The Horseshoe Map

Take a topological square $S$, with boundary $\partial S$ consisting of four arcs or curves with vertices $a, b, c,$ and $d$. Now take an injective homeomorphism $f: S \to {\mathbb{R}}^2$ such that the arcs $[a, d]$ and $[b, c]$ do not intersect $f (S)$. The map $f$ is an orientation-preserving homeomorphism sending clockwise-ordered co-ordinates on the boundary to clockwise-ordered co-ordinates. It sends two horizontal bands across the rectangle to two vertical bands which make up $S \cap f (S)$, and $\{ z \in S \vert f (z), f^2 (z) \in S \}$ consists of four horizontal bands, while $\{ z \in S \vert f^{-1} (z), f^{-2} (z) \in S \}$ consists of four vertical bands. A map such as this one is called a topological horseshoe map. Later on we shall define this map more specifically (linear on some bands), and then we shall be able to say more. But for now, let us consider this topological map.

Let ${\Lambda}_n = \{ z \in S \vert f^i (z) \in S \}$ for $i = -n, \ldots, n \}$ and $\Lambda = {\bigcap}_{i \geq 0} {\Lambda}_n$. $\Lambda \not= \emptyset$, as all the $\Lambda_n$ are nested and compact. This invariant set $\Lambda$ is a Cantor set. That means it is compact, perfect (each point is an accumulation point), and it is totally disconnected (each point is its own connected component).

We can create a map $h: \Lambda \to \{ Le, Ri \}^{\mathbb{Z}} := \Sigma_2$ with $z \mapsto ( \ldots a_{-2}, a_{-1}, a_0, a_1, a_2, \ldots )$ where $a_i (z) = Le$ if $f^i (z)$ is in the left vertical strip and $a_i (z) = Ri$ if it is in the right. This is well-defined. For continuity we need a topology on $\Sigma_2$ and define it with the metric $d ( \underline{a}, \underline{b} ) = \sum_{i \in \mathbb{Z}} \vert a_i - b_i \vert / (2 ^{\vert i \vert})$ where $\vert a_i - b_i \vert = 0$ if $a_i = b_i$ and $\vert a_i - b_i \vert = 1$ otherwise. This makes $h$ continuous by continuity of $f$. It is also clearly surjective, but is it injective?

The map $h$ is a semi-conjugacy from $f: \Lambda \to \Lambda$ to the two-sided shift $\sigma$ on two symbols. Take a periodic symbol $\underline{a}$ of period $n$ (with the pattern of $Le$'s and $Ri$'s repeating every $n$ terms).

The following proposition tells us that every such periodic point of $\sigma$ corresponds with a peridic point of $f$.

Proposition 3.1   Let $\underline{a} = (\ldots, a_1, a_2, a_3, \ldots, a_n, \ldots)\in\Sigma_2$ be a periodic point with period $n$. Let the rectangle $\tilde{S}\subset S$ be the preimage of all the points $\{(\ldots, a_1, a_2, \ldots, a_n, \ldots)\}$.

Then the map $f^n$ has a fixed point in the rectangle $\tilde{S}$.

Proof First note that $f^n$ maps $\tilde{S}$ across itself (see figure 1).

Figure 1: There is a vertical strip in $S$ mapped across itself by $f^n$

Suppose $f^n$ has no fixed point on $\tilde{S}$. Then the map $\phi: S \to S^1$ given by $\phi (z) = (f^n (z) - z) / \vert\vert f^n (z) - z \vert\vert$ (imagine $S$ embedded in a vector space) is well-defined. Take a closed curve ${\gamma}_0 : S^1 \to \delta \tilde{S}$ on the boundary of $\tilde{S}$. The composition map ${\phi}_{{\gamma}_0} = \phi \circ {\gamma}_0 : S^1 \to S^1$ is a continuous circle map of degree $1$. Define a lift $F_{{\gamma}_0} : \mathbb{R}\to \mathbb{R}$ with $F_{{\gamma}_0} (1) - F_{{\gamma}_0} (0) = 1$ (or $-1$). Take ${\gamma}_s$ a continuous deformation of ${\gamma}_0$. This also has degree $1$ since degree is a continuous property which must take an integer value. Let this new curve shrink down towards a single point as $s$ tends to $1$. We have ${\phi}_{{\gamma}_0} (t) = (f^n ({\gamma}_s (t)) - {\gamma}_s (t)) / \vert\vert f^n ({\gamma}_s (t)) - {\gamma}_s (t) \vert\vert$ which tends to a constant as $s \to 1$. This implies degree $0$, a contradiction. There must have been a fixed point of $f^n$ after all. $\square$

Now consider $R$ to be a rectangle on the surface of the sphere $S^2$ by adding the `point at infinity', which may be considered to be a source. For all $x \neq \infty$, there exists $n (x) \in \mathbb{N}$ such that $f^n (x) \in D$, where $D$ is the union of $R$ and two half-discs glued onto it in order to encompass all of $f (D)$ in $D$ itself. See figure 2.

Figure 2: The Smale-Horseshoe in its 'original habitat', the sphere