= 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}^{∞}
p p
(ℝ ,ℝ )
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 11.1.4 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 11.1.4 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
( ) ( )
d ˜f,Ω,xij = d f,Ω,xij
d(g,K ,y) = d(˜g,K ,y)
i i
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
∑ ∑ ∑ ( ( ( i))) ( ( ))
= sgn det D ˜g xj sgn det D ˜f (z)
i j z∈˜f−1(xij)
∑ ∑ ( ( ( i))) ∑ ( ( ˜ ))
= sgn det D ˜g xj 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 i
(xj)
}sgn
( ( ))
det D ˜f (z)
≡ d
( )
˜f,Ω, xi
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}
n n
(ℝ ,ℝ )
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 11.6.3Let Ω be a bounded open set in ℝ^{p}, f ∈ C
(Ω;ℝp)
, and suppose
{Ωi}
_{i=1}^{∞}are disjointopen 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 11.3.4 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.
|--------------------------p------------|
|----bounded-components of ℝ-∖-f (∂K-)---|
|------------------ℋ--------------------|
|-----ℒH--sets of ℒ-contained-in H-∈ ℋ----|
|ℋ1 those sets of ℋ which intersect a set of ℒ|
----------------------------------------
Theorem 11.6.4(Jordan separation theorem)Let f be a homeomorphism of C and f
(C )
where Cis a compact set in ℝ^{p}. Then ℝ^{p}∖C and ℝ^{p}∖f
(C )
have the same 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 ¯ )
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 , (11.8)
H ∈ℋ
(11.8)
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
(f−1,H, y)
= 0. Thus in 11.8, 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,
∑ ∑ (¯ ) ( -−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 = d ¯f,K, L d f−1,L,y = d ¯f,K,L d f−1,L,K (11.9)
L∈ℒ L∈ℒ
(11.9)
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 ∈KL∈ℒ
Similarly,
∑ ∑ ∑ (¯ ) (-−1 )
|ℒ| = 1 = d f,K, L d f ,L,K
L∈ℒ L∈ℒ K∈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 11.6.5Let Ω be an open connected bounded set in ℝ^{p},n ≥ 1 such that ℝ^{p}∖∂Ω consistsof two, three if n = 1, connected components. Let f ∈ C
(Ω;ℝp )
be continuous and one to one. Thenf
(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.