23.8. A FUNCTION WITH VALUES IN SMALLER DIMENSIONS 785

From 5. above, this reduces to

1 = d (A,B(0,r) ,AB(0,r))d(A−1,AB(0,r) ,0

)Thus, since the degree is an integer, either both d (A,B(0,r) ,0) ,d

(A−1,AB(0,r) ,0

)are 1

or they are both −1. This would hold if when A is invertible,

d (A,B(0,r) ,0) = sgn(det(A)) = sgn(det(A−1)) .

It would also hold if d (A,B(0,r) ,0) = −sgn(det(A)) . However, the latter of the two al-ternatives is not the one wanted because, doing something like the above,

d(A2,B(0,r) ,0

)= d (A,B(0,r) ,AB(0,r))d (A,AB(0,r) ,0)= d (A,B(0,r) ,0)d (A,AB(0,r) ,0)

If you had for a definition d (A,B(0,r) ,0) =−sgndet(A) , then you would have

−sgndet(A2)=−1 = (−sgndet(A))2 = 1

Hence the only reasonable definition is to let d (A,B(0,r) ,0) = sgn(det(A)). It followsfrom 23.7.15 that

d (g,Ω,y) =n

∑i=1

d (g,Bir,y) =n

∑i=1

sgn(det(Dg(xi))) .

Thus, if the above conditions hold, then in whatever manner you construct the degree, itamounts to the definition given above in this chapter in the sense that for y a regular pointof a smooth function, you get the definition of the chapter.

23.8 A Function With Values In Smaller DimensionsRecall that we have the degree defined d ( f ,Ω,y) for continuous functions on Ω̄ and y /∈f (∂Ω). It had properties as follows.

1. d (id,Ω,y) = 1 if y ∈Ω.

2. If Ωi ⊆Ω,Ωi open, and Ω1∩Ω2 = /0 and if y /∈ f(Ω\ (Ω1∪Ω2)

), then d (f,Ω1,y)+

d (f,Ω2,y) = d (f,Ω,y).

3. If y /∈ f(Ω\Ω1

)and Ω1 is an open subset of Ω, then

d (f,Ω,y) = d (f,Ω1,y) .

4. For y ∈ Rn \ f(∂Ω) , if d (f,Ω,y) ̸= 0 then f−1 (y)∩Ω ̸= /0.

5. If t→ y(t) is continuous h : Ω̄× [0,1]→Rn is continuous and if y(t) /∈ h(∂Ω, t) forall t, then t→ d (h(·, t) ,Ω,y(t)) is constant.