# A Crossing Lemma for Annular Regions and Invariant Sets with an Application to Planar Dynamical Systems

Authors: Anna Pascoletti; Fabio Zanolin
Journal: Journal of Mathematics
 (2013)

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## Abstract

We present a topological result, namedcrossing lemma, dealing with the existence of a continuum which crosses a topological space between a pair of “opposite” sides. This topological
lemma allows us to obtain some fixed point results. In the works of Pascoletti et al., 2008, and Pascoletti and Zanolin, 2010, we have widely exposed the crossing lemma for planar regions homeomorphic to
a square, and we have also presented some possible applications to the theory of topological horseshoes and to the study of chaotic-like dynamics for planar maps. In this work, we move from the framework of the
generalized rectangles to two other settings (annular regions and invariant sets), trying to obtain similar results. An application to a model of
fluid mixing is given.


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## Body

## 1. Introduction

In 1883-1884, Poincaré [1, 2] introduced a generalization of the Bolzano theorem to continuous vector fields defined on a hypercube of ℝN. Such result, usually known as the Poincaré-Miranda theorem [3–5], in dimension two, reads as follows.Theorem 1.
Letℛ:=[a1,b1]×[a2,b2] be a planar rectangle, and let f:=(f1,f2):ℛ→ℝ2 be a continuous vector field such that f1 changes its sign from the left to the right side of ℛ and f2 changes its sign from the lower to the upper side of ℛ. Then, there exists at least one point w∈ℛ such that f(w)=0.A heuristic proof of this result,“pour faire comprendre comment on peut démontrer ce théorème,” quoting Poincaré’s words [2], can be described as follows. (The more formal proof in [2] makes use of Kronecker integral, a forerunner of topological degree theory (see [5, 6]). We also recommend the recent article [7] for a modern perspective about this topic.) The “curve” f2(x1,x2)=0 starts at some point of the left side x1=a1, and it ends at some point of the right side x1=b1. Similarly, the “curve” f1(x1,x2)=0 starts at some point of the lower side x2=a2, and it ends at some point of the upper side x2=b2. Hence, these “curves” must intersect at some point of the rectangle. Clearly, in modern language, using in this context the term “curve” may be a little misleading, although it well explains the geometry of the problem. Yet the argument of the proof is safe if we note that the set f2(x1,x2)=0 contains a continuum joining the left to the right side and, similarly, the set f1(x1,x2)=0 contains a continuum joining the lower to the upper side. The existence of at least one intersection point between these continua follows from a result of plane topology (see Lemma 7).In order to prove the existence of such continua, one can observe that the setf2(x1,x2)=0 crosses any path from the lower to the upper side of the rectangle (and the same happens for the set f1(x1,x2)=0 with respect to the paths connecting the left to the right side of the rectangle). In [8] we called such kind of results as crossing lemmas and proposed new applications to the study of the dynamics of some planar maps. In an abstract setting we can describe these results as follows. We start from a closed set 𝒮 which cuts all the paths connecting two given sets, and we look for a closed connected subset 𝒞 of 𝒮 with the same cutting property. In many applications, it turns out that the set 𝒞 crosses the original domain “from one side to the other.” A typical example in this direction is presented in Lemma 10 in the framework of topological rectangles, which are topological spaces which are homeomorphic to a planar rectangle.Important theorems, where some forms of these crossing properties are considered, appear in dimension theory with the results of Hurewicz and Wallman [9], in topological games [10, 11], and in proofs of the existence of solutions to nonlinear differential equations [12].In some applications, typically the set𝒮 is a set of solutions of a nonlinear equation depending on a parameter (or equivalently a set of fixed points for a family of parameter-dependent operators). From this point of view, results in this direction, in spite of the fact that they may look quite elementary, present a great usefulness in different areas of nonlinear analysis, especially in connection with bifurcation theory (see [12–15]).When the operators whose fixed points correspond to the element of𝒮 have some special symmetries, it is likely that some of these symmetries are inherited by the set 𝒮 itself. In this case, it would be desirable to prove that also the continuum 𝒞⊂𝒮 inherits the symmetries of 𝒮.In the present paper we move from the framework of generalized rectangles already discussed in previous articles [8, 16] to two other settings, namely, the study of annular regions and the case of invariants sets. More precisely, for the former case, we just collect in Section 2 some classical separation results already applied in [8, 16] to prove the existence of fixed points, periodic points, and chaotic dynamics for planar dynamical systems. In Section 3 and Section 4, respectively, we discuss the variants to annular regions and invariant sets, and, as a possible application of our results, we reconsider in Section 5 an example of planar map arising in fluid mixing previously studied by Kennedy and Yorke in [17].We conclude this introduction with some preliminary results which will be used in what follows. Throughout the paper, by apath in a topological space X, we mean a continuous map γ:I→X, with I:=[0,1]. We also set γ-:=γ(I). Any homeomorphic image of I will be referred to as an arc. The sets B(0,r) and B[0,r] are, respectively, the open and the closed balls of center of origin and radius r in ℝN. In the plane ℝ2 (usually identified with ℂ) with the Euclidean norm, we denote by E2:=B[0,1] the closed unit disc and by S1=∂E2 its boundary.Slightly modifying an analogous definition of Berarducci et al. [18, Definition 2.1], we give the following.Definition 2.
LetX be a topological space, and let A,B⊂X be two nonempty disjoint sets. Let also S⊂X. One says that S  cuts the paths between  A   and  B if S∩γ-≠∅, for every path γ:I→X such that γ(0)∈A and γ(1)∈B.In order to simplify the statements of the next results, we writeS:A∤B to express the fact that S cuts the paths between A and B. If A={a} is a singleton, we write S:a∤B instead of S:{a}∤B. Similar notation is used if B is a singleton. In order to have Definition 2 meaningful, we implicitly assume that there exists at least a path γ in X connecting A with B (otherwise, we could take S=∅ or S any subset of X). Clearly, if a set S satisfies the cutting property of Definition 2, then also its closure clS cuts the paths between A and B. Therefore, in the sequel and without loss of generality, we usually assume S is closed. Such an assumption is also well suited for proving the existence of minimal (closed) sets satisfying Definition 2. Indeed, we have the following.Lemma 3.
LetX be a topological spac,e and let A,B⊂X be two nonempty disjoint sets which are connected by at least one path in X. Let S⊂X be a (nonempty) closed set which cuts the paths between A and B. Then there exists a nonempty, closed set C⊂S which is minimal with respect to the property of cutting the paths between A and B.Proof.
The proof is a standard application of Zorn’s lemma. Letℱ be the set of all the nonempty closed subsets F of S such that F:A∤B, with the elements of ℱ ordered by inclusion. Clearly, ℱ is nonempty for at least S∈ℱ. Let (Fα)α∈J be a totally ordered subset of ℱ. We claim that F*:=⋂α∈JFα∈ℱ. Indeed, let γ:I→X be a path such that γ(0)∈A and γ(1)∈B. The family of compact sets (γ-∩Fα)α∈J has the finite intersection property, and, therefore, γ-∩⋂α∈JFα=γ-∩F*≠∅. This proves our claim, and the conclusion follows from Zorn’s lemma.In [19], Dolcher studied a similar minimality problem, dealing with closed sets separating two points. The definition of separation is the standard one; that is, a set S⊂X separates two points (or, in general, two nonempty sets) if the two points belong to different components of the complement X∖S. In this respect, we reconsider the following example from [19]. Let X⊂ℝ2, with the topology of the plane, be defined by
(1)X:={(x,y):x≥0,y=xn,n∈ℕ0}∪(ℝ+×{0}),
with ℝ+:=[0,+∞) and ℕ0:=ℕ∖{0}.LetA:={(0,0)},B:={(2,0)}, and S:={(x,y)∈X:x=1}. Clearly, S:A∤B, and also S separates A and B in X. As shown in [19], there is no subset of S which is minimal for the property of separating A and B in X. On the other hand, C:={(1,0)} is the minimal subset of S which cuts the paths between A and B.In general, a setS satisfying S:A∤B, even if minimal with respect to such property, need not be connected. For an elementary example, one can take X:=S1 (with the topology of the plane), A:={(-1,0)},B:={(1,0)}, and S:={(0,-1),(0,1)}. In this case, S is a closed set, minimal with respect to the property of cutting the paths between A and B, which is not connected. The connectivity of S (or of a minimal subset of it) is, however, an important property for the proof of the existence of fixed points or of zeros for maps in Euclidean spaces. Such connectivity properties have been employed recently in [20–22] in connection with the theory of topological horseshoes. In order to recall some main results from the above quoted papers and to propose some further developments, we introduce, in the next section, some main definitions which play a crucial role in our approach.


