If Ω_{i}⊆ Ω,Ω_{i} open, and Ω_{1}∩ Ω_{2} = ∅ and if y
∈∕
f
(-- )
Ω ∖(Ω1 ∪ Ω2)
, then
d
(f,Ω1,y)
+ d
(f,Ω2,y)
= d
(f,Ω,y )
.
If y
∕∈
f
(-- )
Ω ∖Ω1
and Ω_{1} is an open subset of Ω, then
d(f,Ω,y) = d (f,Ω1,y).
For y ∈ ℝ^{n}∖ f
(∂Ω )
, if d
(f,Ω, y)
≠0 then f^{−1}
(y)
∩ Ω≠∅.
If t → y
(t)
is continuous h : Ω×
[0,1]
→ ℝ^{n} is continuous and if y
(t)
∕∈
h
(∂Ω,t)
for all t, then t → d
(h (⋅,t),Ω,y(t))
is constant.
d
(⋅,Ω, y)
is defined and constant on
{ (-- ) }
g ∈ C Ω;ℝn : ||g− f||∞ < r
where r = dist
(y,f (∂Ω))
.
d
(f,Ω,⋅)
is constant on every connected component of ℝ^{n}∖ f
(∂Ω )
.
d
(g,Ω,y)
= d
(f,Ω,y )
if g|_{∂Ω} = f|_{∂Ω}.
Theorem 21.8.1Let Ω be a bounded open set in ℝ^{n}and let f ∈ C
( )
¯Ω;ℝnm
where ℝ_{m}^{n} =
{x ∈ ℝn : xk = 0 for k > m}
. Thus x concludes with a columnof n − m zeros. Let y ∈ ℝ_{m}^{n}∖
(id− f)
(∂Ω)
. Then d
(id− f,Ω, y)
=
d
( )
(id− f)|--n-,Ω∩ ℝn ,y
Ω∩ℝm m
.
Proof:To save space, let g = id−f. Then there is no loss of generality in assuming
at the outset that y is a regular value for g. Indeed, everything above was reduced
to this case. Then for x ∈ g^{−1}
(y)
and letting x_{m} be the first m variables for
x,
( D g(x) ∗ )
Dg (x) = xm0 I
n−m
Then it follows that
( )
0 ⁄= det(Dg (x )) = det Dxmg (x) ∗
0 In−m
( Dx g (x) 0 )
= det m0 In−m = det(Dxmg (x))
This last is just the determinant of the derivative of the function which results from
restricting g to the first m variables. Now y ∈ ℝ_{m}^{n} and f also is given to have values in
ℝ_{m}^{n} so if g
(x)
= y, then you have x − f
(x)
= y which requires x ∈ ℝ_{m}^{n} also.
Therefore, g^{−1}
(y)
consists of points in ℝ_{m}^{n} only. Thus, y is also a regular value of the
function which results from restricting g to ℝ_{m}^{n}∩ Ω.
d(id− f,Ω,y ) = d(g,Ω,y)
= ∑ sign(det(Dg (x)))
x∈g−1(y)
∑ ( )
= sign(det(Dxmg (x ))) ≡ d (id − f)|Ω-∩ℝnm,Ω ∩ℝnm,y ■
x∈g−1(y)
Recall that for g ∈ C^{2}
(¯Ω;ℝn )
,
∫
d(g,Ω,y) ≡ lim ϕ (g (x) − y)detDg (x)dx
ε→0 Ω ε
In fact, it can be shown that the degree is unique based on its Properties, 1,2,5 above. It
involves reducing to linear maps and then some complicated arguments involving linear
algebra. It is done in [?]. Here we will be a little less ambitious. The following
lemma will be useful when extending the degree to finite dimensional normed
linear spaces and from there to Banach spaces. It is motivated by the following
diagram.
−1
θ−1(y) θ← y
↑ θ−1 ∘g∘ θ ↑ g
θ−1(Ω) θ→ Ω
Lemma 21.8.2Lety
∕∈
g
(∂ Ω)
and let θ be an isomorphism of ℝ^{n}. That is, θ is one toone onto and linear. Then
( −1 − 1 −1 )
d θ ∘ g∘ θ,θ (Ω ),θ y = d (g,Ω, y)
Proof:It suffices to consider g ∈ C^{2}
(¯ n)
Ω; ℝ
for which y is a regular value because
you can get such a ĝ with
∥ˆg− g∥
_{∞}< δ where
B (y,2δ)∩ g (∂Ω ) = ∅.
