Induced mappings on the hyperspace of convergent sequences

Topology and its Applications 229 (2017) 85–105

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Topology and its Applications www.elsevier.com/locate/topol

Induced mappings on the hyperspace of convergent sequences David Maya a,∗ , Patricia Pellicer-Covarrubias b , Roberto Pichardo-Mendoza b a

Universidad Autónoma del Estado de México, Facultad de Ciencias, Instituto Literiario 100, Col. Centro, Toluca, CP 50000, Mexico b Departamento de Matemáticas, Facultad de Ciencias, Circuito ext. s/n, Ciudad Universitaria, C.P. 04510, D.F., Mexico

a r t i c l e

i n f o

Article history: Received 6 February 2017 Received in revised form 13 July 2017 Accepted 19 July 2017 Available online 24 July 2017 MSC: 54A20 54B20 54C10

a b s t r a c t The symbol Sc (X) denotes the hyperspace of all nontrivial convergent sequences in a Hausdorff space X. This hyperspace is endowed with the Vietoris topology. For a given mapping between Hausdorff spaces f : X → Y , define the induced mapping Sc (f ) by Sc (f )(A) = f [A] (the image of A under f ) for every A ∈ Sc (X). In the current paper, we study under which conditions the fact that f belongs to a given class of mappings M implies that Sc (f ) belongs to M, and vice versa. © 2017 Elsevier B.V. All rights reserved.

Keywords: Hyperspace of nontrivial convergent sequences Induced mapping Open map Almost-open map Pseudo-open map Quotient map Closed map Surjective map Finite-to-1 map Homeomorphism Monotone map Strong light map Light map Sequence covering map 1-Sequence covering map

* Corresponding author. E-mail addresses: [email protected], [email protected] (D. Maya), [email protected] (P. Pellicer-Covarrubias), [email protected] (R. Pichardo-Mendoza). URL: http://www.matematicas.unam.mx/pmr (R. Pichardo-Mendoza). http://dx.doi.org/10.1016/j.topol.2017.07.008 0166-8641/© 2017 Elsevier B.V. All rights reserved.

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1. Introduction Convergence of sequences is an important tool to determine topological properties in Hausdorff spaces. On the other hand, the study of hyperspaces can provide information about the topological behavior of the original space and vice versa. In connection with both concepts, the hyperspace consisting of all nontrivial convergent sequences Sc (X), of a metric space X without isolated points, was introduced in [18]. Interesting properties of this hyperspace are presented in [32] and [33] where the study was extended to Hausdorff spaces. Our main purpose in this paper is to study the following class of continuous functions between hyperspaces. Given a hyperspace H(X) of a Hausdorff space X and a mapping f from X into a Hausdorff space Y satisfying that {f (a) : a ∈ A} ∈ H(Y ) for every A ∈ H(X), we define the H-induced mapping H(f ) : H(X) → H(Y ) by H(f )(A) = {f (a) : a ∈ A}. Now, let M be a class of mappings between topological spaces. A general problem for a fixed hyperspace H(X) is to find all possible relationships among the following statements: (a) f ∈ M, and (b) H(f ) ∈ M. This problem has been of interest for many authors (see [1–14,20–26,28–31]). The aim of this paper is to study the interrelations between the statements (a) and (b) when H(X) = Sc (X) and M is each one of the following classes of mappings: open, almost-open, pseudo-open, quotient, closed, surjective, finite-to-1, homeomorphism, monotone, strong light, light, sequence-covering and 1-sequence-covering (see definitions below). More precisely, we prove that condition (a) is implied by (b) for the following classes of mappings: sequence-covering, surjective, quotient, pseudo-open, almost-open, 1-sequence-covering and finite-to-1; we show the equivalence between (a) and (b) for the classes of mappings: monotone, one-to-one and homeomorphism. We also characterize the mappings for which the induced mapping is open, and finally, we find a condition on the domain of the ground function under which (b) implies (a) for the class of open mappings and for the class of closed mappings. All these results will be presented in Section 4. Examples in Section 5 show that the converse of most of the results in Section 4 do not hold and also that the hypotheses in some theorems of that section are essential. 2. Preliminaries, definitions and basic results All topological notions and all set-theoretic notions whose definition is not included here should be understood as in [15] and [27], respectively. The symbol ω denotes both, the first infinite ordinal and the first infinite cardinal. In particular, we consider all nonnegative integers as ordinals too; thus, n ∈ ω implies that n = {0, . . . , n − 1} and ω \ n = {k ∈ ω : k ≥ n}. The successor of ω is the ordinal ω + 1 = ω ∪ {ω}. The set ω \ {0} is denoted by N. If X is a set, |X| will represent the cardinality of X, [X]<ω is the collection of all finite subsets of X and [X]
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A convergent sequence in a topological space X is a function f from ω into X for which there is x ∈ X in such a way that for each U ∈ τX with x ∈ U there exists n ∈ ω with f [ω \ n] ⊆ U . In this case, we will say that either f converges to x or that x is the limit of f , and this fact will be denoted by lim f (n) = x. n→∞

We shall write (f (n))n∈ω to refer to f . If | ran(f )| = ω, we say that f is nontrivial. In connection with this concept, in this paper, a subset S of a space X will be called a nontrivial convergent sequence in X if S is countably infinite and there is x ∈ S in such a way that S \ U ∈ [X]<ω for each U ∈ τX with x ∈ U (see [33]). When this happens, the point x is called the limit point of S and we will say that S converges to x and write lim S = x. Throughout this paper, the reader will be able to identify from the context what is the intended meaning of nontrivial convergent sequence in the discussion. For a space X, let CL(X) = {A ⊆ X : A is closed in X and A = ∅}, K(X) = {A ∈ CL(X) : A is compact} and Sc (X) = {S ∈ K(X) : S is a nontrivial convergent sequence in X}. Given a family U of subsets of X, we define    U = A ∈ CL(X) : A ⊆ U ∧ ∀ U ∈ U (A ∩ U = ∅) . The Vietoris topology is the topology on CL(X) generated by the base consisting of all sets of the form U , where U ∈ [τX ]<ω (see [34, Proposition 2.1, p. 155]). The hyperspaces K(X) and Sc (X) will be considered as subspaces of CL(X). In particular, a base for the topology of Sc (X) consists of all sets of the form U c = U ∩ Sc (X), where U ∈ [τX ]<ω . For each n ∈ N, we will denote by Fn (X) the subspace [X]
if for all B ⊆ X and for each x ∈ clX B there is a sequence (xn )n∈ω contained in B whose limit is x. Every Fréchet–Urysohn space is a sequential space (see [15, Theorem 1.6.14, p. 53]). From [17, Proposition 7.3, p. 55], it follows that each sequential space X such that |LX | < ω is Fréchet–Urysohn. Proposition 2.1 ([33, Proposition 3.2]). For an arbitrary space X, { U c : U ∈ C(X)} is a base for Sc (X). Lemma 2.2 ([32, Lemma 3.4]). Let X be a space and let U ∈ τX . If x ∈ U ∩ LX , then Uc+ ∩ Sc (X, x) = ∅. Let X and Y be topological spaces and let f : X → Y be a mapping. Then, f is called: 1. strong light if f −1 (y) is discrete for every y ∈ Y ; 2. light if all nonempty fibers f −1 (y) are hereditarily disconnected;

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finite-to-1 provided that {|f −1 (y)| : y ∈ Y } ⊆ ω; monotone if each fiber f −1 (y) is connected; open provided that {f [U ] : U ∈ τX } ⊆ τY ;  almost-open if for each y ∈ Y there exists x ∈ f −1 (y) such that y ∈ {int f [U ] : x ∈ U ∈ τX };  pseudo-open if y ∈ {int f [U ] : f −1 (y) ⊆ U ∈ τX } for every y ∈ Y ; quotient if f is onto and τY = {V ⊆ Y : f −1 [V ] ∈ τX }; sequence-covering if for each sequence (yn )n∈ω converging to y in Y , there exists a convergent sequence (xn )n∈ω in X such that xn ∈ f −1 (yn ) for each n ∈ ω (see [35, p. 147]); 10. 1-sequence-covering if for each y ∈ Y there exists x ∈ f −1 (y) satisfying that for each sequence (yn )n∈ω in Y converging to y there exists a sequence (xn )n∈ω in X converging to x such that xn ∈ f −1 (yn ) for each n ∈ ω (see [29]); 11. closed provided that {f [F ] : F ∈ CL(X)} ⊆ CL(Y ). 3. 4. 5. 6. 7. 8. 9.

Notice that each strong light mapping is light. All open onto mappings are almost-open. Every almostopen mapping is pseudo-open and every pseudo-open mapping is quotient. Clearly, each 1-sequence-covering mapping is sequence-covering. Let X and Y be spaces and let f : X → Y be a mapping. Define the K-induced mapping K(f ) : K(X) → K(Y ) by K(f )(A) = f [A] for each A ∈ K(X). Also, set Sc (f ) = K(f )  Sc (X) and Fn (f ) = K(f )  Fn (X), for any n ∈ N. It is known that K(f ) is continuous (see [34, 5.10.1 of Theorem 5.10, p. 170]) and that ran(Fn (f )) ⊆ Fn (Y ), whenever n ∈ N. For the rest of this paper, we will assume that X is a sequential space such that Sc (X) = ∅, Y is a space and f : X → Y is a mapping. 3. Auxiliary results The following result gives necessary and sufficient conditions under which the induced mapping Sc(f ) satisfies that ran(Sc (f )) ⊆ Sc (Y ). Theorem 3.1 ([33, Corollary 7.4]). The following statements are equivalent. 1. f is a strong light mapping. 2. ran(Sc (f )) ⊆ Sc (Y ). Proposition 3.2. If f is a strong light mapping, then f [LX ] ⊆ LY . Proof. Let x ∈ LX . A consequence of Theorem 3.1 and the continuity of f is that {f [S] : S ∈ Sc (X, x)} ⊆ Sc (Y, f (x)). Hence, f (x) ∈ LY . 2 Lemma 3.3. If f is strong light and S ∈ Sc (X), then S ∩ f −1 (y) is finite for every y ∈ Y . Proof. Let y ∈ Y be arbitrary and set H = S ∩ f −1 (y). Observe that H is a closed subset of the compact space S. Then, H is a compact subset of S. Now, since f −1 (y) is discrete, we get that H must be finite. 2 Lemma 3.4. For an arbitrary sequence (yn )n∈ω in Y converging to yω ∈ Y , the following statements are true.

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1. If f is surjective and one of the following conditions is satisfied, then for each x ∈ f −1 (yω ) there exists a sequence (xn )n∈ω in X converging to x such that xn ∈ f −1 (yn ) for every n ∈ ω. (a) |{yn : n ∈ ω}| < ω; (b) yω ∈ Y \ LY . 2. If f is strong light and, for some S ∈ Sc (X), {yn : n ∈ ω + 1} is an adequate enumeration of f [S] (recall Theorem 3.1), then there exists {xn : n ∈ ω + 1} ∈ Sc+ with f (xn ) = yn for each n ∈ ω + 1. Proof. Let us prove (1). Let x ∈ f −1 (yω ) be arbitrary. First, observe that condition (b) implies (a), so it suffices to suppose that (a) is satisfied. A consequence of the convergence of the sequence (yn )n∈ω is the existence of m ∈ ω satisfying that {yn : n ∈ ω \ m} = {yω }. Now, for each n < m choose xn ∈ f −1 (yn ) and let xn = x for every n ∈ ω \ m. Observe that (xn )n∈ω is a sequence in X converging to x and f (xn ) = yn for every n ∈ ω. In order to show (2), set Q = {yn : n ∈ ω + 1} and let xω = lim S. Now, for each n ∈ ω, let xn ∈ S ∩ f −1 (yn ). So, the fact that |Q| = ω guarantees that |{xn : n ∈ ω + 1}| = ω. Thus, {xn : n ∈ ω + 1} ∈ Sc+ and f (xn ) = yn for every n ∈ ω. 2 Definition 3.5. For each y ∈ Y , let F (y) = {x ∈ f −1 (y) : Sc (Y, y) ⊆ Sc (f )[Sc (X, x)]}. Lemma 3.6. If f is a surjective strong light map and y ∈ Y , then x ∈ F (y) if and only if x ∈ f −1 (y) and for each sequence (yn )n∈ω in Y converging to y there exists a sequence (xn )n∈ω in X converging to x such that xn ∈ f −1 (yn ) for every n ∈ ω. Proof. First, assume that x ∈ F (y). Let (yn )n∈ω be a sequence in Y converging to y. Set Q = {yn : n ∈ ω} ∪ {y}. In light of (1) of Lemma 3.4, it suffices to suppose that Q ∈ Sc (Y, y). The definition of F (y) ensures the existence of S ∈ Sc (X, x) such that Sc (f )(S) = Q. So, (2) of Lemma 3.4 guarantees that there is {xn : n ∈ ω} ∪ {x} ∈ Sc+ such that f (xn ) = yn for every n ∈ ω. So, (xn )n∈ω is the required sequence. Now, in order to show the converse, let Q ∈ Sc (Y, y) and let {qn : n ∈ ω + 1} be an adequate enumeration of Q. Our assumption on x implies the existence of a sequence (xn )n∈ω in X converging to x such that xn ∈ f −1 (qn ) for every n ∈ ω. Set S = {xn : n ∈ ω} ∪ {x}. Since |Q| = ω, we have that |S| = ω. Therefore, S ∈ Sc (X, x) and Sc (f )(S) = Q. So, x ∈ F (y). 2 The next result characterizes 1-sequence-covering mappings in terms of the Sc -induced mappings and its proof follows from (1) of Lemma 3.4 and Lemma 3.6. / {F (y) : Proposition 3.7. If f is strong light and surjective, then f is 1-sequence-covering if and only if ∅ ∈ y ∈ LY }. 4. Main theorems A class of mappings M is said to be preserved by the correspondence f → Sc (f ) if the condition f ∈ M implies that Sc (f ) ∈ M and the class M is reversed by the correspondence f → Sc (f ) provided that the fact that f ∈ M is implied by Sc (f ) ∈ M. In the current section we will analyze which of the classes of mappings mentioned in the introduction are preserved or reversed by the correspondence f → Sc (f ). Theorem 3.1 will be used repeatedly without mentioning it explicitly throughout this section to assume that Sc (f ) is a mapping from Sc (X) into Sc (Y ) when f is a strong light map from X into Y . Let us start by proving that if f is strong light, then Sc (f ) is light. To this end, we will introduce some notation: for each Q ∈ Sc (Y ) and for each x ∈ f −1 [Q], let G(Q, x) = {S ∈ Sc (X) : x ∈ S ∧ f [S] = Q}.

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Lemma 4.1. If f is strong light, Q ∈ Sc (Y ) and x ∈ f −1 [Q \ {lim Q}], then G(Q, x) is a closed open subset of Sc (f )−1 (Q). −1 Proof. First, the equality G(Q, x) = Sc (f )−1 (Q)∩{x}− (Q). c implies that G(Q, x) is a closed subset of Sc (f ) −1 −1 Now, we shall show that G(Q, x) is an open subset of Sc (f ) (Q). Since f (f (x)) is discrete and f (x) ∈ Q \{lim Q}, there exist U ∈ τX and V ∈ τY satisfying that U ∩f −1 (f (x)) = {x} and V ∩Q = {f (x)}. Set W = U ∩ f −1 [V ]. Notice that x ∈ W ∈ τX . Then, Wc− ∩ Sc (f )−1 (Q) is an open subset of Sc (f )−1 (Q). Finally, let us argue that Wc− ∩ Sc (f )−1 (Q) = G(Q, x). First, the fact that x ∈ W and the definition of G(Q, x) guarantee that G(Q, x) ⊆ Wc− ∩ Sc (f )−1 (Q). Now, let S ∈ Wc− ∩ Sc (f )−1 (Q) be arbitrary. So, Sc (f )(S) = Q and there exists y ∈ S ∩ W . It follows that f (y) ∈ f [S] ∩ f [W ] ⊆ Q ∩ V = {f (x)} and, since W ⊆ U , we obtain that y ∈ f −1 (f (x)) ∩ U = {x}. Thus, x = y and x ∈ S. Hence, S ∈ G(Q, x). 2

Lemma 4.2. If f is strong light, Q ∈ Sc (Y ) and C is a component of Sc (f )−1 (Q), then, for every S, R ∈ C, S \ f −1 (lim Q) = R \ f −1 (lim Q) and lim S = lim R. Proof. Set q = lim Q. Lemma 4.1 ensures that G(Q, x) is a closed open subset of Sc (f )−1 (Q) for every x ∈ f −1 [Q \ {q}]. So, for each P ∈ C and x ∈ P \ f −1 (q), since P ∈ G(Q, x), we have that C ⊆ G(Q, x).   Hence, if S, R ∈ C, then S ∈ {G(Q, x) : x ∈ R \ f −1 (q)} and R ∈ {G(Q, x) : x ∈ S \ f −1 (q)}. This implies that R \ f −1 (q) ⊆ S and S \ f −1 (q) ⊆ R. Therefore, S \ f −1 (q) = R \ f −1 (q). Finally, invoke Lemma 3.3 to see that R ∩ f −1 (q) ∈ [X]<ω . Thus, from the equality S \ f −1 (q) = R \ f −1 (q), we deduce that clX (R \ f −1 (q)) ∈ Sc+ ∩ Rc+ , and so lim S = lim clX (R \ f −1 (q)) = lim R. 2 Theorem 4.3. If f is strong light, then Sc (f ) is light. Proof. Fix Q ∈ Sc (Y ) with Sc (f )−1 (Q) = ∅ and let C be a component of Sc (f )−1 (Q). We shall prove that C is degenerate. Fix S ∈ C and set a = lim S and q = lim Q. We are going to verify that {S} is a closed open subset of C. Since Sc (X) is T2 , {S} is a closed subset of C. In order to see that {S} is an open subset of C, set B = (S ∩ f −1 (q)) \ {a} and A = S \ B. From Lemma 3.3, it follows that B ∈ [X]<ω and so A ∈ Sc (X). Now, in order to prove that there exists an open subset U of Sc (X) such that {S} = U ∩ C, we will consider two cases. Case 1. B = ∅. Then, there exist V ∈ τX and W ∈ C(X) satisfying the following properties: 1. 2. 3. 4.

A ⊆ V ⊆ X \ (f −1 (q) \ {a}),  V ∩ W = ∅,  B = f −1 (q) ∩ W and |W ∩ B| = 1 for every W ∈ W.

Set U = {V } ∪ W. Conditions (1)–(4) ensure that U ∈ C(X) and S ∈ U c . Let us prove that {S} = U c ∩ C. Observe that S ∈ U c ∩ C. Now, in order to show that U c ∩ C ⊆ {S}, let T ∈ U c ∩ C be arbitrary. First, Lemma 4.2 and the fact that S, T ∈ C imply that A = (T \ f −1 (q)) ∪ {a} and lim T = lim S = a. Then, T = A ∪ ((T ∩ f −1 (q)) \ {a}). Now, let z ∈ B. Condition (3) guarantees the existence of W ∈ W such that z ∈ W . From the fact that T ∈ U c , it follows that T ∩ W = ∅. On the other hand, notice that T = A ∪ (T ∩ f −1 (q)) and hence, from conditions (1)–(4), we infer that T ∩ W = T ∩ f −1 (q) ∩ W ⊆ T ∩ {z}. Thus, T ∩ {z} = ∅ and we can conclude that z ∈ T . This proves that B ⊆ T and so S = A ∪ B ⊆ T .

