496 APPENDIX A. HOMOLOGICAL METHODS∗

Corollary A.13.4 Let Ω ⊆ Rn, n ≥ 2 be a bounded open connected set such that ∂ΩC

has two components, a bounded and an unbounded component. Suppose f : ∂Ω→ C ≡f (∂Ω)⊆Rn is a homeomorphism. Then CC also has exactly two components, one boundedand one unbounded.

As an application, here is a very interesting result about orientation and the invarianceof domain theorem which says that a one to one continuous function maps open sets toopen sets.

Proposition A.13.5 Let Ω be an open connected bounded set in Rn,n ≥ 2 such that Rn \∂Ω consists of two connected components. Let f ∈ C

(Ω;Rn

)be continuous and one to

one. Then f (Ω) is the bounded component of Rn \ f (∂Ω) and for y ∈ f (Ω) , d ( f ,Ω,y)either equals 1 or −1.

Proof: By the Jordan separation theorem, Corollary A.13.4, Rn \ f (∂Ω) consists oftwo components, a bounded component B and an unbounded component U . Using theTietze extention theorem, there exists g defined on Rn 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 each y∈ Band the above formula shows this equals either 1 or−1 because the degree is an integer. ■

This shows how to generalize orientation. It is just the degree. One could use this todescribe an orientable manifold without any direct reference to differentiability.

In the case of f(Sn−1

)for f one to one and continuous, one wants to verify that this is

the boundary of both components, the bounded one and the unbounded one.

Theorem A.13.6 Let Sn−1 be the unit sphere in Rn,n ≥ 2. Suppose γ : Sn−1 → Γ ⊆ Rn

is one to one onto and continuous. Then Rn \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 Rn \Γ is Γ and Γ has empty interior.

Proof: γ−1 is continuous since Sn−1 is compact and γ is one to one. By the Jordanseparation theorem, Rn \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 = Rn.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 Sn−1and is a proper subset

of Sn−1. It is obvious that taking a relatively open set from Sn−1 results in a compact setwhose complement in Rn is an open connected set. By Proposition A.13.1, Rn \ Γ̂ is alsoan open connected set. Start with x ∈Ui and consider a continuous curve which goes fromx to y ∈Uo which is contained in Rn \ 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, then