## 2. Generalized Rectangles

Definition 4.
A topological spaceX is called a generalized rectangle if it is homeomorphic to the unit square Q:=I2.For a generalized rectangleX and a homeomorphism η:Q→η(Q)=X, the set η(∂Q) is independent of the choice of the homeomorphism η. We call the set η(∂Q) the contour ofX and denote it by ϑX.  Clearly, for a generalized rectangle X embedded in ℝ2, the contour of X coincides with the boundary of X.Definition 5.
Anoriented rectangle is a pair X~:=(X,X-) such that X is a generalized rectangle and X-⊂ϑX is the union of two disjoint arcs Xl- and Xr-. One says that Xl- and Xr- are, respectively, the left- and the right-hand sides of the (generalized) rectangle X. Sometimes one will also write X:=|X~|.Remark 6.
Given an oriented rectangle,X~=(X,X-), we can choose a homeomorphism η:Q→η(Q)=X in such a manner that
(2)Xl-=η({0}×I),Xr-=η({1}×I)
(this can be proved using the Schoenflies theorem [23]). Then we can also introduce the sets
(3)Xb+=η(I×{0}),Xt+=η(I×{1}),
which are called the bottom and the top sides of X~ and define
(4)X+:=Xb+∪Xt+.
Observe that, for any homeomorphism η1:Q→η1(Q)=X such that η1({0,1}×I)=X-, it follows that η1(I×{0,1})=X+. Thus, the choice of X- determines also the choice of X+ (which is also the closure of ϑX∖X-). Given a pair (X-,X+), the order in which we choose to call as “right” and “left” to the two components of X- (as well as to label the two components of X+) is completely irrelevant for what follows. As a consequence of such remark, we have that, to each oriented rectangle X~, a “dual” oriented rectangle X~′ is associated with |X~|=|X~′|=X and with X+ as the [·]--set for X~′.


### 2.1. Crossing Properties for Generalized Rectangles

The next (classical) result guarantees the fact that continua connecting opposite sides of an oriented rectangle must cross each other. It can be proved as a consequence of the Jordan curve theorem, and the proof is omitted (see [16] for the details and also [24] for an application to ordinary differential equations).Lemma 7.
LetX~=(X,X-) be an oriented rectangle, and let C1,C2⊂X be closed connected sets such that
(5)C1∩Xl-≠∅≠C1∩Xr-,C2∩Xb+≠∅≠C2∩Xt+.
Then C1∩C2≠∅.The connectedness ofC1 and C2 is not enough to guarantee the existence of a nonempty intersection (see [25, Example 10.4]).We present now some results about sets separating the opposite faces of an oriented rectangle. We have shown their role in the proof of the existence of fixed points and periodic points for continuous maps in [8, 16, 26]. Some of these results can be extended to higher dimension using the topological degree, the fixed point index, or other index theories (see [22] and the references therein). Since the applications in the present paper will be all related to planar maps, we prefer to confine ourselves to the use of a more direct tool: Alexander’s lemma. Such result, named after Alexander [27, 28], as shown both in Newman’s book [29] and in Sanderson’s article [30], is quite useful in proving a broad range of theorems of plane topology. Quoting Smith [31] “this lemma, the proof of which requires but a few lines, is shown[⋯] to be one of the sharpest tools in the theory of separation, if skilfully handled.”Results based on applications of Alexander’s lemma or to other related theorems in [29] have been fruitfully applied to differential equations by Hastings [32–34], McLeod and Serrin [35], Turner [36], and others. For more recent applications see [15, 37, 38].The following version of Alexander’s lemma will be used in our next results. The proof requires only an elementary modification of the standard one [39] and, therefore, is omitted.Lemma 8.
LetX~=(X,X-) be an oriented rectangle, and let K1,K2 be closed disjoint subsets of X. Assume that there exist two paths γ1,γ2:I→X, with γ1(0),γ2(0)∈Xl- and γ1(1),γ2(1)∈Xr- such that
(6)γ-1∩K1=∅,γ-2∩K2=∅.
Then, there exists a path γ:I→X, with γ(0)∈Xl- and γ(1)∈Xr- such that γ-∩(K1∪K2)=∅.Remark 9.
The original version of Alexander’s lemma is a variant of Lemma8 in which we assume that all the paths have the same initial point p∈Xl- and the same end point q∈Xr-.The next result is perhaps one of the most classical and useful consequences of the previous lemma (see [29]) and is usually expressed by the fact that “if a closed set separates the plane, then some component of this set separates the plane” [34, page 131].Lemma 10.
LetX~=(X,X-) be an oriented rectangle, and let S⊂X be a closed set such that S:Xl-∤Xr-. Then, there exists a compact, connected set C⊂S such that C:Xl-∤Xr-.Proof.
By Lemma3 there exists a closed set C⊂S such that C:Xl-∤Xr-, with C minimal with respect to the cutting property. Suppose, by contradiction, that C is not connected, and let C1,C2⊂C be two closed nonempty disjoint sets with C1∪C2=C. Since C is minimal and C1,C2 are proper subsets of C, there exist two paths γ1,γ2 in X which connects Xl- to Xr-, such that γi avoids Ci (for i=1,2). Then, by Lemma 8, there exists a path γ:I→X with γ(0)∈Xl- and γ(1)∈Xr- with γ-∩C=∅, contradicting the assumption that C:Xl-∤Xr-.The cutting property forC obtained in Lemma 10 can be equivalently expressed as follows.Lemma 11.
LetX~=(X,X-) be an oriented rectangle, and let C⊂X be a closed connected set. Then, C:Xl-∤Xr- if and only if C∩Xb+≠∅≠C∩Xt+.Proof.
IfC:Xl-∤Xr-, then, necessarily, C must cut the upper and the lower sides of X~ which are the images of particular paths connecting Xl- to Xr-. On the other hand, if γ:I→X is any path with γ(0)∈Xl- and γ(1)∈Xr-, then γ- and C are two continua connecting the opposite sides of the oriented rectangle and, therefore, γ-∩C≠∅ by Lemma 7. This proves that C:Xl-∤Xr-.Lemma11, as well as Lemma 7, can be seen as a continuous version of the so-called Hex theorem asserting that Hex cannot end in a tie [11].The combination of Lemma10 and Lemma 11 gives the so-called Crossing Lemma asserting the fact that if a closed set intersects all the paths from the left to the right side in an oriented rectangle, then it contains a continuum connecting the two other sides. Note also that such a continuum can be taken as irreducible betweenXb+  and  Xt+ (by using some classical results from [13, 40]). See [16] for a recent survey on this subject and its connections with various different results of plane topology, as well as for a different proof based on Whyburn’s lemma.


## 2.1. Crossing Properties for Generalized Rectangles

The next (classical) result guarantees the fact that continua connecting opposite sides of an oriented rectangle must cross each other. It can be proved as a consequence of the Jordan curve theorem, and the proof is omitted (see [16] for the details and also [24] for an application to ordinary differential equations).Lemma 7.
LetX~=(X,X-) be an oriented rectangle, and let C1,C2⊂X be closed connected sets such that
(5)C1∩Xl-≠∅≠C1∩Xr-,C2∩Xb+≠∅≠C2∩Xt+.
Then C1∩C2≠∅.The connectedness ofC1 and C2 is not enough to guarantee the existence of a nonempty intersection (see [25, Example 10.4]).We present now some results about sets separating the opposite faces of an oriented rectangle. We have shown their role in the proof of the existence of fixed points and periodic points for continuous maps in [8, 16, 26]. Some of these results can be extended to higher dimension using the topological degree, the fixed point index, or other index theories (see [22] and the references therein). Since the applications in the present paper will be all related to planar maps, we prefer to confine ourselves to the use of a more direct tool: Alexander’s lemma. Such result, named after Alexander [27, 28], as shown both in Newman’s book [29] and in Sanderson’s article [30], is quite useful in proving a broad range of theorems of plane topology. Quoting Smith [31] “this lemma, the proof of which requires but a few lines, is shown[⋯] to be one of the sharpest tools in the theory of separation, if skilfully handled.”Results based on applications of Alexander’s lemma or to other related theorems in [29] have been fruitfully applied to differential equations by Hastings [32–34], McLeod and Serrin [35], Turner [36], and others. For more recent applications see [15, 37, 38].The following version of Alexander’s lemma will be used in our next results. The proof requires only an elementary modification of the standard one [39] and, therefore, is omitted.Lemma 8.
LetX~=(X,X-) be an oriented rectangle, and let K1,K2 be closed disjoint subsets of X. Assume that there exist two paths γ1,γ2:I→X, with γ1(0),γ2(0)∈Xl- and γ1(1),γ2(1)∈Xr- such that
(6)γ-1∩K1=∅,γ-2∩K2=∅.
Then, there exists a path γ:I→X, with γ(0)∈Xl- and γ(1)∈Xr- such that γ-∩(K1∪K2)=∅.Remark 9.
The original version of Alexander’s lemma is a variant of Lemma8 in which we assume that all the paths have the same initial point p∈Xl- and the same end point q∈Xr-.The next result is perhaps one of the most classical and useful consequences of the previous lemma (see [29]) and is usually expressed by the fact that “if a closed set separates the plane, then some component of this set separates the plane” [34, page 131].Lemma 10.
LetX~=(X,X-) be an oriented rectangle, and let S⊂X be a closed set such that S:Xl-∤Xr-. Then, there exists a compact, connected set C⊂S such that C:Xl-∤Xr-.Proof.
By Lemma3 there exists a closed set C⊂S such that C:Xl-∤Xr-, with C minimal with respect to the cutting property. Suppose, by contradiction, that C is not connected, and let C1,C2⊂C be two closed nonempty disjoint sets with C1∪C2=C. Since C is minimal and C1,C2 are proper subsets of C, there exist two paths γ1,γ2 in X which connects Xl- to Xr-, such that γi avoids Ci (for i=1,2). Then, by Lemma 8, there exists a path γ:I→X with γ(0)∈Xl- and γ(1)∈Xr- with γ-∩C=∅, contradicting the assumption that C:Xl-∤Xr-.The cutting property forC obtained in Lemma 10 can be equivalently expressed as follows.Lemma 11.
LetX~=(X,X-) be an oriented rectangle, and let C⊂X be a closed connected set. Then, C:Xl-∤Xr- if and only if C∩Xb+≠∅≠C∩Xt+.Proof.
IfC:Xl-∤Xr-, then, necessarily, C must cut the upper and the lower sides of X~ which are the images of particular paths connecting Xl- to Xr-. On the other hand, if γ:I→X is any path with γ(0)∈Xl- and γ(1)∈Xr-, then γ- and C are two continua connecting the opposite sides of the oriented rectangle and, therefore, γ-∩C≠∅ by Lemma 7. This proves that C:Xl-∤Xr-.Lemma11, as well as Lemma 7, can be seen as a continuous version of the so-called Hex theorem asserting that Hex cannot end in a tie [11].The combination of Lemma10 and Lemma 11 gives the so-called Crossing Lemma asserting the fact that if a closed set intersects all the paths from the left to the right side in an oriented rectangle, then it contains a continuum connecting the two other sides. Note also that such a continuum can be taken as irreducible betweenXb+  and  Xt+ (by using some classical results from [13, 40]). See [16] for a recent survey on this subject and its connections with various different results of plane topology, as well as for a different proof based on Whyburn’s lemma.


