OOPEN MAPS BETWEEN SHIFT SPACES
UIJIN JUNG
Abstract.
Given a code from a shift space to an irreducible sofic shift, any twoof the following three conditions – open, constant-to-one, (right or left) closing– imply the third. If the range is not sofic, then the same result holds whenbi-closingness replaces closingness. Properties of open mappings between shiftspaces are investigated in detail. In particular, we show that a closing open (orconstant-to-one) extension preserves the structure of a sofic shift. Introduction and Preliminaries
A map between topological spaces is open if images of open sets are open. Insymbolic dynamics, the openness of a sliding block code was first considered in [7]and has been proved useful, especially in cellular automata [9]. Hedlund showed thatan endomorphism of a full shift is open if and only if it is constant-to-one, i.e., thecardinality of the preimage of every point is same. In [11], Nasu extended this resultto the finite-to-one codes between irreducible shifts of finite type and showed furtherthat in that category, these conditions are also equivalent to the bi-closingness of acode. In the sofic cases, the situation is different and any single condition does notimply the others in general. So far, the structure of open codes between sofic shiftsare not well understood. In this paper, we investigate the properties of such codesand extend the result of Nasu to the class of sofic shifts. The main result is statedas follows.
Theorem 1.1.
Let φ be a code from a shift space X to an irreducible sofic shift Y .Then any two of the following conditions imply the third: (1) φ is open. (2) φ is constant-to-one. (3) φ is right closing (or left closing).In the case where any two (hence all) of the above conditions hold, X is a nonwan-dering sofic shift. When Y is not necessarily sofic, the same result still holds if we replace (3) withthe condition that φ is bi-closing (see Theorem 4.4). It will also be shown that whenthe domain in Theorem 1.1 is of finite type and φ is finite-to-one, then the code isopen if and only if it is constant-to-one (see Theorem 4.2). Key words and phrases. sofic shift, shift of finite type, open, constant-to-one, bi-closing.2000
Mathematics Subject Classification.
Primary 37B10; Secondary 37B15, 37B40, 54H20.This work was supported by the second stage of the Brain Korea 21 Project, The DevelopmentProject of Human Resources in Mathematics, KAIST in 2008. a r X i v : . [ m a t h . D S ] O c t UIJIN JUNG
In the early of 1990’s, Blanchard and Hansel proved that an irreducible soficconstant-to-one extension of an irreducible shift of finite type is also of finite type[1]. If either soficity or irreducibility is not assumed, then the statement is false [6]. Itturns out, however, that we need not assume the extension to be sofic and irreducibleif the extension map is closing, i.e., right closing or left closing. We will prove that aconstant-to-one closing extension of an irreducible shift of finite type is also of finitetype (see Corollary 4.3). This result can be viewed as a generalization of [1], sincethe proof in [1] is essentially showing that the extension map is closing. We give ananalogue of the above result for right closing open extensions. Indeed, assumed tobe closing, an open code and a constant-to-one code have similar properties. Someother interesting properties of open codes are provided.We assume that the reader is familiar with elementary symbolic dynamics. Foran introduction, see [8] or [10]. For a general theory of topological dynamics, see [9]or [12].Let A be a finite set with discrete topology and consider the set A Z with theproduct topology. It is compact and metrizable, and a typical element is x = ( x i ) i ∈ Z ,with x i ∈ A . Together with a homeomorphism σ , called a shift map , defined by σ ( x ) i = x i +1 , it becomes a topological dynamical system called a full shift . A shiftspace or subshift is a σ -invariant closed subset X of a full shift, together with arestriction of σ to X . We may also refer to X as a shift space.If X is a shift space, then denote by B n ( X ) the set of all words of length n appearing in the points of X and B ( X ) = (cid:83) n B n ( X ). A shift space X is called nonwandering if for all u ∈ B ( X ), we can find a word w such that uwu ∈ B ( X ). Itis called irreducible if for all u, v ∈ B ( X ), there is a word w with uwv ∈ B ( X ). So X is irreducible if and only if there is a point in X in which every word in B ( X )occurs infinitely often to the left and to the right. We call such a point doublytransitive . Indeed, if X is irreducible, then such points are dense in X and ( X, σ ) istopologically transitive.For u ∈ B ( X ), let l [ u ] denote the open and closed set { x ∈ X : x [ l,l + | u |− = u } ,which we call a cylinder . If l ≥ | u | = 2 l + 1, then − l [ u ] is called a central l + 1 cylinder . When u ∈ B ( X ) and l = 0, we usually discard the subscript 0.A code φ : X → Y is a continuous σ -commuting map between shift spaces. Anycode can be recoded to a 1-block code. A code φ is called a factor ( conjugacy , resp.)if it is onto (bijective, resp.) If φ is a conjugacy, then X and Y are called conjugate .Let d be the metric on X by letting d ( x, y ) = 2 − k , where k is the maximal numberwith x [ − k,k ] = y [ − k,k ] if x (cid:54) = y , and 0 otherwise. Two points x and ¯ x in X are leftasymptotic if d ( σ − n ( x ) , σ − n (¯ x )) → n → ∞ . A code φ is called right closing if itnever collapses two distinct left asymptotic points. A 1-block code φ is called rightresolving if whenever ab , ac are in B ( X ) and φ ( ab ) = φ ( ac ), then we have b = c . Aright closing code can be recoded to a right resolving code. Right asymptotic points,and left closing (resolving, resp.) properties are defined similarly. We call φ closing if it is left or right closing. If φ is both left and right closing (resolving, resp.), thenit is called bi-closing ( bi-resolving, resp.) We call φ finite-to-one if φ − ( y ) is a finiteset for all y ∈ Y . Closing codes are finite-to-one. A finite-to-one code is called PEN MAPS BETWEEN SHIFT SPACES 3 constant-to-one if | φ − ( y ) | is independent of y . Note that a constant-to-one code isa factor code.A subshift X is called a shift of finite type if there is a finite set F of words suchthat X consists of all points in which there’s no occurrence of words in F . If A isa nonnegative integral matrix, then a directed graph G A (with the vertex set V andthe edge set E ) and a shift space X A whose elements are the set of all bi-infinitetrips in G A ⊂ E Z are naturally associated to A . We call X A an edge shift . It is wellknown that any shift of finite type is conjugate to an edge shift. On the vertex set V of a graph G A , define an equivalence relation by setting v ∼ w if there is a pathfrom v to w and vice versa. The vertices in each equivalence class, together withall edges whose endpoints are in this equivalence class, form a natural subgraph of G A , called an (irreducible) component of G A . For each component of G A , a subshiftof X A is naturally defined. If it is not empty, we call this subshift an (irreducible)component of X A . It is irreducible and there is no subshift of X A which is irreducibleand contains a component of X A properly. A component X of X A is called a sinkcomponent if every point left asymptotic to X is also in X . A source component is defined similarly. All concepts defined on edge shifts naturally extend to theshifts of finite type by conjugacies. In general, an irreducible subshift X of X is an (irreducible) component of X if there is no irreducible subshift of X which properlycontains X .A shift space is called sofic if it is a factor of a shift of finite type. If X is anirreducible sofic shift, then there exist an irreducible shift of finite type X A and aright resolving factor code π : X A → X with the condition that any right closingfactor code from an irreducible shift of finite type to X factors through π . This pairis unique up to conjugacy. We call ( X A , π ) the canonical cover of X .Let h ( X ) denote the topological entropy of a topological dynamical system ( X, σ ).By the Perron-Frobenius theory, if X is irreducible sofic and Z is a proper subshiftof X , then h ( Z ) < h ( X ). For a shift space X , let Ω( X ) be a (unique) maximalnonwandering subshift of X . If X is of finite type, then Ω( X ) is equal to the(disjoint) union of its components. By the variational principle in ergodic theory, if Z is a subshift of X and Ω( X ) = Ω( Z ), then h ( X ) = h ( Z ) = h (Ω( X )). Finite-to-one codes preserve entropy in the sense that h ( X ) = h ( φ ( X )). A component X ofa shift space X is called maximal if h ( X ) = h ( X ).In this paper, our focus is mainly on open codes between shift spaces. A code φ : X → Y is called an open code if images of open sets are open. It is easy to seethe following equivalent condition of openness of a code. Lemma 1.2.
A code φ : X → Y between shift spaces is open if and only if for each l ∈ N , there is k ∈ N such that whenever x ∈ X, y ∈ Y and φ ( x ) [ − k,k ] = y [ − k,k ] , wecan find ¯ x ∈ X with ¯ x [ − l,l ] = x [ − l,l ] and φ ( x ) = y . So if φ is open, then for any l ∈ N we can find k ∈ N such that the image ofa central (2 l + 1) cylinder in X consists of central (2 k + 1) cylinders in Y . In thefollowing we will often use these observations. UIJIN JUNG Properties of open codes
In this section, we give some properties of open codes which are used in subsequentsections. We show that the image of a shift of finite type by an open code is alsoof finite type. Openness of a code is preserved by the fiber product. If combinedwith constant-to-one property, an open code separates fibers and becomes a localhomeomorphism.
Lemma 2.1.
Let X and Y be shift spaces with Y irreducible. If φ : X → Y is anopen code, then it is onto.Proof. Note that φ ( X ) is a nonempty open and closed σ -invariant subset of Y . Since( Y, σ ) is topologically transitive, any nonempty open σ -invariant subset of Y is densein Y [12] and we get φ ( X ) = φ ( X ) = Y . Thus φ is onto. (cid:3) Lemma 2.2. [5]
Let X be an irreducible shift of finite type and φ a constant-to-onecode from X to a shift space Y . Then Y is of finite type. This result on constant-to-one codes is useful. Indeed, the code in the lemma isopen (see Theorem 4.1).
Proposition 2.3.
Let X be a shift of finite type and φ an open factor code from X to a shift space Y . Then Y is of finite type.Proof. By recoding, we can assume that X is an edge shift and φ is 1-block. Choose l ≥ a ∈ B ( X ), φ ([ a ]) consists of central 2 l + 1 cylinders.Let uw and wv be in B ( Y ) with | w | = 2 l + 1. Then there are points y (1) and y (2) such that y (1)[ − l −| u | ,l ] = uw and y (2)[ − l,l + | v | ] = wv . Take x ∈ φ − ( y (1) ) and consider φ ([ x ]). It contains y (1) and y (1)[ − l,l ] = y (2)[ − l,l ] = w . Hence y (2) ∈ φ ([ x ]) and there is apoint z ∈ [ x ] with φ ( z ) = y (2) . Now define ¯ x by ¯ x i = x i for i ≤ x i = z i for i ≥
0. Since φ (¯ x ) [ − l −| u | ,l + | v | ] = uwv , we have uwv ∈ B ( Y ). Thus Y is a (2 l + 1)-stepshift of finite type. (cid:3) Let X , Y , and Z be shift spaces and φ : X → Z , φ : Y → Z the codes. Then the fiber product of ( φ , φ ) is the triple (Σ , ψ , ψ ) whereΣ = { ( x, y ) ∈ X × Y : φ ( x ) = φ ( y ) } and ψ : Σ → X is defined by ψ ( x, y ) = x ; similarly for ψ .Fiber product construction is useful since it lifts over or pulls down the propertiesof given codes to the opposite codes. More specifically, if φ is one-to-one, onto, rightclosing, right resolving, or finite-to-one, then ψ also satisfies the correspondingcondition. In the case φ is assumed to be onto, if ψ has any of the propertiesmentioned above, then so does φ [10]. It can be easily shown that constant-to-oneproperty is also preserved in this way. Now we show that openness is also preservedby the fiber product. Lemma 2.4.
