= 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.
∑ ∑ ∑ ( ( ( ))) ( ( ))
= sgn det D ˜g xij sgn det D˜f (z)
i j z∈˜f−1(xij)
∑ ∑ ( ( ( i))) ∑ ( ( ))
= sgn detD ˜g x j 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
(˜ i)
f,Ω,x j
= 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 21.6.3Let 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∖ ∪∞ Ω
j=1 j
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 21.3.4 and letting O = ∪_{j=n}^{∞}Ω_{j},
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 21.6.4(Jordan separation theorem)Let f be a homeomorphism of Cand f
(C )
where C is a compact set in ℝ^{p}. Then ℝ^{p}∖ C and ℝ^{p}∖ f
(C )
have thesame number 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
---( )
y ∕∈ f− 1 ¯f (∂K)
Since f^{−1}∘f equals the identity I on ∂K, it follows from the properties of the degree
that
( --- )
1 = d (I,K, y) = d f− 1 ∘¯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 ¯ ) ∑ (¯ ) (-−1 )
1 = d f ∘f,K,y = d f,K, H d f ,H,y , (21.6.9)
H∈ ℋ
(21.6.9)
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,
(--- )
d f−1,H,y = d(f−1,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
(−1 )
f ,H, y
= 0. Thus in 21.6.9, 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
∑ (--- ) ∑ ∑ (--- )
d(¯f,K,H )d f−1,H, y = d (¯f,K, H) d f−1,L, y
H∈ ℋ1 H∈ℋ1 L∈ℒH
∑ ∑ (¯ ) (-−1 )
= d f,K,H d f ,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,
∑ ∑ d(¯f,K,H )d (f−1,L,y) = ∑ ∑ d(¯f,K,L )d(f−1,L,y)
H∈ℋ L∈ℒ H∈ℋ L∈ℒ
1 H 1 H
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 (21.6.10)
L∈ℒ L∈ℒ
(21.6.10)
Let
|K |
denote the number of components in K and similarly,
|ℒ|
denotes the number of
components in ℒ. Thus
∑ ∑ ∑ (¯ ) ( −-1 )
|K| = 1 = d f,K, L d f ,L,K
K∈K K∈K L∈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
∑ ∑ (¯ ) (-−1 )
|K| = d f,K,L d f ,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 21.6.5Let 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)
and for 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
1 = d (id,Ω,g(y)) = d (g ∘f,Ω,g (y))
= d (g,B, g(y))d(f,Ω,B) +d (f,Ω,U )d(g,U,g (y ))
= d (g,B, g(y))d(f,Ω,B)
The reduction happens because d
(f,Ω,U )
= 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.