Topology and its Applications 163 (2014) 66–76

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Topology and its Applications

www.elsevier.com/locate/topol

Induced mappings between quotient spaces of symmetricproducts of continua

Enrique Castañeda-Alvarado, Fernando Orozco-Zitli, Javier Sánchez-Martínez ∗

Universidad Autónoma del Estado de México, Facultad de Ciencias, Instituto Literario No. 100,Col. Centro, C.P. 50000, Toluca, Estado de México, Mexico

a r t i c l e i n f o a b s t r a c t

MSC:primary 54C05, 54C10, 54B20secondary 54B15

Keywords:ContinuumHyperspaceInduced mappingsQuotient spaceSymmetric product

Given a continuum X and n ∈ N. Let H(X) ∈ {2X , C(X), Fn(X)} be a hyperspaceof X, where 2X , C(X) and Fn(X) are the hyperspaces of all nonempty closedsubsets of X, all subcontinua of X and all nonempty subsets of X with at mostn points, respectively, with the Hausdorff metric. For a mapping f : X → Ybetween continua, let H(f) : H(X) → H(Y ) be the induced mapping by f , given byH(f)(A) = f(A). On the other hand, for 1 � m < n, SFn

m(X) denotes the quotientspace Fn(X)/Fm(X) and similarly, let SFn

m(f) denote the natural induced mappingbetween SFn

m(X) and SFnm(Y ). In this paper we prove some relationships between

the mappings f , 2f , C(f), Fn(f) and SFnm(f) for the following classes of mapping:

atomic, confluent, light, monotone, open, OM, weakly confluent, hereditarily weaklyconfluent.

© 2013 Elsevier B.V. All rights reserved.

1. Introduction

A continuum is a nondegenerate compact connected metric space. Given a continuum X. Denote by 2Xthe hyperspace of all nonempty closed subsets of X, by C(X) the hyperspace of subcontinua of X and byFn(X) the hyperspace of all nonempty subsets of X having at most n points, where n is a positive integer.All hyperspaces are considered with the Hausdorff metric (see [20, p. 1]).

Given two positive integers n > m, SFnm(X) denotes the quotient space Fn(X)/Fm(X) obtained by

shrinking Fm(X) to a point in Fn(X), with the quotient topology. The fact that SFnm(X) is a continuum

follows from Theorem 3.10 of [21, p. 40]. A mapping means a continuous function. The quotient mappingfrom Fn(X) onto SFn

m(X) is denoted by ρXn,m.Given a mapping f : X → Y between continua, the mappings 2f : 2X → 2Y given by 2f (A) = f(A),

C(f) : C(X) → C(Y ) given by C(f) = 2f |C(X) and Fn(f) : Fn(X) → Fn(Y ) given by Fn(f) = 2f |Fn(X)are the induced mappings by f . Also, we have an induced mapping SFn

m(f) : SFnm(X) → SFn

m(Y ) given by

SFnm(f)

(ρXn,m(A)

)= ρYn,m

(f(A)

).

* Corresponding author.E-mail addresses: eca@uaemex.mx (E. Castañeda-Alvarado), forozcozitli@gmail.com (F. Orozco-Zitli),

jsanchezm@uaemex.mx (J. Sánchez-Martínez).

0166-8641/$ – see front matter © 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.topol.2013.10.007

E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76 67

By [11, Theorem 4.3, p. 126], SFnm(f) is a mapping and the following diagram

Fn(X)Fn(f)

ρXn,m

Fn(Y )

ρYn,m

SFnm(Y )

SFnm(f)

SFnm(Y )

is commutative.Let M be a class of mappings between continua. A general problem is to find all possible relationships

among the following four statements:

(1) f ∈ M;(2) 2f ∈ M;(3) C(f) ∈ M;(4) Fn(f) ∈ M.

Readers especially interested in this problem are referred to [1–8,12–17].We can also consider the condition:

(5) SFnm(f) ∈ M.

In this paper we study the interrelations among the statements (1)–(5), for the following classes of mappings:open, monotone, confluent, weakly confluent, OM, light and atomic.

2. Preliminaries

Let X be a continuum with metric d. Let δ > 0 and A ⊂ X, we define VXδ (A) = {x ∈ X: there exists

y ∈ A such that d(x, y) < δ}, and we use the symbol Cl(A) to denote the closure of A in X.Given a finite collection K1, . . . ,Kr of subsets of X, 〈K1, . . . ,Kr〉, denotes the following subset of Fn(X):

{A ∈ Fn(X): A ⊂

r⋃i=1

Ki, A ∩Ki �= ∅ for each i ∈ {1, . . . , r}}.