Let (Σ , ψ , ψ ) be the fiber product of φ : X → Z and φ : Y → Z . (1) If φ is open, then so is ψ . (2) Let φ be onto. If ψ is open, then so is φ . PEN MAPS BETWEEN SHIFT SPACES 5
Proof. (1) Suppose φ is open. Let C be the central 2 l + 1 cylinder in Σ, containinga point ( x, y ) with x ∈ X and y ∈ Y . To show ψ is open, it suffices to find an openneighborhood of y , contained in ψ ( C ). Let m be a coding length of φ . Choose n ≥ l so that φ ( − l [ x − l · · · x l ]) is a disjoint union of central 2 n + 1 cylinders. Now,if ¯ y ∈ Y satisfies ¯ y [ − n − m,n + m ] = y [ − n − m,n + m ] , then φ (¯ y ) [ − n,n ] = φ ( y ) [ − n,n ] . Since φ ( − l [ x − l · · · x l ]) contains φ ( y ) and the central 2 n + 1 blocks of y and ¯ y are equal,it follows that φ (¯ y ) ∈ φ ( − l [ x − l · · · x l ]). Hence there is an ¯ x ∈ − l [ x − l · · · x l ] suchthat φ (¯ x ) = φ (¯ y ). Since n ≥ l , we have ¯ y [ − l,l ] = y [ − l,l ] , so (¯ x, ¯ y ) is in C satisfying ψ (¯ x, ¯ y ) = ¯ y . Thus ψ ( C ) contains − l [ y − n − m · · · y n + m ], as desired.(2) Suppose φ is not open. Then there is an open set U of X such that φ ( U )is not open. Hence we can find an x ∈ U and a sequence (cid:8) z ( n ) (cid:9) ∞ n =1 in Z such that z ( n ) / ∈ φ ( U ) for any n and z ( n ) → φ ( x ). Take points y ( n ) ∈ φ − ( z ( n ) ). By taking asubsequence, we may assume y ( n ) converges to a point y ∈ Y . Then φ ( y ) = φ ( x ),so ( x, y ) is in Σ. Now consider an open set V = ( U × Y ) ∩ Σ. Since ( x, y ) ∈ V , itfollows that ψ ( V ) contains y . But y ( n ) is not in ψ ( V ) for any n , since( φ ◦ ψ )( V ) = ( φ ◦ ψ )( V ) = φ ( U ) . Thus ψ ( V ) is not open. Therefore ψ is not an open code. (cid:3) Lemma 2.5.
Let φ be a finite-to-one open code from a shift space X to an irreducibleshift space Y . Then there is d > such that | φ − ( y ) | = d for each doubly transitivepoint y of Y . Furthermore, | φ − ( y ) | ≤ d for all y ∈ Y .Proof. Take a doubly transitive point z of Y and let d = | φ − ( z ) | . Suppose that thereexists a point y ∈ Y such that | φ − ( y ) | > d . Take distinct points u , · · · , u d +1 ∈ φ − ( y ). Since X is a compact metric space, we can take disjoint neighborhoods U , · · · , U d +1 of u , · · · , u d +1 , respectively. Take a sequence { y ( i ) } ∞ i =1 ⊂ { σ k ( z ) } k ∈ Z such that y ( i ) → y .Since | φ − ( y ( i ) ) | = d , for each i there exists j ( i ) such that U j ( i ) ∩ φ − ( y ( i ) ) = ∅ .By taking a subsequence of i such that j ( i ) is constant, without loss of generalitywe may assume that there exist a sequence { y ( i ) } ∞ i =1 and j such that y ( i ) → y and U j ∩ φ − ( y ( i ) ) = ∅ for all i .Consider φ ( U j ) ⊂ Y . Note that y ( i ) → y ∈ φ ( U j ), but y ( i ) is not in φ ( U j ) for any i , hence φ ( U j ) is not open. Thus φ is not an open code and a contradiction comes.Hence | φ − ( y ) | ≤ d for all y in Y . If ¯ z is another transitive points of Y , then aboveargument shows that | φ − (¯ z ) | = | φ − ( z ) | = d . So each doubly transitive point of Y has d preimages. (cid:3) Let φ : X → Y be a factor code between shift spaces. If there is d > Y has d preimages under φ , we call d the degree of φ . If X is an irreducible shift of finite type and φ is finite-to-one, then φ has adegree d and | φ − ( y ) | ≥ d for all y in Y . Hence the above lemma implies that anyfinite-to-one open code has a degree and is d -to-1 if the domain is an irreducibleshift of finite type.We will use the following lemma to prove that any bi-closing factor code also hasa well-defined degree. UIJIN JUNG
Lemma 2.6.
Let φ : X → Y be a code between shift spaces. Then φ is bi-closingif and only if there is an (cid:15) > such that whenever y ∈ Y and x, ¯ x ∈ φ − ( y ) with x (cid:54) = ¯ x , we have d ( x, ¯ x ) ≥ (cid:15) .Proof. First, if φ is not right closing, then there are two left asymptotic points x and z in X such that x (cid:54) = z and φ ( x ) = φ ( z ) = y ∈ Y . Consider σ − n ( y ) for large n . Since d ( σ − n ( x ) , σ − n ( z )) → n → ∞ , we cannot find an (cid:15) > φ is not left closing.Next, suppose that φ is bi-closing. Since φ is right closing, there is an N > x, y ∈ X , φ ( x ) = φ ( y ) and x [ − N , = y [ − N , , then x = y (by an easy compactness argument.) Since φ is left closing, there is an N > N = max { N , N } . It is easy to see that (cid:15) = 2 − N satisfies thecondition in the statement. (cid:3) Remark . Let φ : X → Y be an open code between shift spaces. If φ is bi-closing,then it is a local homeomorphism. For, let (cid:15) > x ∈ X , let U be the ball of radius (cid:15)/ x . Then φ | U is a homeomorphismonto its image. Lemma 2.7.
Let φ be a bi-closing factor code from a shift space X to an irreducibleshift space Y . Then there is a d > such that | φ − ( y ) | = d for each doubly transitivepoint y of Y . Furthermore, | φ − ( y ) | ≥ d for all y ∈ Y .Proof. Let z be a doubly transitive point of Y and y ∈ Y . Then there is a sequence { n k } ∞ k =1 such that σ n k ( z ) → y . Write φ − ( z ) = { z (1) , · · · , z ( d ) } . By compactness, there exist subsequence { n k j } ⊂ { n k } and points x (1) , · · · , x ( d ) suchthat σ n kj ( z ( i ) ) → x ( i ) for each i = 1 , · · · , d. Note that φ ( x ( i ) ) = y for all i . By Lemma 2.6, { σ n kj ( z ( i ) ) } di =1 is (cid:15) -separated foreach j ∈ N . It follows that x ( i ) ’s are distinct and hence | φ − ( y ) | ≥ | φ − ( z ) | = d .If y is also doubly transitive, then by the symmetric argument we get | φ − ( y ) | = | φ − ( z ) | = d . (cid:3) Corollary 2.8.
