Induced maps on n-fold symmetric product suspensions

Topology and its Applications 158 (2011) 1192–1205 Contents lists available at ScienceDirect Topology and its Applications www.elsevier.com/locate/topol Induced maps on n-fold symmetric product suspensions Franco Barragán 1 Facultad de Ciencias Físico Matemáticas, BUAP, Ave. San Claudio y Rio Verde, Ciudad Universitaria, San Manuel Puebla, Pue. C.P. 72570, Mexico a r t i c l e i n f o a b s t r a c t Article history: Received 21 January 2010 Received in revised form 14 April 2011 Accepted 14 April 2011 In a previous paper, we define the n-fold symmetric product suspensions of continua. Now, we investigate the induced maps between these spaces.  2011 Elsevier B.V. All rights reserved. MSC: 54B20 Keywords: Continuum Confluent map Monotone map Open map Quasi-interior map Symmetric product 1. Introduction In 1979 Sam B. Nadler Jr. introduced the hyperspace suspension of a continuum [1]. In 2004 S. Macías, defined the n-fold hyperspace suspension of a continuum [2]. For a continuum X and n  2, in 2009, we define the n-fold symmetric product suspensions of X [3], denoted by SF n ( X ), as the quotient space F n ( X )/ F 1 ( X ), where F n ( X ) is the hyperspace of nonempty subsets of X with at most n points. Given a map f : X → Y between continua and an integer n  2, we let F n ( f ) : F n ( X ) → F n (Y ) and SF n ( f ) : SF n ( X ) → SF n (Y ) denote the corresponding induced maps. Let M be a class of maps between continua. As it was done with hyperspaces (see, for example [4–8]), in this paper we study the interrelations between the following three statements: (1) f ∈ M; (2) F n ( f ) ∈ M; (3) SF n ( f ) ∈ M. The paper consists of ten sections. In Section 2, we give the basic definitions for understanding the paper. In Section 3, we study homeomorphisms. Section 4 is devoted to monotone maps. Section 5 is about open maps. In Section 6, we discuss confluent maps. The light maps are analyzed in Section 7. In Section 8, we consider the class of quasi-interior maps and the class of MO-maps. In Section 9, we prove results concerning to the class of quasi-monotone maps and the class of weakly monotone maps. Finally, in Section 10 we study the class of weakly confluent maps and the class of pseudo-confluent maps. 1 E-mail address: [email protected]. This research is part of the author’s Dissertation under the supervision of Professors Raúl Escobedo and Sergio Macías. 0166-8641/$ – see front matter doi:10.1016/j.topol.2011.04.006  2011 Elsevier B.V. All rights reserved. F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1193 2. Definitions The symbols N and R will denote the set of positive integers and the set of real numbers, respectively. A continuum is a nonempty compact, connected metric space. A subcontinuum is a continuum contained in a space X . A continuum X is said to be irreducible provided that no proper subcontinuum of X contains { p , q} for some p , q ∈ X . If X is a continuum, then given A ⊂ X and ǫ > 0, the open ball about A of radius ǫ is denoted by Vǫ ( A ), the closure of A in X by Cl X ( A ), and the interior of A in X by int X ( A ). A map means a continuous function. A surjective map f : X → Y between continua is said to be: • • • • • • • • • • confluent provided that for each subcontinuum B of Y and for each component C of f −1 ( B ), we have f (C ) = B; light provided that f −1 ( y ) is totally disconnected for each y ∈ Y ; monotone provided that f −1 ( y ) is connected for each y ∈ Y ; open provided f (U ) is open in Y for each open subset U of X ; MO provided that there exist a continuum Z , an open map h : X → Z , and a monotone map g : Z → Y such that f = g ◦ h; pseudo-confluent provided that for each irreducible subcontinuum B of Y , there exists a component C of f −1 ( B ) such that f (C ) = B; quasi-interior provided that for each y ∈ Y if U is an open subset of X containing a component of f −1 ( y ), then y ∈ intY ( f (U )); quasi-monotone provided that for each subcontinuum B of Y having nonempty interior in Y , f −1 ( B ) has only finitely many components and each of these components maps onto B under f ; weakly confluent provided that for each subcontinuum B of Y , there exists a component C of f −1 ( B ) such that f (C ) = B; weakly monotone provided that for each subcontinuum B of Y having nonempty interior in Y , each component of f −1 ( B ) is mapped by f onto B. Given a continuum X and n ∈ N, the product of X with itself n times will be denoted by X n , the symbol F n ( X ) denotes the n-fold symmetric product of X ; that is: F n ( X ) = { A ⊂ X | A has at most n points}, topologized with the Hausdorff metric, which is defined as follows:    H( A , B ) = inf ǫ > 0  A ⊂ Vǫ ( B ) and B ⊂ Vǫ ( A ) , H always denotes the Hausdorff metric. Given a finite collection, U 1 , . . . , U m , of subsets of X , U 1 , . . . , U m n , denotes the following subset of F n ( X ):   m    A ∈ Fn( X ) A ⊂ U k and A ∩ U k = ∅ for each k ∈ {1, . . . , m} . k =1 It is known that the family of all subsets of F n ( X ) of the form U 1 , . . . , U m n , where each U i is an open subset of X , forms a basis for a topology for F n ( X ) (see [4, 0.11]) called the Vietoris topology. The Vietoris topology and the topology induced by the Hausdorff metric coincide [4, 0.13]. Given a continuum X and n ∈ N, with n  2, we define the n-fold symmetric product suspension of the continuum X [3], denoted by SF n ( X ), as the quotient space: SF n ( X ) = F n ( X )/ F 1 ( X ) with the quotient topology. The fact that SF n ( X ) is a continuum follows from 3.10 of [9]. Notation 2.1. Given a continuum X , qnX : F n ( X ) → SF n ( X ) denotes the quotient map. Also, let F nX denote the point qnX ( F 1 ( X )). Remark 2.2. Note that SF n ( X ) \ { F nX } is homeomorphic to F n ( X ) \ F 1 ( X ), using the appropriate restriction of qnX . Given a map f : X → Y between continua and an integer n  2, the function F n ( f ) : F n ( X ) → F n (Y ) given by F n ( f )( A ) = f ( A ) is the induced map by f between the n-fold symmetric products of X and Y . By [10, Corollary 1.8.23] F n ( f ) is continuous. Also, we have an induced map SF n ( f ) : SF n ( X ) → SF n (Y ) called the induced map by f between the n-fold symmetric product suspensions of X and Y , which can be defined by SF n ( f )(χ ) =  qnY ( F n ( f )((qnX )−1 (χ ))), F Yn , if χ = F nX ; if χ = F nX . 