It is known that the family of all subsets of Fn(X) of the form 〈K1, . . . ,Kr〉, where each Ki is an open subsetof X, forms a basis for a topology for Fn(X) (see [20, Theorem 0.11, p. 9]) called the Vietoris Topology.The Vietoris topology and the topology induced by the Hausdorff metric coincide (see [20, Theorem 0.13,p. 10]). If K1, . . . ,Kr are subcontinua of X and r � n, 〈K1, . . . ,Kr〉 is a subcontinuum of Fn(X) (see [19,Lemma 1, p. 230]).

Given two positive integers s > k, we define Fsk(X) = Fs(X) \ Fk(X). Notice that ρXs,k|Fs

k(X) : Fsk(X) →

SFsk(X) \ ρXs,k(Fk(X)) is a homeomorphism.A mapping f : X → Y between continua is said to be:

– atomic if for each subcontinuum K of X such that f(K) is nondegenerate, f−1(f(K)) = K;– confluent if for each subcontinuum K of Y and for each component M of f−1(K), f(M) = K;– light if f−1(y) is totally disconnected for each y ∈ Y ;– monotone if f−1(y) is connected in X for each y ∈ Y ;– open if f(U) is open in Y for each open subset U of X;

68 E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76

– OM if there exist a continuum Z and mappings g : X → Z and h : Z → Y such that f = h ◦ g, g ismonotone and h is open;

– weakly confluent if for each subcontinuum K of Y , there exists a subcontinuum M of X such thatf(M) = K;

– hereditarily weakly confluent if for each nondegenerate subcontinuum M of X, f |M is weakly confluent.

3. Open mappings

Lemma 3.1. Let X be a continuum and n > m. If U ⊂ Fn(X) is such that U ∩ Fm(X) �= ∅ and ρXn,m(U) isan open subset of SFn

m(X), then Fm(X) ⊂ Cl(U).

Proof. Suppose to the contrary that there exists A ∈ Fm(X) such that A /∈ Cl(U). Then, there exists ε > 0such that VFn(X)

ε ({A}) ∩ U = ∅.On the other hand, since U ∩ Fm(X) �= ∅, (ρXn,m)−1(ρXn,m(U)) = U ∪ Fm(X). So, U ∪ Fm(X) is an open

subset of Fn(X).Thus, there exists 0 < δ < ε such that VFn(X)

δ ({A}) ⊂ Fm(X). This contradicts the fact that Fnn−1(X)

is dense in Fn(X). �Theorem 3.2. Let f : X → Y be a mapping between continua, and let n � 2. Then the following conditionsare equivalent.

(1) f is a homeomorphism;(2) 2f is a homeomorphism;(3) C(f) is a homeomorphism;(4) Fn(f) is a homeomorphism;(5) SFn

m(f) is a homeomorphism;(6) SFn

m(f) is open.

Proof. Conditions (1), (2) and (3) are well known to be equivalent (see [16, p. 239] and [20, Theorem 0.52,p. 29]). By Theorem 3.1 of [12, p. 369], (1) and (4) are equivalent. It is easy to prove that (1) ⇒ (5).Notice that (5) ⇒ (6). Finally, we prove that (6) ⇒ (1). Since SFn

m(f) is an open mapping, SFnm(f) is

surjective. Thus, f is surjective. To see that f is one-to-one, assume that there exist x1, x2 ∈ X suchthat x1 �= x2 and f(x1) = f(x2). Take A ∈ Fm+1

m (X) such that {x1, x2} ⊂ A. Consider δ > 0 such thatVFn(X)δ ({A}) ∩ Fm(X) = ∅. Let 0 < ε < δ. Then, ρXn,m(VFm(X)

ε ({A})) is an open set of SFnm(X). So, since

SFnm(f)

(ρXn,m

(VFm(X)ε

({A}

)))= ρYn,m

(Fn(f)

(VFn(X)ε

({A}

))),

ρYn,m(Fn(f)(VFn(X)ε ({A}))) is an open set of SFn

m(Y ). Thus, by Lemma 3.1,

Fm(Y ) ⊂ Cl(Fn(f)

(VFn(X)ε

({A}

))).