## 3. Annular Regions

In this section, we are moving our attention from generalized rectangles to planar annuli, trying to develop analogous results of Section2 in this new setting. First of all, we present some basic definitions.A closed planar annulus (of inner radiusa>0 and outer radius b>a) is the set
(7)A[a,b]:={(x,y)∈ℝ2:a≤∥(x,y)∥≤b}.Definition 12.
A topological spaceX is called a topological annulus if it is homeomorphic to a planar annulus.LetX be a topological annulus, and let η:A[1,2]→η(A[1,2])=X be a homeomorphism. As a consequence of the Schoenflies theorem, the set η(∂A[1,2]) is independent of the choice of the homeomorphism η. We call the set η(∂A[1,2]) the contour ofX and denote it by ϑX. Clearly, for a topological annulus X embedded in ℝ2, the contour of X coincides with the boundary of X. The contour of X consists of two connected components which are closed arcs (the Jordan curves) since they are homeomorphic to S1. We call such closed arcs Xi and Xo. For a planarly embedded topological annulus, they could be chosen as the inner and the outer boundaries of the annulus. In such a special case, the bounded component of ℝ2∖X turns out to be an open simply connected set D=D(X) with ∂D=Xi and clD=D∪Xi homeomorphic to the closed unit disc. On the other hand, in the general setting, speaking of inner and outer boundaries is meaningless, yet we keep this terminology. Finally, we define the interior ofX as
(8)intX:=X∖ϑX.In the sequel, we use the standard covering projection ofℝ×ℝ0+ onto ℝ2∖{0}, defined by the polar coordinates
(9)Π(θ,ρ)=(ρcosθ,ρsinθ).
In the space ℝ×ℝ0+, we consider also the translation
(10)⅁:(θ,ρ)↦(θ+2π,ρ).


