A.13. JORDAN SEPARATION THEOREM 497

x /∈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 n = 2, this theorem is called the Jordan curve theorem.

Corollary A.13.7 Let f : Rn→ Rn be one to one, continuous and let lim|x|→∞ | f (x)|= ∞.Then f is onto.

Proof: From the invariance of domain, Proposition A.13.5, f (Rn) is open. However, iff (xn)→ y, then {xn} must be bounded and so there is a convergent subsequence xnk → x.Therefore, by continuity of f it follows that f

(xnk

)→ y= f (x) and so f (Rn) is also closed.

But Rn = (Rn \ f (Rn))∪ f (Rn) , the union of disjoint open sets. Hence one must be emptysince otherwise Rn would not be connected. ■

Corollary A.13.8 If n > m then there is no continuous one to one function f which mapsRn onto Rm. Thus Rn,Rm are not homeomorphic.

Proof: If there were, then you could let θ : Rm → Rn be given by x→ (x,0) where0 = (0,0, ...,0) . and we would have θ ◦ f is one to one and continuous mapping Rn to Rn

but does not take open sets to open sets. ■

Theorem A.13.9 Let B be a ball in Rn,n≥ 1 and let continuous f : B̄→Rn be odd mean-ing that f (x) =− f (−x) . Suppose 0 /∈ f (∂B). Then d ( f ,B,0) ̸= 0.

Proof: From the above construction of f̂ , this reduces to showing that f̂ ([ĉ]) ̸= 0 where[ĉ] generates Hn (Sn).

First consider the case where n = 1 so the ball is just the interval (−R,R) . Let gdenote the function of the form g(x) = kx where g(R) = f (R) ,g(−R) = f (−R). Thend (t f +(1− t)g,(−R,R) ,0) is constant by the earlier properties of the degree because forall t ∈ [0,1] ,0 /∈ (t f +(1− t)g)(∂ (−R,R)). However, d (g,(−R,R) ,0) =±1 by Proposi-tion A.12.2. Thus the theorem holds in case n = 1. Thus f̂ ([ĉ]) ̸= 0.

This shows what is desired in case n = 1. Assume that d ( f ,B,0) ̸= 0 for f odd forn− 1, n− 1 ≥ 1 so f̂∗ [ĉ] ̸= 0 for [ĉ] generating the homology group in dimension n− 1.Recall that for U and V missing the top and bottom of Sn, Hn−1 (U ∩V )≈Hn−1

(Sn−1

)and

so we have the following Mayer Vietoris sequence in which ∆ is an isomorphism:=Hn(Rn)=0

Hn (U) ⊕=Hn(Rn)=0

Hn (V )g∗→

ZHn (Sn)

∆→ Hn−1(Sn−1)

↓ f̂∗⊕ f̂∗ ↓ f̂∗ ↓ f̂∗Hn(

f̂ (U))⊕Hn

(f̂ (V )

) g∗→ Hn(

f̂ (Sn)) ∆→ Hn−1

(f̂(Sn−1))

h∗→0

Hn−1 (U)⊕0

Hn−1 (V )

↓ f̂∗⊕ f̂∗h∗→ Hn−1

(f̂ (U)

)⊕Hn−1

(f̂ (V )

)In the above, g∗,h∗ mapping to 0 in the bottom line comes from the above observation

that f̂ (U) ⊆ U and f̂ (V ) ⊆ V . Then this implies that both connecting homomorphisms∆ are isomorphisms. Thus, using Lemma A.7.1, the above Mayer Vietoris sequence com-mutes and the following holds for n > 1.

If [ĉ] generates Hn (Sn) , then ∆ [ĉ] generates Hn−1 (U ∩V )≈Hn−1(Sn−1

)and by induc-

tion and Lemma A.7.1, 0 ̸= f̂∗∆ [ĉ] = ∆ f̂∗ [ĉ] and so f̂∗ [ĉ] ̸= 0, so d ( f ,B,0) ̸= 0. ■Since we know that d ( f ,B,0) ̸= 0 for f an odd mapping, this leads to the Borsuk Ulam

theorem.

Theorem A.13.10 Let B be a bounded open ball in Rn centered at 0 and let f : ∂B→V becontinuous where V is an m dimensional subspace of Rn,m ≤ n− 1. Then f (−x) = f (x)for some x ∈ ∂B.