1194 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 We note that, by [11, Theorem 4.3, p. 126], SF n ( f ) is continuous. In addition, the following diagram Fn( X ) Fn ( f ) F n (Y ) qnX SF n ( X ) qnY SF n ( f ) (∗) SF n (Y ) is commutative. Let X be a continuum and let n ∈ N. We denote by f nX : X n → F n ( X ) the map given by f nX ((x1 , . . . , xn )) = {x1 , . . . , xn } (see [12, Lemma 1]), if there is no confusion, we write f n : X n → F n ( X ). Given a map f : X → Y between continua, we denote by f Xn ,Y : X n → Y n the map given by f Xn ,Y ((x1 , . . . , xn )) = ( f (x1 ), . . . , f (xn )), if there is no confusion, we write f n : X n → Y n . In addition, the following diagram Xn f Xn ,Y Yn f nX Fn( X ) (∗∗) f nY Fn ( f ) F n (Y ) is commutative. 3. Homeomorphisms We begin with some simple results. Theorem 3.1. Let f : X → Y be a map between continua, and let n  2 be an integer. Then the following are equivalent: (1) f : X → Y is injective; (2) F n ( f ) : F n ( X ) → F n (Y ) is injective; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is injective. Theorem 3.2. Let f : X → Y be a map between continua, and let n  2 be an integer. Then the following are equivalent: (1) f : X → Y is surjective; (2) F n ( f ) : F n ( X ) → F n (Y ) is surjective; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is surjective. As an easy consequence of Theorems 3.1 and 3.2, we have the following: Theorem 3.3. Let f : X → Y be a map between continua, and let n  2 be an integer. Then the following are equivalent: (1) f : X → Y is a homeomorphism; (2) F n ( f ) : F n ( X ) → F n (Y ) is a homeomorphism; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is a homeomorphism. 4. Monotone maps We prove the equivalence of the monotonicity of all the maps we are considering. Theorem 4.1. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. Then the following are equivalent: (1) f : X → Y is monotone; (2) F n ( f ) : F n ( X ) → F n (Y ) is monotone; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is monotone. Proof. Suppose that f is monotone. Let B ∈ F n (Y ). We assume that B = { y 1 , . . . , y r }, where r  n. Note that Fn( f ) −1 ( B ) = f −1 ( y 1 ), . . . , f −1 ( yr ) n . r To prove this, take A ∈ ( F n ( f ))−1 ( B ). Then f ( A ) = B. Thus, A ⊂ i =1 f −1 ( y i ). For each i ∈ {1, . . . , r }, let xi ∈ A be such that f (xi ) = y i . It follows that for each i ∈ {1, . . . , r }, A ∩ f −1 ( y i ) = ∅. Hence, A ∈  f −1 ( y 1 ), . . . , f −1 ( y r )n . Thus, ( F n ( f ))−1 ( B ) ⊂ F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1195 r  f −1 ( y 1 ), . . . , f −1 ( y r )n . Now, let D ∈  f −1 ( y 1 ), . . . , f −1 ( y r )n . Then D ⊂ i =1 f −1 ( y i ) = f −1 ( B ). It follows that f ( D ) ⊂ B. Let y i ∈ B. Since D ∩ f −1 ( y i ) = ∅, we can find di ∈ D ∩ f −1 ( y i ). Note that f (di ) = y i ∈ f ( D ). Consequently, B ⊂ f ( D ). This implies that F n ( f )( D ) = f ( D ) = B. We have that D ∈ ( F n ( f ))−1 ( B ). Thus,  f −1 ( y 1 ), . . . , f −1 ( y r )n ⊂ ( F n ( f ))−1 ( B ). Since f is monotone, for each i ∈ {1, . . . , r }, f −1 ( y i ) is connected. By [13, Lemma 1], ( F n ( f ))−1 ( B ) is connected. Therefore, (1) implies (2). Note that qnY is a monotone map. If F n ( f ) is monotone, then, by [14, (5.1)], qnY ◦ F n ( f ) is monotone. By (∗), we obtain that SF n ( f ) ◦ qnX is monotone. Hence, by [14, (5.15)], we have that SF n ( f ) is monotone. Thus, (2) implies (3). Finally, we prove that (3) implies (1). Let y ∈ Y . Take a point y ′ ∈ Y \ { y }, and let B = { y , y ′ }. Let x, x′ ∈ X be such that f (x) = y and f (x′ ) = y ′ . Let A = {x, x′ }. It follows that F n ( f )( A ) = f ( A ) = B. Hence, A ∈ ( F n ( f ))−1 ( B ). Since qnX and SF n ( f ) are monotone maps, by [14, (5.1)], we have that SF n ( f ) ◦ qnX is monotone. Thus, by (∗), qnY ◦ F n ( f ) is monotone. Then (qnY ◦ F n ( f ))−1 (qnY ( B )) is connected. Since B ∈ F n ( X ) \ F 1 ( X ), we obtain that (qnY ◦ F n ( f ))−1 (qnY ( B )) = ( F n ( f ))−1 ( B ). This implies that ( F n ( f ))−1 ( B ) is connected. Since f −1 ( y ) and f −1 ( y ′ ) are closed subsets of X and f −1 ( y ) ∩ f −1 ( y ′ ) = ∅, there exists ǫ > 0 such that Vǫ ( f −1 ( y )) ∩ Vǫ ( f −1 ( y ′ )) = ∅. We note that A ∈ ( F n ( f ))−1 ( B ) ⊂ Vǫ ( f −1 ( y )), Vǫ ( f −1 ( y ′ ))n . Then, by [15, Lemma 6.1], it follows that ( ( F n ( f ))−1 ( B )) ∩ Vǫ ( f −1 ( y )) is connected. Since ( F n ( f ))−1 ( B ) = f −1 ( B ) = f −1 ( y ) ∪ f −1 ( y ′ ), we have that − 1 − 1 − 1 − 1 ( ( F n ( f )) ( B )) ∩ Vǫ ( f ( y )) = f ( y ). Hence, f ( y ) is connected. Therefore, f is monotone. ✷ 5. Open maps We study the relationship of openness between the considered maps. Lemma 5.1. Let X be a continuum, let n, r ∈ N be such that r  n, and let U 1 , . . . , U r be open subsets of X . Then an open subset of X . U 1 , . . . , U r n is . . . , U r n such that x ∈ A x . Let J = { j ∈ {1, . . . , r }: x ∈ U j }. Since Proof. Let x ∈ U 1 , . . . , U r n . Thenthere exists A x ∈ U 1 , r x ∈ A x ⊂ i =1 U i , J = ∅, also x ∈ j ∈ J U j . We see that y U ⊂  U , . . . , U  . Let ∈ 1 r n j j∈ J j ∈ J U j , and let A = { y } ∪  ( A x \ {x}). Hence, A ∈ U 1 , . . . , U r n . It follows that y ∈ U 1 , . . . , U r n . Thus, x ∈ j ∈ J U j ⊂ U 1 , . . . , U r n . Therefore, U 1 , . . . , U r n is an open subset of X . ✷ As a consequence of Lemma 5.1, we have the following: Corollary 5.2. Let X be a continuum and n ∈ N. If C is an open subset of F n ( X ), then C is an open subset of X . Theorem 5.3. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is an open map, then f : X → Y is an open map. Proof. Let U be an open subset of X . Then U n is an open subset F n ( X ). It follows that F n ( f )(U n ) is an open subset of F n (Y ). By Corollary 5.2, F n ( f )(U n ) is an open subset of Y . We see that F n ( f )(U n ) = f (U ). Let y ∈ ∪ F n ( f )(U n ). Then there exists B ∈ F n ( f )(U n ) such that y ∈ B. Hence, there exists A ∈ U n such that F n ( f )( A ) = B. Thus, B ⊂ f (U ). It follows that y ∈ f (U ). This implies that F n ( f )(U n ) ⊂ f (U ). Now, let z ∈ f (U ). Take a ∈ U such that f (a) = z. Let C = {a} and D = { z}. It follows that C ∈ U n and F n ( f )(C ) = D. Hence, D ∈ F n ( f )(U n ) and z ∈ D. We have that z ∈ F n ( f )(U n ). This implies that f (U ) ⊂ F n ( f )(U n ). Therefore, F n ( f )(U n ) = f (U ). Hence, we obtain that f (U ) is an open subset of Y . We conclude that f is an open map. ✷ Theorem 5.4. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is an open map, then f : X → Y is a homeomorphism. Proof. It is sufficient to prove that f is injective. Suppose that there exist x1 , x2 ∈ X such that x1 = x2 and f (x1 ) = f (x2 ). Let A = {x1 , x2 } and y = f (x1 ) = f (x2 ). Take a point z in Y different of y. Thus, we can take open subsets V and W of Y such that y ∈ V , z ∈ W and V ∩ W = ∅. It follows that, f −1 ( V ) is an open subset of X . Moreover, x1 , x2 ∈ f −1 ( V ). Let U 1 and U 2 be open subsets of X such that x1 ∈ U 1 ⊂ f −1 ( V ), x2 ∈ U 2 ⊂ f −1 ( V ) and U 1 ∩ U 2 = ∅. Let U = U 1 , U 2 n . Hence, U is an open subset of F n ( X ) such that A ∈ U and U ∩ F 1 ( X ) = ∅. This implies that qnX (U ) is open in SF n ( X ). Now, for each k ∈ N, let zk ∈ Y be such that zk ∈ W , zk = z and such that the sequence { zk }k∞=1 converges to z in Y . For each k ∈ N, let A k = { zk , z}. Hence, the sequence { A k }k∞=1 converges to { z} in F n (Y ). Note that, for each k ∈ N, A k ∈ /  V n . / F n ( f )(U ). This implies that, for each k ∈ N, A k ∈ / [ F 1 (Y ) ∪ Since F n ( f )(U ) ⊂  V n , we have that, for each k ∈ N, A k ∈ F n ( f )(U )]. Since [ F 1 (Y ) ∪ F n ( f )(U )] = (qnY )−1 (qnY ( F n ( f )(U ))), we obtain that, for each k ∈ N, A k ∈ / (qnY )−1 (qnY ( F n ( f )(U ))). n −1 n n −1 n Notice that { z} ∈ (q Y ) (q Y ( F n ( f )(U ))). If (q Y ) (q Y ( F n ( f )(U ))) is an open subset of F n (Y ), there exists N ∈ N such that if k  N, A k ∈ (qnY )−1 (qnY ( F n ( f )(U ))), which is a contradiction. Then (qnY )−1 (qnY ( F n ( f )(U ))) is not an open subset of F n (Y ). 1196 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 It follows that qnY ( F n ( f )(U )) is not an open subset of SF n (Y ). By (∗), we obtain that SF n ( f )(qnX (U )) is not an open subset of SF n (Y ). Therefore, SF n ( f ) is not an open map. ✷ Applying Theorems 5.4 and 3.3, we obtain the following: Theorem 5.5. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is an open map, then F n ( f ) : F n ( X ) → F n (Y ) is a homeomorphism. The following example proves that the converse of Theorem 5.4 is not true. Example 5.6. Let X = [−1, 1], Y = [0, 1], and let f : X → Y be defined by f (x) = |x|. It is clear that f is an open map. If SF n ( f ) : SF n ( X ) → SF n (Y ) were an open map, where n  2, then, by Theorem 5.4, f would be a homeomorphism, which is not true. Hence, SF n ( f ) is not an open map. Theorem 5.7. Let f : X → Y be a surjective map between continua. Consider the following conditions: (1) f : X → Y is open; (2) F 2 ( f ) : F 2 ( X ) → F 2 (Y ) is open; (3) SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is open. Then (1) and (2) are equivalent, (3) implies (1), (3) implies (2), (1) does not imply (3) and (2) does not imply (3). Proof. By Theorem 5.3, we have that (2) implies (1). To prove that (1) implies (2). Consider the map f X2 ,Y : X 2 → Y 2 . Since f is open, we obtain that f X2 ,Y is open. Also, by [12, Lemma 9], we have that f 2Y is open. Hence, f 2Y ◦ f X2 ,Y is open [14, (5.1)]. By (∗∗), it follows that F 2 ( f ) ◦ f 2X is open. Then, by [14, (5.15)], we conclude that F 2 ( f ) is open. Applying Theorem 5.4, we have that (3) implies (1). Since (1) and (2) are equivalent, we obtain that (3) implies (2). By Example 5.6, it follows that (1) does not imply (3). Since (1) and (2) are equivalent, (2) does not imply (3). ✷ It is known that only f 2X is an open map [12, Lemma 9]. So, we prove the following result: Lemma 5.8. Let f : X → Y be a surjective map between continua and let n  2 be an integer. Let U 1 , . . . , U n be open subsets of X such that for each j , l ∈ {1, . . . , n}, U j ∩ U l = ∅ if j = l, and let U = U 1 × · · · × U n . Then f nX (U ) is an open subset of F n ( X ). Proof. Let {b1 , . . . , bk } ∈ f nX (U ), where k  n. Then exists (a1 , . . . , an ) ∈ U such that f nX ((a1 , . . . , an )) = {b1 , . . . , bk }. Thus, {a1 , . . . , an } = {b1 , . . . , bk }. Notice that k = n. Since, for each j ∈ {1, . . . , n}, a j ∈ U j , we have that {b1 , . . . , bn } ∈ U 1 , . . . , U n n . Now, we see that U 1 , . . . , U n n ⊂ f nX (U ). Let {c 1 , . . . , cn } ∈ U 1 , . . . , U n n . Since, for each j , l ∈ {1, . . . , n}, U j ∩ U l = ∅ if j = l, without loss of generality, we assume that, for each j ∈ {1, . . . , n}, c j ∈ U j . Hence, (c 1 , . . . , cn ) ∈ U and f nX ((c 1 , . . . , cn )) = {c 1 , . . . , cn }. Thus, {c 1 , . . . , cn } ∈ f nX (U ). It follows that, U 1 , . . . , U n n ⊂ f nX (U ). Hence, we have that {b1 , . . . , bn } ∈ U 1 , . . . , U n n ⊂ f nX (U ). Therefore, f nX (U ) is an open subset of F n ( X ). ✷ Theorem 5.9. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then F n ( f ) : F n ( X ) → F n (Y ) is an open map if and only if f : X → Y is a homeomorphism. Proof. Suppose that F n ( f ) is an open map. It is enough to prove that f is injective. Suppose, on the contrary, that there exist x1 , x2 ∈ X such that x1 = x2 and f (x1 ) = f (x2 ). Let p 1 = f (x1 ) = f (x2 ). We can take p 2 , . . . , pn−1 ∈ Y \ { p 1 } such that p j = pl if j = l and j , l ∈ {2, . . . , n − 1}. Hence, for each i ∈ {1, . . . , n − 1}, we take an open subset V i of Y such that p i ∈ V i , and V j ∩ V l = ∅ if j = l. Let U 1 and U 2 be open subsets of X such that x1 ∈ U 1 ⊂ f −1 ( V 1 ), x2 ∈ U 2 ⊂ f −1 ( V 1 ) and U 1 ∩ U 2 = ∅. For each i ∈ {2, . . . , n − 1}, let U i +1 = f −1 ( V i ). Notice that, for each j , l ∈ {1, . . . , n}, U j ∩ U l = ∅ if j = l. Let U = U 1 × · · · × U n . It follows that U is an open subset of X n . By Lemma 5.8, f nX (U ) is an open subset of F n ( X ). Since F n ( f ) is an open map, by Theorem 5.3, f is an open map. Hence, f Xn ,Y is an open map. Then f Xn ,Y (U ) = f (U 1 ) × · · · × f (U n ) is an open subset of Y n such that ( p 1 , p 1 , p 2 , . . . , pn−1 ) ∈ f Xn ,Y (U ). Now let { yk }k∞=1 and { zk }k∞=1 be sequences converging to p 2 in Y such that y j = zl for each j , l ∈ N, for each i ∈ {1, . . . , n − 1} and for each k ∈ N, yk = p i and zk = p i . Note that {{ p 1 , yk , zk , p 3 , . . . , pn−1 }}k∞=1 is a sequence converging to the point { p 1 , p 2 , . . . , pn−1 } in F n (Y ). Since ( p 1 , p 1 , p 2 , . . . , pn−1 ) ∈ f Xn ,Y (U ), it follows that { p 1 , p 2 , . . . , pn−1 } ∈ f nY ( f Xn ,Y (U )). Suppose that f nY ( f Xn ,Y (U )) is an open subset of F n (Y ). Then there exists N ∈ N such that { p 1 , y N , z N , p 3 , . . . , pn−1 } ∈ f nY ( f Xn ,Y (U )) and y N , z N ∈ f (U 3 ) = V 2 . Notice that ( f nY )−1 ({ p 1 , y N , z N , p 3 , . . . , pn−1 }) = {( y 1 , . . . , yn ) ∈ Y n : { y 1 , . . . , yn } = F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1197 { p 1 , y N , z N , p 3 , . . . , pn−1 }}. Since the points of { p 1 , y N , z N , p 3 , . . . , pn−1 } are pair wise distinct, any point of the set ( f nY )−1 ({ p 1 , y N , z N , p 3 , . . . , pn−1 }) has two different coordinates in f (U 3 ) ( y