### 3.1. Crossing Properties for Annular Regions

Our aim now is to reconsider the results obtained for topological rectangles in Section2 and adapt them to a form which may be better suited to deal with the new setting of topological annuli. We start with a version of Alexander’s lemma which reads as follows.Lemma 13.
LetX be a topological annulus, and let K1,K2 be closed subsets of X such that K1∩K2 is connected (possibly empty). Assume that there exist two paths γ1,γ2:I→X, with γ1(0),γ2(0)∈Xi and γ1(1),γ2(1)∈Xo such that
(11)γ-1∩K1=∅,γ-2∩K2=∅.
Then, there exists a path γ:I→X, with γ(0)∈Xi and γ(1)∈Xo such that γ-∩(K1∪K2)=∅.This result is an elementary variant of the version of Alexander’s lemma in [29]; see also [38, page 3040] and [37, Corollary 3].We say that a setC⊂X is essentially embedded inX if the inclusion
(12)iC:C⟶X,iC(x)=x,∀x∈C
is not homotopic to a constant map.The next result is a corollary of the Borsuk separation theorem [41, Theorem 6-47] adapted to our situation. We give a proof, for completeness, following [41].Lemma 14.
LetX be a topological annulus, and let S⊂X be a closed set. Then, S is essentially embedded in X if and only if S:Xi∤Xo.Proof.
Up to a homeomorphism defining the annulusX, we can assume X=A[a,b] with
(13)0<a<b<1.
In this case, Xi=∂B(0,a)=aS1 and Xo=∂B(0,b)=bS1.
Suppose thatS:Xi∤Xo, and let 𝒞(0) be the connected component of ℝ2∖S containing the origin. By the assumption, 𝒞(0)∪S is closed and
(14)B[0,a]⊂𝒞(0)∪S⊂B[0,b].
Assume, by contradiction, that S is not essentially embedded in X, and, therefore, the inclusion iS:S→X is homotopic in X to a constant map, say X∋x↦p, for all x∈X, for a suitable point p∈X. From this, it follows immediately that the mapping f:S→S1 defined by x↦x/∥x∥ is homotopic to a constant, which is inessential. Then, by [41, Theorem 4-5], there exists a continuous and inessential extension f~ of f with f~ defined on 𝒞(0)∪S. We define now the map r:E2→S1 by
(15)r(x):={f~(x),forx∈𝒞(0)∪Sx∥x∥,forx∉𝒞(0)∪S
which is continuous. Thus, we are led to a contradiction since r(·) turns out to be a retraction of E2 onto S1.
Suppose now thatS is essentially embedded in X, and also assume, by contradiction, that there exists a path γ:I→X with ∥γ(0)∥=a,∥γ(1)∥=b and that γ(t)∉S, for all t∈I. Without loss of generality (taking possibly an arc inside γ-), we can suppose that  γ is one to one and that a<∥γ(t)∥<b for each t∈]0,1[.
Passing to the covering spaceℋ:=ℝ×[a,b] of X=A[a,b], we have that the path γ lifts to a family of paths γ~n:I→ℋ with γ~n(t)=γ~0(t)+(2nπ,0) and that γ~n(t)∩Π-1(S)=∅, for all t∈I and every n∈ℤ. By the properties of γ we have that γ~0:I→ℋ is one to one and, moreover, γ~0(t′)-γ~0(t′′)∉2πℤ for t′≠t′′ (because Π∘γ~0=γ is one to one). We have thus obtained an arc Γ0:=γ~0(I)⊂ℋ connecting the lines ρ=a and ρ=b and avoiding Π-1(S) such that Γ0 intersects the line {ρ=a} exactly at one point and the same happens with respect to {ρ=b}. Furthermore,
(16)Γ0∩((2πn,0)+Γ0)=∅,∀n∈ℤ∖{0}.
Let Γ0∩{ρ=a}={Pa},Γ0∩{ρ=b}={Pb}, and Γ1:=(2π,0)+Γ0. Note that Γ0∩Γ1=∅. Let 𝒥 be the Jordan curve obtained by joining (in the counterclockwise sense) the point Pa to (2π,0)+Pa along the line {ρ=a}, the point (2π,0)+Pa to (2π,0)+Pb along Γ1, the point (2π,0)+Pb to Pb along the line {ρ=b}, and, finally, the point Pb to Pa along Γ0. The curve 𝒥 is the boundary of an open bounded domain D with clD=D∪𝒥 homeomorphic to the unit square Q. Roughly speaking, clD is the set of all the points of the strip ℋ between Γ0 and Γ1, with the boundary arcs included. Let η:Q→η(Q)=clD be a homeomorphism mapping the left side of Q to Γ0, the lower side of Q to the segment {Pa+(θ,0):θ∈[0,2π]}, the right side of Q to Γ1, and the upper side of Q to the segment {Pb+(θ,0):θ∈[0,2π]}. By the construction of the topological rectangle clD and since Γ0∩Π-1(S)=∅, we have that the set
(17)S′:=Π-1(S)∩η(]0,1[×[0,1])⊂clD
is mapped homeomorphically onto S by the covering projection Π. Now we choose ε∈]0,1/2[ sufficiently small such that
(18)S′⊂Π-1(S)∩η([ε,1-ε]×[0,1]),
and we also introduce the set
(19)B:=Π(η([ε,1-ε]×[0,1])).
By construction, the set B is a topological rectangle contained in A[a,b] and containing the set S. The continuous map
(20)([ε,1-ε]×[0,1])×[0,1]∋(z,λ)↦Π(η((1-λ)z+λ(12,12)))⊂A[a,b]
restricted to S×I provides a homotopy between the identity iS and a constant map. This contradicts the assumption that S is essentially embedded in A[a,b], and the proof is complete.
A variant of the above proof has been suggested by the referee, and we wish to report it here, since it may be more interesting from a geometrical point of view. The argument is as follows.
Having obtained (as above)γ:I→A[a,b]=X which is one to one, avoids the set S, and intersects the inner and the outer boundaries of X only at its extreme ends γ(0) and γ(1), we consider a homeomorphism h of the annulus onto itself such that h(γ(I))={(x,0):x∈[a,b]},h(γ(0))=(a,0), and h(γ(1))=(b,0). Moreover, the segment [a,b]×{0} avoids the set h(S). We lift the segment [a,b]×{0} to sequence of parallel vertical lines ⋃n∈ℤ‍{2nπ}×[a,b]⊂ℋ which avoid the set Π-1(h(S)). Now the topological rectangle clD of the preceding part of the proof turns out to be just the standard rectangle [0,2π]×[a,b], and the set T′:=Π-1(h(S))∩(]0,2π[×[a,b]) (which is mapped homeomorphically by Π onto h(S)) is actually contained in [ε,2π-ε]×[a,b] for some ε>0 and sufficiently small. Now we conclude precisely as before by showing that ih(S) is homotopic to a constant in the rectangle Π([ε,2π-ε]×[a,b])⊂A[a,b].A version of Lemma10 for a topological annulus reads as follows.Lemma 15.
LetX be a topological annulus, and let S⊂X be a closed set such that S:Xi∤Xo. Then, there exists a compact, connected set C⊂S such that C:Xi∤Xo (and, therefore, C is essentially embedded in X).Proof.
The proof follows the same argument of the one of Lemma10. We give the details for completeness. By Lemma 3, there exists a closed set C⊂S such that C:Xi∤Xo, with C minimal with respect to the cutting property. Suppose, by contradiction, that C is not connected, and let C1,C2⊂C be two closed nonempty disjoint sets with C1∪C2=C. Since C is minimal and C1,C2 are proper subsets of C, there exist two paths γ1,γ2 in X which connect Xi to Xo such that γi avoids Ci (for i=1,2). Then, by Lemma 13, there exists a path γ:I→X with γ(0)∈Xi and γ(1)∈Xo with γ-∩C=∅, contradicting the assumption that C:Xi∤Xo. The continuum C is also essentially embedded in X by Lemma 14.The result in Lemma15 has been proved using a minimality argument. In some cases, the minimality of the set C may be useful for the proof of some topological properties of the continuum. An example in this direction is given in the next lemma.Lemma 16.
LetX be a topological annulus, and let C⊂X be a compact connected set which is minimal with respect to the property of cutting all the paths in X from Xi to Xo. Let f:C→ℝ be a nonconstant continuous function. Then, for every k∈]minf(C),maxf(C)[ there exist at least two points w,z∈C with w≠z such that f(w)=f(z)=k.Proof.
Without loss of generality, we can assume thatk=0 and f changes sign on C. The existence of at least a zero for f∣C follows from Bolzano’s theorem. Suppose, by contradiction, that there is only one point say z∈C such that f(z)=0. Consider the two nonempty compact sets K1:={x∈C:f(x)≤0} and K2:={x∈C:f(x)≥0}. By the assumption, we have that {z}=K1∩K2 and K1≠C as well as K2≠C. By the assumption of minimality of C, it follows that there exists a path γ1 connecting Xi to Xo in X and avoiding K1 and, similarly, there exists a path γ2 connecting Xi to Xo in X and avoiding K2. From Lemma 13 we know that there exists a path γ in X connecting Xi to Xo and avoiding C=K1∪K2. This contradicts the cutting property of C.We observe that the existence of two solutions is not guaranteed if the minimality of the setC is not assumed (see [42, Example 2.8]).In the case of topological rectangles, there is a complete symmetry between the fact that a set cuts the paths between a given pair of opposite sides or it cuts the paths connecting a complementary pair of opposite sides. Thus, once we have achieved a result like Lemma11, also its dual version, involving the other pair of sides, is guaranteed. In the case of topological annuli, the situation is different. We have just proved a result which expresses the fact that a compact set which crosses all the paths from the inner to the outer boundary must contain a continuum which nontrivially winds around the annulus. A dual result should express the fact that if a compact set intersects all the nontrivial loops of the annulus, then it must contain a continuum joining the inner and the outer boundaries of the annulus. This is precisely the content of the next lemma. To this end, we first recall some basic facts from homotopy theory. Let ω:I→X be a loop, which is a continuous path such that ω(0)=ω(1). We say that ω is (homotopically) trivial in X if it is homotopic in X to the constant loop ex0:I→x0 with x0=ω(0)=ω(1). Since a loop in X (up to a change in the parameter) may be also seen as a continuous map ω:S1→X, triviality of ω can be also expressed by the fact that there is a continuous extension α:E2→X with αS1=ω. We say that a loop ω in X is nontrivial if it is not homotopically trivial in X. Then we have the following.Lemma 17.
LetX be a topological annulus, and let S⊂X be a closed set such that S∩ω-≠∅ for each nontrivial loop ω in X. Then, there exists a compact, connected set C⊂S such that
(21)C∩Xi≠∅≠C∩Xo.Proof.
Without loss of generality (up to a homeomorphism), we suppose thatX=A[a,b] with 0<a<b, so that X has inner boundary Xi=aS1 and outer boundary Xo=bS1. By the assumption of crossing the nontrivial loops, we know that Si:=S∩Xi≠∅ and also So:=S∩Xo≠∅.
Suppose, by contradiction, thatS does not contain any compact connected set C satisfying (21). Then, by the Kuratowski-Whyburn lemma [13, 40], it follows that S splits as the disjoint union of two compact sets S′,S′′ with S′⊃Si and S′′⊃So. We pass now to the covering space ℋ:=ℝ×[a,b] of X=A[a,b] and consider the closed subsets of ℋ,
(22)W′:=Π-1(S′)∪(ℝ×{a}),W′′:=Π-1(S′′)∪(ℝ×{b}).
By definition, Π(W′∪W′′)⊃S. Moreover, W′∩W′′=∅, and both such sets are invariant with respect to the translation ⅁ defined in (10). We define
(23)δ≔dist(W′,W′′)=inf{∥w′-w′′∥:w′∈W′,w′′∈W′′}.
It is clear that δ>0, and it is actually a minimum (this follows from a standard compactness argument, using the periodicity). Then, we define the two closed ε-tubular neighborhoods of W′ and W′′ as W′[ε]:={y=(θ,ρ)∈ℋ:dist(y,W′)≤ε} and W′′[ε]:={y=(θ,ρ)∈ℋ:dist(y,W′′)≤ε}, for
(24)0<ε≤δ3.
We modify now a pigeonhole argument used in the proof of Theorem 1 in [10] as follows. Let us fix a positive integer N>(b-a)/ε and consider the rectangle
(25)ℛ:=[0,2Nπ]×[a,b].
The sets
(26)K1:=W′[ε]∩ℛ,K2:=W′′[ε]∩ℛ
are closed and disjoint. The lower edge [0,2Nπ]×{a} is the image of a path connecting the left to the right side of ℛ and avoiding the set K2. Similarly, via the upper edge [0,2Nπ]×{b} we find a path connecting the left to the right side of ℛ and avoiding the set K1. Alexander’s lemma (Lemma 8) guarantees the existence of a path γ=(θ(t),ρ(t)):I→ℛ with θ(0)=0 and θ(1)=2Nπ, such that γ(t)∉K1∪K2 for all t∈I. For each i=0,…,N, let
(27)ti:=min{t∈I:θ(t)=2iπ},
so that
(28)0=t0<t1<⋯<tN≤1
and the N+1-tuple of points (ρ(t0),ρ(t1),…,ρ(tN)) in  ]a,b[ is well defined. Clearly, by the choice of N such that Nε>(b-a), there exists at least a pair of points (tj,tk) with j<k such that |ρ(tj)-ρ(tk)|<ε. Since
(29)mint∈[0,1]{dist(γ(t),W′∪W′′)}>ε,
we conclude that the segment joining the points (2kπ,ρ(tj)) and (2kπ,ρ(tk)) does not intersect the set W′∪W′′ (of course, such statement is trivial if ρ(tj)=ρ(tk)). We can now define the path(30)ω~(s)≔{γ(tj+2s(tk-tj)),for0≤s≤12,(2kπ,ρ(tk)+(2s-1)(ρ(tj)-ρ(tk))),for12≤s≤1,which takes values in ℛ∖(W′∪W′′) and is such that
(31)ω~(0)=(2jπ,ρ(tj)),ω~(1)=(2kπ,ρ(tj)).
Hence, the projection ω:=Π∘ω~:I→X is a nontrivial loop in X (in fact, it corresponds to k-j∈ℤ∖{0} in the fundamental group of X) and, by construction, ω(t)∉S,forallt∈I. This contradicts the hypothesis, and hence, the conclusion follows.Our last result in this section can be seen as a continuous version of the no-tie theorem for the Hex game on the annulus. See [10] for a discrete version of this result.Lemma 18.
LetX be a topological annulus, and let C1,C2⊂X be closed connected sets such that
(32)C1∩Xi≠∅≠C1∩Xo,C2isessentiallyembeddedinX.
Then, C1∩C2≠∅.Proof.
Assume, by contradiction, thatC1∩C2=∅, and let dist(C1,C2)=δ>0. In a δ/2-neighborhood of C1, we can find the image of a path γ:I→X with γ(0)∈Xi and γ(1)∈Xo. By construction, γ(t)∉C2, for all t∈I. This proves that it is not true that C2:Xi∤Xo, and therefore (by Lemma 14) C2 is not essentially embedded in X, thus contradicting one of our assumptions.


