Uncountably many planar embeddings of unimodal inverse limit spaces

For point $x$ in the inverse limit space $X$ with a single unimodal bonding map we construct, with the use of symbolic dynamics, a planar embedding such that $x$ is accessible. It follows that there are uncountably many non-equivalent planar embeddings of $X$.


Introduction
Inverse limit spaces can be often used as a model to construct attractors of some plane diffeomorphisms, see for example [22,23,4]. One of the simplest examples is the Knaster continuum (bucket handle), which is the attractor of Smale's horseshoe map and can be modeled as the inverse limit space of the full tent map T 2 (x) := min{2x, 2(1 − x)} for x ∈ [0, 1], see [1]. The topological unfolding of the Smale's horseshoe has been a topic of ongoing interest, related to the pruning front conjecture for the Hénon family developed by Cvitanović et al. in [10,11] and demonstrated more recently with the work of Boyland, de Carvalho & Hall [6] and Mendoza [17].
Inverse limit spaces with unimodal bonding maps T : [0, 1] → [0, 1] (from now on denoted by X) are chainable. The study of embeddings of chainable continua dates back to 1951 when Bing proved in [5] that every chainable continuum can be embedded in the plane. However, his proof does not offer any insight what such embeddings look like. The first explicit class of embeddings of X was given by Brucks & Diamond in [7]. Later, Bruin [9] extended this result showing that the embedding of X can be made such that the shift-homeomorphism extends to a Lipschitz map on R 2 . Both mentioned results are using symbolic dynamics as the main tool in the description of X.
Locally, inverse limit spaces of unimodal maps roughly resemble Cantor sets of arcs. However, this is not true in general. In [2] Barge, Brucks & Diamond proved that in the tent family T s for a dense G δ set of slopes s ∈ [ Can a complicated X be embedded in R 2 in multiple ways? For example, do there exist embeddings in R 2 of X that are non-equivalent to the standard embeddings constructed in [7] and [9]?
For a special case of the full tent map these two questions were already answered in the affirmative by Mayer [15], Mahavier [14], Schwartz [21] and Dȩbski & Tymchatyn [12].
Let K be a continuum. A composant U ⊂ K of a point x ∈ U is the union of all proper subcontinua of K containing point x. We call K indecomposable if it cannot be expressed as the union of two proper subcontinua. In this case, K has uncountably many composants and every composant is dense in K.
There are two fixed points of the map T : 0 and r. Denote the composant of (. . . , 0, 0) by C. We can decompose X = C ∪ X ′ , where X ′ is the core of X and C compactifies on X, see e.g. [13]. Let R be the arc-component of (. . . , r, r); if X ′ is indecomposable then R is indeed the composant of (. . . , r, r) within X ′ . The standard embeddings given in [7] and [9] make C and R respectively accessible. Definition 1. A point a ∈ X ⊂ R 2 is accessible (i.e., from the complement of X) if there exists an arc A = [x, y] ⊂ R 2 such that a = x and A ∩ X = {a}. We say that a composant U ⊂ X ′ or U = C is accessible, if U contains an accessible point.
In the special case of the full tent map, i.e., with the Knaster continuum X as the inverse limit space, Mayer constructed uncountably many non-equivalent embeddings in [15]. Later, Mahavier showed in [14] that for every composant U ⊂ X, there exists a homeomorphism h : X → R 2 such that each point of h(U) is accessible. Schwartz extended Mahavier's result and proved that embeddings of X which do not make C or R accessible are non-equivalent to the standard embeddings.
Definition 2. Denote two planar embeddings of X by g 1 : X → E 1 ⊂ R 2 and g 2 : X → E 2 ⊂ R 2 . We say that g 1 and g 2 are equivalent embeddings if there exists a homeomorphism h : E 1 → E 2 which can be extended to the homeomorphism of the plane.
In this paper we make use of symbolic dynamics description of X introduced in [7] and [9] and answer the questions of Boyland in the affirmative. We construct embeddings of X by selecting (the itinerary of) the accessible point. For every unimodal map of positive topological entropy, we obtain uncountably many embeddings by making an arbitrary point from X accessible. Theorem 1. For every point a ∈ X there exists an embedding of X in the plane such that a becomes accessible.
By h top (T ) we denote the topological entropy of T . If h top (T ) > 0, then the unimodal inverse limit space X with the bonding map T contains an indecomposable subcontinuum.
