On the struture of homeomorphims of the open annulus
Let $h$ be a without fixed point lift to the plane of a homeomorphism of the open annulus isotopic to the identity and without wandering point. We show that $h$ admits a $h$-invariant dense open set $O$ on which it is conjugate to a translation and we study the action of $h$ on the compactly connected components of the closed and without interior set ${\bf R}^2 \setminus O$.
Bounded recurrent sets for planar homeomorphisms
We answer a question proposed by Barge and Franks in 1993 by showing that a bounded connected set in the plane which is closed or open and meets an infinite number of its iterates under an orientation-preserving fixed point free homeomorphism must meet all of them.
A generalization of the Poincaré-Birkhoff theorem
We substitute Poincare's twist hypothesis by the weakest possible topological one: that the homeomorphism in question not be conjugate to a translation. Let A = S' x [0,1] denote the annulus and B = R x [0,1] its universal cover; let T: B -4 B be the translation T(x, y) = (x + 2wr, y) for (x, y) E R x [0, 1]. Let h: B -4 B be a lifting of a homeomorphism h: A -4 A (i.e. hT = Th). Recall that h is said to be topologically conjugate to T, if there exists a homeomorphism k: B -4 B such that hk = kT, we write h T if such a k exists, h Y T otherwise. The purpose of this note is to prove the THEOREM. Let h: A -4 A be boundary component and orientation preserving; if h: B -4 B is a lifting of h such that h ?(' T, then either h has at least one fixed point or there exists in A a closed, simple, noncontractible curve C such that h(C) n c = 0. In other words, in the Poincare-Birkhoff Theorem we substitute Poincare's twist condition (i.e. that h send the boundary components of B in opposite directions) by the weakest possible condition h YT. Our proof is just a short addendum to Kerekjarto's proof of the Poincare-Birkhoff Theorem using Brouwer's translation theory (see [5]). The example in Figure 1 of [3] shows that, unlike in the area-preserving case, the existence of only one fixed point is best possible here. For other generalizations and references see [3 and 4]. Proof of the Theorem. We first recall BROUWER'S LEMMA (SEE [2, SATz 8; 6, SATZ 9]). Let H: R2 -4 R2 be an orientation-preserving, fixed-point free homeomorphism of the plane. Then, for any point P E R2 , the set {H (P) , n an integer} has no accumulation point in R2. We call an arc a joining the boundary components of B free (with respect to h) if c n h(ae) = 0. LEMMA. If h has a free simple arc and h ?" T, then h has a fixed point. Received by the editors January 19, 1987. 1980 Mathematics Subject Classification (1985 Revi.sion). Primary 55H25, 55M25, 54H20.