## 3.1. Crossing Properties for Annular Regions

Our aim now is to reconsider the results obtained for topological rectangles in Section2 and adapt them to a form which may be better suited to deal with the new setting of topological annuli. We start with a version of Alexander’s lemma which reads as follows.Lemma 13.
LetX be a topological annulus, and let K1,K2 be closed subsets of X such that K1∩K2 is connected (possibly empty). Assume that there exist two paths γ1,γ2:I→X, with γ1(0),γ2(0)∈Xi and γ1(1),γ2(1)∈Xo such that
(11)γ-1∩K1=∅,γ-2∩K2=∅.
Then, there exists a path γ:I→X, with γ(0)∈Xi and γ(1)∈Xo such that γ-∩(K1∪K2)=∅.This result is an elementary variant of the version of Alexander’s lemma in [29]; see also [38, page 3040] and [37, Corollary 3].We say that a setC⊂X is essentially embedded inX if the inclusion
(12)iC:C⟶X,iC(x)=x,∀x∈C
is not homotopic to a constant map.The next result is a corollary of the Borsuk separation theorem [41, Theorem 6-47] adapted to our situation. We give a proof, for completeness, following [41].Lemma 14.
LetX be a topological annulus, and let S⊂X be a closed set. Then, S is essentially embedded in X if and only if S:Xi∤Xo.Proof.
Up to a homeomorphism defining the annulusX, we can assume X=A[a,b] with
(13)0<a<b<1.
In this case, Xi=∂B(0,a)=aS1 and Xo=∂B(0,b)=bS1.
Suppose thatS:Xi∤Xo, and let 𝒞(0) be the connected component of ℝ2∖S containing the origin. By the assumption, 𝒞(0)∪S is closed and
(14)B[0,a]⊂𝒞(0)∪S⊂B[0,b].
Assume, by contradiction, that S is not essentially embedded in X, and, therefore, the inclusion iS:S→X is homotopic in X to a constant map, say X∋x↦p, for all x∈X, for a suitable point p∈X. From this, it follows immediately that the mapping f:S→S1 defined by x↦x/∥x∥ is homotopic to a constant, which is inessential. Then, by [41, Theorem 4-5], there exists a continuous and inessential extension f~ of f with f~ defined on 𝒞(0)∪S. We define now the map r:E2→S1 by
(15)r(x):={f~(x),forx∈𝒞(0)∪Sx∥x∥,forx∉𝒞(0)∪S
which is continuous. Thus, we are led to a contradiction since r(·) turns out to be a retraction of E2 onto S1.
Suppose now thatS is essentially embedded in X, and also assume, by contradiction, that there exists a path γ:I→X with ∥γ(0)∥=a,∥γ(1)∥=b and that γ(t)∉S, for all t∈I. Without loss of generality (taking possibly an arc inside γ-), we can suppose that  γ is one to one and that a<∥γ(t)∥<b for each t∈]0,1[.
Passing to the covering spaceℋ:=ℝ×[a,b] of X=A[a,b], we have that the path γ lifts to a family of paths γ~n:I→ℋ with γ~n(t)=γ~0(t)+(2nπ,0) and that γ~n(t)∩Π-1(S)=∅, for all t∈I and every n∈ℤ. By the properties of γ we have that γ~0:I→ℋ is one to one and, moreover, γ~0(t′)-γ~0(t′′)∉2πℤ for t′≠t′′ (because Π∘γ~0=γ is one to one). We have thus obtained an arc Γ0:=γ~0(I)⊂ℋ connecting the lines ρ=a and ρ=b and avoiding Π-1(S) such that Γ0 intersects the line {ρ=a} exactly at one point and the same happens with respect to {ρ=b}. Furthermore,
(16)Γ0∩((2πn,0)+Γ0)=∅,∀n∈ℤ∖{0}.
Let Γ0∩{ρ=a}={Pa},Γ0∩{ρ=b}={Pb}, and Γ1:=(2π,0)+Γ0. Note that Γ0∩Γ1=∅. Let 𝒥 be the Jordan curve obtained by joining (in the counterclockwise sense) the point Pa to (2π,0)+Pa along the line {ρ=a}, the point (2π,0)+Pa to (2π,0)+Pb along Γ1, the point (2π,0)+Pb to Pb along the line {ρ=b}, and, finally, the point Pb to Pa along Γ0. The curve 𝒥 is the boundary of an open bounded domain D with clD=D∪𝒥 homeomorphic to the unit square Q. Roughly speaking, clD is the set of all the points of the strip ℋ between Γ0 and Γ1, with the boundary arcs included. Let η:Q→η(Q)=clD be a homeomorphism mapping the left side of Q to Γ0, the lower side of Q to the segment {Pa+(θ,0):θ∈[0,2π]}, the right side of Q to Γ1, and the upper side of Q to the segment {Pb+(θ,0):θ∈[0,2π]}. By the construction of the topological rectangle clD and since Γ0∩Π-1(S)=∅, we have that the set
(17)S′:=Π-1(S)∩η(]0,1[×[0,1])⊂clD
is mapped homeomorphically onto S by the covering projection Π. Now we choose ε∈]0,1/2[ sufficiently small such that
(18)S′⊂Π-1(S)∩η([ε,1-ε]×[0,1]),
and we also introduce the set
(19)B:=Π(η([ε,1-ε]×[0,1])).
By construction, the set B is a topological rectangle contained in A[a,b] and containing the set S. The continuous map
(20)([ε,1-ε]×[0,1])×[0,1]∋(z,λ)↦Π(η((1-λ)z+λ(12,12)))⊂A[a,b]
restricted to S×I provides a homotopy between the identity iS and a constant map. This contradicts the assumption that S is essentially embedded in A[a,b], and the proof is complete.
A variant of the above proof has been suggested by the referee, and we wish to report it here, since it may be more interesting from a geometrical point of view. The argument is as follows.
Having obtained (as above)γ:I→A[a,b]=X which is one to one, avoids the set S, and intersects the inner and the outer boundaries of X only at its extreme ends γ(0) and γ(1), we consider a homeomorphism h of the annulus onto itself such that h(γ(I))={(x,0):x∈[a,b]},h(γ(0))=(a,0), and h(γ(1))=(b,0). Moreover, the segment [a,b]×{0} avoids the set h(S). We lift the segment [a,b]×{0} to sequence of parallel vertical lines ⋃n∈ℤ‍{2nπ}×[a,b]⊂ℋ which avoid the set Π-1(h(S)). Now the topological rectangle clD of the preceding part of the proof turns out to be just the standard rectangle [0,2π]×[a,b], and the set T′:=Π-1(h(S))∩(]0,2π[×[a,b]) (which is mapped homeomorphically by Π onto h(S)) is actually contained in [ε,2π-ε]×[a,b] for some ε>0 and sufficiently small. Now we conclude precisely as before by showing that ih(S) is homotopic to a constant in the rectangle Π([ε,2π-ε]×[a,b])⊂A[a,b].A version of Lemma10 for a topological annulus reads as follows.Lemma 15.
LetX be a topological annulus, and let S⊂X be a closed set such that S:Xi∤Xo. Then, there exists a compact, connected set C⊂S such that C:Xi∤Xo (and, therefore, C is essentially embedded in X).Proof.
The proof follows the same argument of the one of Lemma10. We give the details for completeness. By Lemma 3, there exists a closed set C⊂S such that C:Xi∤Xo, with C minimal with respect to the cutting property. Suppose, by contradiction, that C is not connected, and let C1,C2⊂C be two closed nonempty disjoint sets with C1∪C2=C. Since C is minimal and C1,C2 are proper subsets of C, there exist two paths γ1,γ2 in X which connect Xi to Xo such that γi avoids Ci (for i=1,2). Then, by Lemma 13, there exists a path γ:I→X with γ(0)∈Xi and γ(1)∈Xo with γ-∩C=∅, contradicting the assumption that C:Xi∤Xo. The continuum C is also essentially embedded in X by Lemma 14.The result in Lemma15 has been proved using a minimality argument. In some cases, the minimality of the set C may be useful for the proof of some topological properties of the continuum. An example in this direction is given in the next lemma.Lemma 16.
