780 CHAPTER 23. DEGREE THEORY

Now d(f̄,K,H

)= d

(f̄,K,L

)where L ∈LH . This is because L is an open connected

subset of H and z→ d(f̄,K,z

)is constant on H. Therefore,

∑H∈H1

∑L∈LH

d(f̄,K,H

)d(

f−1,L,y)= ∑

H∈H1

∑L∈LH

d(f̄,K,L

)d(

f−1,L,y)

As noted above, there are finitely many H ∈H which are involved.

Rp \ f(C)⊆ Rp \ f(∂K)

and so every L must be contained in some H ∈H1. It follows that the above reduces to

∑L∈L

d(f̄,K,L

)d(

f−1,L,y)

Where this is a finite sum because all but finitely many terms are 0.Thus from 23.6.13,

1 = ∑L∈L

d(f̄,K,L

)d(

f−1,L,y)= ∑

L∈Ld(f̄,K,L

)d(

f−1,L,K)

(23.6.14)

Let |K | denote the number of components in K and similarly, |L | denotes the number ofcomponents in L . Thus

|K |= ∑K∈K

1 = ∑K∈K

∑L∈L

d(f̄,K,L

)d(

f−1,L,K)

Similarly, the argument taken another direction yields

|L |= ∑L∈L

1 = ∑L∈L

∑K∈K

d(f̄,K,L

)d(

f−1,L,K)

If |K |< ∞, then ∑K∈K

1︷ ︸︸ ︷∑

L∈Ld(f̄,K,L

)d(

f−1,L,K)< ∞. The summation which equals 1

is a finite sum and so is the outside sum. Hence we can switch the order of summation andget

|K |= ∑L∈L

∑K∈K

d(f̄,K,L

)d(

f−1,L,K)= |L |

A similar argument applies if |L | < ∞. Thus if one of these numbers |K | , |L | is finite,so is the other and they are equal. This proves the theorem because if p > 1 there is exactlyone unbounded component to both Rp \C and Rp \ f(C) and if p = 1 there are exactly twounbounded components.

As an application, here is a very interesting little result. It has to do with d (f,Ω, f(x))in the case where f is one to one and Ω is connected. You might imagine this should equal 1or −1 based on one dimensional analogies. Recall a one to one map defined on an intervalis either increasing or decreasing. It either preserves or reverses orientation. It is similar inn dimensions and it is a nice application of the Jordan separation theorem and the productformula.