Since ε is an arbitrary element in (0, δ), we conclude that Fm(Y ) = {Fn(f)(A)}, a contradiction.Therefore f is a homeomorphism. �

Theorem 3.3. Let f : X → Y be a surjective mapping between continua. Consider the following conditions:

(1) f is open;(2) 2f is open;

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(3) C(f) is open;(4) F2(f) is open;(5) Fn(f) is open for all n � 3;(6) SFn

m(f) is open for every 1 � m < n;(7) SF2

1(f) is open.

Then (1), (2) and (4) are equivalent, (5), (6) and (7) are equivalent, (3) implies (1), (7) implies (3) and (4).

Proof. (1) and (2) are equivalent (see [16, Theorem 4.3, p. 243]). (1) and (4) are equivalent (see [12,Theorem 3.5, p. 370]; compare with [1, Theorem 5.7, p. 1196]). By Theorem 3.6 of [12, p. 371] (comparewith [1, Theorem 5.9, p. 1196]) and by Theorem 3.2, we have (5), (6) and (7) are equivalent. (3) ⇒ (1)follows from Theorem 4.3 of [16, p. 243]. Using Theorem 3.2, we have (7) ⇒ (3) and (7) ⇒ (4). �

Concerning this theorem it is known that (1) does not imply (3) (see [16, p. 243]). Using the example of[16, p. 243] and Theorem 3.5 of [12, p. 370], we obtain that (4) does not imply (3).

Example 3.4. Define f : [0, 1] × [0, 1] → [0, 1] by f(x, y) = x. Clearly, f is open. It is known that F2(f)and C(f) are open mappings (see [12, Theorem 3.5, p. 370] and [8, Corollary 19, p. 73], respectively). ByTheorem 3.2, Fn(f), SFn

m(f), SF21(f) are not open, when n � 3.

Theorem 3.5. Let f : X → Y be a surjective mapping between continua and let n � 2. If X is hereditarilylocally connected, then the following conditions are equivalent.

(1) C(f) is open;(2) SFn

m(f) is open.

Proof. Follows from Theorem 3.2 and Corollary 2 of [10, p. 3730]. �4. Monotone mappings

Proposition 4.1. Let n � 1. Let f : X → Y be a mapping between continua and let y ∈ Y . Then f−1(y) isconnected in X if and only if Fn(f)−1({y}) is connected in Fn(X).

Proof. If Fn(f)−1({y}) is connected in Fn(X), by Lemma 1.49 of [20, p. 102], f−1(y) =⋃

Fn(f)−1({y}) isconnected.

On the other hand, suppose that f−1(y) is connected in X. Since 〈f−1(y)〉 = Fn(f)−1({y}), Fn(f)−1({y})is connected. This finishes the proof. �Proposition 4.2. Let f : X → Y be a surjective mapping between continua and let n � 2. If SFn

m(f) is amonotone mapping, then Fn(f)−1(A) is connected for each A ∈ Fn

m(Y ).

Proof. This proposition follows from the fact that

Fn(f)−1(A) =(ρXn,m

)−1(SFnm(f)−1(ρYn,m(A)

))for each A ∈ Fn

m(Y ). �Theorem 4.3. Let f : X → Y be a surjective mapping between continua and let n � 2. Then the followingconditions are equivalent.

70 E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76

(1) f is monotone;(2) 2f is monotone;(3) C(f) is monotone;(4) Fn(f) is monotone;(5) SFn

m(f) is monotone.

Proof. It is known that (1)–(3) are equivalent (see [16, Theorem 3.2, p. 241]). Also, (1) and (4) are equivalent(see [12, Theorem 3.3, p. 370]; compare with [1, Theorem 4.1, p. 1194]). To prove that (4) ⇒ (5), letA ∈ Fn(Y ). Since Fn(f) and ρYn,m are monotone mappings and

ρXn,m(Fn(f)−1((ρYn,m)−1(

ρYn,m(A))))

= SFnm(f)−1(ρYn,m(A)

),

SFnm(f)−1(ρYn,m(A)) is connected. Hence, SFn

m(f) is monotone.Now, we show that (5) ⇒ (1). Let y ∈ Y . Suppose to the contrary that there exist two nonempty closed

subsets K and L of X such that f−1(y) = K ∪ L and K ∩ L = ∅. We can take B ∈ Fm+1m (Y ) such that

y ∈ B. Let

K ={G ∈ Fn(f)−1(B): G ∩K �= ∅

},

L ={G ∈ Fn(f)−1(B): G ∩ f−1(y) ⊂ L

}.