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 Next, let y ∈ (T ∩ f −1 (q)) \ {a} be arbitrary. Since T ⊆ U, there exists U ∈ U such that y ∈ U . A consequence of the fact that y ∈ f −1 (q) \ {a} and condition (1) is that U = V and so U ∈ W. Then, from condition (3), we deduce that y ∈ U ∩ f −1 (q) ⊆ B. Thus, y ∈ B. Therefore, T = A ∪ ((T ∩ f −1 (q)) \ {a}) ⊆ A ∪ B = S. So, T = S and {S} = U c ∩ C. Case 2. B = ∅. First, there exists U ∈ τX satisfying that U ∩ f −1 (q) = {a}. We may assume that there exists U ∈ C(X)  such that U ∈ U, S ∈ U c and f −1 (q) ∩ (U \ {U }) = ∅. Since S ∈ C ∩ U c , it only remains to show that if T ∈ U c ∩C, then T = S. Lemma 4.2 ensures that S = (T \f −1 (q)) ∪{a}. Next, if t ∈ T ∩f −1 (q), then t ∈ V for some V ∈ U. Thus, t ∈ V ∩ f −1 (q). Hence, the choice of U implies that V = U . So, t ∈ U ∩ f −1 (q) = {a}. This proves that T ∩ f −1 (q) = {a}. It follows that T = (T \ f −1 (q)) ∪ {a} = S. In conclusion, {S} is an open subset of C. This finishes the proof of the fact that {S} is a closed open subset of C. So, the connectedness of C guarantees that C = {S}. Therefore, Sc (f ) is light. 2 In Section 5, we give an example of a strong light map whose Sc -induced mapping is not strong light (see Example 5.1). The following result characterizes, in the class of strong light mappings, sequence-covering mappings in terms of Sc -induced mappings. Theorem 4.4. If f is strong light, then f is sequence-covering if and only if ran(Sc (f )) = Sc (Y ). Proof. Assume that f is sequence-covering. In light of Theorem 3.1, it suffices to prove that Sc (Y ) ⊆ ran(Sc (f )). Let Q ∈ Sc (Y ) be arbitrary and let {qn : n ∈ ω + 1} be an adequate enumeration of Q. Our assumption on f ensures the existence of a convergent sequence (xn )n∈ω in X such that f (xn ) = qn for every n ∈ ω. Let xω be the limit point of the sequence (xn )n∈ω and set S = {xn : n ∈ ω + 1}. The fact that |Q| = ω implies that |S| = ω. So, S ∈ Sc (X) and Sc (f )(S) = Q. This proves that ran(Sc (f )) = Sc (Y ). In order to prove the reverse implication, we shall prove first that f is surjective. Let y ∈ Y be arbitrary and fix S ∈ Sc (X). The fact that f is strong light guarantees that f [S] ∈ Sc (Y ). So, f [S] ∪ {y} ∈ Sc (Y ). The assumption ran(Sc (f )) = Sc (Y ) implies that there exists T ∈ Sc (X) satisfying that Sc (f )(T ) = f [S] ∪ {y}. Hence, f −1 (y) = ∅. Finally, let (yn )n∈ω be a sequence in Y converging to yω and set Q = {yn : n ∈ ω + 1}. By (1) of Lemma 3.4, it suffices to assume that Q ∈ Sc (Y ). The equality ran(Sc (f )) = Sc (Y ) ensures the existence of S ∈ Sc (X) such that Sc (f )(S) = Q. Hence, from (2) of Lemma 3.4, it follows that there exists {xn : n ∈ ω + 1} ∈ Sc+ such that f (xn ) = yn for every n ∈ ω. Therefore, (xn )n∈ω is the required sequence and so f is sequence-covering. 2 The proof of each one of the corollaries below follows from the fact that all sequence covering mappings are surjective and Theorem 4.4. Corollary 4.5. If f is strong light and Sc (f ) is sequence-covering, then f is sequence-covering. Corollary 4.6. If f is strong light and ran(Sc (f )) = Sc (Y ), then f is a surjection. The converses of these two corollaries are false, in Section 5 we provide suitable examples (see Examples 5.2 and 5.3).

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Lemma 4.7. If f is strong light and V is a subset of Y , then  − 1. Sc (f )−1 [Vc− ] = f −1 [V ] c ,  + 2. Sc (f )−1 [Vc+ ] = f −1 [V ] c . Proof. Observe that {S ∈ Sc (X) : f [S] ∩ V = ∅} = {S ∈ Sc (X) : S ∩ f −1 [V ] = ∅} and {S ∈ Sc (X) : f [S] ⊆ V } = {S ∈ Sc (X) : S ⊆ f −1 [V ]}. 2 Theorem 4.8. If f is a strong light mapping and Sc (f ) is quotient, then f is quotient. Proof. Let V be a subset of Y such that f −1 [V ] ∈ τX and let y ∈ V be arbitrary. In order to prove that y ∈ intY V , we will consider two cases. Case 1. LY \ {y} = ∅. In this case there exist V1 , V2 ∈ τY satisfying that y ∈ V1 , V1 ∩ V2 = ∅ and V2 ∩ LY = ∅. Now, the continuity of f implies that f −1 [V1 ] ∈ τX . So, f −1 [V ∩ V1 ] = f −1 [V ] ∩ f −1 [V1 ] ∈ τX . Hence, since −1 −1 Sc (f )−1 [(V ∩ V1 )− [V ∩ V1 ])− [(V ∩ V1 )− c ] = (f c (see (1) of Lemma 4.7), we obtain that Sc (f ) c ] is an open − subset of Sc (X). Our assumption on Sc (f ) ensures that (V ∩ V1 )c is an open subset of Sc (Y ). On the other hand, the choice of V2 and Lemma 2.2 guarantee the existence of Q ∈ (V2 )+ c . Notice that − Q ∪{y} ∈ (V ∩V1 )− . Proposition 2.1 gives the existence of W ∈ C(Y ) such that Q ∪{y} ∈ W

c ⊆ (V ∩V1 )c . c Choose W ∈ W satisfying that y ∈ W . To show that y ∈ intY V we will argue that W ∩ V1 ⊆ V . Let z ∈ W ∩ V1 be arbitrary. Observe that Q ∪ {z} ∈ W c . Thus, we obtain that Q ∪ {z} ∈ (V ∩ V1 )− c , or equivalently, (Q ∪ {z}) ∩ (V ∩ V1 ) = ∅. From the fact that Q ∩ V1 = ∅, it follows that z ∈ V . Case 2. LY ⊆ {y}. Since f is strong light and Sc (X) = ∅, we have that Sc (Y ) = ∅ and so LY = ∅. Thus, we get that LY = {y}. −1 From (2) of Lemma 4.7, we obtain that Sc (f )−1 [Vc+ ] = (f −1 [V ])+ [V ] ∈ τX and that c . Given that f + Sc (f ) is a quotient map, we deduce that Vc is an open subset of Sc (Y ). Now, observe that LX ⊆ f −1 [V ] because y ∈ V (see Proposition 3.2). Then, a consequence of Lemma 2.2 and the assumption Sc (X) = ∅ is −1 that (f −1 [V ])+ [Vc+ ] = ∅ and so Vc+ = ∅. Choose Q ∈ Vc+ . Notice that lim Q = y. Next, c = ∅. Hence, Sc (f )  Proposition 2.1 ensures the existence of U ∈ C(Y ) satisfying that Q ∈ U c ⊆ Vc+ . Then, y ∈ U ⊆ V and  U ∈ τY . In conclusion, y ∈ intY V and so V ∈ τX .

2

The class of quotient mappings is not preserved by the correspondence f → Sc (f ) (see Example 5.6). Corollary 4.9. If f is strong light, Sc (f ) is quotient and |LY | < ω, then Y is Fréchet–Urysohn. Proof. Each quotient map is surjective. Now, Corollary 4.6 and Theorem 4.8 guarantee that f is a surjective quotient mapping. So, from [16, Proposition 1.2, p. 109], we obtain that Y is sequential. Finally, since |LY | < ω, [17, Proposition 7.3, p. 55] implies that Y is Fréchet–Urysohn. 2 The proof of the next result follows from Theorem 4.4 and [35, theorems 4.1 and 4.2, pp. 151–152]. Lemma 4.10. If f is strong light, Y is Fréchet–Urysohn and ran(Sc (f )) = Sc (Y ), then f is pseudo-open. Theorem 4.11. If f is strong light and Sc (f ) is pseudo-open, then f is pseudo-open.

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Proof. Let y ∈ Y be arbitrary and let U ∈ τX be such that f −1 (y) ⊆ U . We shall prove that y ∈ intY f [U ]. Consider two cases. Case 1. LY \ {y} = ∅. In this case, fix V1 , V2 ∈ τY satisfying that y ∈ V1 , V1 ∩ V2 = ∅ and V2 ∩ LY = ∅. The continuity of f guarantees that f −1 [V1 ] ∈ τX . Set W = U ∩ f −1 [V1 ]. Notice that f −1 (y) ⊆ W ∈ τX and f [W ] ∩ V2 = ∅. −1 Next, use Lemma 2.2 to fix some Q ∈ (V2 )+ (Q ∪ {y}) ⊆ Wc− . Each element c . Let us argue that Sc (f ) −1 −1 −1 P ∈ Sc (f ) (Q ∪ {y}) satisfies that P ∩ f (y) = ∅ and so, since f (y) ⊆ W , we have that P ∩ W = ∅. From the previous paragraph and our assumption on Sc(f ) we obtain that Q ∪ {y} ∈ int Sc (f )[Wc− ]. Hence, by Proposition 2.1, there exists V ∈ C(Y ) such that Q ∪ {y} ∈ V c ⊆ Sc (f )[Wc− ]. Let V ∈ V be such that y ∈ V . So, since V ∈ τY , it suffices to show that V ⊆ f [U ]. Let z ∈ V . Notice that Q ∪ {z} ∈ V c ⊆ Sc (f )[Wc− ]. Then, there exists S ∈ Wc− satisfying that Sc (f )(S) = Q ∪ {z}. The fact that S ∩ W = ∅ ensures that (Q ∪ {z}) ∩ f [W ] = f [S] ∩ f [W ] = ∅; thus, the inclusions Q ⊆ V2 ⊆ Y \ f [W ] imply that z ∈ f [W ] ⊆ f [U ]. Therefore, y ∈ int f [U ]. Case 2. LY ⊆ {y}. First, a consequence of the assumption Sc (X) = ∅ and the fact that f is strong light is that Sc (Y ) = ∅. Hence, we infer that LY = ∅. Then LY = {y}. Now, the assumption Sc (f ) is pseudo-open implies that Sc (f ) is a quotient mapping. Then, Corollary 4.9 and Lemma 4.10 imply that f is pseudo-open. 2 Example 5.6 in Section 5 shows that the converse of Theorem 4.11 does not hold. Next we turn our attention to almost open maps; we will prove that, under the assumption that f is strong light, if Sc (f ) is almost open, then so is f . In order to do this, we introduce three lemmas. Lemma 4.12. Assume that f is surjective. If x ∈ LX , S ∈ Sc (X, x) and R ∈ Sc (Y ) satisfy that |R\f [S]| < ω, then there exists Q ∈ Sc (X, x) such that f [Q] = R. Proof. Set P = S ∩ f −1 [R]. Observe that P is a closed subset of S. Our assumption |R \ f [S]| < ω ensures that |P | = ω. Hence, P ∈ Sc+ . Let A ∈ [X]<ω be such that f [A] = R \ f [S]. Set Q = P ∪ A. Then, Q ∈ Sc (X, x) and f [Q] = R. 2 Lemma 4.13. Assume that f is strong light. If x ∈ LX and S ∈ Sc (X, x), then there exists V ∈ τX satisfying that S ∈ Vc+ ∩ Sc (X, f −1 (f (x))) ⊆ Sc (X, S ∩ f −1 (f (x))). Proof. Set y = f (x). Since f −1 (y) is discrete, f −1 (y) \ S is closed in f −1 (y). The continuity of f implies that f −1 (y) \ S is closed in X. Set V = X \ (f −1 (y) \ S). Let us show that V fulfills all our requirements. Notice that V ∈ τX and S ∈ Vc+ ∩ Sc (X, f −1 (y)). In addition, if Q ∈ Vc+ ∩ Sc (X, f −1 (y)) and q = lim Q, then q ∈ V ∩ f −1 (y) = f −1 (y) \ (f −1 (y) \ S) = f −1 (y) ∩ S. In conclusion, Vc+ ∩ Sc (X, f −1 (y)) ⊆ Sc (X, S ∩ f −1 (y)). 2 In our next result, we will use the set F (y) as in Definition 3.5. Lemma 4.14. Assume that f is strong light and surjective. If y ∈ LY , Q ∈ Sc (Y, y), S ∈ Sc (f )−1 (Q) and V ∈ τX satisfy that S ∈ Vc+ ∩ Sc (X, f −1 (y)) ⊆ Sc (X, S ∩ f −1 (y))

