= 0. Thus the only components which give a nonzero term in the sum
are those which intersect both g^{−1}
(y )
and f
(Ω)
and as just observed, there are
finitely many of these. The unbounded component K contributes 0 to the sum
because it contains points which cannot be in f
(Ω-)
and the degree is constant
on components. Thus d
(f,Ω,K )
= 0. From now on, only consider the bounded
components.
Let
˜g
be in C_{c}^{∞}
(ℝp,ℝp)
and let
∥˜g− g∥
_{∞} be so small that
d (g ∘f,Ω,y) = d (˜g ∘f,Ω,y )
d(g,Ki,y) = d (˜g,Ki, y)
and such that y is a regular value. This is from Lemma 16.1.5 and the observation that
∂K_{i}⊆ f
(∂Ω )
. This is done as before, by creating a homotopy between the two functions
which misses g
(f (∂Ω))
or ∂K_{i}⊆ f
(∂Ω)
. Let
˜g−1(y)∩ Ki ≡ {xij}mi
j=1
There are finitely many because y is a regular value of
˜g
and we only consider
the bounded components. Now use Lemma 16.1.5 again to get
˜
f
in C^{∞}
(- n)
Ω, ℝ
such that each x_{j}^{i} is a regular value of
˜f
and also
∥ ∥
∥∥˜f − f∥∥
_{∞} is very small, so small
that
d (˜g ∘f,Ω,y) = d(˜g ∘˜f,Ω,y)
and
( ) ( )
d ˜f,Ω, xij = d f,Ω,xij
for each i,j. It is the same process done earlier only you adjust smallness for each of finitely
many x_{j}^{i}. Thus, from the above,
( )
d(g∘ f,Ω,y) = d ˜g ∘˜f,Ω,y
(˜ i) ( i)
d f,Ω, xj = d f,Ω,xj
d(g,Ki,y) = d(˜g,Ki,y)
and y is a regular value for
˜g
∘
˜f
and each x_{j}^{i} is a regular value for
˜f
. Thus, from the
definition of the degree, the right side equals
∑ ∑ ∑ ( ( ))
= sgn(det(D˜g (xij)))sgn det D ˜f (z)
i j z∈˜f−1xi
∑ ∑ (( j)( ( ))) ∑ ( ( ))
= sgn det D ˜g xij sgn det D ˜f (z)
i j z∈˜f− 1(xij)
∑ ( )
= d(˜g,Ki,y)d ˜f,Ω,Ki
i
To explain the last step, ∑_{z∈˜f
−1(xij)
}sgn
( ( ))
det D ˜f (z)
≡ d
( )
˜f,Ω,xij
= d
( )
˜f,Ω, Ki
. This
proves the product formula because
˜g
and
˜f
were chosen close enough to f,g respectively
that
∑ ( ) ∑
d (˜g,Ki,y)d ˜f,Ω,Ki = d(g,Ki, y)d(f,Ω,Ki)■
i i
Note that g ∈ C_{c}
(ℝn,ℝn)
but it is not really necessary to assume the function has
compact support because everything of interest happens on the compact set f
(∂Ω )
. The
consideration of the more general case involves only technical considerations such as
multiplication by a suitable cutoff function and noting that nothing of interest in the formula
changes because degree is determined by the values of the function on the boundary of
compact sets.
The following theorem is the Jordan separation theorem, a major result. A homeomorphism is
a function which is one to one onto and continuous having continuous inverse. Before the
theorem, here is a helpful lemma.
Lemma 16.6.3Let Ω be a bounded open set in ℝ^{p}, f ∈ C
(- )
Ω; ℝp
, and suppose
{Ωi}
_{i=1}^{∞}are disjoint open sets contained in Ω such that
(- ∞ )
y ∕∈ f Ω ∖ ∪j=1Ωj
Then
∑∞
d(f,Ω, y) = d (f,Ωj,y)
j=1
where the sum has all but finitely many terms equal to 0.