Notice that K and L are disjoint closed subsets of Fn(f)−1(B) and K ∪ L = Fn(f)−1(B). Since K and L

are nonempty subsets of f−1(y), we have K and L are nonempty sets. So, Fn(f)−1(B) is not connected,this contradicts Proposition 4.2. Hence, f is monotone. �5. Confluent and weakly confluent mappings

Proposition 5.1. Let f : X → Y be a mapping between continua and let 1 � r � n. Let K1, . . . ,Kr benonempty disjoint closed subsets of Y . For each i ∈ {1, . . . , r}, let Mi be a component of f−1(Ki). Then

(1) 〈M1, . . . ,Mr〉 is a component of Fn(f)−1(〈K1, . . . ,Kr〉);(2) if M is a component of f−1(Ki) such that M �= Mi and r < n, then 〈M1, . . . ,Mr,M〉 is a component

of Fn(f)−1(〈K1, . . . ,Kr〉).

Proof. Notice that 〈M1, . . . ,Mr〉 is connected (see [19, Lemma 1, p. 230]) and it is contained inFn(f)−1(〈K1, . . . ,Kr〉). Let C be the component of Fn(f)−1(〈K1, . . . ,Kr〉) containing 〈M1, . . . ,Mr〉. Weneed to prove that M1, . . . ,Mr are the components of

⋃C. Clearly,

⋃ri=1 Mi ⊂

⋃C ⊂

⋃ri=1 f

−1(Ki). Now,let P be a component of

⋃C. By Lemma 3.1 of [16, p. 241], P ∩ Mj �= ∅ for some j ∈ {1, . . . , r}. Thus,

P = Mj .Hence, using Lemma 3.1 of [16, p. 241], it follows that C = 〈M1, . . . ,Mr〉.(2) can be proved in a similar way. �

Proposition 5.2. Let f : X → Y be a surjective mapping between continua.

(1) If SFnm(f) is confluent, then for each subcontinuum A ⊂ Fn

m(Y ) and each component B of Fn(f)−1(A),Fn(B) = A.

(2) If SFnm(f) is weakly confluent, then for each subcontinuum A ⊂ Fn

m(Y ), there exists B a subcontinuumof Fn(X) such that Fn(B) = A.

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Proof. The proposition follows from the fact that ρXn,m|Fnm(X), ρYn,m|Fn

m(Y ) are homeomorphisms and thefact that

Fn(f)|Fn(f)−1(Fnm(Y )) =

(ρYn,m

∣∣Fn

m(Y )

)−1 ◦ SFnm(f) ◦ ρXn,m

∣∣Fn

m(X). �Theorem 5.3. Let f : X → Y be a surjective mapping between continua and let n � 2. Consider the followingconditions:

(1) f is confluent;(2) 2f is confluent;(3) C(f) is confluent;(4) Fn(f) is confluent;(5) SFn

m(f) is confluent for all m < n;(6) there exists m � 1 such that SFn

m(f) is confluent;(7) SFn

1 (f) is confluent.

Then (1) is implied by each one of the conditions (2)–(7) and the implications (4) ⇒ (5) ⇒ (6), (5) ⇒ (7)hold.

Proof. By Theorem 6.3 of [16, p. 246], (1) is implied by (2) or (3). We will prove that (4) ⇒ (5). Since Fn(f)is confluent and ρYn,m ◦ Fn(f) = SFn

m(f) ◦ ρXn,m, by Proposition 13.27 of [21, p. 290], SFnm(f) is confluent.

Clearly, (5) ⇒ (6) and (5) ⇒ (7).To see that (6) ⇒ (1), let C be a proper subcontinuum of Y and let K be a component of f−1(C).

Consider y1, . . . , ym ∈ Y \ C such that yi �= yj if i �= j. For each i ∈ {1, . . . ,m}, let Mi be a component off−1(yi). By Proposition 5.1, 〈M1, . . . ,Mm,K〉 is a component of Fn(f)−1(〈{y1}, . . . , {ym}, C〉). Notice that〈{y1}, . . . , {ym}, C〉 ⊂ Fn

m(Y ). Thus, using Proposition 5.2, Fn(f)(〈M1, . . . ,Mm,K〉) = 〈{y1}, . . . , {ym}, C〉.Hence f(K) = C.