N and z N ). Since f (U 3 ) ∩ f (U i ) = ∅, for each i ∈ {1, . . . , n}\{3}, we obtain that ( f nY )−1 ({ p 1 , y N , z N , p 3 , . . . , pn−1 })∩ f Xn ,Y (U ) = ∅. This implies that { p 1 , y N , z N , p 3 , . . . , pn−1 } ∈ / f nY ( f Xn ,Y (U )), which is a contradiction. Thus, f nY ( f Xn ,Y (U )) is not an open subset of F n (Y ). By (∗∗), it follows that F n ( f )( f nX (U )) is not an open subset of F n (Y ). Hence, F n ( f ) is not open map, which is a contradiction. Therefore, f is injective. By Theorem 3.3, the reverse implication follows. ✷ To see that the converse of Theorem 5.3 is not true, we have the following: Example 5.10. Let X = [−1, 1], Y = [0, 1], and f : X → Y defined by f (x) = |x|. It is clear that f is an open map. If F n ( f ) : F n ( X ) → F n (Y ) were an open map, where n  3, by Theorem 5.9, f would be a homeomorphism, which is not true. Hence, F n ( f ) is not an open map. Theorem 5.11. Let f : X → Y be a surjective map between continua, and let n  3 be a positive integer. Consider the following conditions: (1) f : X → Y is open; (2) F n ( f ) : F n ( X ) → F n (Y ) is open; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is open. Then (2) and (3) are equivalent, (2) implies (1), (3) implies (1), (1) does not implies (3) and (1) does not implies (2). Proof. Applying Theorems 5.3, 5.4, and 5.5, we have that (2) implies (1), (3) implies (1) and (3) implies (2), respectively. Now, if F n ( f ) is open, then, by Theorem 5.9, f is a homeomorphism. Hence, by Theorem 3.3, SF n ( f ) is a homeomorphism, in particular SF n ( f ) is open. Thus, (2) implies (3). By Example 5.6, (1) does not imply (3). Moreover, applying Example 5.10, we obtain that (1) does not imply (2). ✷ 6. Confluent maps Lemma 6.1. Let f : X → Y be a surjective map between continua, and let n  2 be an integer, and let B be a subcontinuum of Y . If C is a component of f −1 ( B ), then F n (C ) is a component of ( F n ( f ))−1 ( F n ( B )). Proof. We note that, by [16, p. 877], F n (C ) is connected. Furthermore, F n (C ) ⊂ ( F n ( f ))−1 ( F n ( B )). Let C the component of ( F n ( f ))−1 ( F n ( B )) such that F n (C ) ⊂ C . By [15, Lemma 2.2], C is a subcontinuum of X such that C ⊂ C ⊂ f −1 ( B ). Hence, C = C . We see that C = F n (C ). Let A ∈ C . Thus, A ⊂ C . This implies that A ∈ F n (C ). Then, C ⊂ F n (C ). Therefore, C = F n (C ). ✷ Theorem 6.2. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a confluent map, then f : X → Y is a confluent map. Proof. Let B be a subcontinuum of Y , and let C be a component of f −1 ( B ). By Lemma 6.1, we have that F n (C ) is a component of ( F n ( f ))−1 ( F n ( B )). It follows that F n ( f )( F n (C )) = F n ( B ). Now, we prove that B ⊂ f (C ). Let b ∈ B. Then there exists H ∈ F n (C ) such that f ( H ) = {b}. This implies that b ∈ f (C ). Thus, we have that B ⊂ f (C ). Hence, we obtain that f (C ) = B. Therefore, f is a confluent map. ✷ Theorem 6.3. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a confluent map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a confluent map. Proof. Since qnY is monotone, by [9, Theorem 13.15], qnY is confluent. Then, by [17, 2.6], we have that qnY ◦ F n ( f ) is confluent. Hence, by (∗), SF n ( f ) ◦ qnX is confluent. By [14, 5.16], we obtain that SF n ( f ) is confluent. ✷ The proof of the following theorem is similar to the proof [6, Theorem 18]. We include the details for the convenience of the reader. Theorem 6.4. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then F n ( f ) : F n ( X ) → F n (Y ) is a confluent map if and only if f : X → Y is a monotone map. 1198 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 Proof. If f is a monotone map, then, by Theorem 4.1, F n ( f ) is a monotone map. Hence, by [9, Theorem 13.15], F n ( f ) is a confluent map. Now, we assume that f is not monotone. Then there exists y ∈ Y such that f −1 ( y ) is not a connected subset of X . Let P and Q be different components of f −1 ( y ), and let p ∈ P and q ∈ Q . We take a point z ∈ Y \ { y }. Let R be a subcontinuum of Y such that { z}  R ⊂ Y \ { y } [9, Corollary 5.5]. Let   K = { y } ∪ K : K ∈ F n −1 ( R ) . It is clear that K is a subcontinuum of F n (Y ). We take a point s ∈ X such that f (s) = z. Let S be the component of f −1 ( R ) such that s ∈ S. Let C be the component of ( F n ( f ))−1 (K) such that { p , q, s} ∈ C . Let Z = C . Then, by [15, Lemma 2.2], Z ∈ C n ( X ). We note that P ∪ Q ∪ S ⊂ Z . We see that P is a component of Z . Let P ′ be the component of Z such that P ⊂ P ′ . We assume that P  P ′ . Let w ∈ P ′ \ f −1 ( y ). Then f ( w ) ∈ f ( P ′ ) ∩ R. We note that y ∈ { y } ∩ f ( P ′ ) and f ( P ′ ) ⊂ { y } ∪ R. Thus, we have that f ( P ′ ) is not connected, a contradiction. Hence, P is component of Z . Similarly, Q is component of Z . Now, we prove that F n ( f )(C ) ⊂ {{ y } ∪ K : K ∈ F n−2 ( R )}. Let A ∈ C . Since C is a connected subset of 2 Z , by [18, Lemma 3.1, p. 241], A ∩ P = ∅ and A ∩ Q = ∅. Thus, f ( A ) ∈ {{ y } ∪ K : K ∈ F n−2 ( R )}. Since {{ y } ∪ K : K ∈ F n−2 ( R )}  K, we have that F n ( f )(C )  K. Therefore, F n ( f ) is not confluent. ✷ As a consequence of the proof of Theorem 6.4, we have the following: Theorem 6.5. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then SF n ( f ) : SF n ( X ) → SF n (Y ) is a confluent map if and only if f : X → Y is a monotone map. Applying Theorems 4.1, 6.4 and 6.5, we obtain the following: Theorem 6.6. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then the following are equivalent: (1) (2) (3) (4) (5) f : X → Y is monotone; F n ( f ) : F n ( X ) → F n (Y ) is monotone; SF n ( f ) : SF n ( X ) → SF n (Y ) is monotone; F n ( f ) : F n ( X ) → F n (Y ) is confluent; SF n ( f ) : SF n ( X ) → SF n (Y ) is confluent. The following example proves that the converse of Theorem 6.2 is not true. Example 6.7. Let X = [−1, 1], Y = [0, 1], and f : X → Y defined by f (x) = |x|. It is readily seen that f is confluent, but f is not monotone. By Theorem 6.6, F n ( f ) : F n ( X ) → F n (Y ), where n  3, is not confluent. Theorem 6.8. Let f : X → Y be a surjective map between continua. If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a confluent map, then f : X → Y is a confluent map. Proof. Let B be a subcontinuum of Y , and let C be a component of f −1 ( B ). If B = Y , then C = X and f (C ) = B. We assume that there exists a point y ∈ Y \ B. Let   K = { y} ∪ K : K ∈ F 1 ( B ) . We note that K is a subcontinuum of F 2 (Y ) such that K ∩ F 1 (Y ) = ∅. Hence, q2Y (K) is a subcontinuum of SF 2 (Y ). Now let z ∈ f −1 ( y ), and