A.12. THE DEGREE ON OPEN SUBSETS OF Rn 491

Lemma A.11.15 g−1 (p) has empty intersection with all but finitely many of the Ki.

The next theorem is a major result called the product formula. Recall that by Propos-tion A.11.12, d ( f ,Ω, p) is constant for p ∈ Ki and so we denote this common value asd ( f ,Ω,Ki) .

Theorem A.11.16 Let g : Sn→ Sn and let f : Ω→ Sn. Suppose p /∈ g( f (∂Ω)) . Then forKi the components of f (∂Ω)C ,

d (g◦ f ,Ω, p) = ∑i

d ( f ,Ω,Ki)d (g,Ki, p) (1.19)

and the sum is finite.

Proof: By the above Lemma A.11.15, we will consider only finitely many i for whichg−1 (p) intersects Ki. Otherwise d (g,Ki, p) = 0 and the term in the sum is 0. Considerf−1 (Ki)∩Ω, disjoint open sets because the Ki are disjoint. Also we are assuming thatg−1 (p)⊆ ∪iKi so

(g◦ f )−1 (p)⊆ ∪i f−1 (Ki)∩Ω⊆Ω.

and p /∈ (g◦ f )(Ω̄\∪i f−1 (Ki)

). It follows from Corollary A.11.11 and Proposition A.11.9

thatd (g◦ f ,Ω, p) = ∑

id(g◦ f , f−1 (Ki)∩Ω, p

)Now letting f̂i be an appropriate extension of f , on f−1 (Ki)∩Ω and [c] a generator ofHn (Sn), d

(f , f−1 (Ki)∩Ω,qi

)[c] = f̂i∗ ([c]). Therefore,

d(g◦ f , f−1 (Ki)∩Ω, p

)[c] = g∗ f̂i∗ ([c]) = g∗d

(f , f−1 (Ki)∩Ω,qi

)[c]

= g∗d ( f ,Ω,qi) [c] = d ( f ,Ω,Ki)g∗ [c] = d ( f ,Ω,Ki)d (g,Ki, p) [c]

Note that on the top line, f restricted to f−1 (Ki)∩Ω has the properties that qi is not inf(Ω̄\ f−1 (Ki)∩Ω

)and so Corollary A.11.11 applies. Thus we obtain the product formula

1.19. ■

A.12 The Degree on Open Subsets of Rn

I am not all that interested in spheres. I am much more interested in what can be said aboutopen bounded subsets of Rn. However, this is essentially included in the above. It will bebased on the following mapping θ illustrated in this picture.

(⃗0,2) = q

(⃗0,1)x

θ(x)

Rn

Note that if Ω is an open bounded subset of Rn, then θΩ̄ = θΩ is a proper subset of Sn

which does not contain q. Letting p /∈ f (∂Ω) d ( f ,Ω, p)≡ d(θ ◦ f ◦θ

−1,θΩ,θ p).

Definition A.12.1 Let Ω be a bounded open set in Rnand f : Ω̄→ Rn is continuous. Letp /∈ f (∂Ω). Then for θ the homeomorphism which maps Rn∪{∞} to all of Sn, p = θ (∞) ,

A.12. THE DEGREE ON OPEN SUBSETS OF R” 491Lemma A.11.15 g~! (p) has empty intersection with all but finitely many of the K;.The next theorem is a major result called the product formula. Recall that by Propos-tion A.11.12, d(f,Q, p) is constant for p € K; and so we denote this common value asd(f,Q,Ki).Theorem A.11.16 Let g: S" > S" and let f :Q— S". Suppose p ¢ g(f (0Q)). Then forK; the components of f (0Q)° ,d(gof,Q,p) = Yid(f,2,Ki)d(8,Ki,p) (1.19)and the sum is finite.Proof: By the above Lemma A.11.15, we will consider only finitely many i for whichg | (p) intersects K;. Otherwise d(g,Kj,p) = and the term in the sum is 0. Consider~! (K;) VQ, disjoint open sets because the K; are disjoint. Also we are assuming thatg | (p) CU;K; so(gof) | (p) CUif | (Ki)N@CQ.and p ¢ (go f) (Q\U;f | (Ki) . It follows from Corollary A.11.11 and Proposition A.11.9thatd(gof,Q,p)=Yid(gof.f ' (Ki) NQ,p)Now letting f; be an appropriate extension of f, on f~!(K;) MQ and [c] a generator ofHn (S"), d (f,f-' (Ki) 0, qi) [ce] = fis ([c]). Therefore,d(gof,f—! (Ki) Q, p) [c] = gfix ([c]) = gd (f,.f' (Ki) Q, 41) [el]= g.d(f,Q,qi) [c] =d(f,Q,Ki) gx [ce] = d(f,Q,Ki)d(g, Ki, p) [c]Note that on the top line, f restricted to f ~!(K;) Q has the properties that q; is not inf (Q\ f~' (Ki) NQ) and so Corollary A.11.11 applies. Thus we obtain the product formula1.19.A.12 The Degree on Open Subsets of IR”I am not all that interested in spheres. Iam much more interested in what can be said aboutopen bounded subsets of IR”. However, this is essentially included in the above. It will bebased on the following mapping @ illustrated in this picture.(6,2) =¢q. R”Note that if Q is an open bounded subset of R”, then 9Q = OQ is a proper subset of S”which does not contain g. Letting p ¢ f (OQ) d(f,Q,p)=d (0 o fo! AQ, Op) .Definition A.12.1 Let Q be a bounded open set in R’and f :Q— R" is continuous. Letp ¢ f (AQ). Then for @ the homeomorphism which maps R" U {0°} to all of S", p = 0 (%),