The implication (7) ⇒ (1) follows from Theorems 6.5 and 6.8 of [1, p. 1198]. �Concerning Theorem 5.3 it is known that (1)–(3) are equivalent when Y is locally connected (see [16,

Theorem 6.3, p. 246]), (4) ⇒ (1) (see [12, Theorem 3.19]; compare with [1, Theorem 6.2]), the implications(1) ⇒ (2) and (1) ⇒ (3) do not hold (see [16, p. 247]), (3) does not imply (2) (see [9, Example 4.12, p. 138]),(1) ⇒ (4) and (1) ⇒ (7) do not hold (see [12, Example 3.18, p. 376] and [1, Example 10.7]), we do not knowanything about the implication (2) ⇒ (3) (see [9, Question 4.25, p. 144]).

The following result is known (see [1, Theorem 6.6, p. 1198]).

Proposition 5.4. Let f : X → Y be a surjective mapping between continua and let n � 3. Then the followingconditions are equivalent.

(1) f is monotone;(2) Fn(f) is monotone;(3) SFn

1 (f) is monotone;(4) Fn(f) is confluent;(5) SFn

1 (f) is confluent.

Theorem 5.5. Let f : X → Y be a surjective mapping between continua and let n � 4. Then the followingconditions are equivalent.

(1) f is monotone;(2) there exists 2 � m � n− 2 such that SFn

m(f) is confluent.

72 E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76

Proof. (1) ⇒ (2) follows from Theorem 4.3 and Theorem 13.15 of [21, p. 285].To see that (2) ⇒ (1), suppose to the contrary that there exists y1 ∈ X such that f−1(y1) is not

connected. Consider M0 and M1 different components of f−1(y1). By Corollary 5.5 of [21, p. 74], thereexists a nondegenerate subcontinuum K of Y such that y1 /∈ K. Consider y2, . . . , ym ∈ Y \K ∪ {y1}. Foreach i ∈ {2, . . . ,m}, let Mi be a component of f−1(yi). Let Mm+1 be a component of f−1(K). Noticethat 〈{y1}, {y2}, . . . , {ym},K〉 intersects Fn

n−1(Y ) and it is connected (see [19, Lemma 1, p. 230]). ByProposition 5.1, 〈M0, . . . ,Mm+1〉 is a component of Fn(f)−1(〈{y1}, . . . , {ym},K〉). Then, by Proposition 5.2,Fn(f)(〈M0, . . . ,Mm+1〉) = 〈{y1}, . . . , {ym},K〉. Thus, 〈{y1}, {y2}, . . . , {ym},K〉 is contained in Fn−1(Y ).This contradiction proves the implication (2) ⇒ (1). �Question 5.6. Let f : X → Y be a surjective mapping between continua.

1. If SF32(f) is a confluent mapping, then is SF2

1(f) a confluent mapping?2. If SF2

1(f) is a confluent mapping, then is SF32(f) a confluent mapping?

3. If SF32(f) is a confluent mapping, then is F3(f) a confluent mapping?

4. If SF21(f) is a confluent mapping, then is F2(f) a confluent mapping?

5. If f is a confluent mapping, then is SF32(f) is a confluent mapping?

Theorem 5.7. Let f : X → Y be a surjective mapping between continua and let n � 2. Consider the followingconditions:

(1) f is weakly confluent;(2) 2f is weakly confluent;(3) C(f) is weakly confluent;(4) Fn(f) is weakly confluent;(5) SFn

m(f) is weakly confluent for all m < n;(6) there exists m < n such that SFn

m(f) is weakly confluent;(7) SFn

1 is weakly confluent.

Then, (1) is implied by each one of the conditions (2)–(7), and the implications (4) ⇒ (5) ⇒ (6), (5) ⇒ (7)hold.

Proof. (2) ⇒ (1) follows from Proposition 6.10 of [9, p. 150]. If C(f) is weakly confluent, in particular C(f)is a surjective mapping, so (3) ⇒ (1). Clearly (5) ⇒ (6) and (5) ⇒ (7). We will prove that (4) ⇒ (5). SinceρYn,m is monotone, by Theorem 5.4 of [18, p. 29], ρYn,m ◦ Fn(f) is weakly confluent. So, SFn

m(f) ◦ ρXn,m isweakly confluent. Hence, by Theorem 5.16 of [18, p. 32], SFn

m(f) is weakly confluent.To see that (6) ⇒ (1), let C be a proper subcontinuum of Y . We take y1, . . . , yn−1 ∈ Y \ C such

that yi �= yj if i �= j. By Lemma 5.2, there exists a component L of Fn(f)−1(〈{y1}, . . . , {yn−1}, C〉)such that Fn(f)(L) = 〈{y1}, . . . , {yn−1}, C〉. Since L ⊂ Fn

n−1(Y ), we can find subsets M1, . . . ,Mn of X

such that for each i ∈ {1, . . . , n − 1}, Mi is a component of f−1(yi), Mn is a component of f−1(C) andL∩〈M1, . . . ,Mn〉 �= ∅. Notice that 〈M1, . . . ,Mn〉 ⊂ Fn(f)−1(〈{y1}, . . . , {yn−1}, C〉). So, by Proposition 5.1,L = 〈M1, . . . ,Mn〉. Thus, Fn(f)(〈M1, . . . ,Mn〉) = 〈{y1}, . . . , {yn−1}, C〉. Hence, f(Mn) = C.