we define A = {{ z} ∪ A: A ∈ F 1 (C )}. Then A is a connected subset of ( F 2 ( f ))−1 (K). Let C be the component of ( F 2 ( f ))−1 (K) such that A ⊂ C . Thus, q2X (C ) is a component of (SF 2 ( f ))−1 (q2Y (K)). Since SF 2 ( f ) is a confluent map, we have that SF 2 ( f )(q2X (C )) = q2Y (K). By (∗), q2Y ( F 2 ( f )(C )) = q2Y (K). Since K ∩ F 1 (Y ) = ∅, F 2 ( f )(C ) = K. On the other hand, since C ⊂ f −1 ( y ) ∪ f −1 ( B ), C is not connected. Moreover, by [15, Lemma 2.2], C = C 1 ∪ C 2 , where C 1 and C 2 are the components of C . Without loss of generality, we assume that C 1 ⊂ f −1 ( y ) and C 2 ⊂ f −1 ( B ). It follows that C 2 = C . We prove that B ⊂ f (C ). Let b ∈ B. Then { y , b} ∈ K. Thus, there exists D ∈ C such that F 2 ( f )( D ) = { y , b}. Hence, b ∈ f ( D ) ⊂ f (C 1 ) ∪ f (C ). If b ∈ f (C 1 ), then b = y, a contradiction. This implies that b ∈ f (C ). Then B ⊂ f (C ). Therefore, f is a confluent map. ✷ Lemma 6.9. If f : X → Y is a confluent map between continua and Y is locally connected, then f is a quasi-interior map. Proof. Let y ∈ Y , let C be a component of f −1 ( y ) and let U be an open subset ∞ of X such that C ⊂ U . Since Y is locally connected, there exists a sequence { B i }∞ of subcontinua of Y such that i =1 B i = { y } and, for each i, B i +1 ⊂ B i and i =1 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1199 y ∈ int( B i ). For each i, let C i be the component of f −1 ( B i ) such thatC ⊂ C i . Since, for each i, f −1 ( B i +1 ) ⊂ f −1 ( B i ), we ∞ have that C ⊂ C i +1 ⊂ f −1 ( B i ). Hence, for each i, C i +1 ⊂ C i . Let K = i =1 C i . By [9, Theorem 1.8], K is a continuum. We  ∞ note that C ⊂ K . Moreover, K ⊂ f −1 ( y ), because if x ∈ K , then f (x) ∈ i =1 B i = { y }, thus x ∈ f −1 ( y ). It follows that C = K . By [9, Proposition 1.7], there exists a positive integer N such that for each i  N, C i ⊂ U . Let j  N. Hence, f (C j ) ⊂ f (U ). Since f is a confluent map, we have that B j ⊂ f (U ). Since y ∈ int( B j ), we conclude that y ∈ int( f (U )). Therefore, f is a quasi-interior. ✷ Theorem 6.10. Let f : X → Y be a surjective map between continua, and let Y be locally connected. Then the following are equivalent: (1) f : X → Y is confluent; (2) F 2 ( f ) : F 2 ( X ) → F 2 (Y ) is confluent; (3) SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is confluent. Proof. If f is a confluent map then, by Lemma 6.9, f is a quasi-interior map. By [17, 3.1], there exist a continuum Z , a monotone map h : X → Z , and an open map g : Z → Y such that f = g ◦ h. Hence, by Theorems 4.1 and 5.7, it follows that F 2 (h) is a monotone map and F 2 ( g ) is an open map, respectively. Furthermore, F 2 ( f ) = F 2 ( g ) ◦ F 2 (h). Then, by [17, 3.1], F 2 ( f ) is a quasi-interior map. By [17, 2.7], SF 2 ( f ) is a confluent map. Thus (1) implies (2). By Theorems 6.3 and 6.8, it follows that (2) implies (3) and (3) implies (1), respectively. ✷ Question 6.11. Let f : X → Y be a surjective map between continua such that Y is not locally connected. If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a confluent map, then is F 2 ( f ) : F 2 ( X ) → F 2 (Y ) a confluent map? 7. Light maps It is easy to verify that if f : X → Y is a light map between continua and A is a subcontinuum of X , then f | A : A → f ( A ) is a light map. Therefore, we have the following: Lemma 7.1. Let f : X → Y be a surjective map between continua, and let n ∈ N. If F n ( f ) : F n ( X ) → F n (Y ) is a light map, then F m ( f ) : F m ( X ) → F m (Y ) is a light map, for each m ∈ N with m  n. Lemma 7.2. Let X be a continuum, and let n  2 be an integer. If C 1 , . . . , C k , where k  n, are totally disconnected closed subsets of X and C i ∩ C j = ∅, for each i = j, then C 1 , . . . , C k n is a totally disconnected subset of F n ( X ). Proof. Suppose that there exists a nondegenerate component C of C 1 , . . . , C k n . By [15, Lemma 2.2], C has at most n components, we say that C = is=1 K i , where K i is a component of C and s  n. If K i is degenerate, for each i ∈ {1, . . . , s}, then C is degenerate, which is a contradiction. Hence, there exists j ∈ {1, . . . , s} such that K j is not degenerate. k Since C ⊂ i =1 C i , there exists l ∈ {1, . . . , k} such that K j ⊂ C l , which is a contradiction. Therefore, C 1 , . . . , C k n is a totally disconnected subset of F n ( X ). ✷ Theorem 7.3. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. Then F n ( f ) : F n ( X ) → F n (Y ) is a light map if and only if f : X → Y is a light map. Proof. Suppose that F n ( f ) : F n ( X ) → F n (Y ) is a light map. Applying Lemma 7.1, we have that F 1 ( f ) : F 1 ( X ) → F 1 (Y ) is a light map. On the other hand, since X and Y are homeomorphic to F 1 ( X ) and F 1 (Y ), respectively, we can consider two homeomorphisms h : X → F 1 ( X ) and k : Y → F 1 (Y ) such that k ◦ f = F 1 ( f ) ◦ h. It follows that f is a light map. Now assume that f : X → Y is a light map. Let B ∈ F n (Y ). Let B = {b1 , . . . , bk }, where k  n and b i = b j , for each i = j. It is easy to see that ( F n ( f ))−1 ( B ) =  f −1 (b1 ), . . . , f −1 (bk ). Hence, by Lemma 7.2, we conclude that ( F n ( f ))−1 ( B ) is a totally disconnected subset of F n ( X ). Thus, F n ( f ) is a light map. ✷ Theorem 7.4. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is a light map, then F n ( f ) : F n ( X ) → F n (Y ) is a light map. Proof. Let B ∈ F n (Y ). Then we have the following cases: Case (1). B ∈ F 1 (Y ). It follows that q Y ( B ) = F nY . Since SF n ( f ) is light, we obtain that (SF n ( f ))−1 ( F Yn ) is a totally disconnected subset of SF n ( X ). Suppose that there exists a nondegenerate component C of ( F n ( f ))−1 ( B ). Then, by [15, Lemma 2.2], C has at most n components, say C 1 , . . . , C r , where r  n. We note that there exists j ∈ {1, . . . , r } such that C j is nondegenerate. Hence, F n (C j ) is a nondegenerate subcontinuum of F n ( X ) such that F n (C j ) ∩ ( F n ( X ) \ F 1 ( X )) = ∅. Since F n (C j ) ⊂ ( F n ( f ))−1 ( B ) ⊂ 1200 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 (qnX )−1 ((SF n ( f ))−1 ( F Yn )), it follows that qnX ( F n (C j )) is a nondegenerate subcontinuum of (SF n ( f ))−1 ( F Yn ). Which is a contradiction. Thus, C is degenerate. This implies that ( F n ( f ))−1 ( B ) is a totally disconnected subset of F n ( X ). Case (2). B ∈ F n (Y ) \ F 1 (Y ). Then q Y ( B ) ∈ SF n ( X ) \ { F Yn }. Hence (SF n ( f ))−1 (q Y ( B )) is totally disconnected and (SF n ( f ))−1 (q Y ( B )) ⊂ SF n ( X ) \ { F nX }. It follows that (q X )−1 ((SF n ( f ))−1 (q Y ( B ))) is totally disconnected. Since (q X )−1 ((SF n ( f ))−1 (q Y ( B ))) = ( F n ( f ))−1 ( B ), we conclude that ( F n ( f ))−1 ( B ) is totally disconnected. Therefore, F n ( f ) is a light map. ✷ Applying Theorems 7.3 and 7.4, we obtain the following: Theorem 7.5. