8.7. HARD TOPOLOGY THEOREMS 209

first curve from a to b. Then there exists a point of intersection of these two paths. That is,there is s ∈ [−1,1] , t ∈ [−1,1] such that (u(s) ,v(s)) = ( f (t) ,g(t)) .

Proof: Suppose this is not true, then G : [−1,1]× [−1,1]→ [−1,1]× [−1,1] is contin-uous where G(s, t)≡

≡(

f (t)−u(s)max{| f (t)−u(s)| , |v(s)−g(t)|}

,v(s)−g(t)

max{| f (t)−u(s)| , |v(s)−g(t)|}

)Both components are in [−1,1] and one of them is ±1.Thus G maps [−1,1]× [−1,1] to∂ ([−1,1]× [−1,1]) , and is continuous. The only fixed points possible are of the form(±1, t) ,(s,±1). Then letting (≤ 0, t̂) denote an ordered pair in which the first componentis non-positive and other uses of this notation similar,

G(1, t) = (≤ 0, t̂) ̸= (1, t) ,G(−1, t) = (≥ 0, t̂) ̸= (−1, t) ,G(s,1) = (ŝ, ≤ 0) ̸= (s,1) ,G(s,−1) = (ŝ,≥ 0) ̸= (s,−1) .

Thus G has no fixed point contrary to Brouwer fixed point theorem. It follows that G(s, t)=0 for some (s, t) and this says there is a point of intersection of these two curves. ■

A Jordan arc will be the continuous one to one image of a closed interval. Then theconclusion of the above lemma implies the following easier to use proposition.

Corollary 8.7.19 Let J1,J2 be two oriented Jordan arcs which lie in [a,b]× [c,d] andsuppose the first component of J1 includes both a and b and the second component of J2includes both c and d, then there is a point on the intersection of these Jordan arcs.

Proof: This follows from the above lemma by changing the parametrization γ i to haveγ1

1 (−1) = a,γ11 (1) = b,γ2

2 (−1) = c,γ22 (1) = d. Then apply the above Lemma to these

functions restricted to [−1,1]. ■

Proposition 8.7.20 Let y→ α (y) ,β (y) be non-negative continuous functions and letU consist of (x,y) such that a−α (y) ≤ x ≤ b+β (y) and y ∈ [c,d] . Let J1 and J2 be twooriented Jordan arcs such that some first component of J1 equals a−a(y) for some y andsome first component of J1 equals b+ β (y) for some y and some second component ofpoints on J2 equals c while some second component of J2 equals d. Then the two Jordanarcs intersect.

Proof: Let σ be the midpoint of [a,b]. Let f : [a,b]× [c,d]→U be defined as follows.For x ≥ σ , f (x,y) ≡

(x+( x−σ

b−σ

)β (y) ,y

)and for x ≤ σ , f (x,y) ≡

(x−(

σ−xσ−a

)α (y) ,y

).

Then f is one to one onto and continuous. Also the left side of [a,b]× [c,d] is mapped tothe left side of U while the right side of [a,b]× [c,d] is mapped to the right side of U. Nowif Ji are as described, then f−1 (Ji) satisfy the conditions of the above corollary and so theseintersect at some point (x0,y0) . Then f (x0,y0) is a point of intersection of J1 and J2. ■

Theorem 8.7.21 Let J be a Jordan curve in the plane, J = γ(S1)

where γ is contin-uous and one to one. Then JC consists of a bounded component Ui called the inside, and anunbounded component Uo called the outside and J = ∂Ui = ∂Uo. That is, J is the commonboundary of both Ui and Uo.