(†)

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and for each sequence (Qm )m∈ω in Sc (Y ) converging to Q there exists k ∈ ω such that Sc (f )−1 (Qk )∩Vc+ = ∅, then S ∩ F (y) = ∅. Proof. Suppose to the contrary that for each z ∈ S ∩ f −1 (y) there exists Rz ∈ Sc (Y, y) \ Sc (f )[Sc (X, z)].  Set R = {Rz : z ∈ S ∩ f −1 (y)}. By Lemma 3.3, it follows that S ∩ f −1 (y) ∈ [X]<ω . This implies that R ∈ Sc (Y, y). Let {rn : n ∈ ω + 1} be an adequate enumeration of R. For each m ∈ ω, set Qm = Q ∪ {rn : n ≥ m}. Observe that (Qm )m∈ω is a sequence in Sc (Y ) converging to Q. So, from our assumptions, it follows the existence of k ∈ ω such that Sc (f )−1 (Qk ) ∩ Vc+ = ∅. Let P ∈ Vc+ be such that Sc (f )(P ) = Qk . Set p = lim P . Then, f (p) = f (lim P ) = lim f [P ] = lim Qk = y and so P ∈ Sc (X, f −1 (y)). Hence, the inclusion (†) guarantees that p ∈ S ∩ f −1 (y). Finally, since Sc (f )(P ) = Qk and |Rp \ Qk | < ω, from Lemma 4.12, we obtain that Rp ∈ Sc (f )[Sc (X, p)]. This is a contradiction. Therefore, S ∩ F (y) = ∅. 2 Theorem 4.15. If f is strong light and Sc (f ) is almost-open, then f is almost-open. Proof. Let y ∈ Y . We will consider two cases. Case 1. LY \ {y} = ∅. −1 In this case, there exists Q ∈ (Y \ {y})+ (Q ∪ {y}) c . Since Sc (f ) is almost-open, there exists S ∈ Sc (f ) such that Q ∪ {y} ∈ int Sc (f )[U] for every open subset U of Sc (X) with S ∈ U. Let x ∈ f −1 (y) ∩ S be  arbitrary. In order to show that y ∈ {intY f [U ] : x ∈ U ∈ τX }, let U ∈ τX be such that x ∈ U . Notice that if W = U \ f −1 [Q], then x ∈ W ∈ τX ; hence S ∈ Wc− . Thus Q ∪ {y} ∈ int Sc (f )[Wc− ] and so, by Proposition 2.1, there exists V ∈ C(Y ) satisfying that Q ∪ {y} ∈ V c ⊆ Sc (f )[Wc− ]. Take V ∈ V such that y ∈ V . Let us show that V \ Q ⊆ f [U ]. Let z ∈ V \ Q be arbitrary. Then Q ∪ {z} ∈ V c . So, the inclusion V c ⊆ Sc (f )[Wc− ] guarantees that there exists P ∈ Wc− such that Sc (f )(P ) = Q ∪ {z}. Thus, there exists p ∈ P ∩ W and P ∩ W is a subset of f −1 (z). Therefore p ∈ U and f (p) = z. This implies that V \ Q ⊆ f [U ]. Finally, since V \ Q ∈ τX and y ∈ V \ Q, we conclude that y ∈ int f [U ].

Case 2. LY ⊆ {y}. From Proposition 3.2 and our assumption that Sc (X) = ∅, it follows that LY = ∅. So, LY = {y}. Now, let Q ∈ Sc (Y ) be arbitrary. Then lim Q = y. By assumption there exists S ∈ Sc (f )−1 (Q) satisfying that Q ∈ int Sc (f )[U] for every open subset U of Sc (X) such that S ∈ U. So, we have that f (lim S) = y. Let us  argue that there exists x ∈ f −1 (y) ∩ S satisfying that y ∈ {intY f [V ] : x ∈ V ∈ τX }. Suppose to the contrary that for each z ∈ f −1 (y) ∩ S there exists Vz ∈ τX satisfying that z ∈ Vz and y∈ / intY f [Vz ]. On the other hand, since each almost-open mapping is quotient and surjective, Corollary 4.9 ensures that Y is Fréchet–Urysohn. This and the fact that y ∈ clY (Y \ f [Vz ]) imply the existence of Rz ∈ Sc (Y, y) satisfying that Rz \ {y} ⊆ Y \ f [Vz ]. Observe that if P ∈ Sc (X, z), then P \ Vz is finite and so f [P ] \ f [Vz ] must be finite; thus, we conclude that f [P ] = Rz . Hence, Rz ∈ Sc (Y, y) \ Sc (f )[Sc (X, z)] for every z ∈ f −1 (y) ∩ S.

(4.1)

Now, in light of Lemma 4.13, there exists V ∈ τX satisfying that S ∈ Vc+ ∩ Sc (X, f −1 (y)) ⊆ Sc (X, S ∩ f −1 (y)). Our assumption on S implies that Q ∈ int Sc (f )[Vc+ ]. Thus, for each sequence (Qm )m∈ω in Sc (Y ) / {Sc (f )−1 (Qm ) ∩ converging to Q, there is k ∈ ω satisfying that {Qm : m ∈ ω \k} ⊆ int Sc (f )[Vc+ ] and so ∅ ∈ + Vc : m ∈ ω \k}. Since each almost-open mapping is surjective, by Corollary 4.6, we have that f is surjective. Then, Lemma 4.14 guarantees the existence of x ∈ f −1 (y) ∩ S such that Sc (Y, y) ⊆ Sc (f )[Sc (X, x)], a contradiction to (4.1). From the two cases above, we conclude that f is an almost-open mapping. 2 Theorem 4.16. If f is strong light and Sc (f ) is 1-sequence-covering, then f is 1-sequence-covering.

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Proof. From the fact that each 1-sequence-covering mapping is surjective and Corollary 4.6, it follows that f is surjective. So, in light of Proposition 3.7, it suffices to prove that ∅ ∈ / {F (y) : y ∈ LY }. Let y ∈ LY be arbitrary. The definition of LY guarantees the existence of Q ∈ Sc (Y, y). By assumption there exists S ∈ Sc (f )−1 (Q) such that for each sequence (Qm )m∈ω in Sc (Y ) converging to Q there exists a sequence (Sm )m∈ω in Sc (X) converging to S such that Sc (f )(Sm ) = Qm for every m ∈ ω. We will employ Lemma 4.14 to prove that F (y) = ∅. The existence of V ∈ τX satisfying that S ∈ Vc+ ∩ Sc (X, f −1 (y)) ⊆ Sc (X, S ∩ f −1 (y)) is guaranteed by Lemma 4.13. Now, if (Qm )m∈ω is a convergent sequence in Sc (Y ) whose limit point is Q, our assumption on S ensures that there exists a sequence (Sm )m∈ω in Sc (X) converging to S and such that Sc (f )(Sm ) = Qm for every m ∈ ω. Then, there exists k ∈ ω such that {Sm : m ∈ ω \ k} ⊆ Vc+ and so, Sm ∈ Sc (f )−1 (Qm ) ∩ Vc+ for each m ∈ ω \ k. Thus, from Lemma 4.14, we get that F (y) = ∅. In conclusion, f is 1-sequence-covering. 2 Example 5.3 shows that the converses of Theorems 4.15 and 4.16 fail. Remark 4.17. Observe that if x ∈ X is not an isolated point of X, then x ∈ clX (X \ {x}) and since X is sequential (recall the assumptions we made in the last paragraph of Section 2), there is a sequence (xn )n∈ω in X \ {x} which converges to x; now, the fact that X is T2 implies that {xn : n ∈ ω} is infinite and so, x ∈ LX . In other words, for each x ∈ X \ LX , {x} ∈ τX . Hence, we infer that LX ∈ CL(X). Also notice that if w ∈ LX , there exists S ∈ Sc (X, w) and since |S| = ω, we conclude that {w} ∈ / τX . Both results below are auxiliary to study the openness of the induced mapping Sc (f ). Lemma 4.18. Assume that f is strong light and surjective. If f −1 [LY ] \ LX = ∅ and {{f (x)} : x ∈ X \ LX and Sc (f )[{x}− c ] \ Sc (Y, f (x)) = ∅} ⊆ τY ,

