In this section is an important theorem which can be used to verify that d
(f,Ω,y)
≠0. This is
significant because when this is known, it follows from Theorem 16.3.4 that f^{−1}
(y)
≠∅. In
other words there exists x ∈ Ω such that f
(x)
= y.
Definition 16.4.1A bounded open set, Ω is symmetric if −Ω = Ω. Acontinuous function, f : Ω→ ℝ^{p}is odd if f
(− x )
= −f
(x)
.
Suppose Ω is symmetric and g ∈ C^{∞}
( -- p)
Ω;ℝ
is an odd map for which 0 is a regular
value. Then the chain rule implies Dg
(− x )
= Dg
(x)
and so d
(g,Ω,0 )
must equal an odd
integer because if x ∈ g^{−1}
(0)
, it follows that −x∈ g^{−1}
(0)
also and since Dg
(− x)
= Dg
(x )
,
it follows the overall contribution to the degree from x and −x must be an even integer. Also
0 ∈ g^{−1}
(0)
and so the degree equals an even integer added to sgn
(det Dg (0))
, an
odd integer, either −1 or 1. It seems reasonable to expect that something like this
would hold for an arbitrary continuous odd function defined on symmetric Ω. In fact
this is the case and this is next. The following lemma is the key result used. This
approach is due to Gromes [14]. See also Deimling [6] which is where I found this
argument.
The idea is to start with a smooth odd map and approximate it with a smooth odd map
which also has 0 a regular value.
Lemma 16.4.2Let g ∈ C^{∞}
( -- p)
Ω;ℝ
be an odd map. Then for every ε > 0, thereexists h ∈ C^{∞}
( -- p)
Ω;ℝ
such that h is also an odd map,
||h− g||
_{∞}< ε, and 0 is aregular value ofh.Here Ω is a symmetric bounded open set. In addition, d
(g,Ω, 0)
isan odd integer.
Proof:In this argument η > 0 will be a small positive number. Let h_{0}
(x)
= g
(x)
+ ηx
where η is chosen such that detDh_{0}
(0)
≠0. Note that h_{0} is odd and 0 is a value of h_{0} thanks
to h_{0}
(0)
= 0. This has taken care of 0. Now we need to take care of the other points.
Let
∑p ∑0
h (x) ≡ h0(x)− yjx3j, ≡ 0.
j=1 j=1
The y^{j} will each be small. Note that h
(0)
= 0 and det
(Dh (0))
= det
(Dh0 (0))
≠0. We want
to choose small y^{j},
∥∥yj∥∥
< η in such a way that 0 is a regular value for h for each x≠0 such
that x ∈ h^{−1}
(0)
. Let
Ωi ≡ {x ∈ Ω : xi ⁄= 0}, so ∪pj=1 Ωj = {x ∈ ℝp : x ⁄= 0} .
Let Ω_{0}≡
{0}
. Then the theorem will be proved if for each k ≤ p, there exist y^{i},
∥∥yi∥∥
< η and
0 is a regular value of
∑k j 3 k
hk (x) ≡ h0 (x) − y xj for x ∈ ∪i=0Ωi
j=1
For k = 0 this is done. Suppose then that this holds for k − 1, k ≥ 1. Thus 0 is a regular value
for
k−1
h (x) ≡ h (x) −∑ yjx3on ∪k−1 Ω
k−1 0 j=1 j i=0 i
What should y^{k} be? Keeping y^{1},
⋅⋅⋅
,y^{k−1}, it follows that if x_{k} = 0, h_{k}
(x )
= h_{k−1}
(x)
so for
x_{k}∈ Ω_{k},
k
h (x) ≡ h (x)− ∑ yjx3 = 0
k 0 j=1 j
if and only if
∑k −1 j 3
h0-(x-)−---j=1-y-xj = yk
x3k
So let y^{k} be a regular value of h_{k−1}
(x)
≡
h0(x)−-∑kj−=11 yjx3j
x3k
on Ω_{k} and also
∥∥ k∥∥
y
< η. Then for
such x, and using the quotient rule,
Ω_{k}, then x_{k} = 0 and so, because of the
power of 3 in x_{k}^{3}, Dh_{k}
(x)
= Dh_{k−1}
(x)
which has nonzero determinant by induction.
Thus for each k ≤ n, there is a function h_{k}of the form described above, such that
for x ∈∪_{j=1}^{k}Ω_{k}, with h_{k}
(x)
= 0, it follows that det
(Dhk (x))
≠0. Let h ≡ h_{n}.
Then
-- p
∥h − g∥∞,Ω ≤ η (diam (Ω)+ n diam (Ω) ) < ε < dist(g(∂Ω ),0)
provided η was chosen sufficiently small to begin with.
So what is d
(h, Ω,0)
? Since 0 is a regular value and h is odd,
h−1 (0) = {x1,⋅⋅⋅,xr,− x1,⋅⋅⋅,− xr,0}.
So consider Dh
(x)
and Dh
(− x)
.
Dh (− x)u + o(u) = h (− x + u)− h(− x)
= − h(x+ (− u))+ h (x)
= − (Dh (x)(− u))+ o (− u)
= Dh (x)(u) +o (u)
Hence Dh
(x)
= Dh
(− x)
and so the determinants of these two are the same. It follows that
∑r ∑r
d (h,Ω, 0) = sgn(det(Dh (xi)))+ sgn(det(Dh (− xi)))
i=1 i=1
+ sgn (det(Dh (0)))
= 2m ± 1 some integer m
an odd integer. However, d
(g,Ω,0)
= d
(h,Ω, 0)
because for x ∈ ∂Ω,t ∈
[0,1]
∥g(x)+ t(h(x) − g (x))∥ ≥ ∥g(x)− 0∥ − ε > 0 ■
Theorem 16.4.3(Borsuk) Let f ∈ C
(-- )
Ω; ℝp
be odd and let Ω besymmetricwith0
∕∈
f
(∂Ω )
. Then d
(f,Ω, 0)
equals an odd integer.
Proof:Let ψ_{n} be a mollifier which is symmetric, ψ
(− x)
= ψ
(x)
. Also recall that f is the
restriction to Ω of a continuous function, still denoted as f which is defined on all of ℝ^{p}. Let g
be the odd part of this function. That is,
1
g (x) ≡ 2 (f (x)− f (− x))
Since f is odd, g = f on Ω. Then
∫
gn(− x) ≡ g∗ ψn(− x ) = Ωg (− x − y)ψn(y)dy
∫ ∫
= − g (x + y)ψ (y) dy = − g (x− (− y))ψ (− y)dy = − g (x)
Ω n Ω n n
Thus g_{n} is odd and is infinitely differentiable. Let 3δ = dist