13.6. PRODUCT FORMULA, SEPARATION THEOREM 319

By Lemma 13.1.5, there exists g̃ such that y is a regular value of g̃ in addition to 13.9and 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 13.1.5 g̃−1 (y) is countable. As discussed there,

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

j

}mi

j=1,mi ≤ ∞. mi could only

be ∞ on the unbounded component.Now use Lemma 13.1.5 again to get f̃ in C∞

(Ω;Rp

)such that each xi

j is a regular

value of 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)

and d(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) .

Then f̃ (z) = xij 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? As observed, the sum of 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 13.8 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 xij. 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

and so, it suffices to only consider bounded components in what follows and the sum makessense because there are finitely many xi

j in bounded Ki. This also shows that if there are