Multivariable Advanced Calculus

280 BROUWER DEGREEand y /∈ C. Since f −1 ◦ f equals the identity id on ∂K, it follows from the properties ofthe degree that()1 = d (id, K, y) = d f −1 ◦ f, K, y .Let H denote the set of bounded components of R n \ f (∂K).By the product formula,()1 = d f −1 ◦ f, K, y = ∑ d ( f, K, H ) ( )d f −1 , H, y , (10.19)H∈Hthe sum being a finite sum from the product formula.following diagram.R n \ Cf→f −1←R n \ f (C)KKy ∈ K H, H 1HL HLR n \ f (∂K)It might help to consult theNow letting x ∈ L ∈ L, if S is a connected set containing x and contained in R n \ f (C),then it follows S is contained in R n \ f (∂K) because ∂K ⊆ C. Therefore, every set ofL is contained in some set of H. Furthermore, if any L ∈ L has nonempty intersectionwith H ∈ H then it must be contained in H. This is because(L = (L ∩ H) ∪ (L ∩ ∂H) ∪ L ∩ H C) .Now by Lemma 10.5.6,L ∩ ∂H ⊆ L ∩ f (∂K) ⊆ L ∩ f (C) = ∅.Since L is connected, L∩H C = ∅. Letting L H denote those sets of L which are containedin H equivalently having nonempty intersection with H, if p ∈ H \ ∪L H = H \ ∪L,then p ∈ H ∩ f (C) and soH = (∪L H ) ∪ (H ∩ f (C)) (10.20)Claim 1:H \ ∪L H ⊆ f (C) .Proof of the claim: Suppose p ∈ H \ ∪L H but p /∈ f (C). Then p ∈ L ∈ L.It must be the case that L has nonempty intersection with H since otherwise p couldnot be in H. However, as shown above, this requires L ⊆ H and now by 10.20 andp /∈ ∪L H , it follows p ∈ f (C) after all. This proves the claim.Claim 2: y /∈ f ( ) −1 H \ ∪L H . Recall y ∈ K ∈ K the bounded components ofR n \ C.Proof of the claim: If not, then f −1 (z) = y where z ∈ H \ ∪L H ⊆ f (C) andso z = f (w) for some w ∈ C and so y = f −1 (f (w)) = w ∈ C contrary to y ∈ K, acomponent of R n \ C.Now every set of L is contained in some set of H. What about those sets of H whichcontain no set of L so that L H = ∅? From 10.20 it follows H ⊆ f (C). Therefore,( )d f −1 , H, y = d ( f −1 , H, y ) = 0