Corollary 1. Let T be a unimodal interval map with h top (T ) > 0. Then there are uncountably many non-equivalent embeddings of X in the plane.
The short outline of the paper is as follows. Section 2 provides the basic set-up as introduced in [7] and [9]. Next, we construct specific representations of X in the plane in Section 3. In Section 4 we prove that the representations given in Section 3 are indeed embeddings and prove the main results.
In the construction of planar embeddings of spaces X we recall a well-known symbolic description introduced in [7]. The space X will be represented by the quotient space Σ adm /∼, where Σ adm ⊆ {0, 1} Z is equipped with the product topology. We first need to recall the kneading theory for unimodal maps. To every x ∈ [0, 1] we assign its itinerary: Note that if ν i (x) = * for some i ∈ N 0 , then ν i+1 (x)ν i+2 (x) . . . = I(T (c)). The sequence ν := I(T (c)) is called the kneading sequence of T and is denoted by ν = c 1 c 2 . . ., where c i := ν i (T (c)) ∈ {0, * , 1} for every i ∈ N. Observe that if * appears in the kneading sequence, then c is periodic under T , i.e., there exists n > 0 such that T n (c) = c and the kneading sequence is of the form ν = (c 1 . . . c n−1 * ) ∞ . In this case we adjust the kneading sequence by taking the smallest of (c 1 . . . c n−1 0) ∞ and (c 1 . . . c n−1 1) ∞ in the parity-lexicographical ordering defined below.
By # 1 (a 1 . . . a n ) we denote the number of ones in a finite word a 1 . . . a n ∈ {0, 1} n ; it can be either even or odd.
In the same way we modify itinerary of an arbitrary point x ∈ [0, 1]. If ν i (x) = * and i is the smallest positive integer with this property then we replace ν i+1 (x)ν i+2 (x) . . . with the modified kneading sequence. Thus * can appear only once in the modified itinerary of an arbitrary point x ∈ [0, 1].
From now onwards we assume that the itineraries of points from [0, 1] are modified.
It is a well-known fact (see [18]) that a kneading sequence completely characterizes the dynamics of unimodal map in the sense of the following proposition:  Conversely, assume s 0 s 1 . . . ∈ {0, * , 1} N satisfies (1). If there exists j ∈ N 0 such that s j+1 s j+2 . . . = ν, and j is minimal with this property, assume additionally that s j = * .
Next we show how to expand the above construction to X. Take x = (. . . , Define the itinerary of x as a two-sided infinite sequencē We make the same modifications as above. If * appears for the first time at ν k (x) for some k ∈ Z, then ν k+1 (x)ν k+2 (x) . . . = ν. If there is no such minimal k, then the kneading sequence is periodic with a period n ∈ N, In this way * can appear at most once in every itinerary. Now we are ready to identify the inverse limit space with a quotient of a space of two-sided sequences consisting of two symbols. Let Σ := {0, 1} Z be the space of two-sided sequences equipped with the metric By Σ adm ⊆ Σ we denote all s ∈ Σ such that either (a) s k s k+1 . . . is admissible for every k ∈ Z, or (b) there exists k ∈ Z such that s k+1 s k+2 . . . = ν and s k−i . . . s k−1 * s k+1 s k+2 . . . is admissible for every i ∈ N.
We abuse notation and call the two-sided sequences in Σ adm also admissible.
Let us define an equivalence relation on the space Σ adm . For sequences s = (s i ) i∈Z , t = (t i ) i∈Z ∈ Σ adm we define the relation It is not difficult to see that this is indeed an equivalence relation on the space Σ adm . Furthermore, every itinerary is identified with at most one different itinerary and the quotient space Σ adm / ∼ of Σ adm is well defined. It was also shown that Σ adm / ∼ is homeomorphic to X. So in order to embed X in the plane it is enough to embed Σ adm /∼ in the plane. For all observations in this paragraph we refer to the paper [ Lemma 1] it was observed that A( ← − s ) is indeed an arc (possibly degenerate). For every basic arc we define two quantities as follows: These definitions first appeared in [9] in order to study the number of endpoints of inverse limit spaces X. We now adapt two lemmas from [9] that we will use later in the paper.
Example. Take the unimodal map with the kneading sequence ν = (101) ∞ . Then ← − s = (011) ∞ 010. and , and by Lemma 1, A( ← − s ) and A( ← − t ) have a common boundary point which is projected to T 3 (c), see Figure 1. Note that in this example both τ L and τ R agree for ← − s and ← − t , which need not be the case in general.
A((011) ∞ 010.) Figure 1. Example of two basic arcs having a boundary point in common.

Representation in the plane
This section is the first step towards embedding X in the plane so that an arbitrary point a ∈ X becomes accessible. We denote the symbolic representation of a by . . . l −2 l −1 .l 0 l 1 . . . :=Ī(a), so a ∈ A(. . . l −2 l −1 .). We present the following ordering on {0, 1} −N depending on some L = . . . l −2 l −1 . and we work with this ordering from now onwards.
Note that such ordering is well-defined and the left infinite tail L is the largest sequence.
Proof. If n = 1 the statement follows easily so let us assume that n ≥ 2. Assume that there exists k < n such that u −k = s −k and take k the smallest natural number with this property. Assume without loss of generality that (−1) # 1 (s −(k−1) ...s −1 ) = (−1) # 1 (l −(k−1) ...l −1 ) (the proof follows similarly when the parities are different). Since Let C ⊂ [0, 1] be the middle-third Cantor set, Points in C are coded by the left-infinite sequences of zeros and ones. We embed basic arcs in the plane as horizontal lines along the Cantor set and then join corresponding endpoints with semi-circles as in Figure 1. The ordering has to be defined in a way that semi-circles neither cross horizontal lines nor each other.
Example. For L = 1 ∞ ., points in C are coded as in Figure 2 (a). Note that this is the same ordering as in the paper by Bruin [9]. The ordering obtained by L = 0 ∞ 1. is the ordering from the paper by Brucks & Diamond [7]. In Figure 2 1.
(b) From now onwards let d e denote the Euclidean distance in R 2 .
Now we represent X as the quotient space of the subset of I × C adm for I := [0, 1]. To every point x = (. . . , x −2 , x −1 , x 0 ) ∈ X we will assign either a point or two points in I ×C adm by rule (2) below. From now on, write . . . s −3 s −2 s −1 . := . . . ν −3 (x)ν −2 (x)ν −1 (x). Let ϕ : X → I × C adm be defined in the following way: Set Y := ϕ(X) ⊂ I × C adm . The next step is to identify points in Y in the same way as they are identified in the symbolic representation of X. For a, b ∈ Y : a ∼ b if there exists x ∈ X such that a, b ∈ ϕ(x).
If a = b ∼ a we writeã := b. Ifã = b and x ∈ X is such that a, b ∈ ϕ(x) and s −i = * we say that a and b are joined at level i.
Note that ϕ : X → Y /∼ is a well-defined map. Equip Y with the Euclidean topology and Y /∼ with the standard quotient topology. Let π C : I × C → C and π I : I × C → I denote the natural projections. The next proposition is an analogue of Proposition 4 from [9]. We prove it here for the sake of completeness.
Proof. We first prove that Y /∼ is a Hausdorff space and because X is compact it is enough to check that ϕ is a continuous bijection to obtain a homeomorphism between X and Y /∼, see e.g. Theorem 26.6. in [19].
Now we prove that ϕ is continuous. It is enough to prove that for a ∈ X and a sequence (x n ) n∈N ⊂ X such that lim n→∞ x n = a it holds that lim n→∞ ϕ(x n ) = ϕ(a).
Case I: For every i ∈ N, ν −i (a) = * . If there exists K ∈ N such that for every n ≥ K it follows that ν −j (x n ) = * for every j ∈ N, then there is N ′ ≥ K such that ϕ(x n ) ∈ U for every n ≥ N ′ . Now assume that there exists an increasing sequence (n i ) i∈N ⊂ N such that ν −j (x n i ) = * for some j ∈ N.
Then there exist open sets U n i 1 , U n i 2 ⊂ Y such that ϕ ′ (x n i ) ∈ U n i 1 and ϕ ′′ (x n i ) ∈ U n i 2 and ϕ −1 (U) = U n i 1 ∪ U n i 2 for every i ∈ N. Because x n → a as n → ∞, by the definition of ϕ it follows that ϕ ′ (x n i ) → ϕ(a) and ϕ ′′ (x n i ) → ϕ(a) as i → ∞. Thus we again conclude that there exists N ′ ∈ N such that for every n ≥ N ′ it follows that ϕ(x n ) ∈ U.