(7) ⇒ (1) follows from Theorem 10.1 of [1, p. 1202]. �Concerning Theorem 5.7, it is known that the implications (1) ⇒ (2) and (1) ⇒ (3) do not hold (see

[9, Example 3.8, p. 149]), (1) ⇒ (4) and (1) ⇒ (7) do not hold (see [12, Example 3.18, p. 376] and[1, Example 10.7]). Additionally, we do not know anything about the implications (2) ⇒ (3) and (3) ⇒ (2)(see [9, p. 150]).

E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76 73

Question 5.8. Let f : X → Y be a surjective mapping between continua.

1. If f is weakly confluent and n � 3, then is SFnm(f) weakly confluent for all m < n?

2. If SFnm(f) is weakly confluent for some m < n, then is Fn(f) weakly confluent?

Theorem 5.9. Let f : X → Y be a surjective mapping between continua and n � 3. Consider the followingconditions:

(1) f is hereditarily weakly confluent;(2) 2f is hereditarily weakly confluent;(3) C(f) is hereditarily weakly confluent;(4) Fn(f) is hereditarily weakly confluent;(5) SFn

m(f) is hereditarily weakly confluent for some 1 � m < n.

Then (2)–(4) are equivalent and (1) is implied by each one of the conditions (2)–(5).

Proof. By Theorem 3.11 of [3, p. 199] and Theorem 3.40 of [12, p. 384], the conditions (2)–(4) are equiva-lent and each one of them implies (1). To see that (5) ⇒ (1), let C be a subcontinuum of X. If f(C) = Y ,we consider the weakly confluent mapping SFn

m(f)|SFnm(C) : SFn

m(C) → SFnm(Y ), then by Theorem 5.7

we conclude that f |C is weakly confluent. Suppose that f(C) is a proper subcontinuum of Y and let B

be a subcontinuum of f(C). We take {y1, . . . , yn−1} ∈ Y \ C such that yi �= yj if i �= j. Let Mi be acomponent of f−1(yi), for each i ∈ {1, . . . , n − 1}. Since ρXn,m(〈M1, . . . ,Mn−1, C〉) is a subcontinuum ofSFn

m(X) and ρYn,m(〈{y1}, . . . , {yn−1}, B〉) is a subcontinuum of SFnm(f)(ρXn,m(〈M1, . . . ,Mn−1, C〉)), there

exists a subcontinuum K of ρXn,m(〈M1, . . . ,Mn−1, C〉) such that SFnm(f)(K) = ρYn,m(〈{y1}, . . . , {yn−1}, B〉).

Let K′ = (ρXn,m)−1(K). Notice that K′ is a subcontinuum of 〈M1, . . . ,Mn−1, C〉 and Fn(f)(K′) =〈{y1}, . . . , {yn−1}, B〉. By Lemma 3.1 of [16, p. 241], K =

⋃K′ ∩C is a nonempty connected set. Moreover,

f(K) = B. �6. OM-mappings

Definition 6.1. Given a sequence {Am}∞m=1 of subsets of X define lim supm→∞ Am as the set of points x ∈ X

such that there exists a sequence of positive numbers m1 < m2 < · · · and there exist points xmk∈ Amk

such that lim xmk= x.

The following characterization of OM mappings was showed in (4.6) of [18, p. 15].

Lemma 6.2. A mapping f : X → Y is OM if and only if, for every sequence {ym}∞m=1 in Y that convergesto a point y ∈ Y , we have that lim supm→∞ f−1(ym) meets every component of f−1(y).

Theorem 6.3. Let f : X → Y be a surjective mapping between continua and let n � 2. Consider the followingconditions:

(1) f is an OM-mapping;(2) 2f is an OM-mapping;(3) C(f) is an OM-mapping;(4) F2(f) is an OM-mapping;(5) Fn(f) is an OM-mapping;(6) SFn

m(f) is an OM-mapping.