LetX be a topological annulus, and let C⊂X be a compact connected set which is minimal with respect to the property of cutting all the paths in X from Xi to Xo. Let f:C→ℝ be a nonconstant continuous function. Then, for every k∈]minf(C),maxf(C)[ there exist at least two points w,z∈C with w≠z such that f(w)=f(z)=k.Proof.
Without loss of generality, we can assume thatk=0 and f changes sign on C. The existence of at least a zero for f∣C follows from Bolzano’s theorem. Suppose, by contradiction, that there is only one point say z∈C such that f(z)=0. Consider the two nonempty compact sets K1:={x∈C:f(x)≤0} and K2:={x∈C:f(x)≥0}. By the assumption, we have that {z}=K1∩K2 and K1≠C as well as K2≠C. By the assumption of minimality of C, it follows that there exists a path γ1 connecting Xi to Xo in X and avoiding K1 and, similarly, there exists a path γ2 connecting Xi to Xo in X and avoiding K2. From Lemma 13 we know that there exists a path γ in X connecting Xi to Xo and avoiding C=K1∪K2. This contradicts the cutting property of C.We observe that the existence of two solutions is not guaranteed if the minimality of the setC is not assumed (see [42, Example 2.8]).In the case of topological rectangles, there is a complete symmetry between the fact that a set cuts the paths between a given pair of opposite sides or it cuts the paths connecting a complementary pair of opposite sides. Thus, once we have achieved a result like Lemma11, also its dual version, involving the other pair of sides, is guaranteed. In the case of topological annuli, the situation is different. We have just proved a result which expresses the fact that a compact set which crosses all the paths from the inner to the outer boundary must contain a continuum which nontrivially winds around the annulus. A dual result should express the fact that if a compact set intersects all the nontrivial loops of the annulus, then it must contain a continuum joining the inner and the outer boundaries of the annulus. This is precisely the content of the next lemma. To this end, we first recall some basic facts from homotopy theory. Let ω:I→X be a loop, which is a continuous path such that ω(0)=ω(1). We say that ω is (homotopically) trivial in X if it is homotopic in X to the constant loop ex0:I→x0 with x0=ω(0)=ω(1). Since a loop in X (up to a change in the parameter) may be also seen as a continuous map ω:S1→X, triviality of ω can be also expressed by the fact that there is a continuous extension α:E2→X with αS1=ω. We say that a loop ω in X is nontrivial if it is not homotopically trivial in X. Then we have the following.Lemma 17.
LetX be a topological annulus, and let S⊂X be a closed set such that S∩ω-≠∅ for each nontrivial loop ω in X. Then, there exists a compact, connected set C⊂S such that
(21)C∩Xi≠∅≠C∩Xo.Proof.
Without loss of generality (up to a homeomorphism), we suppose thatX=A[a,b] with 0<a<b, so that X has inner boundary Xi=aS1 and outer boundary Xo=bS1. By the assumption of crossing the nontrivial loops, we know that Si:=S∩Xi≠∅ and also So:=S∩Xo≠∅.
Suppose, by contradiction, thatS does not contain any compact connected set C satisfying (21). Then, by the Kuratowski-Whyburn lemma [13, 40], it follows that S splits as the disjoint union of two compact sets S′,S′′ with S′⊃Si and S′′⊃So. We pass now to the covering space ℋ:=ℝ×[a,b] of X=A[a,b] and consider the closed subsets of ℋ,
(22)W′:=Π-1(S′)∪(ℝ×{a}),W′′:=Π-1(S′′)∪(ℝ×{b}).
By definition, Π(W′∪W′′)⊃S. Moreover, W′∩W′′=∅, and both such sets are invariant with respect to the translation ⅁ defined in (10). We define
(23)δ≔dist(W′,W′′)=inf{∥w′-w′′∥:w′∈W′,w′′∈W′′}.
It is clear that δ>0, and it is actually a minimum (this follows from a standard compactness argument, using the periodicity). Then, we define the two closed ε-tubular neighborhoods of W′ and W′′ as W′[ε]:={y=(θ,ρ)∈ℋ:dist(y,W′)≤ε} and W′′[ε]:={y=(θ,ρ)∈ℋ:dist(y,W′′)≤ε}, for
(24)0<ε≤δ3.
We modify now a pigeonhole argument used in the proof of Theorem 1 in [10] as follows. Let us fix a positive integer N>(b-a)/ε and consider the rectangle
(25)ℛ:=[0,2Nπ]×[a,b].
The sets
(26)K1:=W′[ε]∩ℛ,K2:=W′′[ε]∩ℛ
are closed and disjoint. The lower edge [0,2Nπ]×{a} is the image of a path connecting the left to the right side of ℛ and avoiding the set K2. Similarly, via the upper edge [0,2Nπ]×{b} we find a path connecting the left to the right side of ℛ and avoiding the set K1. Alexander’s lemma (Lemma 8) guarantees the existence of a path γ=(θ(t),ρ(t)):I→ℛ with θ(0)=0 and θ(1)=2Nπ, such that γ(t)∉K1∪K2 for all t∈I. For each i=0,…,N, let
(27)ti:=min{t∈I:θ(t)=2iπ},
so that
(28)0=t0<t1<⋯<tN≤1
and the N+1-tuple of points (ρ(t0),ρ(t1),…,ρ(tN)) in  ]a,b[ is well defined. Clearly, by the choice of N such that Nε>(b-a), there exists at least a pair of points (tj,tk) with j<k such that |ρ(tj)-ρ(tk)|<ε. Since
(29)mint∈[0,1]{dist(γ(t),W′∪W′′)}>ε,
we conclude that the segment joining the points (2kπ,ρ(tj)) and (2kπ,ρ(tk)) does not intersect the set W′∪W′′ (of course, such statement is trivial if ρ(tj)=ρ(tk)). We can now define the path(30)ω~(s)≔{γ(tj+2s(tk-tj)),for0≤s≤12,(2kπ,ρ(tk)+(2s-1)(ρ(tj)-ρ(tk))),for12≤s≤1,which takes values in ℛ∖(W′∪W′′) and is such that
(31)ω~(0)=(2jπ,ρ(tj)),ω~(1)=(2kπ,ρ(tj)).
Hence, the projection ω:=Π∘ω~:I→X is a nontrivial loop in X (in fact, it corresponds to k-j∈ℤ∖{0} in the fundamental group of X) and, by construction, ω(t)∉S,forallt∈I. This contradicts the hypothesis, and hence, the conclusion follows.Our last result in this section can be seen as a continuous version of the no-tie theorem for the Hex game on the annulus. See [10] for a discrete version of this result.Lemma 18.
LetX be a topological annulus, and let C1,C2⊂X be closed connected sets such that
(32)C1∩Xi≠∅≠C1∩Xo,C2isessentiallyembeddedinX.
Then, C1∩C2≠∅.Proof.
Assume, by contradiction, thatC1∩C2=∅, and let dist(C1,C2)=δ>0. In a δ/2-neighborhood of C1, we can find the image of a path γ:I→X with γ(0)∈Xi and γ(1)∈Xo. By construction, γ(t)∉C2, for all t∈I. This proves that it is not true that C2:Xi∤Xo, and therefore (by Lemma 14) C2 is not essentially embedded in X, thus contradicting one of our assumptions.