Case II: Let K ∈ N be such that ν K (a) = * and thus ϕ(a) = ϕ ′ (a) ∪ ϕ ′′ (a). Take M > K so that ν −M (a) . . . ν M (a) = ν −M (x n ) . . . ν −K (x n ) . . . ν K (x n ) . . . ν M (x n ) for every n ≥ N, and so ϕ(x n ) = ϕ ′ (x n ) ∪ ϕ ′′ (x n ). Thus there exist open sets U 1 , U 2 ⊂ Y such that ϕ ′ (a) ∈ U 1 , ϕ ′′ (a) ∈ U 2 and ϕ −1 (U) = U 1 ∪ U 2 . It follows that there exists N ′ > N such that for every n > N ′ it holds that ϕ ′ (x n ) ∈ U 1 and ϕ ′′ (x n ) ∈ U 2 and thus ϕ(x n ) ∈ U. Now we are ready to represent X in the plane. This is still not an embedding but it is the first step towards it. Connect identified points in I × C adm with semi-circles. Suppose a = b ∈ Y are joined at level n. By Lemma 1, points a and b are both endpoints of basic arcs in I × C adm and are both right or left endpoints. If # 1 (c 1 . . . c n−1 ) is even (odd), a and b are right (left) endpoints and we join them with a semi-circle on the right (left), see Figure 1. Translated to symbolics, this means that there exist n ∈ N and ← − s is even. By Lemma 3, u −(n−1) . . . u −1 = c 1 . . . c n−1 . By Lemma 2 it follows that sup{π I (A)} ≤ T n (c), and thus an intersection between the arc A and a semi-circle cannot occur.
Case II: Assume that we have a crossing of two semi-circles which project to the same point in I. (See Figure 4.)  Thus our ordering gives a representation Y ∪ {semi-circles} of X in the plane. Figure 5 and Figure 6 give two examples of these planar representations.

Embeddings
In this section we show that representations of X constructed in the previous section are indeed embeddings.
Lemma 4. Let U ⊂ R 2 be homeomorphic to the open unit disk, and let W ⊂ R be a closed set such that W × J ⊂ U for some closed interval J. There exists a continuous function f : Note that g(a, ·) is injective for every a ∈ [0, 2], g(0, x) = 0 for all x ∈ [−1/2, 1/2], and where d e (x, W ) = inf w∈W {d e (x, w)}. Note that x → d e (x, W ) is continuous, sof is continuous. Also,f (w, y) = (w, g(0, y)) = (w, 0) for (w, y) ∈ W × J andf is injective otherwise. Also note thatf is the identity on the boundary of [−1, 2] × [−1, 1], sof can be extended continuously to the map f : R 2 → R 2 such that f | R 2 \U is the identity.
Define W n ⊂ R 2 to be the set consisting of all semi-circles that join pairs of points at level n. Note that there exists a set W ⊂ R such that W n is homeomorphic to W × J. Observe that W is closed. Indeed, if for a sequence ( ← − s k ) k∈N ⊂ {0, 1} N there exists    Let M n denote the midpoint of A n . If # 1 (c 1 . . . c n ) is odd, let be the closed left semi-disc centered around (T n (c), M n ). Similarly, if # 1 (c 1 . . . c n ) is even, let be the closed right semi-disc centered around (T n (c), M n ). Note that see Figure 8. For every n ∈ N, the open set contains W n , because otherwise there exists an increasing sequence (i k ) k∈N ⊂ N so that points x i k ∈ {T i k (c)} × G i k and x := lim k x i k ∈ W n . Since x i k ∈ {T i k (c)} × G i k , the corresponding itinerary satisfies τ R ( ← − s i k ) = i k , but because i k → ∞ as k → ∞ this implies that the corresponding itinerary ← − s of x satisfies τ R ( ← − s ) = ∞, a contradiction. Figure 8. Sets constructed in the proof of Lemma 4.
Now define a continuous function f n : R 2 → R 2 as in Lemma 4 replacing U with U n and W with W n . Let F n : R 2 → R 2 be defined as F n := f n • . . . • f 1 for every n ∈ N. We need to show that F := lim n→∞ F n exists and is continuous. It is enough to show the following: Lemma 6. Sequence (F n ) n∈N is uniformly Cauchy.