Proof: By assumption, the compact set f^{−1}
(y)
≡
{x ∈ Ω-: f (x) = y}
has empty
intersection with
-- ∞
Ω ∖∪ j=1Ωj
and so this compact set is covered by finitely many of the Ω_{j}, say
{Ω1,⋅⋅⋅,Ωn−1}
and
( )
y ∕∈ f ∪∞j=nΩj .
By Theorem 16.3.4 and letting O = ∪_{j=n}^{∞}Ω_{j},
n∑−1 ∑∞
d (f,Ω,y ) = d(f,Ωj,y)+ d(f,O,y) = d (f,Ωj,y)
j=1 j=1
because d
(f,O,y)
= 0 as is d
(f,Ωj,y)
for every j ≥ n. ■
I will give a shorter version of the proof and a longer version. First is the shorter
version which leaves out a few details which may or may not be clear. Sometimes,
when you put in too many details, you lose the forest by stumbling around hitting
trees. It may still have too many details. To help remember some symbols, here is a
short diagram. I have a little trouble remembering what is what when I read the
proof.
|---------------------------------------|
|----bounded-components-of ℝp-∖f-(∂K-)----|
|------------------ℋ--------------------|
|-----ℒH-sets of ℒ-contained-in-H-∈-ℋ-----|
-ℋ1-those sets of ℋ-which-intersect-a set of ℒ
Theorem 16.6.4(Jordan separation theorem)Let f be a homeomorphism ofC and f
(C)
where C is a compact set in ℝ^{p}. Then ℝ^{p}∖C and ℝ^{p}∖f
(C )
have the samenumber of connected components.
Proof: Denote by K the bounded components of ℝ^{p}∖ C and denote by ℒ,
the bounded components of ℝ^{p}∖ f
(C )
. Also, using the Tietze extension theorem
on components, there exists f an extension of f to all of ℝ^{p} which maps into a
bounded set and let f^{−1} be an extension of f^{−1} to all of ℝ^{p} which also maps into a
bounded set. We can assume these extensions have compact support by multiplying
by a suitable cutoff function. Pick K ∈K and take y ∈ K. Then ∂K ⊆ C and
so
-−1( )
y∈∕f ¯f (∂K )
Since f^{−1}∘f equals the identity I on ∂K, it follows from the properties of the degree
that
(-−1 )
1 = d(I,K,y ) = d f ∘¯f,K,y .
Recall that if two functions agree on the boundary, then they have the same degree. Let ℋ
denote the set of bounded components of ℝ^{p}∖f
(∂K )
. These will be as large as those in ℒ and
if a set in ℒ intersects one of these larger H ∈ℋ then H contains the component in ℒ.By
the product formula,
(--- ) ∑ ( ) (--- )
1 = d f−1 ∘¯f,K,y = d ¯f,K,H d f−1,H,y , (16.9)
H∈ℋ
(16.9)
the sum being a finite sum from the product formula. That is, there are finitely many H
involved in the sum, the other terms being zero.
What about those sets of ℋ which contain no set of ℒ? These sets also have empty
intersection with all sets of ℒ as just explained. Therefore, for H one of these, H ⊆ f
(C)
.
Therefore,
( --- ) ( −1 )
d f− 1,H, y = d f ,H,y = 0
because y ∈ K a component of ℝ^{p}∖ C, but for u ∈ H ⊆ f
(C )
,f^{−1}
(u)
∈ C so f^{−1}
(u)
≠y
implying that d
(f−1,H, y)
= 0. Thus in 16.9, all such terms are zero. Then letting ℋ_{1} be
those sets of ℋ which contain (intersect) some sets of ℒ, the above sum reduces to
∑ (¯ ) (-−1 ) ∑ (¯ ) ∑ (-−1 )
d f,K,H d f ,H, y = d f,K,H d f ,L,y
H∈ℋ1 H∑∈ℋ1 ∑ ( L∈ℒH) (--- )
= d ¯f,K,H d f−1,L,y
H ∈ℋ1L∈ℒH
where ℒ_{H} are those sets of ℒ contained in H. If ℒ_{H} = ∅, the above shows that the second sum
is 0 with the convention that ∑_{∅} = 0. Now d
(¯f,K, H )
= d
(¯f,K,L )
where L ∈ℒ_{H}.
