322 CHAPTER 13. DEGREE THEORY

Proof: By the Jordan separation theorem, Corollary 13.6.6, Rp \ f (∂Ω) consists oftwo components, a bounded component B and an unbounded component U . Using theTietze extention theorem, there exists g defined on Rp such that g = f−1 on f

(Ω). Thus

on ∂Ω,g ◦f = id. It follows from this and the product formula that

1 = d (id,Ω,g (y)) = d (g ◦f ,Ω,g (y)) = d (g,B,g (y))d (f ,Ω,B)

Therefore, d (f ,Ω,B) ̸= 0 and so for every z ∈ B, it follows z ∈ f (Ω) . Thus B ⊆ f (Ω) .On the other hand, f (Ω) cannot have points in both U and B because it is a connected set.Therefore f (Ω) ⊆ B and this shows B = f (Ω). Thus d (f ,Ω,B) = d (f ,Ω,y) for eachy ∈ B and the above formula shows this equals either 1 or −1 because the degree is aninteger. ■

The one dimensional case also fits into this although it is easier to do by more elemen-tary means. In the case where n = 1, the argument is essentially the same. There is one andonly one bounded component for R\ f ({a,b}) . This shows how to generalize orientation.It is just the degree. One could use this to describe an orientable manifold without anydirect reference to differentiability.

In the case of f(Sp−1

)one wants to verify that this is the is the boundary of both

components, the bounded one and the unbounded one.

Theorem 13.6.8 Let Sp−1 be the unit sphere in Rp, p≥ 2. Suppose γ : Sp−1→ Γ⊆Rp is one to one onto and continuous. Then Rp \Γ consists of two components, a boundedcomponent (called the inside) Ui and an unbounded component (called the outside), Uo.Also the boundary of each of these two components of Rp \Γ is Γ and Γ has empty interior.

Proof: γ−1 is continuous since Sp−1 is compact and γ is one to one. By the Jordanseparation theorem, Rp \Γ =Uo∪Ui where these on the right are the connected compo-nents of the set on the left, both open sets. Only Ui is bounded. Thus Γ∪Ui ∪Uo = Rp.Since both Ui,Uo are open, ∂U ≡U \U for U either Uo or Ui. If x ∈ Γ, and is not a limitpoint of Ui, then there is B(x,r) which contains no points of Ui. Let S be those points x ofΓ for which, B(x,r) contains no points of Ui for some r > 0. This S is open in Γ. Let Γ̂ beΓ \ S. Then if Ĉ = γ−1

(Γ̂), it follows that Ĉ is a closed set in Sp−1and is a proper subset

of Sp−1. It is obvious that taking a relatively open set from Sp−1 results in a compact setwhose complement in Rp is an open connected set. By Proposition 13.6.4, Rp \ Γ̂ is also anopen connected set. Start with x ∈Ui and consider a continuous curve which goes from xto y ∈Uo which is contained in Rp \ Γ̂ . Thus the curve contains no points of Γ̂. However,it must contain points of Γ which can only be in S. The first point of Γ intersected by thiscurve is a point in Ui and so this point of intersection is not in S after all because every ballcontaining it must contain points of Ui. Thus S = /0 and every point of Γ is in Ui. Similarly,every point of Γ is in Uo. Thus Γ ⊆Ui \Ui and Γ ⊆Uo \Uo. However, if x ∈Ui \Ui, thenx /∈Uo because it is a limit point of Ui and so x ∈ Γ. It is similar with Uo. Thus Γ =Ui \Uiand Γ = Uo \Uo. This could not happen if Γ had an interior point. Such a point would bein Γ but would fail to be in either ∂Ui or ∂Uo. ■

When p = 2, this theorem is called the Jordan curve theorem.What if γ maps B̄ toRp instead of γ only being defined on Sp−1? Obviously, one should

be able to say a little more.

Corollary 13.6.9 Let B be an open ball and let γ : B̄→ Rp be one to one and contin-uous. Let Ui,Uo be as in the above theorem, the bounded and unbounded components ofγ (∂B)C. Then Ui = γ (B).