Let φ : X → Y be an open code between shift spaces. If φ isbi-closing and Y irreducible, then φ is constant-to-one.Proof. This follows from Lemma 2.5 and Lemma 2.7. (cid:3)
Proposition 2.9.
Let φ : X → Y be an open code between shift spaces. If φ isconstant-to-one, then it is bi-closing.Proof. Let φ be constant-to-one and d the number of preimages. If φ is not bi-closing, then by Lemma 2.6 for each n ∈ N , there is a point y ( n ) ∈ Y such that φ − ( y ( n ) ) = (cid:8) x n, , · · · , x n,d (cid:9) with x n, − n,n ] = x n, − n,n ] . By choosing a subsequence, wecan assume that y ( n ) → y , x n, → x (2) , and x n,i → x ( i ) for i = 2 , · · · , d .Since φ is d -to-1, there is z ∈ X such that φ ( z ) = y and z (cid:54) = x ( i ) for any i = 2 , · · · , d . Also there is a neighborhood U of z such that x n,i / ∈ U for any PEN MAPS BETWEEN SHIFT SPACES 7 i = 1 , · · · , d and for any n ∈ N . Note that y ( n ) → y ∈ φ ( U ) but y ( n ) is not in φ ( U )for any n . Hence φ ( U ) is not open, which is a contradiction. So φ is bi-closing. (cid:3) Extension by open codes
In [11], Nasu considered the constant-to-one extensions of irreducible shifts offinite type for the case when the extension is assumed to be of finite type, andobtained the following result.
Proposition 3.1. [11]
Let X be a shift of finite type, Y an irreducible shift offinite type, and φ : X → Y a constant-to-one code. Then X is nonwandering, allcomponents are maximal, and the restriction of φ to any component is constant-to-one. We will show a similar structural result for the finite-to-one open extensions ofirreducible sofic shifts. First we recall some definitions.Let φ be a finite-to-one 1-block factor code from an irreducible shift of finite type X to a shift space Y . Given a word w ∈ B ( Y ), define d ( w ) = min ≤ t ≤| w | (cid:12)(cid:12)(cid:8) a ∈ B ( X ) : ∃ u ∈ φ − ( w ) with u t = a (cid:9)(cid:12)(cid:12) and d = min w ∈B ( Y ) d ( w ). If a word w satisfies d ( w ) = d , then it is called a magicword . In this case, a coordinate t where the minimum occurs is called a magiccoordinate . For the properties of magic words, see [8] or [10]. A word v ∈ B ( X ) is intrinsically synchronizing if whenever uv and vw are in B ( X ), we have uvw ∈ B ( X ).Every irreducible sofic shift has an intrinsically synchronizing word.In [11], Nasu proved the following result for the case of shifts of finite type byusing the notions of compatible and complete sets. We generalize this to sofic shifts. Lemma 3.2.
Let X be a sofic shift, Y an irreducible sofic shift, and φ : X → Y afinite-to-one factor code. If y is a doubly transitive point in Y , then φ − ( y ) ⊂ Ω( X ) . Proof.
First, we prove the case where X is of finite type. Without loss of generality,we can assume X is an edge shift and φ is 1-block. Suppose there is a point x in X \ Ω( X ) with φ ( x ) = y . Then there are components X , X of X such that x isleft asymptotic to X and right asymptotic to X . Hence there are k, l ∈ Z , k < l ,such that x i ∈ B ( X ) for all i ≤ k and x i ∈ B ( X ) for all i ≥ l . Since y is doublytransitive and φ is finite-to-one, it follows that x is left transitive in X and righttransitive in X . Hence φ | X and φ | X are onto.Note that there is an intrinsically synchronizing word w ∈ B ( Y ) which is amagic word for both φ | X , φ | X . (Just find three words satisfying each conditionand glue them using the irreducibility of Y .) Since y is doubly transitive, there are i , i , j , j ∈ Z such that i < i < k < l < j < j and y ( i ,i ] = y ( j ,j ] = w .Let ¯ y = ( y ( i ,j ] ) ∞ . Then ¯ y ∈ Y , since w is intrinsically synchronizing. Note that φ ( x ( i ,j ] ) = ¯ y ( i ,j ] .Let m and n denote magic coordinates of w for φ | X and φ | X , respectively. Thenthere is a unique point x (1) ∈ X such that φ ( x (1) ) = ¯ y and x (1) i + m = x i + m . Also, UIJIN JUNG there is a unique point x (2) ∈ X such that φ ( x (2) ) = ¯ y and x (2) j + n = x j + n . Nowdefine ¯ x ∈ X by(3.1) ¯ x i = x (1) i if i ≤ i + mx i if i + m ≤ i ≤ j + nx (2) i if i ≥ j + n Note that ¯ x is well defined and φ (¯ x ) = ¯ y . Also ¯ x is not periodic. Since ¯ y is periodic,it follows that φ is not finite-to-one, which is a contradiction. Thus φ − ( y ) ⊂ Ω( X ).Next, we prove the case where X is sofic. Let y be a doubly transitive pointin Y . Take the canonical cover π : X A → X . By applying the previous result to φ ◦ π , we get x ∈ Ω( X A ) for all x ∈ π − ( φ − ( y )). Since the factor code preservesnonwandering points, it follows that π ( x ) ∈ Ω( X ). Thus φ − ( y ) ⊂ Ω( X ). (cid:3) Proposition 3.3.