Therefore,
∑ ∑ (¯ ) ( −-1 ) ∑ ∑ (¯ ) (-−1 )
d f,K,H d f ,L,y = d f,K,L d f ,L,y
H ∈ℋ1L∈ℒH H ∈ℋ1L∈ℒH
As noted above, there are finitely many H ∈ℋ which are involved. ℝ^{p}∖ f
(C )
⊆ ℝ^{p}∖ f
(∂K )
and so every L must be contained in some H ∈ℋ_{1}. It follows that the above reduces
to
∑ (¯ ) (-−1 ) ∑ (¯ ) (-−1 )
1 = d f,K, L d f ,L,y = d f,K,L d f ,L,K (16.10)
L∈ℒ L∈ℒ
(16.10)
Let
|K |
denote the number of components in K and similarly,
|ℒ|
denotes the number of
components in ℒ. Thus
∑ ∑ ∑ ( ) (--- )
|K | = 1 = d ¯f,K, L d f−1,L,K
K∈K K ∈KL∈ℒ
Similarly,
∑ ∑ ∑ ( ) ( --- )
|ℒ| = 1 = d ¯f,K, L d f− 1,L, K
L∈ℒ L∈ℒK ∈K
If
|K|
< ∞, then ∑_{K∈K}
◜----------1◞◟----------◝
∑ d (¯f,K, L)d (f−1,L, K)
L∈ℒ
< ∞. The summation which equals 1 is
a finite sum and so is the outside sum. Hence we can switch the order of summation and
get
∑ ∑ ( ) (--- )
|K| = d ¯f,K, L d f−1,L,K = |ℒ|
L∈ℒ K∈K
A similar argument applies if
|ℒ|
< ∞. Thus if one of these numbers is finite, so is the other
and they are equal. It follows that
|ℒ|
=
|K |
.This proves the theorem because if n > 1 there
is exactly one unbounded component to both ℝ^{p}∖ C and ℝ^{p}∖ f
(C )
and if n = 1 there are
exactly two unbounded components. ■
As an application, here is a very interesting little result. It has to do with d
(f,Ω,f (x))
in
the case where f is one to one and Ω is connected. You might imagine this should equal 1 or
−1 based on one dimensional analogies. In fact this is the case and it is a nice application of
the Jordan separation theorem and the product formula.
Proposition 16.6.5Let Ω be an open connected bounded set in ℝ^{p},n ≥ 1 suchthat ℝ^{p}∖ ∂Ω consists of two, three if n = 1, connected components. Let f ∈ C
(Ω;ℝp )
be continuous and one to one. Then f
(Ω)
is the bounded component of ℝ^{p}∖ f
(∂Ω)
andfor y ∈ f
(Ω)
, d
(f,Ω,y )
either equals 1 or −1.
Proof: First suppose n ≥ 2. By the Jordan separation theorem, ℝ^{p}∖f
(∂Ω)
consists of two
components, a bounded component B and an unbounded component U. Using
the Tietze extention theorem, there exists g defined on ℝ^{p} such that g = f^{−1} on
f
(--)
Ω
. Thus on ∂Ω,g ∘ f = id. It follows from this and the product formula that
= 0 as explained above. Since U is unbounded,
there are points in U which cannot be in the compact set f
( )
Ω¯
. For such, the degree is 0 but
the degree is constant on U, one of the components of f
(∂Ω )
. Therefore, d
(f,Ω,B)
≠0 and so
for every z ∈ B, it follows z ∈ f
(Ω)
. Thus B ⊆ f
(Ω)
. On the other hand, f
(Ω )
cannot
have points in both U and B because it is a connected set. Therefore f
(Ω )
⊆ B
and this shows B = f
(Ω )
. Thus d
(f,Ω, B)
= d
(f,Ω, y)
for each y ∈ B and the
above formula shows this equals either 1 or −1 because the degree is an integer. In
the case where n = 1, the argument is similar but here you have 3 components in
ℝ^{1}∖ f
(∂Ω )
so there are more terms in the above sum although two of them give 0.
■
Here is another nice application of the Jordan separation theorem to the Jordan curve
theorem.