## 4. A Crossing Lemma for Invariant Sets

LetX be a topological space, and let h:X→X be a homeomorphism. Our first result is a version of Lemma 3 for h-invariant sets. Indeed, we have the following.Lemma 19.
LetA,B⊂X be two nonempty disjoint sets which are connected by at least one path in X. Let S⊂X be a closed set which satisfies S:A∤B and is invariant for h. Then, there exists a nonempty, closed set C⊂S which is minimal with respect to the property of cutting the paths between A and B and is invariant for h.Proof.
Letℱ be the set of all the nonempty closed subsets F of S such that F:A∤B and h(F)=F, with the elements of ℱ ordered by inclusion. Let (Fα)α∈J be a totally ordered subset of ℱ, and define F*:=⋂α∈JFα. From h(Fα)=Fα for all α∈J, it follows that h(F*)=F*. The proof that F*:A∤B is the same as that of Lemma 3. Thus, we obtain F*∈ℱ, and the conclusion follows from Zorn’s lemma.Our goal now is to develop a result analog of Lemma10 in the frame of h-invariant sets. Note that if in Lemma 10 we have a continuum C:Xl-∤Xr- which is also h-invariant, then by Lemma 11 we also have an h-invariant continuum intersecting Xb+ and Xt+.In order to fix the ideas and before moving to the class of generalized and oriented rectangles, we consider for a moment a planar rectangleℛ=[a,b]×[c,d] oriented in the standard manner and suppose that h:ℛ→ℛ is a homeomorphism. Let S⊂ℛ also be a compact set which intersects all the paths in ℛ joining ℛl- to ℛr- such that h(S)=S. We are looking for the existence of a continuum C⊂S with C:ℛl-∤ℛr- and h(C)=C. It is not difficult to see that, in general, the answer is negative, as shown by the following elementary example.Example 20.
Letℛ=[-2,2]×[0,1] and S={(±1,y):y∈[0,1]}. Clearly, S has the property of intersecting all the paths in ℛ joining the left edge to the right edge. Moreover, consistently with Lemma 10, S contains two continua {-1}×[0,1] and {1}×[0,1] connecting the lower and the upper sides of ℛ. However, if we take as a homeomorphism h(x,y)=(-x,y), that is, the symmetry with respect to the y-axis, we have that h(S)=S, but there is no connected subset of S which is invariant under h.Observe that in Example20 the set S cannot be split as the union of two disjoint closed (nonempty) subsets which are invariant. This property is also expressed by saying that S, (although not connected) is invariantly connected (see [43, Definition 4.2]). If we are allowed to replace the word “connected” with “invariantly connected,” we can get a full extension of Lemma 10, as follows.Lemma 21.
LetX~=(X,X-) be an oriented rectangle, let h:X→X be a homeomorphism, and let S⊂X be a closed set such that
(33)S:Xl-∤Xr-,h(S)=S.
Then, there exists a compact, invariantly connected set C⊂S such that
(34)C:Xl-∤Xr-.Proof.
By Lemma19 there exists a closed set C⊂S such that C:Xl-∤Xr-, with C invariant for h and minimal with respect to the cutting property. Suppose, by contradiction, that C is not invariantly connected, and let C1,C2⊂C be two closed nonempty disjoint sets with C1∪C2=C and h(Ci)=Ci for i=1,2. Now we conclude as in the proof of Lemma 10. Indeed, since C is minimal and C1,C2 are proper subsets of C, there exist two paths γ1,γ2 in X which connect Xl- to Xr-, such that γi avoids Ci (for i=1,2). Then, by Lemma 8, there exists a path γ:I→X with γ(0)∈Xl- and γ(1)∈Xr- with γ-∩C=∅, contradicting the assumption that C:Xl-∤Xr-.In order to have the connectedness of the setC, we propose a partial extension of Lemma 10 by adding a further assumption on h which prevents the possibility of a situation like the one described in Example 20. For simplicity in the exposition, we confine ourselves to the case of a planar rectangle ℛ=[a,b]×[c,d]. Here, we do not require h to be a homeomorphism.Lemma 22 (h-invariant crossing lemma).
Letℛ=[a,b]×[c,d], and let h:ℛ→ℛ be a continuous map such that
(35)h(ℛb+)⊂ℛb+,h(ℛt+)⊂ℛt+.
Suppose that there exists a path σ:I→ℛ with σ(0)∈ℛl- and σ(1)∈ℛr- such that ∀t∈I,∃s≥t:h(σ(t))=σ(s).
Assume that there exists a compact setS⊂ℛ which cuts the paths between ℛl- and ℛr- and satisfies
(36)h(S)⊂S.
Then, there exists a compact connected set C⊂S such that (i)
h(C)=C,(ii)
C∩ℛb+≠∅≠C∩ℛt+,(iii)
C cuts the paths between ℛl- and ℛr-.Proof.
As a first step, we apply Lemmas10 and 11 which guarantee the existence of a continuum contained in S which joins the lower and the upper sides of the rectangle and cuts the paths from the left to the right side in ℛ. We call such a continuum C0. Note that C0 is not necessarily invariant.
In order to obtain an invariant set, we define a sequence of continua by naming(37)Ci+1=h(Ci),∀i≥0.
Since h(S)⊂S, we also know that Ci⊂S,foralli≥0. By the cutting property of C0, we have that
(38)C0∩σ-≠∅,
so that there exists t0∈I with σ(t0)∈C0. Clearly, h(σ(t0))∈C1. On the other hand, by the hypothesis on σ, there exists t1∈I such that t1≥t0 and h(σ(t0))=σ(t1)∈C1. Proceeding by induction and using step by step the hypothesis on σ, we obtain a monotone sequence t0≤t1≤⋯ti≤ti+1≤⋯ in [0,1] such that σ(ti)∈Ci,foralli≥0. Let ti↗t*∈[0,1]. Then, passing to the limit in σ(·) and h∘σ, we obtain
(39)h(σ(t*))=σ(t*)∈LiCi,
where LiCi is the set of all the points x such that for every open neighborhood U of x it holds that U∩Ci≠∅ for each i large enough. Therefore, by a classical result from [40, Theorem 6, Chapter 5, Section 47,II], we have that
(40)C:=LsCi
is a nonempty continuum (where LsCi is the set of all the points x such that for every open neighborhood U of x it holds that U∩Ci≠∅ for infinitely many i). Recall that z∈C if and only if there exists a sequence zk with zk∈Cik for (ik)k an increasing sequence of indexes such that zk→z. Then, h(zk)→h(z), with h(zk)∈C1+ik, and, therefore, h(z)∈C. Thus, we have proved that h(C)⊂C. Conversely, for z∈C and (zk)k→z, as above, let us fix k≥2, and, from zk∈Cik=h(C-1+ik), let us take yk∈C-1+ik such that h(yk)=zk. By compactness, (yk)k has a convergent subsequence. Let us set (ykn)n→w. By definition, we have that w∈C. On the other hand, h(w)=limh(ykn)=limzkn=z. This proves that h(C)=C.
Since we know thatC0∩ℛb+≠∅, we have
(41)∅≠h(C0∩ℛb+)⊂h(C0)∩h(ℛb+)⊂C1∩ℛb+.
From this, by induction, we obtain that
(42)Ci∩ℛb+≠∅,∀i≥0,
and then, by compactness, we conclude that C∩ℛb+≠∅. The fact that C∩ℛt+≠∅ is proved in the same manner. Having proved that C is a continuum intersecting the horizontal edges of the rectangle, we conclude that it cuts all the paths between ℛl- and ℛr-.Remark 23.
Observe that the same result holds true (with an obvious modification in the proof) if we replace condition (35) with
(43)h(ℛb+)⊂ℛt+,h(ℛt+)⊂ℛb+.Remark 24.
By the assumptionh(ℛb+)⊂ℛb+ in (35), we have that
(44)h(x,c)=(f(x),c),∀x∈[a,b],
where f:[a,b]→[a,b] is a suitable continuous function. It easily follows that if f is monotone nondecreasing, then the path σ(t):=(a+t(b-a),c) satisfies the hypothesis of Lemma 22. A similar observation holds for ℛt+. Hence, we easily obtain the following.Corollary 25.
Letℛ be as in Lemma 22, and let h:ℛ→ℛ be a continuous map satisfying (35), and suppose also that at least one between h∣ℛb+ and h∣ℛt+ is monotone nondecreasing. Assume that there exists a compact set S⊂ℛ which cuts the paths between ℛl- and ℛr- and satisfies h(S)⊂S. Then, the same conclusion of Lemma 22 holds.Notice that in Example20 the function h is decreasing along both horizontal edges of the rectangle.A trivial case of a continuous map which is monotone nondecreasing along the horizontal lines is the identity. In such a case, Corollary25 reduces to the crossing lemma in Section 2.A useful property of the continua connecting two compact disjoint sets is the minimality, often named as irreducibility [13]. Indeed, a stronger version of Lemma 10 holds, guaranteeing the existence of compact connected set C⊂S which is irreducible between Xl- and Xr- (see [13, Proposition 3]).Then, as a next step, we look for the existence of irreducible invariant continua in this new setting. Such a result, indeed, is of general nature and independent of the fact that in the present paper we consider only the case of planar continua.Lemma 26.
LetX be a compact Hausdorff space, and let h:X→X be a homeomorphism. Assume that A,B⊂X are closed disjoint sets which are invariant for h, and let C⊂X be a continuum such that
(45)h(C)=C,C∩A≠∅≠C∩B.
Then, there exists ℰ⊂C satisfying (45) and minimal with respect to such property.Proof.
Letℱ be the family of all the continua 𝒞⊂C which are invariant for the homeomorphism h and intersect both A and B, with ℱ ordered by inclusion. The family ℱ is nonempty since C∈ℱ. The existence of a minimal subcontinuum of C satisfying (45) will be ensured by Zorn’s lemma.
Given any chain(𝒞j)j∈ℒ in ℱ, we observe that ⋂j∈𝒥𝒞j≠∅ for every finite subset of indices 𝒥⊂ℒ. Hence, by compactness, C*:=⋂j∈ℒCj is nonempty. Moreover C* is compact and intersects both A and B. Using the fact that h is a homeomorphism, we obtain the invariance of C*. Thus, if we prove that C* is connected, we have that C*∈ℱ, and Zorn’s lemma allows to conclude the proof.
Assume, by contradiction, thatC* is not connected. Then, there exist C′,C′′ nonempty compact sets such that C*=C′∪C′′ and C′∩C′′=∅. Then, there are also two open disjoint sets A′,A′′ with A′⊃C′ and A′′⊃C′′.
We claim that there exists an indexj* such that the set Cj*⊂A′∪A′′. Otherwise, it would happen that Cj⊄A′∪A′′ which implies that Dj:=Cj-(A′∪A′′)≠∅,∀j∈ℒ. The family (Dj)j∈ℒ is a family of closed sets with the finite-intersection property, and, therefore, we obtain
(46)∃x^∈⋂j∈ℒ‍Dj⊂X-(A′∪A′′).
This is a contradiction to ⋂j∈ℒ‍Dj⊂⋂j∈ℒ‍Cj=C*⊂A′∪A′′.
To conclude, we observe thatC*⊂Cj*⊂A′∪A′′ with Cj* a connected set. Since C*=C′∪C′′ with C′⊂A′ and C′′⊂A′′, we have that Cj*∩A′≠∅ and Cj*∩A′′≠∅, which contradicts the fact that Cj* is connected (in fact, A′ and A′′ are disjoint).


