764 CHAPTER 23. DEGREE THEORY

3. d (I,Ω,y) = 1 if y ∈Ω.

4. d (f,Ω, ·) is continuous and constant on every connected component of Rp \ f(∂Ω).

5. d (g,Ω,y) = d (f,Ω,y) if g|∂Ω

= f|∂Ω

.

6. If y /∈ f(∂Ω), and if d (f,Ω,y) ̸= 0, then there exists x ∈Ω such that f(x) = y.

Proof: That the degree is well defined follows from Lemma 23.1.12.Consider 1., the first property about homotopy. This follows from Theorem 23.2.1

applied to H (x, t)≡ h(x, t)−y(t).Consider 2. where y /∈ f

(Ω\ (Ω1∪Ω2)

). Note that

dist(y, f(Ω\ (Ω1∪Ω2)

))≤ dist(y, f(∂Ω))

Then let g be in C(Ω;Rp

)and

∥g− f∥∞

< dist(y, f(Ω\ (Ω1∪Ω2)

))≤ min(dist(y, f(∂Ω1)) ,dist(y, f(∂Ω2)) ,dist(y, f(∂Ω)))

where y is a regular value of g. Then by definition,

d (f,Ω,y)≡∑{

det(Dg(x)) : x ∈ g−1 (y)}

= ∑{

det(Dg(x)) : x ∈ g−1 (y) ,x ∈Ω1}

+∑{

det(Dg(x)) : x ∈ g−1 (y) ,x ∈Ω2}

≡ d (f,Ω1,y)+d (f,Ω2,y)

It is of course obvious that this can be extended by induction to any finite number of disjointopen sets Ωi.

Note that 3. is obvious because I (x) = x and so if y∈Ω, then I−1 (y) = y and DI (x) = Ifor any x so the definition gives 3.

Now consider 4. Let U be a connected component of Rp \ f(∂Ω) . This is open as wellas connected and arc wise connected by Theorem 7.13.10. Hence, if u,v ∈U, there is acontinuous function y(t) which is in U such that y(0) = u and y(1) = v. By homotopyinvariance, it follows d (f,Ω,y(t)) is constant. Thus d (f,Ω,u) = d (f,Ω,v).

Next consider 5. When f = g on ∂Ω, it follows that if y /∈ f(∂Ω) , then y /∈ f(x) +t (g(x)− f(x)) for t ∈ [0,1] and x ∈ ∂Ω so

d (f+ t (g− f) ,Ω,y)

is constant for t ∈ [0,1] by homotopy invariance in part 1. Therefore, let t = 0 and thent = 1 to obtain 5.

Claim 6. follows from Lemma 23.1.12 which says that if y /∈ f(Ω), then d (f,Ω,y) = 0.

From the above, there is an easy corollary which gives related properties of the degree.