774 CHAPTER 23. DEGREE THEORY

Since ∂Ki ⊆ f(∂Ω), both conditions are obtained by letting

∥g− g̃∥∞,f(Ω) < dist(y,g(f(∂Ω)))

By Lemma 23.1.5, there exists g̃ such that y is a regular value of g̃ in addition to 23.5.12and g̃−1 (y)∩ f(∂Ω) = /0. Then g̃−1 (y) is contained in the union of the Ki along with theunbounded component(s) and by Lemma 23.1.5 g̃−1 (y) is countable. As discussed there,

g̃−1 (y)∩Ki is finite if Ki is bounded. Let g̃−1 (y)∩Ki ={

xij

}mi

j=1,mi ≤ ∞. mi could only

be ∞ on the unbounded component.Now use Lemma 23.1.5 again to get f̃ in C∞

(Ω;Rp

)such that each xi

j is a regular valueof f̃ on Ω and also

∥∥f̃− f∥∥

∞is very small, so small that

d(g̃◦ f̃,Ω,y

)= d (g̃◦ f,Ω,y) = d (g◦ f,Ω,y)

andd(f̃,Ω,xi

j)= d

(f,Ω,xi

j)

for each i, j.Thus, from the above,

d (g◦ f,Ω,y) = d(g̃◦ f̃,Ω,y

)d(f̃,Ω,xi

j)

= d(f,Ω,xi

j)= d (f,Ω,Ki)

d (g̃,Ki,y) = d (g,Ki,y)

Is y a regular value for g̃◦ f̃ on Ω? Suppose z ∈Ω and y = g̃◦ f̃(z) so f̃(z) ∈ g̃−1 (y) . Thenf̃(z) = xi

j for some i, j and Df̃(z)−1 exists. Hence D(g̃◦ f̃

)(z) = Dg̃

(xi

j

)Df̃(z) , both

linear transformations invertible. Thus y is a regular value of g̃◦ f̃ on Ω.What of xi

j in Ki where Ki is unbounded? Then as observed above, the sum of the terms

sgn(detDf̃(z)

)for z ∈ f̃−1

(xi

j

)is d

(f̃,Ω,xi

j

)and is 0 because the degree is constant on

Ki which is unbounded.From the definition of the degree, the left side of 23.5.11 d (g◦ f,Ω,y) equals

∑{

sgn(detDg̃

(f̃(z)

))sgn(detDf̃(z)

): z ∈ f̃−1 (g̃−1 (y)

)}The g̃−1 (y) are the xi

j. Thus the above is of the form

= ∑i

∑j

∑z∈f̃−1

(xi

j

)sgn(det(Dg̃(xi

j)))

sgn(det(Df̃(z)

))

As mentioned, if xij ∈ Ki an unbounded component, then

∑z∈f̃−1

(xi

j

)sgn(det(Dg̃(xi

j)))

sgn(det(Df̃(z)

))= 0