Denote by Z := Y ∪ {semi-circles} ⊂ R 2 . We want to argue that F (Z) ⊂ R 2 is homeomorphic to Y /∼. Since F : Z → F (Z) is continuous, it follows from [20,Theorem 3.21], that {F −1 (y) : y ∈ F (Z)} is a decomposition of Z homeomorphic to F (Z). Note that this decomposition is exactly Y /∼. Proof of Theorem 1. Assume that the symbolic representation of a = (. . . , a −2 , a −1 , a 0 ) ∈ X is given byĪ(a) = . . . l −2 l −1 .l 0 l 1 . . .. Consider the planar representation Z of X obtained by the ordering on C making L = . . . l −2 l −1 the largest. The point a is represented as (a 0 , 1). Take the arc A = {(a 0 , t + 1), t ∈ [0, 1]} which is a vertical interval in the plane (see Figure 9). Note that A ∩ Z = {a}. Then F (A) is an arc such that F (A) ∩ F (Z) = {F (a)} which concludes the proof.
From now onwards we denote the core inverse limit space lim ← − ([T 2 (c), T (c)], T ) by X ′ . Lemma 7. Let T be a unimodal map such that s := exp(h top (T )) > √ 2. Then the core inverse limit space X ′ is indecomposable.
Proof. The map T is semiconjugate to the tent map T s , and if T is locally eventually onto (leo), then the semiconjugacy h is in fact a conjugacy. In this case, X ′ is indecomposable, see [13]. We give the argument if T is not leo (which also works in the general case). Let p be the orientation reversing fixed point of T , so h(p) = s s+1 is the fixed point of T s . Let J ∋ p be a neighbourhood such that h(J) is a non-degenerate neighbourhood of s s+1 .
Suppose now by contradiction that X ′ = A ∪ B for some proper subcontinua A and B of X ′ . Hence there exists n 0 ∈ N such that the projections π n 0 (A) = [T 2 (c), T (c)] = π n 0 (B). Take n 1 = n 0 + N. Since π n 1 (A) and π n 1 (B) are intervals and π n 1 (A) ∪ π n 1 (B) = [T 2 (c), T (c)], at least one of them, say π n 1 (A), contains at least one of J − or J + . But then π n 0 (A) ⊃ T N (J − ) ∩ T N (J + ) = [T 2 (c), T (c)], contradicting the definition of n 0 . This completes the proof.
Proof of Corollary 1. First assume that exp(h top (T )) > √ 2. By Lemma 7, X ′ is indecomposable and thus it has uncountably many pairwise disjoint composants which are dense in X ′ . By Proposition 2 from [8] every subcontinuum H ⊂ X ′ contains a ray which is dense in H. Therefore, every composant U of X ′ contains a non-degenerate basic arc in X ′ . We embed X so that this non-degenerate basic arc is the largest. Such embedding of X makes a non-degenerate arc of U accessible. Assume that g 1 : X → E 1 and g 2 : X → E 2 are equivalent embeddings, so there exists a homeomorphismh : R 2 → R 2 such thath(E 1 ) = E 2 . It was proven in [3] that every homeomorphism h : X → X is pseudo-isotopic to σ R for some R ∈ Z. This means that h permutes the composants of X ′ in the same way as σ R does. We apply this tõ h = g 2 • h • g −1 1 . Thus if p ∈ E 1 is an accessible point, then q := g 2 • h • g −1 1 (p) is also accessible, and g 2 • σ R • g −1 1 (p) belongs to the same composant of X ′ as q. Hence h maps a composant of X ′ with an accessible non-degenerate arc to a composant of X ′ with an accessible non-degenerate arc. Mazurkiewicz [16] proved that every indecomposable planar continuum contains at most countably many composants with an accessible non-degenerate arc. Combining this with the fact that there are uncountably many composants in X ′ that are not shifts of one another, we finish the proof in case when exp(h top (T )) > √ 2. Now assume that √ 2 ≥ exp(h top (T )) > 1. The core X ′ is decomposable and there exists an indecomposable subcontinuum of X ′ which is homeomorphic to the inverse limit space of a unimodal map with entropy greater than log √ 2. It follows from the arguments above that we can embed this indecomposable subcontinuum in uncountably many non-equivalent ways; therefore we obtain uncountably many non-equivalent embeddings of X.

Remark 4.
Mazurkiewicz' result is perhaps too strong a tool to apply, but to complete the argument we need to know which points are accessible besides those in (the composant of ) the arc which is made the largest. Frequently, this composant is indeed the only accessible composant, but there are many exceptions. Determining which points are accessible is a nontrivial matter, to which we plan to come back in a forthcoming paper.