23.5. PRODUCT FORMULA, JORDAN SEPARATION THEOREM 773

23.5 Product Formula, Jordan Separation TheoremThis section is on the product formula for the degree which is used to prove the Jordan sep-aration theorem. To begin with is a significant observation which is used without commentbelow. Recall that the connected components of an open set are open. The formula is allabout the composition of continuous functions.

Ωf→ f(Ω)⊆ Rp g→ Rp

Lemma 23.5.1 Let {Ki}∞

i=1 be the connected components of Rp \C where C is a closedset. Then ∂Ki ⊆C.

Proof: Since Ki is a connected component of an open set, it is itself open. See Theorem7.13.10. Thus ∂Ki consists of all limit points of Ki which are not in Ki. Let p be such apoint. If it is not in C then it must be in some other K j which is impossible because theseare disjoint open sets. Thus if x is a point in U it cannot be a limit point of V for V disjointfrom U .

Definition 23.5.2 Let the connected components of Rp \ f(∂Ω) be denoted by Ki. Fromthe properties of the degree listed in Theorem 23.2.2, d (f,Ω, ·) is constant on each of thesecomponents. Denote by d (f,Ω,Ki) the constant value on the component Ki.

The following is the product formula. Note that if K is an unbounded component off(∂Ω)C , then d (f,Ω,y) = 0 for all y∈K by homotopy invariance and the fact that for largeenough ∥y∥ , f−1 (y) = /0 since f

(Ω)

is compact.

Theorem 23.5.3 (product formula)Let {Ki}∞

i=1 be the bounded components of Rp \ f(∂Ω)for f ∈ C

(Ω;Rp

), let g ∈ C (Rp,Rp), and suppose that y /∈ g(f(∂Ω)) or in other words,

g−1 (y)∩ f(∂Ω) = /0. Then

d (g◦ f,Ω,y) =∞

∑i=1

d (f,Ω,Ki)d (g,Ki,y) . (23.5.11)

All but finitely many terms in the sum are zero. If there are no bounded components off(∂Ω)C , then d (g◦ f,Ω,y) = 0.

Proof: The compact set f(Ω)∩ g−1 (y) is contained in Rp \ f(∂Ω) and so, f

(Ω)∩

g−1 (y) is covered by finitely many of the components K j one of which may be the un-bounded component. Since these components are disjoint, the other components fail tointersect f

(Ω)∩g−1 (y). Thus, if Ki is one of these others, either it fails to intersect g−1 (y)

or Ki fails to intersect f(Ω). Thus either d (f,Ω,Ki) = 0 because Ki fails to intersect f

(Ω)

or d (g,Ki,y) = 0 if Ki fails to intersect g−1 (y). Thus the sum is always a finite sum. I amusing Theorem 23.2.2, the part which says that if y /∈ h

(Ω), then d (h,Ω,y) = 0. Note that

∂Ki ⊆ f(∂Ω) so y /∈ g(∂Ki).Let g̃ be in C∞ (Rp,Rp) and let ∥g̃−g∥

∞be so small that for each of the finitely many

Ki intersecting f(Ω)∩g−1 (y) ,

d (g◦ f,Ω,y) = d (g̃◦ f,Ω,y)d (g,Ki,y) = d (g̃,Ki,y) (23.5.12)