## 5. Applications

We conclude this paper by presenting a possible application of the above-described classical separation results to an example which is inspired by a model arising from the theory of fluid mixing previously considered by Kennedy and Yorke in [17]. Following [17, Section 2], we consider a planar map Φ:ℝ2→ℝ2 which is the composition of a squeeze map Jλ and a rotation map Rθ0. In [17], these two maps are defined as follows:
(47)Jλ(x,y)=(λx,yλ),withλ>1.
In order to define the map Rθ0 we pass to the polar coordinates (θ,ρ) and require that Rθ0 is a counterclockwise rotation which leaves all the concentric circumferences ρ>0 invariant  and keeps all the points of the plane with ρ≥1 still, and, moreover,
(48)limρ→0+Θ(θ,ρ)=θ0>0,
where Θ(θ,ρ) is the angular displacement performed by Rθ0 on the point (ρcosθ,ρsinθ).In [17], under the hypotheses that Rθ0 is a diffeomorphism and that θ0>π/2, with θ0 not an odd multiple of π/2, the authors prove the existence of a Smale horseshoe if λ>0 is sufficiently large. In particular, there exists an invariant Cantor set on which the map F:=Rθ0∘Jλ is conjugate to an m-shift.We are going to prove a result which, although not so sharp like that in [17], makes use of weaker conditions. To be more specific, from now on, the following assumptions are made.LetJ:ℝ2→ℝ2 be a continuous map with
(49)J(x,y)=(J1(x,y),J2(x,y))
which satisfies the following properties:(i)
J(0)=0,J(z)≠0 for z≠0, and J(Qi)⊂Qi for i=1,…,4, where with Qi we denote the ith closed quadrant of the plane,(ii)
J2(x,0)=0 and J1(x,0)=a(x)x, with a(x)>1 for x≠0,(iii)
J1(0,y)=0 and J2(0,y)=b(y)y, with 0<b(y)<1 for y≠0.Plainly speaking,J leaves the quadrants invariant and moves the points of the x-axis away from the origin, while it pushes the points of the y-axis toward the origin. In the sequel it will be convenient to express the map J (restricted to ℝ2∖{0}) via its lifting to the covering space ℝ×ℝ0+ as
(50)J~:(θ,ρ)↦(θ′,ρ′),θ′=θ+ΘJ(θ,ρ),ρ′=ℛJ(θ,ρ),
with ΘJ and ℛJ continuous functions which are 2π-periodic in the θ-variable. Note that the assumption J(Qi)⊂Qi, for i=1,…,4, reflects the fact that
(51)|ΘJ(θ,ρ)|≤π2.
As a second map, we consider a continuous counterclockwise rotation R around the origin such that R(0)=0 and R(z)≠0 for z≠0. We also express R (restricted to ℝ2∖{0}) by means of polar coordinates through its lifting
(52)R~:(θ,ρ)↦(θ′,ρ′),θ′=θ+ΘR(θ,ρ),ρ′=ℛR(θ,ρ),
with ΘR and ℛR continuous functions which are 2π-periodic in the θ-variable, and assume the following conditions.There existr0∈]0,1[ and θ0 such that (i)
ΘR(θ,r0)≥θ0,ΘR(θ,1)=0,forallθ∈ℝ,(ii)
ℛR(θ,ρ)=ρ,forallθ∈ℝandρ∈[r0,1].According to the above hypotheses, the mapR leaves the circumferences of center the origin and radius ρ∈[r0,1] invariant. Moreover, the points with ρ=r0 are rotated in the counterclockwise sense by an angle larger or equal to θ0, while the points with ρ=1 are kept still.Under the previous assumptions onJ and R, the following result holds.Theorem 27.
LetR be a homeomorphism of the annulus 𝒜:=A[r0,1] onto itself, and suppose also that θ0>2π+π/2. Then, the map Ψ:=J∘R has at least four fixed points in the interior of the annulus. Such result is stable with respect to small continuous perturbations of the map Ψ.Proof.
Let us denote by𝒜i and 𝒜o the inner and outer boundaries of 𝒜. We also restrict the map Ψ to the annulus 𝒜 and consider its lifting Ψ~ the covering space ℝ×[r0,1] as
(53)Ψ~(θ,ρ)↦(θ′′,ρ′′),θ′′=θ+ΘΨ(θ,ρ),ρ′′=ℛΨ(θ,ρ),
with ΘΨ and ℛΨ continuous functions which are 2π-periodic in the θ-variable. By the above positions for J~ and R~ we have that
(54)θ′′=θ′+ΘJ(θ′,ρ′),withθ′=θ+ΘR(θ,ρ),ρ′=ℛR(θ,ρ).
We also introduce a set S⊂𝒜 defined as
(55)S:=Π({(θ,ρ):ΘΨ(θ,ρ)=2π}),
where Π is the standard covering projection associated with the polar coordinates. The set S is a compact subset of the annulus consisting of the points which are rotated by an angle of exactly 2π under the action of Ψ.
Suppose thatγ:I→𝒜 is a path with γ(0)∈𝒜i and γ(1)∈𝒜o. We express the points of γ(t) in polar coordinates as
(56)γ(t)=(ρ(t)cosθ(t),ρ(t)sinθ(t))
and consider the angular displacement for the map R along the points of γ(t), using the function
(57)ωγ(t):I∋t↦ΘR(θ(t),ρ(t)).
By the assumptions, ωγ(0)>2π+π/2 and ωγ(1)=0. Now we apply the map J to the points of R(γ(t)). Using condition (51), we have that the angular displacement ΘΨ along the curve γ(t) can be expressed as
(58)ΘΨ(θ(t),ρ(t))=ωγ(t)+Δ(t),
where Δ(t) is a continuous function satisfying
(59)|Δ(t)|≤π2,∀t∈[0,1].
Recalling the properties of ωγ, we find
(60)ΘΨ(θ(0),ρ(0))>2π,ΘΨ(θ(1),ρ(1))≤π2<2π,
and, therefore, by the continuity of γ, we can conclude that γ-∩S≠∅. Now we can apply Lemma 15 which ensures the existence of a compact connected set C⊂S which is essentially embedded into 𝒜. It is also clear that C (as well as S) is contained in the interior of 𝒜.
Let us consider now the intersection of𝒜 with the first quadrant Q1. The boundary of such intersection consists of two segments L1:=[r0,1]×{0},L2:={0}×[r0,1] and two arcs C1:=r0S1∩Q1,C2:=S1∩Q1. We also define
(61)ℬ:=R-1(𝒜∩Q1).
The set ℬ is a topological rectangle for which we give an orientation by setting
(62)ℬl-:=R-1(L1),ℬr-≔R-1(L2),ℬt+≔R-1(C1),ℬb+:=R-1(C2).
By the assumptions, we see that
(63)∥Ψ(z)∥>∥z∥,∀z∈ℬl-,∥Ψ(z)∥<∥z∥,∀z∈ℬr-.
Hence, on each path with values in ℬ connecting ℬl- with ℬr- there is some point where ∥Ψ(z)∥=∥z∥. Lemma 11 ensures the existence of a continuum C′⊂ℬ with C′∩ℬb+≠∅,C′∩ℬt+≠∅, and ∥Ψ(z)∥=∥z∥, for each z∈C.
We are now in a position to apply Lemma18 which guarantees that C′∩C≠∅. By definition of C and C′ we conclude that any point w∈C′∩C is a fixed point for Ψ with Ψ(w)=w∈int(Q1∩𝒜). Repeating the same argument for the other quadrants, we find the remaining three fixed points.
We end the proof with a final comment about the “stability” of the existence result with respect to small perturbations. By this assertion, we mean that the existence of at least four fixed points is preserved if, insteadΨ, we take a continuous map which is sufficiently near to Ψ in the ∥·∥∞-norm on 𝒜. The reason for this fact comes from the strict inequalities that we have in (60) and (63). More precisely, the map Θψ(z)-2π is strictly positive on 𝒜i and strictly negative on 𝒜o. Similarly, ∥Ψ(z)∥-∥z∥, is strictly positive on one side of ℬ and strictly negative on an opposite side of ℬ. Hence, the maps Θψ(z)-2π and ∥Ψ(z)∥-∥z∥ can be seen as the two components of a vector field, and such components change their sign passing from one side to the opposite side of a topological rectangle. In such a case, we can prove that the topological degree (the Brouwer degree) associated with such a vector field is nonzero (this is exactly what happens when we give a proof of the Poincaré-Miranda theorem using the Brouwer degree and assume that the vector field does not vanish on the boundary of the rectangle in order to have the topological degree well defined). In case of nonzero degree (which, one could prove, is our situation) it is known that the results about the existence of zeros (or, resp., fixed points) are stable with respect to small perturbations of the vector field. This is our case when the above strict inequalities hold.Following the proof, it is clear that ifθ0>2jπ+π/2, for some positive integer j, then there are at least 4j fixed points.For a different application of these results to the existence of fixed points and periodic points to planar maps arising from ordinary differential equations, we refer also to [42].



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