Then (1)–(4) are equivalent, (1) is implied by each one of the conditions (2)–(6), and (5) implies (6).

74 E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76

Proof. (1)–(3) are equivalent (see [16, Theorem 5.2, p. 244]). (1) and (4) are equivalent (see [12, Theo-rem 3.11, p. 373]). By Theorem 3.10 of [12, p. 372], (5) ⇒ (1).

To see (6) ⇒ (1), we will use Lemma 6.2. Let {ys}∞s=1 be a sequence in Y such that {ys}∞s=1 convergesto an element y ∈ Y , and let C be a component of f−1(y). Take z1, . . . , zn−1 ∈ Y \ ({y} ∪ {ys ∈ Y :s � 1}). For each i ∈ {1, . . . , n − 1}, let Mi be a component of f−1(zi). Then, by Proposition 5.1,〈M1, . . . ,Mn−1, C〉 is a component of Fn(f)−1(〈{z1}, . . . , {zn−1}, {y}〉). So ρXn,m(〈M1, . . . ,Mn−1, C〉) is acomponent of SFn

m(f)−1(ρYn,m({z1, . . . , zn−1, y})). Thus, by hypothesis,

ρXn,m(〈M1, . . . ,Mn−1, C〉

)∩ lim sup

s→∞SFn

m(f)−1(ρYn,m({z1, . . . , zn−1, ys}

))�= ∅.

Then, there exists a subsequence {ρXn,m(Ask)}∞k=1 of {ρXn,m(As)}∞s=1 such that for each positive integer k,

ρXn,m(Ask) ∈ SFnm(f)−1(ρYn,m(

{z1, . . . , zn−1, ysk}))

and limk→∞ ρXn,m(Ask) = ρXn,m(A) for some A ∈ 〈M1, . . . ,Mn−1, C〉.Now, let a ∈ A∩C. Since {Ask}∞k=1 converges to A, there exists a sequence {ask}∞k=1 such that converges

to a and ask ∈ Ask for each k � 1. Since f(Ask) = {z1, . . . , zn−1, ysk} for each k � 1, there exists a positiveinteger l0 such that f(ask) = ysk for each k � l0. Hence a ∈ A ∩ C ∩ lim sups→∞ f−1(ys).

We will prove that (5) ⇒ (6). Since ρYn,m is monotone and ρYn,m◦Fn(f) = SFnm(f)◦ρXn,m, we have SFn

m(f)is an OM-mapping (see [18, 5.5, p. 29] and [18, 5.20, p. 33]). �

Concerning Theorem 6.3, (1) does not imply (5) (see [12, Example 3.13, p. 373]).

Question 6.4. Let f : X → Y be a surjective mapping between continua. If SFnm(f) is an OM-mapping,

then is Fn(f) an OM-mapping?

Theorem 6.5. Let f : X → Y be a surjective mapping between continua and let n � 3. Then the followingconditions are equivalent.

(1) f is monotone;(2) C(f) is monotone;(3) 2f is monotone;(4) Fn(f) is an OM-mapping;(5) SFn

m(f) is an OM-mapping, 2 � m � n− 2.

Proof. (1)–(3) are equivalent (see Theorem 4.3). (1) ⇒ (4) follows from Theorem 4.3 and (4.1) of [18, p. 15].By Theorem 6.3, (4) ⇒ (5). We will prove that (5) ⇒ (1). By (4.9) of [18, p. 16], SFn

m(f) is a confluentmapping. So, by Theorem 5.5, f is monotone. �Example 6.6. Define f(t) = |t| for t ∈ [−1, 1]. The mapping f : [−1, 1] → [0, 1] is open and then is anOM-mapping. By Theorem 6.5, SFn

m(f) is not an OM-mapping for each n � 3 and 1 � m � n− 2.

Question 6.7. Let f : X → Y be a surjective mapping between continua. If f is an OM-mapping, then isSF2

1(f) an OM-mapping?

7. Light mappings

Theorem 7.1. Let f : X → Y be a surjective mapping between continua and let n � 2. Consider the followingconditions:

E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76 75

(1) f is light;(2) C(f) is light;(3) 2f is light;(4) Fn(f) is light;(5) SFn

m(f) is light.

Then (1) and (4) are equivalent and the implications (3) ⇒ (2) ⇒ (1) and (5) ⇒ (1) hold.