Let X be a shift of finite type, Y an irreducible sofic shift, and φ : X → Y a finite-to-one open code. Then X is nonwandering, all components aremaximal, and the restriction of φ to any component is open.Proof. We can assume X = X A is an edge shift and φ is a 1-block code. Then wecan write Ω( X ) = ˙ (cid:83) X A i , where X A i ’s are components of X A .Let X A i be a sink component. We claim that φ | X Ai is onto. If not, then there is aword v ∈ B ( Y ) \B ( φ ( X A i )). If there is no transition edge to X A i , then X A i is open in X A . But φ ( X A i ) is a proper closed σ -invariant subset of Y and it is not open in Y since Y is topologically transitive, thus we get a contradiction. So there is an edge e ∈ G A whose terminal vertex lies in G A i , but e / ∈ G A i . Consider φ ([ e ]). Choose l ≥ φ ([ e ]) is a union of central 2 l + 1 cylinders, one of which is − l [ u ].Since Y is irreducible, there is a word w with uwv ∈ B ( Y ). Take a point y ∈ Y with y [ − l,l + | w | + | v | ] = uwv . Then y ∈ − l [ u ]. But if x ∈ B ( X A ) with x = e , then x k ∈ B ( X A i ) for all k > v cannot occur in φ ( x ) [1 , ∞ ) . It follows that y / ∈ φ ([ e ]), which is a contradiction. Hence φ | X Ai is onto for each sink component.Similarly φ | X Aj is onto for each source component X A j .Suppose that Ω( X A ) (cid:54) = X A . Then there is a point x ∈ X A \ Ω( X A ) such that x is leftasymptotic to some source component and right asymptotic to some sink component.Furthermore, we can assume that x is left transitive in the source component, andright transitive in the sink component. Then φ ( x ) is a doubly transitive point in Y , which contradicts Lemma 3.2. Hence X A is nonwandering. Now, since eachcomponent X A i is open and closed, the restriction φ | X Ai must be open, hence ontoby Lemma 2.1. Consequently, all components are maximal. (cid:3) We are now ready to prove the main part of Theorem 1.1. The heart of the prooflies in the next two propositions. For a 0-1 matrix A , denote ˆ X A by the shift spaceconsisting of ( x i ) i ∈ Z with A x i x i +1 = 1. There is a natural conjugacy π : X A → ˆ X A [10]. Proposition 3.4.
Let φ : Σ → Y be a right closing open code from a shift space Σ to an irreducible shift of finite type Y . Then Σ is a nonwandering shift of finitetype. PEN MAPS BETWEEN SHIFT SPACES 9
Proof.
By recoding, we can assume that Y is an edge shift and φ is a 1-block rightresolving code. Define a 0-1 matrix A , indexed by symbols of Σ, by A ij = 1 if ij ∈ B (Σ) and A ij = 0 otherwise. We can extend φ to a right resolving code¯ φ : ˆ X A → Y by letting ¯ φ ( x ) i = φ ( x i ) for i ∈ Z . Since Y is an edge shift, ¯ φ is welldefined [4, Theorem 4.12]. The existence of ¯ φ implies that every maximal componentof ˆ X A , hence of X A , is a sink component [8, Lemma 5.1.4]. By precomposing theconjugacy π : X A → ˆ X A , we can assume that ¯ φ is a code from X A to Y which is anextension of φ .Let X A i be a maximal component of X A . If Σ ∩ X A i (cid:40) X A i , then take words v ∈ B ( Y ) \ B ( φ ( X A i ∩ Σ)) and e ∈ B ( X A i ). As in the proof of Proposition 3.3, φ ([ e ]) is a union of 2 l + 1 cylinders, one of which is − l [ u ]. Take a point y ∈ Y with y [ − l,l + | w | + | v | ] = uwv for some w ∈ B ( Y ). Then y ∈ − l [ u ], but it cannot be in φ ([ e ]),hence we get a contradiction. Thus Σ ∩ X A i = X A i for each maximal component.Consider the subgraph G ¯ A of G A consisting of all nonmaximal components of G A and transition edges between these components. If X ¯ A is nonempty, then sinceevery component of G ¯ A is not maximal, φ ( X ¯ A ) is not equal to Y . Take words v ∈ B ( Y ) \ B ( φ ( X ¯ A )) and e ∈ B ( X ¯ A ). As above, φ ([ e ]) does not contain a pointof the form · · · v · · · .φ ( e ) · · · , so φ cannot be open. Hence X ¯ A is empty and eachmaximal component is also a source component. Thus, Σ = ˙ (cid:83) X A i is a nonwanderingshift of finite type. (cid:3) Proposition 3.5.
Let φ : Σ → Y be a right closing constant-to-one code from ashift space Σ to an irreducible shift of finite type Y . Then Σ is a nonwandering shiftof finite type.Proof. By proceeding as in the proof of Proposition 3.4, we obtain a shift of finitetype X A and ¯ φ : X A → Y where each maximal component of X A is a sink and ¯ φ is aright resolving extension code of φ .Write X A = ( ˙ (cid:91) X A i ) ˙ ∪ X ¯ A ˙ ∪ S where X ¯ A is described as in the proof of Proposition 3.4 and S is the set of all pointswhich are left asymptotic to X ¯ A and right asymptotic to some maximal component X A i . Also write Σ = ( ˙ (cid:91) Σ i ) ˙ ∪ ˜Σ ˙ ∪ ˜ S where Σ i = Σ ∩ X A i , ˜Σ = Σ ∩ X ¯ A and ˜ S = Σ ∩ S . Let p be the number of maximalcomponents in X A .Since φ is onto, there is at least one i satisfying Σ i = X A i . For, if there is nosuch i , then since ˜ S ⊂ Σ \ Ω(Σ) and the topological entropy is concentrated on thenonwandering set, we have h (Σ) = max { h (Σ ) , · · · , h (Σ p ) , h ( ˜Σ) } < h ( X A ) , which is a contradiction. Without loss of generality, we can assume the set of such i ’s are { , , · · · , q } with q ≤ p . Since h ( X ¯ A ) < h ( Y ), ¯ φ | X ¯ A is not onto. Hence thereexists a doubly transitive point y in Y \ ¯ φ ( X ¯ A ). Since ¯ φ − ( y ) ⊂ Ω( X A ) by Lemma3.2, it follows that ¯ φ ( X A \ (cid:83) pi =1 X A i ) (cid:54) = Y and hence φ (Σ \ (cid:83) pi =1 Σ i ) (cid:54) = Y . Since h (Σ i ) < h ( Y ) for each i = q + 1 , · · · , p , it follows that φ − ( y ) ∩ Σ i = ∅ . Thus we get φ (Σ \ (cid:83) qi =1 Σ i ) (cid:54) = X .For each i = 1 , · · · , q , φ | Σ i is a finite-to-one factor code between irreducible shiftsof finite type. Let d i be the degree for φ | Σ i and d = (cid:80) qi =1 d i . If Σ \ (cid:83) qi =1 Σ i (cid:54) = ∅ ,then take a doubly transitive point z in Y \ φ ( (cid:83) i>q Σ i ∪ ˜Σ). By Lemma 3.2, we get¯ φ − ( z ) ⊂ Ω( X A ), hence φ − ( z ) ⊂ ( (cid:83) pi =1 Σ i ) ∪ ˜Σ. By the condition on z , φ − ( z ) lies in (cid:83) qi =1 Σ i , so | φ − ( z ) | = d and φ is d -to-1 everywhere. But if we take x ∈ φ (Σ \ (cid:83) qi =1 Σ i ),then by letting φ i = φ | Σ i and ˜ φ = φ | Σ \ S qi =1 Σ i , it follows that | φ − ( x ) | = q (cid:88) i =1 | φ i − ( x ) | + | ˜ φ − ( x ) | ≥ d + 1 , which is a contradiction. Hence Σ \ (cid:83) qi =1 Σ i = ∅ . ThusΣ = q (cid:91) i =1 Σ i = q (cid:91) i =1 X A i so that Σ is a nonwandering shift of finite type. (cid:3) Remark . If φ is assumed to be bi-closing in the above propositions, then theproofs can be organized to be simpler. First, in the proof of Proposition 3.4, if weassume φ is bi-resolving by recoding, then ¯ φ is bi-resolving. Hence each maximalcomponent X A i is a sink and a source, so an open and closed set in X A . Hence X A = ( ˙ (cid:83) X A i ) ˙ ∪ X ¯ A . Since φ (Σ ∩ X A i ) is open, we get Σ ∩ X A i = X A i for each i . Since φ (Σ ∩ X ¯ A ) is open, we get X ¯ A = ∅ , so that Σ = (cid:83) X A i .Next, in the proof of Proposition 3.5, we have a representation X A = ( ˙ (cid:83) pi =1 X A i ) ˙ ∪ X ¯ A ,where X A i ’s are maximal components and h ( X ¯ A ) < h ( X ). Let Σ = ( ˙ (cid:83) pi =1 Σ i ) ˙ ∪ ˜Σ asin the proof. Since φ is onto, Σ i = X A i for some i . By reordering, we can assume i = 1. Now take Σ (1) = ( (cid:83) pi =2 Σ i ) ∪ ˜Σ. If Σ (1) is empty, we are done. Otherwise, φ | Σ is constant-to-one, since it is a bi-closing code between two irreducible shifts offinite type. It follows that φ | Σ (1) is also constant-to-one, so it is onto and Σ i = X A i for some i ≥
2, which we can assume to be 2. Now consider Σ (2) = ( (cid:83) pi =3 Σ i ) ∪ ˜Σ,and so on. By processing, we terminate after at most p steps and get Σ = (cid:83) qi =1 X A i for some q ≤ p . 4. Main Theorems
As we have seen, there are relations among open, constant-to-one, and bi-closingcodes. Some of such relations have been considered in [5, 7, 11]. Especially, in thecategory of irreducible shifts of finite type, they are all equivalent. As we shall see,in the case of general shift spaces, any two of the above properties imply the third.When the range is sofic, the code forces the domain to be sofic and nonwandering.It turns out that the existence of a cross section of a code is useful.We first give the well-known result of Nasu.
PEN MAPS BETWEEN SHIFT SPACES 11
Theorem 4.1. [11]
Let X and Y be irreducible shifts of finite type with equal entropyand φ : X → Y a factor code. Then the following are equivalent. (1) φ is open. (2) φ is constant-to-one. (3) φ is bi-closing.Proof. If φ is open, then it is constant-to-one by the remark following Lemma 2.5.Suppose φ is constantly d -to-1. Without loss of generality, we can assume that X is an edge shift and φ is 1-block. It is well known that the preimage of each pointin Y contains a set of d mutually separated points [10]. Since | φ − ( y ) | = d , φ − ( y )is 1-separated set for all y ∈ Y . So by Lemma 2.6, φ is bi-closing. Finally, if φ is bi-closing, then by recoding it is conjugate to a bi-resolving factor code. By thePerron-Frobenius theory, it is a bi-covering code [10, Theorem 8.2.2], which is clearlyopen. (cid:3) These conditions do not coincide when we are in the sofic category. For the casewhen Y is sofic, the following theorem holds. For the case when X is irreducible,the result appears in [5] without proof. Theorem 4.2.
Let φ be a finite-to-one code from a shift of finite type X to anirreducible sofic shift Y . Then φ is open if and only if it is constant-to-one. If φ isopen, or equivalently, constant-to-one, then φ is bi-closing.Proof. First, suppose that φ is constant-to-one and let π : X B → Y be the canonicalcover so that X B is irreducible. Consider the fiber product (Σ , ψ , ψ ) of ( X, φ ) and( X B , π ). Then ψ is constant-to-one. Since X and X B are of finite type, Σ is alsoof finite type. Now by Proposition 3.1, the restriction of ψ to each component isconstant-to-one, hence is open and bi-closing by Theorem 4.1. So ψ itself is openand bi-closing. Thus φ is also open and bi-closing by Lemma 2.4.Conversely, suppose φ is open. By Lemma 3.3, X is nonwandering and the re-striction of φ to each component is open. By Proposition 2.3, Y is of finite type.Thus Theorem 4.1 applies and φ is constant-to-one. (cid:3) We prove the main theorem of this paper.
Proof of Theorem 1.1.
First, suppose φ is open and right closing. Let π : X B → Y be the canonical cover of Y so that X B is irreducible. Consider the fiber product of(Σ , ψ , ψ ) of ( X, φ ) and ( X B , π ). Σ X ψ < X B ψ > Y π < φ > By Lemma 2.4, ψ is open and right closing, and by Proposition 3.4, Σ is indeed ashift of finite type. Now by Proposition 3.3, Σ is nonwandering, and the restriction of ψ to each component is open, hence constant-to-one and bi-closing by Theorem4.1. So ψ itself is constant-to-one and bi-closing. Hence φ is also constant-to-oneand bi-closing.Next, if φ is constant-to-one and right closing, then the proof is similarly estab-lished, by using Propositions 3.5, 3.1, and Lemma 2.4.The case where φ is open and constant-to-one holds by Proposition 2.9.Finally, note that ψ is onto, since π is onto. Hence if all of these conditions hold,then X is a factor of a nonwandering shift of finite type Σ, so it is nonwanderingsofic. (cid:3) A shift space X is called almost Markov if it is a bi-closing factor of a shift offinite type. We call X a strictly almost Markov shift if it is almost Markov but notof finite type. An almost of finite type shift (AFT) is an irreducible almost Markovshift [3]. It is known that an irreducible sofic shift is an AFT if and only if itscanonical cover is also left closing [2]. Corollary 4.3.
