786 CHAPTER 23. DEGREE THEORY

6. d (·,Ω,y) is defined and constant on{g ∈C

(Ω;Rn) : ||g− f||

∞< r}

where r = dist(y, f(∂Ω)).

7. d (f,Ω, ·) is constant on every connected component of Rn \ f(∂Ω).

8. d (g,Ω,y) = d (f,Ω,y) if g|∂Ω

= f|∂Ω

.

Theorem 23.8.1 Let Ω be a bounded open set in Rn and let f ∈ C(Ω̄;Rn

m)

where Rnm =

{x ∈ Rn : xk = 0 for k > m} . Thus x concludes with a column of n−m zeros. Let y ∈ Rnm \

(id−f)(∂Ω) . Then d (id−f,Ω,y) = d((id−f) |

Ω∩Rnm,Ω∩Rn

m,y).

Proof: To save space, let g = id−f. Then there is no loss of generality in assuming atthe outset that y is a regular value for g. Indeed, everything above was reduced to this case.Then for x ∈ g−1 (y) and letting xm be the first m variables for x,

Dg(x) =(

Dxmg(x) ∗0 In−m

)Then it follows that

0 ̸= det(Dg(x)) = det(

Dxmg(x) ∗0 In−m

)= det

(Dxmg(x) 0

0 In−m

)= det(Dxmg(x))

This last is just the determinant of the derivative of the function which results from restrict-ing g to the first m variables. Now y ∈ Rn

m and f also is given to have values in Rnm so

if g(x) = y, then you have x− f(x) = y which requires x ∈ Rnm also. Therefore, g−1 (y)

consists of points in Rnm only. Thus, y is also a regular value of the function which results

from restricting g to Rnm∩Ω.

d (id−f,Ω,y) = d (g,Ω,y)

= ∑x∈g−1(y)

sign(det(Dg(x)))

= ∑x∈g−1(y)

sign(det(Dxmg(x)))≡ d((id−f) |

Ω∩Rnm,Ω∩Rn

m,y)

Recall that for g ∈C2(Ω̄;Rn

),

d (g,Ω,y)≡ limε→0

∫Ω

φ ε (g(x)−y)detDg(x)dx

In fact, it can be shown that the degree is unique based on its Properties, 1,2,5 above. Itinvolves reducing to linear maps and then some complicated arguments involving linear