Proof. (1) and (4) are equivalent (see [12, Theorem 3.26, p. 380]). The implications (3) ⇒ (2) ⇒ (1) hold(see [4, Theorem 3.10, p. 184]). To see that (5) ⇒ (1), suppose to the contrary that there exists y ∈ Y suchthat f−1(y) is not totally disconnected. Let C be a nondegenerate component of f−1(y). Take y1, . . . , yn−1 ∈Y \ {y} such that yi �= yi if i �= j. For i ∈ {1, . . . , n− 1}, let xi ∈ f−1(yi). Then, 〈{x1}, . . . , {xn−1}, C〉 is anondegenerate subcontinuum of Fn(f)−1({y1, . . . , yn−1, y}). Thus, since 〈{x1}, . . . , {xn−1}, C〉 ⊂ Fn

n−1(X),ρXn,m(〈{x1}, . . . , {xn−1}, C〉) is a nondegenerate subcontinuum of SFn

m(f)−1(ρYn,m({y1, . . . , yn−1, y})), a con-tradiction. �

Concerning Theorem 7.1, it is known that the implications (1) ⇒ (2) and (1) ⇒ (3) do not hold (see [4,Example 3.8, p. 184]) and (2) does not imply (3) (see, [4, Example 3.9, p. 184]).

Example 7.2. Let X = [−1, 1], Y = [0, 1] and f : X → Y be defined by f(x) = |x|. For eachy ∈ Y , f−1(y) is finite and then is totally disconnected. This implies that f and Fn(f) are light. SinceSFn

m(f)−1(ρYn,m(Fm(Y ))) is not totally disconnected, SFnm(f) is not light.

8. Atomic mappings

Lemma 8.1. If f : X → Y is a surjective mapping between continua and y ∈ Y , then exists a subcontinuum A

of X such that f(A) is a nondegenerate set and y /∈ f(A).

Proof. Take x ∈ X such that f(x) �= y. By Exercise 5.25 of [21, p. 84], there exists an order arcα : [0, 1] → C(X) such that α(0) = {x} and α(1) = X. Let g : C(Y ) → [0,∞) defined by g(K) = d(y,K)for all K ∈ C(Y ), where d(y,K) = glb{d(y, b): b ∈ K}. So, since (g ◦ C(f) ◦ α)(0) = d(f(x), y) and(g ◦ C(f) ◦ α)(1) = 0, there exists t ∈ [0, 1] such that (g ◦ C(f) ◦ α)(t) = d(f(x),y)

2 . It is easy to check thatα(t) has the required properties. �Theorem 8.2. Let f : X → Y be a surjective mapping between continua and let n � 3. If SFn

m(f) is atomic,then f is one-to-one.

Proof. Suppose to the contrary that there exist a, b ∈ X such that a �= b and f(a) = f(b). ByLemma 8.1, there exists a subcontinuum C of X such that f(C) is a nondegenerate subcontinuumof Y and f(a) /∈ f(C). Let x2, . . . , xn−1 ∈ X such that {f(x2), . . . , f(xn−1)} ∈ Fn−2

n−3 (Y ) and eachf(xi) /∈ f(C) ∪ {f(a)}. Let K = ρXn,m(〈{a}, {x2}, . . . , {xn−1}, C〉). Notice that K is a subcontinuum ofSFn

m(X) (see [19, Lemma 1, p. 230]) and SFnm(f)(K) = ρYn,m(〈{f(a)}, {f(x2)}, . . . , {f(xn−1)}, f(C)〉). Then,

SFnm(f)(K) is a nondegenerate subcontinuum of SFn

m(Y ). Thus, K = SFnm(f)−1(SFn

m(f)(K)). Hence,since ρXn,m(〈{b}, {x2}, . . . , {xn−1}, C〉) ⊂ SFn

m(f)−1(SFnm(f)(K)), we have 〈{b}, {x2}, . . . , {xn−1}, C〉 ⊂

〈{a}, {x2}, . . . , {xn−1}, C〉, a contradiction. �Theorem 8.3. Let f : X → Y be a surjective mapping between continua and let n � 2. The followingconditions are equivalent:

76 E. Castañeda-Alvarado et al. / Topology and its Applications 163 (2014) 66–76

(1) f is a homeomorphism;(2) C(f) is atomic;(3) 2f is atomic;(4) Fn(f) is atomic;(5) SFn

m(f) is atomic.

Proof. (1)–(3) are equivalent (see [3, Theorem 3.11, p. 199]). (1) and (4) are equivalent (see [12, Theo-rem 3.33, p. 382]). The equivalence between (1) and (5) follows from Theorem 8.2. �Acknowledgement

The authors wish to thank the referee for his/her useful suggestions.

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