The following statements hold. (1)
A right closing open extension of an irreducible shift of finite type is of finitetype. (2)
A right closing open extension of an irreducible strictly almost Markov shiftis strictly almost Markov. (3)
A right closing open extension of an irreducible non-AFT sofic shift is soficand not almost Markov.Each extension is nonwandering and all irreducible components in the extensionsare maximal. The same results hold for right closing constant-to-one extensions.Proof. (1) is merely a restatement of Proposition 3.4.(2) If φ : X → Y is a right closing open code from a shift space to an AFT, then π in the proof of Theorem 1.1 is bi-closing. Hence ψ is bi-closing, so that X = ψ (Σ)is an almost Markov shift. Next, if X is of finite type, then Y is also of finite typeby Proposition 2.3, which is a contradiction. So X is strictly sofic.(3) Let φ : X → Y is a right closing open code from a shift space to an irreduciblenon-AFT sofic shift. Note that the extension shift is sofic and φ is bi-closing byTheorem 1.1. If X is almost Markov, then Y has a bi-closing cover, namely, thebi-closing cover of X followed by φ , which is a contradiction. Hence X is not almostMarkov.The remainder of the statement follows from Theorem 1.1. (cid:3) As stated in the introduction, the above result can be viewed as a generalizationof [1]. In a sense, right closing open (or constant-to-one) extension preserves thestructure of a shift space.
Remark . The code constructed in [6] is a constant-to-one code from a nonsoficcoded system to an irreducible shift of finite type. By Corollary 4.3 it is not closing.Applying Theorem 1.1 it is not open.Now we consider codes between general shift spaces. If we weaken the condition(3) in Theorem 1.1, then the following generalization holds. In contrast to the sofic
PEN MAPS BETWEEN SHIFT SPACES 13 case, we cannot use the theorem of Nasu, since there is no way to reduce to thefinite type situation.
Theorem 4.4.
Let φ be a code from a shift space X to an irreducible shift space Y .Then any two of the following conditions imply the third: (1) φ is open. (2) φ is constant-to-one. (3) φ is bi-closing.Proof. The theorem follows from Corollary 2.8, Proposition 2.9, and the followingproposition 4.5. (cid:3)
Let φ : X → Y be a continuous map between topological spaces. A continuousmap f : Y → X is called a cross section of φ if φ ( f ( y )) = y for all y in Y . Anyopen factor code between shift spaces has a cross section [7]. We say φ has d disjointcross sections if there exist d cross sections f i : Y → X such that f i ( Y ) ∩ f j ( Y ) = ∅ for all i (cid:54) = j . We adopt the ideas from [9] to prove the following result. Proposition 4.5.
Let φ : X → Y be a d -to- code between shift spaces. Then thefollowing are equivalent. (1) φ is open. (2) φ is bi-closing. (3) φ has d disjoint cross sections such that the union of their images is X . (4) For any x ∈ X , there exists a cross section f such that x ∈ f ( Y ) .Proof. (1) ⇒ (2): This holds by Proposition 2.9.(2) ⇒ (3): We can assume that φ is a 1-block code. Since φ is bi-closing, we canfind an (cid:15) > p ∈ N with 2 − p < (cid:15) .For each n ≥ p and w ∈ B n +1 ( Y ), define D ( w ) = (cid:8) x [ − p,p ] : x ∈ X, φ ( x [ − n,n ] ) = w (cid:9) ⊂ B p +1 ( X )and d ( w ) = |D ( w ) | . Note that d ( w ) ≥ d for all w by Lemma 2.6. Now define Y n = (cid:8) y ∈ Y : d ( y [ − n,n ] ) > d (cid:9) . Each Y n is a closed set and the family of Y n ’s is nested. If Y n (cid:54) = ∅ for all n , thenany y ∈ (cid:84) Y n has at least d + 1 preimages by compactness and Lemma 2.6, whichis a contradiction. So there is an n ≥ p with Y n = ∅ . Hence d ( w ) = d for all w ∈ B n +1 ( Y ).By the choice of n , we can define g , · · · , g d : B n +1 ( Y ) → B p +1 ( X ) satisfying { g ( w ) , · · · , g d ( w ) } = D ( w ) . Finally, we define cross sections f , · · · , f d : Y → X as follows: For each i =1 , · · · , d , f i ( y ) is the unique point in φ − ( y ) such that f i ( y ) [ − p,p ] = g i ( y [ − n,n ] ) . Thenall conditions are clearly satisfied except continuity.Fix i and define an open and closed set V i = (cid:91) (cid:8) − n [ u ] : u ∈ B n +1 ( X ) , ( g i ◦ φ )( u ) = u [ n +1 − p,n +1+ p ] (cid:9) . Then x ∈ V i if and only if x = ( f i ◦ φ )( x ). So f − i = φ | V i : V i → Y is a bijectivecontinuous map between compact metric spaces. Hence f i is continuous, as desired.It is clear that (3) implies (4).(4) ⇒ (1): Suppose that φ is not open. Then there is an open set U of X suchthat φ ( U ) is not open. Hence we can find a point x ∈ U and a sequence (cid:8) y ( n ) (cid:9) ∞ n =1 in Y such that y ( n ) / ∈ φ ( U ) for any n ∈ N and y ( n ) → φ ( x ). Find a cross section f with x ∈ f ( Y ). Then x = ( f ◦ φ )( x ). Let z ( n ) = f ( y ( n ) ) for each n . Then bythe continuity of f , we have z ( n ) → x . But since U is open, z ( n ) ∈ U , and hence y ( n ) ∈ φ ( U ) for all large n , which is a contradiction. So φ is open. (cid:3) Remark . In Theorem 4.4, the irreducibility of Y is needed only to prove thatwhen φ is open and bi-closing, it is constant-to-one. If Y is not irreducible, thenthere is a simple counterexample: X = { (01) ∞ , (10) ∞ , ∞ } , Y = { a ∞ , b ∞ } , and φ ((01) ∞ ) = φ ((10) ∞ ) = a ∞ , φ (2 ∞ ) = b ∞ .5. Examples
In sofic category, all the three conditions in Theorem 4.1 are different, that is, thereis no implication between any two of those conditions. We give three examples, ineach of which either X or Y (or both) is strictly sofic. Example 5.1.
