)
H(t, x) =
,
|Ax + tu(x)|
Ax
Ax + u(x)
is homotopic to the map
.
we can conclude that the map
|Ax + u(x)|
|Ax|
In the last homework, we proved that GLn (R) has two (path-)connected components
determined by the sign of a determinant. Thus every matrix A ∈ GLn (R) is homotopic
to the identity I (if det A > 0) or a reflection R = diag(−1, 1, · · · , 1) (if det A < 0)
Ax
x
Rx
via a path inside GLn (R). In ∂B(0, ), we can find a homotopy from |Ax|
to |x|
or |Rx|
.
(Case I) f preserves orientation at 0.
det A = det df0 > 0, so A is homotopic to I.
Ax + u(x)
k−1
W (∂f, 0) = deg
: ∂B(0, ) → S
|Ax + u(x)|
x
k−1
: ∂B(0, ) → S
= deg
|x|
1
k−1
= deg
I : ∂B(0, ) → S
= +1
since 1 I is bijective and orientation-preserving.
(Case II) f reverses orientation at 0.
Similarly, A is homotopic to R and
1
W (∂f, 0) = deg( R) = −1
1
since R is bijective and orientation-reversing.
1
2
GEUNHO GIM
2. Because z is a regular value, there are only finitely many preimages of z in int(B).
Call them {x1 , · · · , xt } and choose small balls B(xi , ) ⊆ int(B) for each i. Note
f (x) − z
that u(x) =
: ∂B q (∪(∂Bi )) = ∂(B − ∪ int(Bi )) → S k−1 extends to
|f (x) − z|
B − ∪ int(Bi ). Since B − ∪ int(Bi ) is compact, we know that I(∂u, {y}) = 0 for a
regular value y of u. Thus we can conclude that
f (x) − z
k−1
: ∂B → S
W (∂f, z) = deg
|f (x) − z|
f (x) − z
k−1
: ∪(∂Bi ) → S
= deg
|f (x) − z|
X
f (x) − z
k−1
=
deg
: ∂Bi → S
,
|f (x) − z|
i
which equals the number of preimages of z counted with orientation convention(+1 if
dfxi is orientation-preserving, −1 otherwise) by Exercise 1.
3. (I guess this statement is true, but couldn’t construct an extension explicitly. Instead,
I proved a weaker statement which is enough for further arguments)
Claim : Let B = B(0, 1) be a closed ball in Rk , and let f : Rk − int(B) → Y be a
smooth map. If ∂f : ∂B → Y is homotopic to a constant, then there exists a smooth
map F : Rk → Y such that F = f on Rk − int(B(0, 2)).
(Proof) Let H : ∂B × [0, 1] → Y be a homotopy such that H1 = ∂f and H0 is a
constant map. Choose a smooth increasing function ρ : R → R such that

 0, if s ≤ 31
1, if 21 ≤ s ≤ 32
ρ(s) =

s, if s ≥ 2
By using ρ, we can define a smooth map F : Rk → Y by
  f ρ(|x|) x ,
if |x| ≥ 1
|x|
F (x) =
 H x , ρ(|x|) , if |x| ≤ 1
|x|
Note that F is constant on the set {x | |x| ≤ 13 , 21 ≤ |x| ≤ 32 }, and F = f is on |x| ≥ 2.
4. Suppose f : S l → Rl+1 − {0} has W (f, 0) = 0. Then, deg(u(x) := |ff (x)
: Sl → Sl) =
(x)|
W (f, 0) = 0 by definition. The special case implies that u is homotopic to a constant
map. On the other hand, H : S l × [0, 1] → Rl+1 − {0} gives a homotopy between f
and u by
f (x)
H(x, t) = (1 − t)f (x) + t
(6= 0 for any t, x).
|f (x)|
Therefore, f is homotopic to a constant map.
(FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL
< HOPF DEGREE THEOREM >
3
5. Take a large ball B = B(0, r) so that it contains every preimage of 0. Then, W (∂f :
∂B → Rk − {0}, 0) = 0 by Exercise 2. Let’s assume that the special case holds when
l = k − 1, and that r = 1 by suitable scaling. By Exercise 4, Corollary holds for
l = k − 1, thus ∂f is homotopic to a constant. By Exercise 3, we can construct
g : Rk → Rk − {0} such that g = f on Rk − int(B(0, 2)). Note that g does not assume
the value 0 because f |Rk −int(B(0,1)) and the homotopy (between f and a constant) don’t
assume it.
6. Firstly, we need the following lemma.
Lemma 1. Let f, g : S 1 → S 1 be smooth maps. Then, f and g are homotopic if and
only if deg f = deg g.
Proof. ( ⇒ ) Obvious.
( ⇐ ) Suppose deg f = d = deg g. There exists a map h : R → R such that f (eiθ ) =
eih(θ) . Moreover, h(θ + 2π) − h(θ) is a continuous map which takes values in a discrete
set 2πZ. Thus h(θ + 2π) − h(θ) = 2πm for some fixed m ∈ Z.
