492 APPENDIX A. HOMOLOGICAL METHODS∗

θΩ is an open set in Sn. Then(θ ◦ f ◦θ

−1) : θΩ→ Sn does not include θ p along witha closed ball Ā and we can define d ( f ,Ω, p) ≡ d

(θ ◦ f ◦θ

−1,θΩ,θ p)

as in DefinitionA.11.3. This will be d f̂ where f̂ is the extention of θ ◦ f ◦θ

−1 off of θΩ described by lettingf̂ = θ ◦ f̄ ◦θ

−1 off Ω̄ and θ ◦ f ◦θ−1on θΩ̄.

The earlier material on spheres yield the following proposition.

Proposition A.12.2 The degree has the following properties:

1. Let continuous h : [0,1]×Ω→ Rn and let t → p(t) also be continuous with p(t) /∈h(t,∂Ω) for each t ∈ [0,1]. Then t → d (h(t, ·) ,Ω, p(t)) is constant. Also p →d ( f ,Ω, p) is constant on each component of f (∂Ω)C .

2. The identity map id, satisfies d (id,Ω,x) = 1 if x ∈Ω and d (id,Ω,x) = 0 if x /∈Ω. IfΩ is a ball centered at 0, then d (− id,Ω,x) = (−1)n for all x ∈Ω.

3. If p /∈ f(Ω̄\ Ω̂

)for Ω̂ an open subset of Ω, then d

(f ,Ω̂, p

)= d ( f ,Ω, p) .

4. Also if Ω is an open bounded subset of Rn equal to the union of disjoint open sets Ωi,then for f : Ω̄→ Rn continuous, p /∈ f (∂Ω) , d ( f ,Ω, p) = ∑i d ( f ,Ωi, p). The sumwill be finite.

5. If f = g on ∂Ω, p /∈ f (∂Ω) , then d ( f ,Ω, p) = d (g,Ω, p) .

6. If Ω is an open ball centered at 0 in Rn and f : Ω→ Rn is given by f (x1, ...,xn) =(k1x1, ...,knxn) where none of the ki are 0, then d ( f ,Ω,0) = (−1)m where m is thenumber of negative constants ki.

7. If Ω is an open ball centered at 0 in Rn and f : Ω→Rn is given by f (x1, ...,xn) = Axwhere A−1 exists, then d ( f ,Ω,0) = sgn(det(A)) .

Proof: Consider 2. the one about − id. The points on Sn,(x1, ...,xn,xn+1) with the sumof the squares of the components equal to 1 are obtained by letting f̂ correspond to thefollowing on the sphere Sn : (x1, ...,xn,xn+1)→ (−x1, ...,−xn,xn+1) and so from TheoremA.10.5 and those leading to this theorem, this would be (−1)n because we changed sign inn components.

For part 6., consider first the case where all the ki > 0. From 1., d (λ f +(1−λ ) id,B,0)is constant in λ ∈ [0,1] since 0 /∈ (λ f +(1−λ ) id)(∂B) . When λ = 0 this is 1 and so whenλ = 1 it is also 1, this by part 2. Now if you compose such an f with a map which changesthe sign of m of the entries, then similar to 2., the result will be (−1)m because of LemmaA.10.4 about the degree of a composition being the product of the degrees.

For part 7. I will consider other elementary operations. Also regard each of these invert-ible elementary operations E as taking ∞ to ∞. Thus xn→∞ if and only if Exn→∞, similarfor an invertible A. Let Li j (x)≡ (x1, ...,xi−1,xi + x j,xi+1, ...,xn)

T which comes from an el-ementary matrix in which the ith row is replaced with the jth row added to the ith row. Thenfor x ∈ ∂Ω and t ∈ [0,1] ,0 /∈ (tLi j +(1− t) id)∂Ω. Indeed, the matrix for the mappingtLi j +(1− t) id is just the identity with t added into the i jth position which is invertible andso cannot send anything in ∂B to 0. Therefore, from 1., d (Li j,Ω,0)= d (id,Ω,0)= 1. SinceA−1 exists, A is the composition of finitely many elementary operations. For the operationwhich switches two components, the degree is −1 as is the determinant of the elementary