(∗)

then LY = f [f −1 [LY ] \ LX ] and |LY | = 1. Proof. Notice that f [f −1 [LY ] \ LX ] ⊆ LY . Now, we shall prove that LY ⊆ f [f −1 [LY ] \ LX ]. Let b ∈ LY be arbitrary and choose x ∈ f −1 (b). If x ∈ / LX , then x ∈ f −1 [LY ] \ LX and so b ∈ f [f −1 [LY ] \ LX ]. Assume that x ∈ LX and fix S0 ∈ Sc (X, x) and a ∈ f −1 [LY ] \ LX . By letting S = S0 ∪ {a} we obtain that S ∈ {a}− c ∩ Sc (X, x). Let us argue that lim Sc (f )(S) = f (a). To this end suppose to the contrary that lim Sc (f )(S) = f (a). Then Sc (f )(S) ∈ Sc (f )[{a}− c ] \ Sc (Y, f (a)). By (∗) we obtain that {f (a)} ∈ τY . This and Remark 4.17 contradict the fact that f (a) ∈ LY . Then b = f (x) = lim Sc (f )(S) = f (a). From both cases, we can conclude that LY = f [f −1 [LY ] \ LX ]. Finally, in order to see that LY consists of a unique element, let w, z ∈ LY . Take x, y ∈ f −1 [LY ] \ LX − satisfying that f (x) = w and f (y) = z. Next, use the assumption Sc (X) = ∅ to choose Q ∈ {x}− c ∩ {y}c . So, − − since {w}, {z} ∈ / τY , condition (∗) ensures that Sc (f )[{x}c ] ⊆ Sc (Y, w) and Sc (f )[{y}c ] ⊆ Sc (Y, z). Hence, Sc (f )(Q) ∈ Sc (Y, w) ∩ Sc (Y, z) and thus w = z. 2 Lemma 4.19. Assume that f is a strong light surjective mapping which satisfies condition (∗) of Lemma 4.18. + −1 If U ∈ τX is such that U ∩ LX = ∅ and Q ∈ (f [U ])+ (lim Q)) = ∅. c , then Uc ∩ Sc (X, f Proof. In the case that f −1 [LY ] \ LX = ∅, by Lemma 4.18, we obtain that LY = {lim Q}, and so Uc+ ⊆ Sc (X, f −1 (lim Q)). Now, apply Lemma 2.2 to get that Uc+ = ∅. Thus, Uc+ ∩ Sc (X, f −1 (lim Q)) = ∅. Now, assume that f −1 [LY ] \ LX = ∅. By Proposition 3.2, the equality f −1 [LY ] = LX holds. Next, choose p ∈ U such that f (p) = lim Q. Then p ∈ LX . So, a consequence of Lemma 2.2 and the inclusion Sc (X, p) ⊆ Sc (X, f −1 (lim Q)) is that Uc+ ∩ Sc (X, f −1 (lim Q)) = ∅. 2

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The purpose of introducing the following notion is to get a useful characterization of the mappings f for which the induced mapping Sc (f ) is open. The mapping f is called hereditarily sequence-covering provided that whenever U ∈ τX has nonempty intersection with LX , we obtain the inclusion + (f [U ])+ c ∩ ran(Sc (f )) ⊆ Sc (f )[Uc ].

Applying Theorem 4.4, for each strong light f such that f is sequence-covering and hereditarily sequencecovering, the mapping f  U : U → f [U ] is sequence covering provided that U ∈ τX is such that U ∩ LX = ∅. Our next result is the above mentioned characterization of the openness of Sc (f ). Theorem 4.20. If f is strong light and surjective, then Sc (f ) is open if and only if each one of the following conditions is satisfied: 1. 2. 3. 4.

{f [U ] : U ∈ τX ∧ U ∩ LX = ∅} ⊆ τY , condition (∗) of Lemma 4.18, ran(Sc (f )) is an open subset of Sc (Y ), and f is hereditarily sequence-covering.

Proof. Assume that Sc (f ) is open. With the idea of proving (1), let U ∈ τX be such that U ∩ LX = ∅. In order to prove that f [U ] ∈ τY , let a ∈ U be arbitrary. Lemma 2.2 guarantees the existence of S ∈ Uc+ . Then S ∪ {a} ∈ Uc+ . By assumption Sc (f )[Uc+ ] is open and so there exists V ∈ [τY ]<ω satisfying that   Sc (f )(S ∪ {a}) ∈ V c ⊆ Sc (f )[Uc+ ]. Observe that f (a) ∈ V ∈ τY . Next, let us argue that V ⊆ f [U ].  Let b ∈ V be arbitrary. Notice that Sc (f )(S ∪ {a}) ∪ {b} ∈ V c . This implies that there exists P ∈ Uc+ such that Sc (f )(P ) = Sc (f )(S ∪ {a}) ∪ {b}. Hence, P ∩ f −1 (b) = ∅ and so b ∈ f [U ]. This finishes the proof that (1) holds. Now, suppose that x ∈ X \ LX and that there exists R ∈ {x}− c such that lim Sc (f )(R) = f (x). First, notice that Remark 4.17 guarantees that {x} ∈ τX . Now, since Sc (f ) is open, by Proposition 2.1, there exists W ∈ C(Y ) satisfying that Sc (f )(R) ∈ W c ⊆ Sc (f )[{x}− c ]. Let W ∈ W be such that f (x) ∈ W . We will show that W = {f (x)}. Consider w ∈ W . Notice that (Sc (f )(R) \ {f (x)}) ∪ {w} ∈ W c . Then, there exists T ∈ {x}− c such that Sc (f )(T ) = (Sc (f )(R) \ {f (x)}) ∪ {w}. Thus, the fact that x ∈ T ensures that f (x) ∈ (Sc (f )(R) \ {f (x)}) ∪ {w}. Therefore, w = f (x) and so we can conclude that (2) is satisfied. Condition (3) follows easily from our assumption. It only remains to verify that f is hereditarily sequence covering. Let U ∈ τX be such that U ∩ LX = + ∅ and let Q ∈ (f [U ])+ c ∩ ran(Sc (f )) be arbitrary. Lemma 4.19 guarantees that there exists R ∈ Uc ∩ Sc (X, f −1 (lim Q)). Since Sc (f ) is open, there exists V ∈ C(Y ) such that Sc (f )(R) ∈ V c ⊆ Sc (f )[Uc+ ] (see Proposition 2.1). Choose V ∈ V satisfying that lim Sc (f )(R) = lim Q ∈ V . Then, |Q \ V | < ω and so Q ∩ V ∈ Sc (Y, lim Q). Hence, Sc (f )(R) ∪ (Q ∩ V ) ∈ V c . This implies that there exists P ∈ Uc+ such that Sc (f )(P ) = Sc (f )(R) ∪ (Q ∩ V ). Thus, there exists T ∈ Sc (X) satisfying that T ⊆ P and Sc (f )(T ) = Q ∩ V . Finally, let A ∈ [U ]<ω be such that f [A] = Q \ V . Then, T ∪ A ∈ Uc+ and Sc (f )(T ∪ A) = Q. This shows that (4) holds. Next, assume that (1)–(4) are true. By Proposition 2.1 and condition (3), in order to see that Sc (f ) is open, it suffices to show that for every U ∈ C(X) and for each S ∈ U c there exists V ∈ [τY ]<ω such that Sc (f )(S) ∈ V c ∩ ran(Sc (f )) ⊆ Sc (f )[ U c ].  Let U and S be as described above. Set U ∗ = {U ∈ U : U ∩ LX = ∅} and O = U ∗ . Choose W ∈ U ∗ satisfying that lim S ∈ W . Then, W ⊆ O ∈ τX and so |S \ O| < ω. Define V = {f [U ] : U ∈ U ∗ } ∪ {{f (x)} : x ∈ S \ (f −1 [LY ] ∪ O)}.

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Note that if x ∈ S \ f −1 [LY ], then f [S] ∈ Sc (f )[{x}− c ] \ Sc (Y, f (x)) and so, we may apply (1), (2) and the inequality |V| ≤ |U| + |S \ O| to deduce that V ∈ [τY ]<ω . Now, we will verify that Sc (f )(S) ∈ V c ∩ ran(Sc (f )) ⊆ Sc (f )[ U c ].   In order to show that Sc (f )(S) ⊆ V, let a ∈ S be arbitrary. Since S ⊆ U, there exists U ∈ U such that  a ∈ U . If U ∈ U ∗ , then f (a) ∈ f [U ] ∈ V and so f (a) ∈ V. In the case that a ∈ S \ (f −1 [LY ] ∪ O), we have  that {f (a)} ∈ V and thus f (a) ∈ V. Otherwise, we get that U ∈ / U ∗ and a ∈ f −1 [LY ] ∪ O. The fact that −1 U ∈ C(X) ensures that U ∩ O = ∅. This guarantees that a ∈ f [LY ] \ LX . From Lemma 4.18 and condition  (2), it follows that LY = {f (a)}. Then f (a) = f (lim S) ∈ f [W ] ∈ V. This proves that Sc (f )(S) ⊆ V. Now, the fact that ∅ ∈ / {S ∩ U : U ∈ U} implies that ∅ ∈ / {Sc (f )(S) ∩ V : V ∈ V}. In conclusion, Sc (f )(S) ∈ V c ∩ ran(Sc (f )). It only remains to show that V c ∩ ran(Sc (f )) ⊆ Sc (f )[ U c ]. Let Q ∈ V c ∩ ran(Sc (f )) be arbitrary. Then, there exists G ∈ U ∗ such that lim Q ∈ f [G] ∈ τY . Thus, |Q \ f [G]| < ω and hence we deduce that + Q ∩ f [G] ∈ (f [G])+ c . Condition (4) and the assumption Q ∈ ran(Sc (f )) guarantee the existence of P ∈ Gc satisfying that Sc (f )(P ) = Q ∩ f [G]. Define the family A by the following formula: A ∈ A if and only if either (i) there are U ∈ U ∗ and y ∈ Q ∩ f [U ] \ f [G] such that A = U ∩ f −1 (y) or (ii) for some U ∈ U ∗ satisfying ∅ = Q ∩ f [U ] ⊆ f [G], we get A = U ∩ f −1 [Q]. / A. Let h : A → Observe that |A| < ω because lim Q ∈ f [G] ∈ τY . Also, ∅ ∈ h(A) ∈ A for every A ∈ A. Set



