Note that finitely many of these sets Q cover ∂Ω because ∂Ω is compact. The following
picture describes the situation.
PICT
Define P1 : ℝn→ ℝn−1 by
P1u ≡ (u2,⋅⋅⋅,un)
and Σ : ℝn→ ℝn given by
Σu ≡ u− g (P u)e
1 1
≡ u− g (u2,⋅⋅⋅,un)e1
≡ (u1 − g(u2,⋅⋅⋅,un),u2,⋅⋅⋅,un)
Thus Σ is invertible and
Σ−1u = u+ g (P1u )e1
≡ (u1 + g(u2,⋅⋅⋅,un),u2,⋅⋅⋅,un)
For x ∈ ∂Ω ∩ Q, it follows the first component of Rx is g
(P1 (Rx))
. Now define
R :W → ℝ≤n as
u ≡ Rx ≡ Rx − g(P1 (Rx))e1 ≡ ΣRx
and so it follows
R −1 = R ∗Σ−1.
These mappings R involve first a rotation followed by a variable sheer in the direction of
the u1 axis.
Since ∂Ω is compact, there are finitely many of these open sets, Q1,
⋅⋅⋅
,Qp which
cover ∂Ω. Let the orthogonal transformations and other quantities described above also
be indexed by k for k = 1,
⋅⋅⋅
,p. Also let Q0 be an open set with Q0⊆ Ω and Ω is
covered by Q0,Q1,
⋅⋅⋅
,Qp. Let u ≡ R0x ≡ x − ke1 where k is large enough that
R0Q0⊆ ℝ<n. Thus in this case, the orthogonal transformation R0 equals I and
Σ0x ≡ x − ke1. I claim Ω is an oriented manifold with boundary and the charts are
(Wi, Ri)
.
Letting A be an open set contained in Ri
(Wi ∩ Wj )
such that ∂A has measure 0 and
∂A separates ℝn into two components, consider
d(u,A,Rj ∘R −i1),u ∕∈ Rj ∘R −i1(∂A)
By convolving g with a mollifier, there exists a sequence of infinitely differentiable
functions gε which converge uniformly to g on all of ℝn−1 as ε → 0. Therefore, letting Σε
be the corresponding functions defined above with g replaced with gε, it follows the Σε
will converge uniformly to Σ and Σε−1 will converge uniformly to Σ−1. Thus from the
above descriptions of Rj−1, it follows Rjε−1 converges uniformly to Rj−1 for each j.
Therefore, if ε is small enough, u
∈∕
(tR ∘R −1 +(1 − t)R ∘R −1)
jε iε j i
(∂A )
and so from
properties of the degree, the mappings Rj and Ri−1 can be replaced with smooth ones in
computing the degree. To save on notation, I will drop the ε. The mapping involved
is
ΣjRjR ∗Σ−1
i i
and it is a one to one mapping. What is the determinant of its derivative? By the chain
rule,
D (ΣjRjR ∗iΣ −i1) = D Σj(RjR ∗iΣ−i1)DRj (R∗iΣ−i1)DR ∗i (Σ−i1)D Σ−i1
However,
( )
det(DΣj ) = 1 = det D Σ−j1
and det
(Ri)
= det
(R ∗i)
= 1 by assumption. Therefore, if u ∈
(Rj ∘ R−i 1)
(A)
, the
above degree is 1 and if u is not in this set, the above degree is 0 or undefined if u is on
(Rj ∘ R−i 1)
(∂A)
. By Definition 28.1.5 Ω is indeed an oriented manifold.