20 0. Prequel

Proof. Identify consecutive vertices v1,v2,v3 of P , and form the triangle

∆ determined by them. Impose coordinates on

R2

with v1,v3 on the x-axis

and v2 below it. Let K denote the complement in P of the two segments

v1v2 and v2v3. If no point of K meets ∆, then the segment determined by

v = v1 and w = v3 obviously has the desired property. Otherwise, choose

w to have least y-coordinate among the finitely many vertices of K {v2}

touching the 2-cell bounded by ∆ (see Figure 0.4); in this case the segment

formed by v = v2 and w

works.2

x v3

v2

v1

w

y

K

Figure 0.4. The interior of segment v2w misses P

Theorem 0.11.4. Every polygonal simple closed curve P in

R2

bounds a

PL 2-cell.

Proof. By induction on the number n of vertices in P , with the initial

case n = 3 being trivial. Inductively assume the result for all polygons of

fewer than n vertices, and consider a polygon P having n vertices. Apply

Lemma 0.11.3 to form two polygonal simple closed curves J1,J2, each having

fewer than n vertices, with J1 ∪J2 = P ∪vw and J1 ∩J2 = vw; by induction,

each Ji bounds a PL 2-cell Di (i = 1, 2). If D1 ∩ D2 = vw, then D1 ∪ D2 is

a disk and ∂(D1 ∪ D2) = P . Otherwise, one of the Ji misses the interior of

the other 2-cell Dj, for if J1 ∩ Int D2 = ∅, then J2 ∩ Int D1 = ∅, as D1 ⊂ D2

(see Figure 0.5). Let us say J1 ∩ Int D2 = ∅ for definiteness. Let C denote

the closure of the bounded component of

R2

P . It is left to the reader

to check that there is an elementary shelling (see Rourke and Sanderson

(1972), p. 40) of D1 to C across D2. Hence, C

∼

= D1 is a 2-cell.

Proof of Theorem 0.11.2. The theorem is an immediate corollary of The-

orem 0.11.4 and the Disc Theorem of Rourke and Sanderson (1972, Theo-

rem 3.34).

2J.

W. Cannon deserves credit for this elegant argument.