A be a function such that

R = P ∪ ran(h) ∪ (S \ (f −1 [LY ] ∪ O)) ∪ (S ∩ f −1 (f (lim S)) ∩ f −1 [Q]). Invoke Lemma 3.3 to show that S ∩ f −1 (f (lim S)) ∩ f −1 [Q] ∈ [X]<ω . Hence, since ran(h), S \ (f −1 [LY ] ∪ O) ∈ [X]<ω and P ∈ Sc (X), we deduce that R ∈ Sc (X). To end the proof, it suffices to argue that R ∈ U c ∩ Sc (f )−1 (Q). Clearly, f [P ], f [ran(h)] and f [S ∩ f −1 (f (lim S)) ∩ f −1 [Q]] are subsets of Q. Now, for the inclusion f [S \ (f −1 [LY ] ∪ O)] ⊆ Q recall the definition of V and the fact that Q ∈ V c . Thus, f [R] ⊆ Q. For the  reverse inclusion, let q ∈ Q be arbitrary. The fact that Q ∈ V c implies that q ∈ V. Hence, if there exists x ∈ S \ (f −1 [LY ] ∪ O) satisfying that q ∈ {f (x)}, then x ∈ R and q ∈ f [R]. Otherwise, there exists U ∈ U ∗ such that q ∈ f [U ]. In the case that q ∈ f [G], we have that q ∈ Q ∩ f [G] = f [P ] ⊆ f [R]. Now, assume that q ∈ / f [G]. This implies that q ∈ Q ∩ f [U ] \ f [G] and so, A = U ∩ f −1 (q) ∈ A; hence, h(A) ∈ A ∩ R and therefore, q = f (h(A)) ∈ f [R].  Finally, let z ∈ R. When z ∈ / ran(h), we just need to recall that P ⊆ G ∈ U and that S ⊆ U to get that  z ∈ U. In the case that z ∈ ran(h), then there exist A ∈ A and U ∈ U such that h(A) = z and A ⊆ U ;   since z ∈ A, we obtain that z ∈ U. In conclusion, R ⊆ U. Next, we shall show that ∅ ∈ / {R ∩ U : U ∈ U}. Let U ∈ U be arbitrary. Observe that if U ∩ (S \ (f −1 [LY ] ∪ O)) = ∅, then U ∩ R = ∅. Now, assume that U ∩ (S \ (f −1 [LY ] ∪ O)) = ∅. In the case that U ∈ U ∗ , we obtain that f [U ] ∈ V and so Q ∩ f [U ] = ∅; then there exists A ∈ A such that h(A) ∈ R ∩ U . When U ∈ / U ∗ , we have that U ∩ (S \ O) = U ∩ S (recall that U ∈ C(X)). Then U ∩ (S \ f −1 [LY ]) = ∅. This and the fact that U ∩ LX = ∅ guarantee that S ∩ U ⊆ f −1 [LY ] \ LX . So, from Lemma 4.18, it follows that LY = f [f −1 [LY ] \ LX ] and |LY | = 1. Then LY = {f (lim S)} = {lim Q} and f [S ∩ U ] ⊆ LY . Hence, we infer that S ∩ U ⊆ S ∩ f −1 (f (lim S)) ∩ f −1 [Q] ∩ U ⊆ R ∩ U . A consequence of the last inclusion and the fact that S ∩ U = ∅ is that R ∩ U = ∅. In conclusion, R ∩ U = ∅ for all U ∈ U. This completes the proof that R ∈ U c . Therefore, Sc (f ) is open. 2 Corollary 4.21. If LX is dense in X, f is strong light and Sc (f ) is open, then so is f .

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Proof. The result follows from (1) of Theorem 4.20. 2 Corollary 4.22. If f is strong light, () {x ∈ X \ LX : Sc (f )[{x}− c ] \ Sc (Y, f (x)) = ∅} = X \ LX and Sc (f ) is open, then f is open. Proof. Let U ∈ τX . If U ∩ LX = ∅, since Sc (f ) is open, by (1) of Theorem 4.20, we have that f [U ] ∈ τY . Now, assume that U ∩ LX = ∅. Let y ∈ U . Then, y ∈ X \ LX . So, condition () and (2) of Theorem 4.20 ensure that {f (y)} ∈ τY . 2 In the next section, Example 5.4 shows that the condition clX LX = X in Corollary 4.21 and condition () in Corollary 4.22 can not be omitted. Theorem 4.23. Assume that f is strong light, open and hereditarily sequence-covering. If ran(Sc (f )) is an open subset of Sc (Y ), then Sc (f ) is open. Proof. In light of Theorem 4.20, it suffices to verify that conditions (1) and (2) of Theorem 4.20 hold. Clearly, (1) follows from the fact that f is open. Now, regarding (2), we apply Remark 4.17 to get {{x} : x ∈ X \ LX } ⊆ τX and so {{f (x)} : x ∈ X \ LX } ⊆ τY as needed. 2 The proof of the next result follows from Corollary 4.22 and Theorem 4.23. Corollary 4.24. Suppose that f is strong light and hereditarily sequence covering. If ran(Sc (f )) is an open subset of Sc (Y ) and condition () of Corollary 4.22 holds, then Sc (f ) is open if and only if f is open. Our next theorem shows that in the class of strong light mappings being one-to-one and being monotone are equivalent. Also, it proves that the class of one-to-one mappings and the class of monotone mappings are preserved and reversed by the correspondence f → Sc (f ). Theorem 4.25. If f is strong light, then the following statements are equivalent. (a) (b) (c) (d)

f is one-to-one. Sc (f ) is one-to-one. f is monotone. Sc (f ) is monotone.

Proof. Notice that (a) implies both (b) and (c), and (b) implies (d). Now, from Theorem 4.3 and the fact that f is strong light, it follows that Sc (f ) and f are light. Hence, (c) implies (a) and (d) implies (b). Thus, it only remains to prove that (a) is implied by (b). Let x, y ∈ X be such that x = y. Fix S ∈ Sc (X) satisfying that {x, y} \ S = ∅. Thus, we obtain that S ∪ {x} = S ∪ {y}. Hence, condition (b) ensures that Sc (f )(S ∪ {x}) = Sc (f )(S ∪ {y}). This means that f [S] ∪ {f (x)} = f [S] ∪ {f (y)} and so, f (x) = f (y). 2 Lemma 4.26. If f is a strong light mapping and ran(Sc (f )) = Sc (Y ), then f [LX ] = LY . Proof. From Proposition 3.2, we have that f [LX ] ⊆ LY . Now, let y ∈ LY be arbitrary. Then, there exists Q ∈ Sc (Y, y). Our assumption on Sc (f ) implies the existence of S ∈ Sc (X) such that Sc (f )(S) = Q. Thus, the continuity of f and the fact that lim S ∈ LX ensures that y = lim Q = lim Sc (f )(S) = f (lim S) ∈ f [LX ]. 2

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Theorem 4.27. If Sc (f ) is a homeomorphism between Sc (X) and Sc (Y ), then f is a homeomorphism. Proof. From Corollary 4.6 and Theorem 4.25, we deduce that f is a bijection. It only remains to show that f is open. The equality ran(Sc (f )) = Sc (Y ), Lemma 4.26 and the fact that f is one-to-one ensure that f −1 [LY ] = LX and this condition implies that condition () of Corollary 4.22 holds (recall that Sc (X) = ∅). Thus, from Corollary 4.22, it follows that f is open. 2 Theorem 4.28. If f is a homeomorphism, then Sc (f ) is a homeomorphism between Sc (X) and Sc (Y ). Proof. Let g be the inverse function of f . Observe that Sc (g) is a map from Sc (Y ) into Sc (X) which is the inverse function of Sc (f ). Therefore, Sc (f ) is a homeomorphism. 2 From Theorems 4.27 and 4.28 we obtain the following result. Corollary 4.29. If f is a strong light mapping, then f is a homeomorphism if and only if Sc (f ) is a homeomorphism between Sc (X) and Sc (Y ). Theorem 4.30. If Y is Fréchet–Urysohn, f is one-to-one and ran(Sc (f )) = Sc (Y ), then Sc (f ) is a homeomorphism between Sc (X) and Sc (Y ). Proof. In light of Theorem 4.28, it suffices to prove that f is a homeomorphism. By our assumptions on f and Corollary 4.6, we only need to show that f is open. From Lemma 4.10, it follows that f is pseudo-open. Now, let U ∈ τX and let y ∈ f [U ] be arbitrary. Since f is one-to-one, we have that f −1 (y) ⊆ U . Then, the fact that f is pseudo-open implies that y ∈ intY f [U ]. Therefore, f [U ] ∈ τY . 2 Corollary 4.31. If Y is Fréchet–Urysohn and Sc (f ) is bijective, then Sc (f ) is a homeomorphism between Sc (X) and Sc (Y ). For a space Z, a subset A of Z and S ∈ Sc (Z), let N(S, A) = {S ∪ {a} : a ∈ A}. Theorem 4.32. If f is strong light and Sc (f ) is finite-to-1, then f is finite-to-1. Proof. Let y ∈ Y and fix S ∈ Sc (X). Set Q = Sc (f )(S). Observe that N(S, f −1 (y) \ S) ⊆ Sc (f )−1 (Q ∪ {y}). So, our assumption on Sc (f ) implies that |N(S, f −1 (y) \ S)| < ω and, as a consequence, |f −1 (y) \ S| < ω. On the other hand, Lemma 3.3 implies that |S∩f −1 (y)| < ω. Thus, |f −1 (y)| = |f −1 (y)∩S|+|f −1 (y)\S| < ω. 2 Example 5.1 in Section 5 shows that the converse of Theorem 4.32 does not hold. The result below will help us study the closedness of the induced mappings Sc (f ). Lemma 4.33. If S ∈ Sc (X) and A is a subset of X, then N(S, clX A) ⊆ clSc (X) N(S, A) ⊆ {P ∈ (clX A)− c : S ⊆ P ⊆ S ∪ clX A}. Proof. In order to prove the inclusion on the left, let z ∈ clX A be arbitrary and let U ∈ C(X) be such that S ∪ {z} ∈ U c . Choose U ∈ U satisfying that z ∈ U . The fact that z ∈ clX A guarantees the existence of x ∈ U ∩ A. Then we have that S ∪ {x} ∈ U c ∩ N(S, A). This and Proposition 2.1 prove that S ∪ {z} ∈ clSc (X) N(S, A). Now, let P ∈ clSc (X) N(S, A). First, we shall prove that X\P ⊆ X\S. Let y ∈ X\P . Then, P ∈ (X\{y})+ c . The choice of P and the fact that X \{y} ∈ τX imply that there exists a ∈ A such that S ∪{a} ∈ (X \{y})+ c , i.e., S ∪ {a} ⊆ X \ {y}. Thus, y ∈ X \ S. This verifies that S ⊆ P .