Let Y be the even shift with its canonical cover φ : X A → Y , where X A is the edge shift formed by the underlying graph in Figure 5.1 and φ is given bythe labeling. Note that φ is bi-resolving and so Y is an AFT.Figure 5.1: A presentation of Y Since | φ − ( y ) | = 1 for all y ∈ Y except y = b ∞ , it follows that φ is not constant-to-one. Also, by Proposition 2.3, φ is not open. Thus φ is a bi-closing code whichis neither open nor constant-to-one. Example 5.2.
Let X be the sofic shift obtained by identifying two fixed points inthe full 2-shift, Y = { a, b } Z the full 2-shift, and φ : X → Y the subscript droppingcode (see Figure 5.2).Note that φ is a 1-block code and X is strictly sofic. Given u ∈ B ( X ) and l ≥ v = φ ( u ). Then for given y ∈ l [ v ], it is easy to find an x in l [ u ] such that φ ( x ) = y .Hence φ ( l [ u ]) = l [ v ] so that φ is open. Every point except a ∞ has two preimages,while a ∞ has only one preimage. Hence φ is not constant-to-one. Also, it identifiestwo points a ∞ . ( b b ) ∞ and a ∞ . ( b b ) ∞ , so φ is not right closing. Similarly it is notleft closing. Thus φ is an open code which is neither constant-to-one nor closing. PEN MAPS BETWEEN SHIFT SPACES 15
Figure 5.2: A presentation of X .This example also shows that an open extension of an irreducible shift of finite typeneed not be of finite type (see Corollary 4.3 (1)).We define a cross section f as follows: Let y ∈ Y . If y i = b with i ≥
0, replaceit with b if the number of occurrence of b in y [0 ,i ] is odd and otherwise with b . If y i = b with i <
0, replace it to b if the number of occurrence of b in y [ i, is odd andotherwise with b . The resulting point is f ( y ). Define f similarly by exchangingthe roles of b and b . Then these two cross sections cover X and are disjoint exceptone point a ∞ . Hence this example satisfies the condition (4) in Proposition 4.5.Note that the implication (4) ⇒ (1) in that proposition holds for any codes (notnecessarily constant-to-one). Example 5.3.
Let X be a sofic shift defined by the labeling in Figure 5.3, Y theeven shift as shown in Figure 5.1, and φ : X → Y the subscript dropping code. Thisexample appears in [1]. Figure 5.3: A presentation of X .It is easy to see that φ is 2-to-1 everywhere. We will show that φ is neither closingnor open. First, since φ identifies the points ( b b ) ∞ .a i ( b b ) ∞ for i = 1 ,
2, it is notclosing. Now, if φ is open, then φ ([ b ]) = (cid:83) ki =1 C i for cylinders C i defined by central2 l + 1 blocks in Y , for some l ≥
0. Since b ∞ is in φ ([ b ]), there exists i with b ∞ ∈ C i ,so C i is of the form − l [ b · · · b ]. Now, let y = b ∞ .b l +1 a ∞ . Note that y is in C i . Butif x ∈ X with x = b , then the first occurrence of a or a in x [0 , ∞ ) must be in aneven coordinate. So y cannot be in φ ([ b ]), which is a contradiction. Thus φ is notopen.In contrast to the finite type case, where any constant-to-one code has a crosssection, this example shows that a constant-to-one code may have no cross sec-tion in general. Indeed, if f : Y → X is a cross section, we can assume that f ( b ∞ ) = ( b b ) ∞ . Since f must send a ∞ b n +1 .b ∞ to a ∞ i ( b b ) n b . ( b b ) ∞ , it cannotbe continuous, which is a contradiction. Acknowledgment.
I am grateful to my advisor, Sujin Shin, for encouragement andguidance. Her valuable comments have improved an earlier version of this papergreatly.
References
1. F. Blanchard and G. Hansel, Sofic constant-to-one extensions of subshifts of finite type,
Proc.Amer. Math. Soc. (1991), 259–265.2. M. Boyle, B. Kitchens, and B. Marcus, A note on minimal covers for sofic systems,
Proc. Amer.Math. Soc. (1985), 403–411.3. M. Boyle and W. Krieger, Almost Markov and shift equivalent sofic systems, Springer-VerlagLecture Notes in Math. , 1988.4. M. Boyle, B. Marcus, and P. Trow, Resolving Maps and the Dimension Group for Shifts ofFinite Type , Mem. Amer. Math. Soc. , 1987.5. E. Coven and M. Paul, Finite procedures for sofic systems,
Monatsh. Math. (1977), 265–278.6. D. Fiebig, Constant-to-one extensions of shifts of finite type, Proc. Amer. Math. Soc. (1996), 2917–2922.7. G. A. Hedlund, Endomorphisms and automorphisms of the shift dynamical system,
Math.Systems Theory (1969), 320–375.8. B. Kitchens, Symbolic Dynamics: One-sided, Two-sided and Countable State Markov Shifts ,Springer-Verlag, 1998.9. P. K˚urka,
Topological and Symbolic Dynamics,
Cours Sp´ecialis´es , Soci´et´e Math´ematique deFrance, Paris, 2003.10. D. Lind and B. Marcus, An Introduction to Symbolic Dynamics and Coding , Cambridge Univ.Press, 1995.11. M. Nasu, Constant-to-one and onto global maps of homomorphisms between strongly con-nected graphs,
Ergod. Th. & Dynam. Sys. (1983), 387–413.12. P. Walters, An Introduction to Ergodic Theory , Springer-Verlag, 1982.
Department of Mathematical Sciences, Korea Advanced Institute of Science andTechnology, Daejeon 305-701, Korea
E-mail address ::