30.5. GREEN’S THEOREM 1075
Define P1 : Rn→ Rn−1 byP1u≡ (u2, · · · ,un)
and Σ : Rn→ Rn given byΣu≡ u−g(P1u)e1
≡ u−g(u2, · · · ,un)e1
≡ (u1−g(u2, · · · ,un) ,u2, · · · ,un)
Thus Σ is invertible andΣ−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 →Rn≤
asu≡ Rx≡ Rx−g(P1 (Rx))e1 ≡ ΣRx
and so it followsR−1 = R∗Σ−1.
These mappings R involve first a rotation followed by a variable sheer in the direction ofthe u1 axis.
Since ∂Ω is compact, there are finitely many of these open sets, Q1, · · · ,Qp whichcover ∂Ω. Let the orthogonal transformations and other quantities described above also beindexed by k for k = 1, · · · , p. Also let Q0 be an open set with Q0 ⊆ Ω and Ω is coveredby Q0,Q1, · · · ,Qp. Let u≡ R0x≡ x− ke1 where k is large enough that R0Q0 ⊆ Rn
<. Thusin this case, the orthogonal transformation R0 equals I and Σ0x ≡ x− ke1. I claim Ω is anoriented 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 ∂Aseparates Rn into two components, consider
d(u,A,R j ◦R−1
i), u /∈ R j ◦R−1
i (∂A)
By convolving g with a mollifier, there exists a sequence of infinitely differentiable func-tions gε which converge uniformly to g on all of Rn−1 as ε → 0. Therefore, letting Σε bethe corresponding functions defined above with g replaced with gε , it follows the Σε willconverge uniformly to Σ and Σ−1
ε will converge uniformly to Σ−1. Thus from the abovedescriptions of R−1
j , it follows R−1jε converges uniformly to R−1
j for each j. Therefore,if ε is small enough, u /∈
(tR jε ◦R−1
iε +(1− t)R j ◦R−1i
)(∂A) and so from properties of
the degree, the mappings R j and R−1i can be replaced with smooth ones in computing the
degree. To save on notation, I will drop the ε . The mapping involved is
Σ jR jR∗i Σ−1i
and it is a one to one mapping. What is the determinant of its derivative? By the chain rule,
D(Σ jR jR∗i Σ
−1i)= DΣ j
(R jR∗i Σ
−1i)
DR j(R∗i Σ
−1i)
DR∗i(Σ−1i)
DΣ−1i