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Next, we are going to argue that P \ S ⊆ clX A. Let x ∈ P \ S and let U ∈ τX be such that x ∈ U . Thus, x ∈ U \ S ∈ τX . Then P ∈ (U \ S)− c . The choice of P ensures the existence of b ∈ A with the property that S ∪ {b} ∈ (U \ S)− . This implies that b ∈ U and so U ∩ A = ∅. Hence, x ∈ clX A and we can conclude that c P ⊆ S ∪ clX A. It only remains to show that P ∈ (clX A)− c . If S ∩ clX A = ∅, then the fact that S ⊆ P implies that + P ∩clX A = ∅. Now, suppose that S∩clX A = ∅. Notice that S ∈ (X \clX A)+ c and (X \clX A)c ∩N(S, A) = ∅. Then, S ∈ / clSc (X) N(S, A). Hence, P \S = ∅. From the inclusion P \S ⊆ clX A, we deduce that P ∩clX A = ∅. Therefore, P ∈ (clX A)− c . 2 Theorem 4.34. Assume that f is strong light and that Sc (f ) is closed. If |f [LX ]| ≥ 2, then f is closed; in particular, if clX LX = X, then f is closed. Proof. Let A ∈ CL(X) and let y ∈ clY f [A]. Our assumption on |f [LX ]| implies that LX ∩ (X \ f −1 (y)) = ∅ and so Lemma 2.2 ensures the existence of S ∈ (X \ f −1 (y))+ c . Set Q = Sc (f )(S). From Lemma 4.33, it follows that Q ∪ {y} ∈ clSc (Y ) N(Q, f [A]). Now, the equality N(Q, f [A]) = Sc (f )[N(S, A)] and the fact that Sc (f ) is closed guarantee that Q ∪{y} ∈ Sc (f )[clSc (X) N(S, A)]. Finally, by Lemma 4.33, there exists P ∈ A− c such that S ⊆ P ⊆ S ∪ A and Sc (f )(P ) = Q ∪ {y}. Thus, since f −1 (y) ∩ P = ∅ and f −1 (y) ∩ S = ∅, we obtain that f −1 (y) ∩ A = ∅. Then, y ∈ f [A]. 2 The converse of Theorem 4.34 is not true (see Example 5.8). 5. Examples and questions The aim of this section is to show that the conclusion of Theorem 4.3 can not be strengthened and that the converses of Theorems 4.8, 4.11, 4.15, 4.16, 4.32 and 4.34 and of Corollaries 4.5 and 4.6 do not hold. We also show that the additional conditions in Corollaries 4.21 and 4.22 and Theorem 4.30 are essential. Finally, we pose some questions related to Theorems 4.23 and 4.34 and an open problem about different classes of mappings from those studied in the current paper. Throughout this section the symbol S 1 denotes the subspace {(x, y) : x2 + y 2 = 1} of the plane R2 . We will consider the exponential mapping exp : R → S 1 defined by exp(t) = (cos(t), sin(t)) for every t ∈ R. Our first example shows that the Sc -induced mappings of strong light mappings are not necessarily strong light, although these mappings are always light (see Theorem 4.3), and that the converse of Theorem 4.32 does not hold. Example 5.1. Define f : [0, 4π) → S 1 as follows: f (t) =

exp(t), if t ∈ [0, 2π], exp(4π − t), if t ∈ [2π, 4π).

Observe that f is continuous and |f −1 (y)| = 2 for every y ∈ S 1 . Thus, f is finite-to-1 and so f is a strong

 light mapping. Set S = {2π} ∪ 2π + πn : n ∈ N . Let us argue that Sc (f )−1 (Sc (f )(S)) is not discrete.

 First, for each n ∈ N, let Qn = S ∪ 2π − nπ . Observe that {Qn : n ∈ N} ⊆ Sc (f )−1 (Sc (f )(S)). Next, in order to prove that Sc (f ) is not a strong light mapping, let U ∈ [τ[0,4π) ]<ω be such that S ∈ U c . We are going to prove that U c ∩ Sc (f )−1 (Sc (f )(S)) is infinite. Take U ∈ U satisfying that 2π ∈ U . Then there exists k ∈ N such that {2π + (−1)i πn : i ∈ {0, 1} and n ≥ k} ⊆ U . Notice that {Qn : n ≥ k} ⊆ U c ∩ Sc (f )−1 (Sc (f )(S)). So, Sc (f )−1 (Sc (f )(S)) is not discrete. Therefore, Sc (f ) is not strong light and is not finite-to-1. The example below verifies that the converse of Corollary 4.6 is not true, even if f is a bijection.

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Example 5.2. Let f be the restriction of exp to [0, 2π) onto S 1 . Observe that f is a strong light surjection

      and {f (0)} ∪ f nπ : n ∈ N ∪ f 2π − nπ : n ∈ N ∈ Sc (S 1 ) \ran(Sc (f )). Therefore, ran(Sc (f )) = Sc (S 1 ). Recall that the symbol α∈I Xα denotes the topological sum of the family {Xα : α ∈ I} of topological spaces (see [15, p. 74]). The following example shows that the converses of Corollary 4.5 and Theorems 4.15 and 4.16 fail. Example 5.3. Let us argue the existence of a first countable space X and a strong light mapping f : X → R such that f is 1-sequence-covering but Sc (f ) is not sequence-covering. First, we are going to define X. For each y ∈ R, let By be a countable base of R with the usual topology at y. Let X(y) be the topological space obtained by endowing R with the topology which has {{x} : x ∈ R \ {y}} ∪ By as a base. Notice that X(y) is a first countable Hausdorff space. Let X = {X(y) : y ∈ R} and let f : X → R be the natural projection. A point z in X will be denoted by (x, y) where f (z) = x and z ∈ X(y). In order to prove that f is a strong light mapping, let w ∈ R and let (w, y) ∈ f −1 (w) be arbitrary. Since X(y) ∈ τX and f −1 (w) ∩ X(y) = {(w, y)}, we obtain that {(w, y)} is an open subset of f −1 (w). This implies that f −1 (w) is discrete. Next, we shall prove that f is 1-sequence-covering. Let q ∈ R. Notice that (q, q) ∈ f −1 (q) and, for each Q ∈ Sc (R, q), we have that S = {(x, q) : x ∈ Q} ∈ Sc (X, (q, q)) and Sc (f )(S) = Q. Hence, Sc (R, q) ⊆ Sc (f )[Sc (X, (q, q))]. So, from Proposition 3.7, it follows that f is 1-sequence-covering. A consequence of the inclusion {{(w, z)} : w, z ∈ R and w = z} ⊆ τX is that for each R ∈ Sc (X) there exists y ∈ R satisfying that lim R = (y, y).

(†)

1  Now, suppose that S c (f ) is sequence-covering. Let Q = {0} ∪ m : m ∈ N and, for each n ∈ N, let √ Qn = q + n2 : q ∈ Q . Notice that Q ∈ Sc (R), (Qn )n∈N is a sequence in Sc (R) converging to Q and √

lim Qn = n2 for each n ∈ N. The assumption that Sc (f ) is sequence-covering implies that there exists a convergent sequence (Pn )n∈N in Sc (X) satisfying that Sc (f )(Pn ) = Qn for each n ∈ N. Let P ∈ Sc (X) be the limit point of the sequence (Pn )n∈N . The continuity of Sc (f ) guarantees that Sc (f )(P ) = Q. Apply (†) to get w ∈ R with lim P = (w, w). Then, w = f (w, w) = lim Sc (f )(P ) = lim Q = 0. Hence, lim P = (0, 0). So, from the fact that X(0) ∈ τX , we have that |P \ X(0)| < ω. Now, set A = {(x, y) ∈ P \ X(0) : x = y}, B = {(x, y) ∈ P \ X(0) : x = y} and U = {X(y) : (y, y) ∈ A ∪ {(0, 0)}} ∪ {{z} : z ∈ B}. Notice that f [A ∪ B] ⊆ Q, U ∈ [τX ]<ω and P ∈ U c . In order to end the proof, we are going to argue that U c ∩ {Pn : n ∈ N} = ∅.  Let n ∈ N. Since Pn ∈ Sc (X), the remark (†) guarantees lim Pn ∈ / {{z} : z ∈ B}. If lim Pn ∈ X(y) for √ 2 some (y, y) ∈ A ∪ {(0, 0)}, then lim P = (y, y) and so y = f (y, y) = lim S (f )(P ) = lim Q = n n n n . Hence,

  c √

√

√

√

√

we obtain that n2 , n2 ∈ A ⊆ P . This implies that n2 = f n2 , n2 ∈ Sc (f )(P ) = Q, a contradiction.  Therefore, lim Pn ∈ / U. Hence, we conclude that Pn ∈ / U c . This contradicts the fact that (Pn )n∈N is a sequence converging to P . In conclusion, Sc (f ) is not sequence-covering. Finally, since each 1-sequence-covering mapping is sequence-covering, we infer that f is sequence-covering and Sc (f ) is not 1-sequence-covering. So, the converses of Corollary 4.5 and Theorem 4.16 are not true. On the other hand, [30, Theorem 2.10, p. 1372] and the fact that R is Fréchet–Urysohn guarantee that f is almost-open. Since X is first countable, by [33, Theorem 6.6], we deduce that Sc (X) is first countable. So, if Sc (f ) were almost-open, by [19, Theorem 2.15, p. 6], Sc (f ) would be 1-sequence-covering, a contradiction. Thus, Sc (f ) is not almost-open and therefore the converse of Theorem 4.15 is not true. Our next example shows that the condition clX LX = X in Corollary 4.21 and the condition () in Corollary 4.22 are essential.

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Example 5.4. Denote by X the topological space which results of considering the ordinal ω + 1 as a linearly ordered topological space and let f : X → X be given by f (0) = ω, f (ω) = ω and f (x) = x − 1 otherwise. Observe that f is a strong light surjective mapping but it is not open. In light of Theorem 4.20, in order to prove that Sc (f ) is open, it suffices to show that each one of the following conditions holds: 1. 2. 3. 4.

{f [U ] : U ∈ τX ∧ U ∩ LX = ∅} ⊆ τX , condition (∗) of Lemma 4.18, ran(Sc (f )) is an open subset of Sc (X), and f is hereditarily sequence-covering.