Note that
(1 − t)h(θ + 2π) + tm(θ + 2π) = (1 − t)h(θ) + tmθ + 2mπ.
By the well-defined homotopy H(θ, t) = exp(i{(1 − t)h(θ) + tmθ}), f is homotopic to
the m-th power map θm : S 1 → S 1 . Since m = deg θm = deg f , m = d. Because both
f and g are homotopic to θd , they are homotopic to each other.
We will use an induction on l to prove Special Case.
First suppose l = 1. By the lemma, any smooth map f : S 1 → S 1 with deg f = 0 is
homotopic to the 0-th power map, i.e., a constant map.
Assume that Special Case is true for l = 1, 2, · · · , k − 1. Let f : S k → S k be a
degree-0 map. Choose two distinct regular values a, b ∈ S k (possible by Sard). Let
f −1 (a) = {a1 , · · · , as }. Take an open set U ∼
= Rk such that U ⊆ S k − f −1 (b). By
isotropy lemma, there exists a diffeomorphism h : S k → S k , isotopic to the identity,
such that h maps a1 , · · · , as into U . Since f = f ◦ id is homotopic to f ◦ h−1 , we can
assume that f −1 (a) ⊆ U . We can view f |U : U → S k − {b} as a map from Rk to Rk
where a is identified with 0 in the latter case. By Exercise 2, the induction hypothesis,
and the fact that U contains every preimage of a, Exercise 5 is applicable in this case.
There exists a map g : U → S k − {a, b} such that g = f outside a compact set in U .
Extend g : S k → S k − {a} by defining g = f on S k − U . g is smoothly extended since
f = g outside a compact set. We can define a homotopy H(x, t) = (1 − t)f (x) + tg(x)
on S k − {b} ∼
= Rk and extend it to S k by setting H(x, t) = f (x) when x ∈ S k − U .
Because g = f outside a compact set in U , and g(x) = f (x) implies H(x, t) = f (x), H
is well-defined smooth homotopy between f and g. On the other hand, g is homotopic
to a constant map since S k − {a} is contractible. Thus, so is f .
4
GEUNHO GIM
7. Suppose W ⊆ RM and f : ∂W → Rk+1 is a smooth map. By -neighborhood theorem,
there is an open set U of points in RM with distance less than from ∂W . Choose a
sufficiently small so that each point x ∈ U possesses a unique closest point wx ∈ ∂W .
Also, there exists a submersion π : U → ∂W which is the identity on ∂W . We can
extend f to U by composing g := f ◦π : U → Rk+1 . Take a smooth increasing function
h : R → R such that
1, x ≤ /3
h(x) =
0, x ≥ 2/3
Extend g smoothly to the whole RM by setting
g(wx + (x − wx )h(d(x, ∂W ))), x ∈ U
g̃(x) =
0,
x ∈ RM − U
where d(x, ∂W ) is the distance between the point x ∈ U and ∂W . In particular, f is
extendable to W .
8. Let W be a compact, connected oriented k + 1 dimensional manifold with boundary,
and let f : ∂W → S k be a smooth map.
If f extends to F : W → S k , with ∂F = f , then clearly deg f = 0.
Conversely, suppose deg f = 0. By Exercise 7, f : ∂W → S k ,→ Rk+1 extends to
G : W → Rk+1 . By taking a homotopy of F (which is invariant on ∂W ) if necessary,
we can assume that 0 is a regular value of G. By a similar argument in Exercise 6,
we can take an open set U ∼
= Rk+1 such that G−1 (0) ⊆ U ⊆ W . Take a closed ball B
G(x)
in U which contains G−1 (0). Note that |G(x)|
is defined on a compact set W − int(B),
which has ∂(W − int(B)) = ∂W − ∂B. Let ∂G = G|∂B , then
G
k
W (∂G, 0) = deg
: ∂B → S
|G|
G
k
= deg
: ∂W → S
|G|
= deg(f ) = 0
k
∼
by assumption. Since ∂B = S , Corollary after Exercise 4 is applicable in this case. ∂G
is homotopic to a constant map, thus by Exercise 3, G|U −int(B) : U − int(B) → Rk+1 −
{0} enduces G̃ : U → Rk+1 − {0} such that G̃ = G on U − int(B 0 ) where B 0 ⊆ U is a
closed ball properly containing B. Also, it extends to W by defining G̃ = G outside
U . We get F = |G̃
: W → S k which extends f .
G̃|
(FALL 2011) 225A - DIFFERENTIAL TOPOLOGY FINAL
< HOPF DEGREE THEOREM >
5
9.
Theorem 1. (The Hopf Degree Theorem) Two maps of a compact, connected, oriented
k-manifold X into S k are homotopic if and only if they have the same degree.
Proof. If g, h : X → S k are homotopic, they have the same degree.
Conversely, suppose deg g = deg h. Let W = X × [0, 1], and define f : ∂W → S k by
f (x, 0) = g(x), f (x, 1) = h(x). Note that deg f = deg h − deg g = 0. By Extension
Theorem, f can be extended to F : W → S k , which gives a homotopy between g and
h.