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is a light map, then f : X → Y is a light map. The following example proves that the converse of Theorems 7.4 and 7.5 are not true. Example 7.6. Let S 1 = {e it : t ∈ R}, f : S 1 → S 1 such that f ( z) = z2 , and let n  2 be an integer. We note that f is a light map. By Theorem 7.3, F n ( f ) is a light map. Let A = {{e it , e i (π +t ) }: t ∈ [0, π2 ]}. Then A is a nondegenerate connected subset of F n ( S 1 ) such that A ∩ F 1 ( S 1 ) = ∅ and F n ( f )(A) ⊂ F 1 ( S 1 ). It follows that qnS 1 (A) is a nondegenerate connected subset of SF n ( S 1 ) such that qnS 1 (A) ⊂ (SF n ( f ))−1 ( F nS1 ). This implies that SF n ( f ) is not a light map. 8. Quasi-interior maps and MO-maps Theorem 8.1. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a quasiinterior map, then f is a quasi-interior map. Proof. Let y ∈ Y , let U be an open subset of X , and let C be a component of f −1 ( y ) such that C ⊂ U . Then, by Lemma 6.1, F n (C ) is a component of ( F n ( f ))−1 ({ y }). Moreover, U n is open in F n ( X ) and F n (C ) ⊂ U n . Hence, { y } ∈int F n (Y ) ( F n ( f )(U n )). Let U 1 , . . . , U m be open subsets of Y such that { y } ∈ U 1 , . . . , U m n ⊂ F n ( f )(U n ). We define m V = i =1 U i . It follows that y ∈ V ⊂ f (U ). Thus, y ∈ intY ( f (U )). Then f is a quasi-interior map. ✷ Theorem 8.2. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a quasiinterior map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a quasi-interior map. Proof. Since qnY is a monotone map, by [17, 2.1], it follows that qnY is a quasi-interior map. Then, by [17, 2.8], qnY ◦ F n ( f ) is a quasi-interior map. By (∗), SF n ( f ) ◦ qnX is a quasi-interior map. Then, by [14, 5.20] and [17, 3.1], SF n ( f ) is a quasi-interior map. ✷ Theorem 8.3. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then the following are equivalent: (1) f : X → Y is monotone; (2) F n ( f ) : F n ( X ) → F n (Y ) is quasi-interior; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is quasi-interior. Proof. If f is a monotone map then, by Theorem 4.1, F n ( f ) is a monotone map. Hence, by [17, 2.1], F n ( f ) is a quasi-interior map. Thus, (1) implies (2). If F n ( f ) is a quasi-interior map, by Theorem 8.2, SF n ( f ) is a quasi-interior map. We obtain that (2) implies (3). Finally, if SF n ( f ) is a quasi-interior map then, by [17, 2.7], SF n ( f ) is a confluent map. By Theorem 6.6, we have that f is a monotone map. ✷ Theorem 8.4. Let f : X → Y be a surjective map between continua. Then the following are equivalent: (1) f : X → Y is quasi-interior; (2) F 2 ( f ) : F 2 ( X ) → F 2 (Y ) is quasi-interior; (3) SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is quasi-interior. Proof. We assume that f is a quasi-interior map. Then, by [17, 3.1], there exist a continuum Z , a monotone map h : X → Z , and an open map g : Z → Y such that f = g ◦ h. Using Theorems 4.1 and 5.7, we have that F 2 (h) is a monotone map and F 2 ( g ) is an open map. Furthermore, F 2 ( f ) = F 2 ( g ) ◦ F 2 (h). Hence, by [17, 3.1], F 2 ( f ) is a quasi-interior map. Thus, (1) implies (2). By Theorem 8.2, it follows that (2) implies (3). F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1201 Now we prove that (3) implies (1). We assume that SF 2 ( f ) is quasi-interior. Let y ∈ Y , let U be an open subset of X , and let C be a component of f −1 ( y ) such that C ⊂ U . If f (U ) = Y , then y ∈ intY ( f (U )) = Y . We suppose that there exists a point z ∈ Y \ f (U ). Let V and W be open subsets of Y such that y ∈ V , z ∈ W and V ∩ W = ∅. We define U 1 = U ∩ f −1 ( V ) and U 2 = f −1 ( W ). Then U 1 and U 2 are open subsets of X such that U 1 ∩ U 2 = ∅. Let U = U 1 , U 2 2 . It follows that U is an open subset of F 2 ( X ) such that U ∩ F 1 ( X ) = ∅. We take a component K of f −1 ( z). Let A = C , K 2 and B = { y , z}. By [13, Lemma 1], A is connected. We note that A ⊂ U and A ⊂ ( F 2 ( f ))−1 ( B ). Next we prove that A is component of ( F 2 ( f ))−1 ( B ). Let C be the component of ( F 2 ( f ))−1 ( B ) such that A ⊂ C . Then C ∪ K ⊂ A ⊂ C . Since C ⊂ f −1 ( y ) ∪ f −1 (z), ( C ) ∩ f −1 ( y ) = ∅ and ( C ) ∩ f −1 (z) = ∅, we have that C is not connected. By [15, Lemma 2.2], C has exactly two components. Let C = C 1 ∪ C 2 , where C 1 and C 2 are the components of C . Without loss of generality, we assume that C 1 ⊂ f −1 ( y ) and C 2 ⊂ f −1 (z). It follows that C 1 = C and C 2 = K . Hence, C = C ∪ K . To see that C ⊂ A, let D ∈ C . Thus, D ⊂ C ∪ K . We suppose that D ∩ K = ∅. This implies that D ⊂ C . Then B = f ( D ) ⊂ f (C ) = { y }, a contradiction. Hence, D ∩ K = ∅. Similarly, D ∩ C = ∅. We obtain that D ∈ A. It follows that C ⊂ A. Therefore, A is component of ( F 2 ( f ))−1 ( B ). We note that q2X (U ) is an open subset of SF 2 ( X ), q2X (A) ⊂ q2X (U ) and q2X (A) is a component of (SF 2 ( f ))−1 (q2Y ( B )). Since SF 2 ( f ) is a quasi-interior map, we have that q2Y ( B ) ∈ intSF 2 (Y ) (SF 2 ( f )(q2X (U ))). By (∗), q2Y ( B ) ∈ intSF 2 (Y ) q2Y ( F 2 ( f )(U )). Since F 2 ( f )(U ) ⊂ F 2 (Y ) \ F 1 (Y ) and q2Y | F 2 (Y )\ F 1 (Y ) : F 2 (Y ) \ F 1 (Y ) → SF 2 (Y ) \ { F Y2 } is a homeomorphism, we obtain that B ∈ int F 2 (Y ) ( F 2 ( f )(U )). Now we prove that y ∈ intY ( f (U )). Let V 1 , . . . , V m be open subsets of Y such that B ∈  V 1 , . . . , V m 2 ⊂ F 2 ( f )(U ). We define I = {i ∈ {1, . . . , m}: y ∈ V i } and G = ( i ∈ I V i ) ∩ V . To see that G ⊂ f (U ), let x ∈ G. We put D = { z, x}. Then D ∈  V 1 , . . . , V m 2 . Let E ∈ U be such that f ( E ) = D. We have that x ∈ f ( E ) ⊂ f (U 1 ) ∪ f (U 2 ). If x ∈ f (U 2 ), then x ∈ V ∩ W , a contradiction. Hence, x ∈ f (U 1 ) ⊂ f (U ). Thus, y ∈ G ⊂ f (U ). This implies that y ∈ intY ( f (U )). Therefore, f is a quasiinterior map. ✷ Theorem 8.5. Let f : X → Y be a surjective map between continua, and let n  3 be an integer. Then the following are equivalent: (1) f : X → Y is monotone; (2) F n ( f ) : F n ( X ) → F n (Y ) is an MO-map; (3) SF n ( f ) : SF n ( X ) → SF n (Y ) is an MO-map. Proof. We assume that f is a monotone map. Then, by Theorem 4.1, F n ( f ) is a monotone map. Hence, F n ( f ) is an MO-map [14, p. 16]. If F n ( f ) is an MO-map then, by [17, 3.2] and [17, 3.1], F n ( f ) is a quasi-interior map. By Theorem 8.3, f is a monotone map. This implies, by Theorem 4.1, that SF n ( f ) is a monotone map. Thus, SF n ( f ) is an MO-map [14, p. 16]. If SF n ( f ) is an MO-map then, by [17, 3.2] and [17, 3.1], SF n ( f ) is a quasi-interior map. By Theorem 8.3, we have that f is a monotone map. ✷ Theorem 8.6. Let f : X → Y be a surjective map between continua. If f is an MO-map, then F 2 ( f ) : F 2 ( X ) → F 2 (Y ) is an MO-map. Proof. We assume that f is an MO-map. Let Z be a continuum, let h : X → Z be an open map, and let g : Z → Y be a monotone map such that f = g ◦ h. By Theorems 5.7 and 4.1, we have that F 2 (h) is an open map and F 2 ( g ) is a monotone map. Furthermore, F 2 ( f ) = F 2 ( g ) ◦ F 2 (h). Hence, F 2 ( f ) is an MO-map. ✷ 9. Quasi-monotone maps and weakly monotone maps Theorem 9.1. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a quasimonotone map, then f is a quasi-monotone map. Proof. Let B be a subcontinuum of Y such that intY ( B ) = ∅. Let y ∈ intY ( B ) and let U be an open subset of Y such that y ∈ U ⊂ B. This implies that { y } ∈ U n ⊂ F n ( B ). Hence, F n ( B ) is a subcontinuum of F n (Y ) such that int F n (Y ) ( F n ( B )) = ∅. Then ( F n ( f ))−1 ( F n ( B )) has only finitely many components, K1 , . . . , Km , and F n ( f )(Ki ) = F n ( B ) for each i ∈ {1, . . . , m}. Now let C be a component of f −1 ( B ). Then, by Lemma 6.1, F n (C ) is a component of ( F n ( f ))−1 ( F n ( B )). Thus, there exists j ∈ {1, . . . , m} such that F n (C ) = K j . We note that, by [15, Lemma 2.2], K j is connected; also C ⊂ K J ⊂ f −1 ( B ). − 1 It follows that C = K j . Hence, f ( B ) has at most m components. Furthermore, if b ∈ B, then there exists D ∈ F n (C ) such that f ( D ) = {b}. Then b ∈ f (C ). Hence, B ⊂ f (C ). Thus, we obtain that f (C ) = B. Therefore, f is a quasi-monotone map. ✷ With similar arguments to the ones given to prove Theorem 9.1, we can verify the following theorem: Theorem 9.2. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a weakly monotone map, then f is a weakly monotone map. 1202 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 Theorem 9.3. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a quasimonotone map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a quasi-monotone map. Proof. We note that qnY is a monotone map. Then, by [9, Proposition 13.18], qnY is a quasi-monotone map. Hence, by [14, Theorem 5.6], qnY ◦ F n ( f ) is a quasi-monotone map. By (∗), SF n ( f ) ◦ qnX is a quasi-monotone map. By [14, 5.19], it follows that SF n ( f ) is a quasi-monotone map. ✷ It is known that the composition of weakly monotone maps may not be weakly monotone [14, 5.9]. However, the following is true. Lemma 9.4. Let f : X → Z be a weakly monotone map between continua and let g : Z → Y be a monotone map between continua. Then g ◦ f : X → Y is a weakly monotone map. Proof. Let B be a subcontinuum of Y such that intY ( B ) = ∅ and let C be a component of ( g ◦ f )−1 ( B ). Since g is a monotone map, by [19, 2.2, p. 138], g −1 ( B ) is a subcontinuum of Z . Furthermore, int Z ( f −1 ( B )) = ∅. Since f is a weakly monotone map, f (C ) = g −1 ( B ). Then ( g ◦ f )(C ) = B. Hence, g ◦ f is a weakly monotone map. ✷ Theorem 9.5. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a weakly monotone map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a weakly monotone map. Proof. Since qnY is a monotone map. Then, by Lemma 9.4, qnY ◦ F n ( f ) is a weakly monotone map. Hence, by (∗), SF n ( f ) ◦ qnX is a weakly monotone map. By [14, 5.19], it follows that SF n ( f ) is a weakly monotone map. ✷ Example 9.6. Let f : [−1, 1] → [0, 1] be defined by f (x) = |x| and let n  3 be an integer. We note that f is a quasimonotone map and, by [14, 4.40], f is a weakly monotone map, but f is not a monotone map. We assume that SF n ( f ) is a weakly monotone map. By [3, Theorem 5.2], SF n ([0, 1]) is locally connected. Then, by [18, Lemma 7.1], SF n ( f ) is a confluent map. Hence, by Theorem 6.5, f is a monotone map, a contradiction. Therefore, SF n ( f ) is not a weakly monotone map. Thus, by Theorem 9.5, F n ( f ) is not a weakly monotone map. By [14, 4.40], it follows that neither SF n ( f ) nor F n ( f ) is a quasimonotone map. We note that, by [9, Proposition 13.18] and [9, Theorem 13.20], the class of weakly monotone maps whose range is locally connected coincide with the class of confluent maps. Thus, we have the following: Question 9.7. Let f : X → Y be a surjective map between continua such that Y is not locally connected. (1) If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a weakly monotone map, then is F 2 ( f ) : F 2 ( X ) → F 2 (Y ) a weakly monotone map? (2) If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a weakly monotone map, then is f a weakly monotone map? Also, by [9, Theorem 13.23], the class of quasi-monotone maps whose domain is locally connected coincide with the class of confluent maps. Thus, we have the following: Question 9.8. Let f : X → Y be a surjective map between continua such that X is not locally connected. (1) If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a quasi-monotone map, then is F 2 ( f ) : F 2 ( X ) → F 2 (Y ) a quasi-monotone map? (2) If SF 2 ( f ) : SF 2 ( X ) → SF 2 (Y ) is a quasi-monotone map, then is f a quasi-monotone map? 10. Weakly confluent and pseudo-confluent maps Theorem 10.1. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a weakly confluent map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a weakly confluent map. Proof. Since qnY is a monotone map, it is clear that qnY is a weakly confluent map. Hence, by [14, 5.4], qnY ◦ F n ( f ) is a weakly confluent map. By (∗), SF n ( f ) ◦ qnX is a weakly confluent map. By [14, 5.16], it follows that SF n ( f ) is a weakly confluent map. ✷ The proof of the following theorem is similar to the one given to prove Theorem 10.1, we need to use [14, 5.9] and [14, 5.4]. Theorem 10.2. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a pseudoconfluent map, then SF n ( f ) : SF n ( X ) → SF n (Y ) is a pseudo-confluent map. F. Barragán / Topology and its Applications 158 (2011) 1192–1205 1203 Theorem 10.3. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is a weakly confluent map, then f : X → Y is a weakly confluent map. Proof. Let B be a subcontinuum of Y . We assume that Y \ B = ∅. Let y 1 , . . . , yn−1 be points in Y \ B such that y i = y j if i = j. Let D = { y 1 , . . . , yn−1 }. We define   K = D ∪ K : K ∈ F 1(B) . Then K is a subcontinuum of F n (Y ) such that K ∩ F 1 (Y ) = ∅. This implies that qnY (K) is a subcontinuum of SF n (Y ) such that F Yn ∈ / C, (qnX )−1 (C) / qnY (K). Then there exists a component C of (SF n ( f ))−1 (qnY (K)) such that SF n ( f )(C) = qnY (K). Since F nX ∈ is a component of (qnX )−1 ((SF n ( f ))−1 (qnY (K))) = ( F n ( f ))−1 (K). We note that F n ( f )((qnX )−1 (C)) = K. On the other hand, we have that (qnX )−1 (C) ⊂ f −1 ( y 1 ) ∪ · · · ∪ f −1 ( yn−1 ) ∪ f −1 ( B ), ( (qnX )−1 (C)) ∩ f −1 ( y i ) = ∅ and ( (qnX )−1 (C)) ∩ f −1 ( B ) = ∅. Since, by [15, Lemma 2.2], (qnX )−1 (C) has at most n components, it follows that (qnX )−1 (C) has exactly n components, C 1 , . . . , C n . Without loss of generality, we assume that C 1 ⊂ f −1 ( y 1 ), . . . , C n−1 ⊂ f −1 ( yn−1 ) and C n ⊂ f −1 ( B ). Let C the component of f −1 ( B ) such that C n ⊂ C . We prove that f (C ) = B. Let b ∈ B and let E = D ∪ {b}. Then E ∈ K. Hence, there exists A ∈ (qnX )−1 (C) such that f ( A ) = E. This implies that b ∈ f ( A ) ⊂ f ( (qnX )−1 (C)) = f (C 1 ) ∪ · · · ∪ f (C n ). If there exists j ∈ {1, . . . , n − 1} such that b ∈ f (C j ), then b = y j , a contradiction. Hence, b ∈ f (C n ). Thus, b ∈ f (C ). It follows that B ⊂ f (C ). Therefore, f is a weakly confluent map. ✷ Given a continuum X , an integer n  2, B a subcontinuum of X and D = {x1 , . . . , xn−1 } ∈ F n ( X ), it is easy to see that K = { D ∪ K : K ∈ F 1 ( B )} is homeomorphic to B. Hence, if B is an irreducible continuum, then K is an irreducible continuum. Therefore, with similar arguments to the ones given to prove Theorem 10.3, we can verify the following: Theorem 10.4. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If SF n ( f ) : SF n ( X ) → SF n (Y ) is a pseudo-confluent map, then f : X → Y is a pseudo-confluent map. By Theorems 10.1 and 10.3, we obtain the following: Theorem 10.5. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a weakly confluent map, then f : X → Y is a weakly confluent map. As a consequence of Theorems 10.2 and 10.4, we have the following theorem: Theorem 10.6. Let f : X → Y be a surjective map between continua, and let n  2 be an integer. If F n ( f ) : F n ( X ) → F n (Y ) is a pseudoconfluent map, then f : X → Y is a pseudo-confluent map. The following example shows that the converse of Theorems 6.2 (for n = 2), 6.8, 10.3, 10.4, 10.5 and 10.6 are not true. Example 10.7. Let A 1 = {(x, y ) ∈ R2 : y = sin( 1x ) − 1, 0 < x  1}, B 1 = {(x, y ) ∈ R2 : x = 0, −3  y  0}, A 2 = {( y , x) ∈ R2 : (x, y ) ∈ A 1 }, B 2 = {( y , x) ∈ R2 : (x, y ) ∈ B 1 }, A 3 = {(−x, y ) ∈ R2 : (x, y ) ∈ A 1 }. We define X = A 1 ∪ B 1 ∪ A 2 ∪ B 2 , Y = A 1 ∪ B 1 ∪ A 3 (see Fig. 1), and f : X → Y such that f (x, y ) =  (x, y ), if (x, y ) ∈ A 1 ∪ B 1 ; (− y , x), if (x, y ) ∈ A 2 ∪ B 2 . Then f is a confluent map (weakly confluent map, pseudo-confluent map). However, F 2 ( f ) is not a pseudo-confluent map (either weakly confluent map or confluent map). We prove that F 2 ( f ) is not a pseudo-confluent map. Define A = {{(0, y − 1), (0, y )} ∈ F 2 ( X ): −2  y  0}, B = {{( y − 1, 0), ( y , 0)} ∈ F 2 ( X ): −2  y  0}. It follows that A and B are subcontinua of F 2 ( X ) such that A ⊂ F 2 ( B 1 ), B ⊂ F 2 ( B 2 ) and A ∩ B = ∅. 1204 F. Barragán / Topology and its Applications 158 (2011) 1192–1205 Fig. 1. Continua X and Y . Let α1 : [0, 1) → A 1 be a homeomorphism. Let α1 (t ) = (x(t ), y (t )). Hence, α2 : [0, 1) → A 2 given by α2 (t ) = ( y (t ), x(t )) is a homeomorphism. We take the maps δ1 : [0, 1) → F 2 ( X ) given by δ1 (t ) = {α1 (t ), (0, y (t ) − 1)} and δ2 : [0, 1) → F 2 ( X ) given by δ2 (t ) = {α2 (t ), ( y (t ) − 1, 0)} Let H = δ1 ([0, 1)) and K = δ2 ([0, 1)). Then H and K are disjoint connected subsets of F 2 ( X ). Moreover, these subsets of F 2 ( X ) do not intersect A (or B ). Since A ⊂ Cl F 2 ( X ) (H), we have that A ∪ H ⊂ Cl F 2 ( X ) (H). On the other hand, we note that Cl F 2 ( X ) (H) ∩ F 2 ( B 1 ) = A and Cl F 2 ( X ) (H) \ H ⊂ F 2 ( B 1 ). This implies that Cl F 2 ( X ) (H) ⊂ A ∪ H. Hence, Cl F 2 ( X ) (H) = A ∪ H. Similarly, Cl F 2 ( X ) (K) = B ∪ K. Therefore, Cl F 2 ( X ) (H) and Cl F 2 ( X ) (K) are disjoint subcontinua of F 2 ( X ). Note each continuum Cl F 2 ( X ) (H) and Cl F 2 ( X ) (K) is homeomorphic to the sin 1x -continuum, with remainder A and B , respectively. Hence, Cl F 2 ( X ) (H) and Cl F 2 ( X ) (K) are irreducible continua. Also, each continuum F 2 ( f )(Cl F 2 ( X ) (H)) and F 2 ( f )(Cl F 2 ( X ) (K)) is homeomorphic to the sin 1x -continuum, with remainder F 2 ( f )(A) and F 2 ( f )(B ), respectively. Let D = F 2 ( f ) Cl F 2 ( X ) (H) ∪ F 2 ( f ) Cl F 2 ( X ) (K) . Since F 2 ( f ) identifies A and B only (F 2 ( f )(A) = F 2 ( f )(B ) = A), it follows that D is an irreducible subcontinuum of F 2 (Y ). We note that D = H ∪ A ∪ F 2 ( f )(K). Let A1 = {{( y − 1, 0), (0, y )} ∈ F 2 ( X ): −2  y  0}, B1 = {{(0, y − 1), ( y , 0)} ∈ F 2 ( X ): −2  y  0}. Then A1 and B2 are disjoint subcontinua of F 2 ( X ) each of which is disjoint to Cl F 2 ( X ) (H) and Cl F 2 ( X ) (K). We note that ( F 2 ( f ))−1 (A) = A ∪ B ∪ A1 ∪ B1 . Now we put H1 = {{α1 (t ), ( y (t ) − 1, 0)} ∈ F 2 ( X ): 0  t < 1}, K1 = {{α2 (t ), (0, y (t ) − 1)} ∈ F 2 ( X ): 0  t < 1}. It follows that H1 and K1 are connected sets such that H1 ∩ K1 = ∅, each of which is disjoint to Cl F 2 ( X ) (H) and Cl F 2 ( X ) (K). Furthermore, ( F 2 ( f ))−1 (H) = H ∪ H1 and ( F 2 ( f ))−1 ( F 2 ( f )(K)) = K ∪ K1 . We note that Cl F 2 ( X ) (H1 ) = H1 ∪ A1 and Cl F 2 ( X ) (K1 ) = K1 ∪ B1 . Hence, Cl F 2 ( X ) (H), Cl F 2 ( X ) (K), Cl F 2 ( X ) (H1 ) and Cl F 2 ( X ) (K1 ) are the components of ( F 2 ( f ))−1 (D ). We observe that neither of these components is mapped onto D under F 2 ( f ). Therefore, F 2 ( f ) is not a pseudo-confluent map. Hence, F 2 ( f ) is not either a weakly confluent map or a confluent map. To see that SF 2 ( f ) is not a pseudo-confluent map, we note that D ∩ F 1 (Y ) = ∅ and ( F 2 ( f ))−1 (D ) ∩ F 1 ( X ) = ∅. Since F 2 (Y ) \ F 1 (Y ) is homeomorphic to SF 2 (Y ) \ { F Y2 }, it follows that q2Y (D ) is an irreducible subcontinuum of SF 2 (Y ). Similarly, we obtain that q2X (Cl F 2 ( X ) (H)), q2X (Cl F 2 ( X ) (H1 )), q2X (Cl F 2 ( X ) (K)) and q2X (Cl F 2 ( X ) (K1 )) are the components of (SF 2 ( f ))−1 (q2Y (D)) neither of which is mapped onto q2Y (D) under SF 2 ( f ). Therefore, SF 2 ( f ) is not a pseudo-confluent map. Hence, SF 2 ( f ) is not either a weakly confluent map or a confluent map. Example 10.7 shows a confluent map f whose range is not locally connected such that neither F 2 ( f ) or SF 2 ( f ) is a confluent map. However, Theorem 6.10 is true. Question 10.8. 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