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.

764 CHAPTER 23. DEGREE THEORY3. d(I,Q,y) =1ify€o.4, d(f,Q,-) is continuous and constant on every connected component of R? \ f (dQ).5. d(g,Q,y) = d(f,Q,y) if Sla0 = E50:6. Ify E£ (dQ), and if d (f,Q,y) 4 0, then there exists x € Q 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 23.2.1applied to H (x,t) =h(x,t) —y(r).Consider 2. where y ¢ f (Q\ (Q; UQ2)) . Note thatdist (y, f (Q\ (Q; UQ2))) < dist (y, f(AQ))Then let g be in C (Q;R?) andIg—f\|., < dist (y,f(Q\ (Q;UQ2)))< min (dist (y,f(AQ,)) , dist (y,f(AQ2)) , dist (y, f (AQ)))where y is a regular value of g. Then by definition,d(f,Q,y) = Y" {det (Dg (x)):xeg '(y)}= PY {det (Dg (x) :x eg"! (y),xe Qi}+)" {det (Dg (x)) :x Eg! (y),x € Qo}= d(f,Q1,y)+d(f,Q2,y)It is of course obvious that this can be extended by induction to any finite number of disjointopen sets Q;.Note that 3. is obvious because / (x) =x and so if y € Q, then J~! (y) =y and DI (x) =/for any x so the definition gives 3.Now consider 4. Let U be a connected component of R? \ f (dQ). 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,Q,y (t)) is constant. Thus d (f,Q,u) = d (f,Q,v).Next consider 5. When f = g on 0Q, it follows that if y ¢ f(0Q), then y ¢ f(x) +t (g(x) —f(x)) fort € [0,1] and x € dQ sod(f+t(g—f),Q,y)is constant for t € [0,1] by homotopy invariance in part 1. Therefore, let t = 0 and thent = | to obtain 5. _Claim 6. follows from Lemma 23.1.12 which says that if y ¢ f (Q) , then d (f,Q,y) =0.|From the above, there is an easy corollary which gives related properties of the degree.