23.3. BORSUK’S THEOREM 765

Corollary 23.2.3 The following additional properties of the degree are also valid.

1. If y /∈ f(Ω\Ω1

)and Ω1 is an open subset of Ω, then d (f,Ω,y) = d (f,Ω1,y) .

2. d (·,Ω,y) is defined and constant on{g ∈C

(Ω;Rp) : ∥g− f∥

∞< r}

where r = dist(y, f(∂Ω)).

3. If dist(y, f(∂Ω))≥ δ and |z−y|< δ , then d (f,Ω,y) = d (f,Ω,z).

Proof: Consider 1. You can take Ω2 = /0 in 2 of Theorem 23.2.2 or you can modifythe proof of 2 slightly. Consider 2. To verify, let h(x, t) = f(x)+ t (g(x)− f(x)) . Thennote that y /∈ h(∂Ω, t) and use Property 1 of Theorem 23.2.2. Finally, consider 3. Lety(t)≡ (1− t)y+ tz. Then for x ∈ ∂Ω

|(1− t)y+ tz− f(x)| = |y− f(x)+ t (z−y)|≥ δ − t |z−y|> δ −δ = 0

Then by 1 of Theorem 23.2.2, d (f,Ω,(1− t)y+ tz) is constant. When t = 0 you getd (f,Ω,y) and when t = 1 you get d (f,Ω,z) .

Another simple observation is that if you have y1, · · · ,yr in Rp \ f(∂Ω) , then if f̃ hasthe property that

∥∥f̃− f∥∥

∞< mini≤r dist(yi, f(∂Ω)) , then

d (f,Ω,yi) = d(f̃,Ω,yi

)for each yi. This follows right away from the above arguments and the homotopy invarianceapplied to each of the finitely many yi. Just consider d

(f+ t

(f̃− f

),Ω,yi

), t ∈ [0,1] . If

x ∈ ∂Ω, f+ t(f̃− f

)(x) ̸= yi and so d

(f+ t

(f̃− f

),Ω,yi

)is constant on [0,1] , this for each

i.

23.3 Borsuk’s TheoremIn this section is an important theorem which can be used to verify that d (f,Ω,y) ̸= 0. Thisis significant because when this is known, it follows from Theorem 23.2.2 that f−1 (y) ̸= /0.In other words there exists x ∈Ω such that f(x) = y.

Definition 23.3.1 A bounded open set, Ω is symmetric if −Ω = Ω. A continuous function,f : Ω→ Rp is odd if f(−x) =−f(x).

Suppose Ω is symmetric and g ∈ C∞(

Ω;Rp)

is an odd map for which 0 is a regularvalue. Then the chain rule implies Dg(−x) = Dg(x) and so d (g,Ω,0) must equal an oddinteger because if x∈ g−1 (0), it follows that−x∈ g−1 (0) also and since Dg(−x) =Dg(x),it follows the overall contribution to the degree from x and −x must be an even integer.Also 0 ∈ g−1 (0) and so the degree equals an even integer added to sgn (detDg(0)), anodd integer, either −1 or 1. It seems reasonable to expect that something like this wouldhold for an arbitrary continuous odd function defined on symmetric Ω. In fact this is the