Thus B
(y,δ)
∩ĝ
(∂Ω)
= ∅ and so d
(g,Ω,y)
= d
(ˆg,Ω,y)
. One can assume similarly that
∥ˆg− g∥
_{∞} is sufficiently small that
d(θ−1g ∘θ,θ−1(Ω),θ−1y) = d(θ−1ˆg∘ θ,θ− 1(Ω),θ−1y)
because both θ and θ^{−1} are continuous. Thus it suffices to consider at the outset
g ∈ C^{2}
(¯Ω;ℝn )
. Then from the definition of degree for C^{2} maps,
What about functions which have values in finite dimensional vector spaces?
Theorem 21.8.3Let Ω be an open bounded set in V a real normed ndimensional vector space. Then there exists a topological degree d
(f,Ω,y)
forf ∈ C
(Ω¯, V)
,y
∕∈
f
(∂Ω )
which satisfies all the properties of the degree for functionshaving values in ℝ^{n}described above,
d
(id,Ω,y)
= 1 if y ∈ Ω.
If Ω_{i}⊆ Ω,Ω_{i}open, and Ω_{1}∩ Ω_{2} = ∅ and if y
∕∈
f
(Ω-∖(Ω ∪ Ω ))
1 2
, thend
(f,Ω ,y)
1
+ d
(f,Ω ,y)
2
= d
(f,Ω, y)
.
If y
∕∈
f
(-- )
Ω ∖Ω1
and Ω_{1}is an open subset of Ω, then
d(f,Ω,y) = d (f,Ω1,y).
For y ∈ ℝ^{n}∖ f
(∂Ω )
, if d
(f,Ω,y)
≠0 then f^{−1}
(y)
∩ Ω≠∅.
If t → y
(t)
is continuous h : Ω×
[0,1]
→ ℝ^{n}is continuous and if y
(t)
∕∈
h
(∂Ω, t)
for all t, then t → d
(h(⋅,t),Ω,y (t))
is constant.
d
(⋅,Ω, y)
is defined and constant on
{ (- ) }
g ∈ C Ω; ℝn : ||g− f||∞ < r
where r = dist
(y,f (∂ Ω))
.
d
(f,Ω,⋅)
is constant on every connected component of ℝ^{n}∖ f
(∂Ω )
.
d
(g,Ω,y)
= d
(f,Ω,y)
if g|_{∂Ω} = f|_{∂Ω}.
Proof:There is an isomorphism θ : ℝ^{n}→ V which also preserves all topological
properties. This follows from the properties of finite dimensional vector spaces. In fact,
every algebraic isomorphism is automatically a homeomorphism preserving all
topological properties. Then it is pretty easy to see what the degree should
be.
d(f,Ω,y) ≡ d (θ− 1 ∘f ∘θ,θ−1(Ω),θ−1y)
Then by standard material on finite dimensional vector spaces, the norm on V is
equivalent to the norm defined by
|v|
≡
||θ−1v||
_{ℝn}. Hence all of those properties hold.
By Lemma 21.8.2 this definition does not depend on the particular isomorphism used. If
ˆθ
is another one, then one would need to verify that
( )
d(θ−1 ∘f ∘ θ,θ−1(Ω),θ−1y) = d ˆθ−1 ∘ f ∘ ˆθ,ˆθ−1 (Ω ),ˆθ−1y
However, you could use that lemma to conclude that
( −1 −1 −1 )
d ˆθ ∘ f ∘ˆθ,ˆθ (Ω ),ˆθ y
( −1 −1 −1 −1 −1 −1 )
= d α ∘ ˆθ ∘ f ∘ ˆθ∘α,α ˆθ (Ω),α ˆθ y
where α is such that
ˆθ
∘ α = θ. Then this verifies the appropriate equation.
■
Next one considers what happens when the function I − f has values in a smaller
dimensional subspace.
Theorem 21.8.4Let Ω be a bounded open set in V an n dimensional normedlinear space and let f ∈ C
( )
¯Ω;Vm
where V_{m}is an m dimensional subspace. Lety ∈ V_{m}∖