488 APPENDIX A. HOMOLOGICAL METHODS∗

Lemma A.11.4 There is an open ball containing p such that d ( f ,Ω, p̂) is the same for allp̂ in this ball.

Proof: By definition of f̂ , there is an open ball A containing p such that f̂−1(Ā)⊆Ω.

Then if p̂ ∈ A, there is a smaller open ball  centered at p̂ with f̂−1(Â)⊆Ω and so, since

this is the same f̂ , the d ( f ,Ω, p) = d ( f ,Ω, p̂) because d ( f ,Ω, p) is defined in terms of f̂∗and the generator of Hn (Sn). ■

Say [c] generates Hn (Sn) and let α be the isomorphism of Hn (Sn) andZ. Then [c] wouldbe either α (1) or else α (−1) = [c] and so either [c] or − [c] will work. But then d stillwould be the same because f̂∗ (− [c]) =− f̂∗ ([c]) =−d [c] = d (− [c]) . Thus the definitionof the degree does not change if [c] is switched to − [c] the other generator of Hn (Sn) .The hard issue is whether two different extensions satisfying the above definition give thesame degree.This is discussed in the following proposition and shows that d ( f ,Ω, p) iswell defined.

The proper open balls A on Sn,n ≥ 1, are just intersections of open balls in Rn+1 withSn and so they are connected and open sets. Each is homeomorphic to a convex subset ofRn. Say α is the name of this homeomorphism so that C = αA where C is convex. Then ifc is a cycle with values in A we have α∗ [c]A = [α#c]C = 0 because C is convex so cyclesand boundaries are the same. Since α∗ is an isomorphism, this implies [c]A = 0.

Proposition A.11.5 Let Ω, f , f̂ be as defined in Definition A.11.3 with respect to someopen ball A for which f̂−1

(Ā)⊆ Ω. If f̂ , f̃ are two extensions of f from Definition A.11.3

and [ĉ] generating Hn (Sn), f̂∗ [ĉ] = f̃∗ [ĉ] and so, for any p ∈ A,d ( f ,Ω, p) is well defined.If f−1

(Ā)= /0 then for f̂ such an extension, f̂∗ [ĉ] = 0 and so for p ∈ A,d ( f ,Ω, p) = 0.

Proof: Letting L,θ be as in Proposition A.11.1, one can consider the following homo-topy of f̂ , f̃

θ(tθ−1 f̂ +(1− t)θ

−1 f̃), t ∈ [0,1] .

The two functions are homotopic and so by Theorem A.3.7 f̂∗ [ĉ] = f̃ [ĉ]. In particular, thedefinition of the degree is well defined.

If f−1(Ā)= /0 then for each simplex φ in ĉ, f̂#φ has values in Sn \A which is homeo-

morphic to a convex subset of Rn and so[

f̂#ĉ]= f̂∗ [ĉ] = 0. ■

Another claim which is easy to get is the following which deals with homeomophismsof Sn.

Lemma A.11.6 From the definition, if α is a homeomorphism on Sn, then d ( f ,Ω, p) =±d (α f ,αΩ,α p).

Proof: Let f̂ and A go with the definition for d ( f ,Ω, p) . Then, since α is a homeo-morphism, α (A) contains A′ a ball centered at α p and α f̂ will serve for the definition ofd (α f ,αΩ,α p) . Then for [ĉ] a generator of Hn (Sn) ,α∗ is an isomorphism of homologygroups and so α∗ [ĉ] is also a generator of Hn (Sn) .

d (α f ,αΩ,α p) [ĉ]≡(α f̂)∗ [ĉ] = α∗ f̂∗ [ĉ] = α∗d ( f ,Ω, p) [ĉ] = d ( f ,Ω, p)α∗ [ĉ]

But, since α∗ is an isomorphism, α∗ [ĉ] is also a generator of Hn (Sn)≈ Z so α∗ [ĉ] is either[ĉ] or − [ĉ]. ■

There is also a generalization of Proposition A.11.5 which replaces f̃ with ĝ where gis sufficiently close to f using the observation that all that mattered in the above argumentwas f̃ (∂Ω)∩ Ā = /0.