First, observe that LX = {ω}. If U ∈ τX satisfies that ω ∈ U , then there exists m ∈ N such that (ω + 1) \ m ⊆ U . Thus, we have that f [U ] = ((ω + 1) \ (m − 1)) ∪ {x − 1 : x ∈ U ∩ m \ {0}} ∈ τX . This proves that (1) holds. − Note that Sc (f )[{0}− c ] ⊆ Sc (X, f (0)). Therefore, if we let x ∈ X \ LX = ω be such that Sc (f )[{x}c ] \ Sc (X, f (x)) = ∅, we get that x = 0 and so, {f (x)} ∈ τX . Hence, (2) is true. Next, since ran(Sc (f )) = Sc (X), condition (3) is satisfied. Finally, in order to prove that (4) holds, let U ∈ τX be such that U ∩ LX = ∅ and let Q ∈ (f [U ])+ c ∩ ran(Sc (f )) be arbitrary. Then, by letting P = {ω} ∪ {q + 1 : q ∈ Q ∩ ω} we obtain that P ∈ Uc+ and Sc (P ) = Q. In conclusion, Sc (f ) is hereditarily sequence covering. A topological property P is called Sc -preserved if whenever a space X has property P , so does Sc (X). Connectedness and local connectedness are Sc -preserved properties but normality, being a Fréchet–Urysohn space and path connectedness are not (see [33] and [18, Example 2.8, p. 800]). Given a cardinal number κ, endow κ with the discrete topology and set S(κ) = ((ω + 1) × κ)/({ω} × κ), where the ordinal ω + 1 is considered as a linearly ordered topological space, i.e., S(κ) is the quotient space of the topological product (ω + 1) × κ which results of collapsing the set {ω} × κ to a single point. Usually, S(κ) is called the sequential fan of κ spines. Now, let Z = (ω × ω) ∪ {∞} where ∞ ∈ / ω × ω. Make Z a topological space by declaring all points of ω × ω as isolated and by letting all sets of the form {∞} ∪ ((ω \ k) × ω), where k ∈ ω, be a local base for Z at ∞. Observe that Z is first countable and thus, it is sequential (see [15, Theorem 1.6.14, p. 53]). Proposition 5.5. Being sequential is not a Sc -preserved property. Proof. Define Y = (ω + 1) ⊕ S(ω) ⊕ Z, where ω + 1 is considered as a linearly ordered space. We shall show that Y is sequential but Sc (Y ) is not sequential. First, the fact that ω + 1, Z and S(ω) are Fréchet–Urysohn and [16, Proposition 2.1, p. 113] imply that Y is Fréchet–Urysohn, in particular, sequential. Notice that S(ω) and Z are closed subsets of Y and (ω + 1) ∩ (S(ω) ∪ Z) = ∅. Then, by [33, Lemma 3.4], F1 (S(ω)) × F1 (Z) embeds as a closed subspace of Sc (Y ). Since all closed subspaces of a sequential space are sequential (see [16, Proposition 1.90, p. 110]) and F1 (S(ω)) × F1 (Z) is homeomorphic to S(ω) × Z, which is not sequential (see Lemma and the argument given in Example 2 of [36, pp. 163, 164]), we deduce that Sc (Y ) is not sequential. 2 The example below shows that the converses of Theorems 4.8 and 4.11 do not hold.

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Example 5.6. There are spaces X, Y and a map f : X → Y in such a way that Sc (X) is first countable, Y is Fréchet–Urysohn, f is strong light and pseudo-open, but Sc (f ) is not quotient. Let Y be as in the proof of Proposition 5.5. As we argued there, Y is Fréchet–Urysohn but Sc (Y ) is not sequential. Now, set A = {S(i) : i ∈ ω} ∪ {Z, ω + 1} and define X = A, where ω + 1 is considered as a linearly ordered space. Notice that X is first countable. So, [33, Corollary 5.8] guarantees that Sc (X) is first countable. In order to define the mapping f : X → Y , let ρ be the identification mapping from (ω + 1) × ω onto S(ω) and, for each i ∈ ω, let gi : S(i) → S(ω) be an embedding such that gi [S(i)] = ρ[(ω + 1) × i]. Define f : X → Y by x, if x ∈ Z ∪ (ω + 1), f (x) = gi (x), if i ∈ ω is such that x ∈ S(i). Notice that f is continuous, strong light, pseudo-open and sequence-covering. Hence, by Theorem 4.4, ran(Sc (f )) = Sc (Y ). Finally, from the fact that each first countable space is sequential and [16, Proposition 1.2, p. 109], it follows that Sc (f ) is not quotient. In Theorem 4.30 the assumption ran(Sc (f )) = Sc (Y ) is essential, as we now show (recall Theorem 4.27). Example 5.7. The restriction f of exp to [0, 2π) is 1-to-1 and onto but it is not open. Observe that {f (0)} ∪ {f (2−n π) : n ∈ ω} ∪ {f (2π − 2−n π) : n ∈ ω} ∈ Sc (S 1 ) \ ran(Sc (f )). Our last example shows that the converse of Theorem 4.34 fails. Example 5.8. If f : [0, 4π] → S 1 is the corresponding restriction of exp, then f is a closed strong light mapping. In order to prove that Sc (f ) is not a closed mapping, let F = [0, 2π]+ c . Observe that F is a closed subset of Sc ([0, 4π]). Now, we will verify that Sc (f )[F] is not a closed subset of Sc (S 1 ). Set Q = {f (2π)} ∪

      f 2π − πn : n ∈ N ∪ f 2π + nπ : n ∈ N . Notice that Q ∈ Sc (S 1 ) \ Sc (f )[F]. We are going to check that  Q ∈ clSc (S 1 ) (Sc (f )[F]). Let U ∈ C(S 1 ) be such that Q ∈ U c (see Proposition 2.1). Let h : U → U be    <ω a function such that h(U ) ∈ U for every U ∈ U. Since f [0, 2π] = S 1 , there exists A ∈ [0, 2π] such

 that f [A] = ran(h). Set S = 2π − πn : n ∈ N ∪ {2π} ∪ A. We have that S ∈ F and Sc (f )(S) ∈ U c . Thus, Sc (f )[F] ∩ U c = ∅. This proves that Sc (f )[F] is not a closed subset of Sc (S 1 ). We finish this paper posing some questions. The first two are related to Theorems 4.23 and 4.34. Question 5.9. Is there a strong light and open mapping f such that Sc (f ) is not open? Question 5.10. Are there spaces X and Y and a strong light mapping f from X into Y such that |f [LX ]| = 1, Sc (f ) is closed and f is not closed? A mapping f is called 2-sequence-covering if for each y ∈ Y and for each x ∈ f −1 (y), whenever (yn )n∈ω is a sequence converging to y in Y , there exists a sequence (xn )n∈ω in X converging to x such that xn ∈ f −1 (yn ) for each n ∈ ω (see [29]). From [19, (2) of Proposition 2.9, Proposition 2.10, Theorem 2.11, pp. 4–5] and [33, Theorem 6.6, p. 16], it follows that if X is first countable, Y is sequential and f is 2-sequence-covering, then Sc (f ) is 2-sequencecovering. The question below arises from this statement. Question 5.11. In general, does the fact that a strong light map f is 2-sequence-covering imply that Sc (f ) is 2-sequence-covering?

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Question 5.12. Under which conditions is a strong light map f 2-sequence-covering provided that Sc (f ) is 2-sequence-covering? A mapping f : X → Y is called: (i) OM if there exist a space Z, a monotone mapping h : X → Z and an open mapping g : Z → Y in such a way that f = g ◦ h; (ii) MO if there exist a space Z, an open mapping h : X → Z and a monotone mapping g : Z → Y satisfying that f = g ◦ h; (iii) confluent provided that for each closed connected subset Q of Y and for each component K of f −1 [Q], we have that f [K] = Q; (iv) weakly confluent if for each closed connected subset Q of Y there exists a component K of f −1 [Q] satisfying the equality f [K] = Q. Problem 5.13. Let M be one of the following classes of mappings: OM, MO, confluent or weakly confluent. Find the positive implications among the statements (a) f ∈ M and (b) Sc (f ) ∈ M. References [1] J.G. Anaya, F. Capulín, D. Maya, F. Orozco-Zitli, Induced mappings on symmetric products of continua, Topol. Appl. 214 (2016) 100–108. [2] J. Camargo, On the semi-open induced mappings, Topol. Proc. 32 (2008) 145–152. [3] J. Camargo, Openness of the induced map Cn (f ), Bol. Mat. 16 (2) (2009) 115–123. [4] J. Camargo, Some relationships between induced mappings, Topol. Appl. 157 (2010) 2038–2047. [5] J. Camargo, Lightness of the induced maps and homeomorphisms, Can. Math. Bull. 54 (4) (2011) 607–618. [6] J. Camargo, On the induced MO-mappings between arcs and simple closed curves, Mat. Enseñ. Univ. 19 (1) (2011) 1–11. [7] J.J. Charatonik, Recent results on induced mappings between hyperspaces of continua, Topol. Proc. 22 (1997) 103–122. [8] J.J. Charatonik, W.J. Charatonik, Hereditarily weakly confluent mappings are homeomorphisms, Colloq. Math. 75 (1998) 195–203. [9] J.J. Charatonik, W.J. Charatonik, Lightness of the induced mappings, Tsukuba J. Math. 22 (1998) 179–192. [10] J.J. Charatonik, W.J. Charatonik, Induced MO-mappings, Tsukuba J. Math. 23 (1999) 245–252. [11] J.J. Charatonik, W.J. Charatonik, Limit properties of induced mappings, Topol. Appl. 100 (2000) 103–118. [12] J.J. Charatonik, W.J. Charatonik, A. Illanes, Openness of induced mappings, Topol. Appl. 98 (1999) 67–80. [13] J.J. Charatonik, A. Illanes, S. Macías, Induced mappings on the hyperspaces Cn (X) of a continuum X, Houst. J. Math. 28 (2002) 781–805. [14] W.J. Charatonik, Openness and monotoneity of induced mappings, Proc. Am. Math. Soc. 127 (1999) 3729–3731. [15] R. Engelking, General Topology, Sigma Series in Pure Mathematics, vol. 6, Heldermann Verlag, Berlin, 1989, translated from Polish by the author. [16] S.P. Franklin, Spaces in which sequences suffice, Fundam. Math. 1 (57) (1965) 107–115. [17] S.P. Franklin, Spaces in which sequences suffice II, Fundam. Math. 61 (1) (1967) 51–56. [18] S. García-Ferreira, Y.F. Ortiz-Castillo, The hyperspace of convergent sequences, Topol. Appl. 196 (2015) 795–804. [19] Y. Ge, Weak forms of open mappings and strong forms of sequence-covering mappings, Mat. Vesn. 59 (1) (2007) 1–8. [20] G. Higuera, A. Illanes, Induced mappings on symmetric products, Topol. Proc. 37 (2011) 367–401. [21] H. Hosokawa, Induced mappings between hyperspaces, Bull. Tokyo Gakugei Univ. 41 (4) (1989) 1–6. [22] H. Hosokawa, Mappings of hyperspaces induced by refinable mappings, Bull. Tokyo Gakugei Univ. 42 (4) (1990) 1–8. [23] H. Hosokawa, Induced mappings between hyperspaces II, Bull. Tokyo Gakugei Univ. 44 (4) (1992) 1–7. [24] H. Hosokawa, Induced mappings on hyperspaces, Tsukuba J. Math. 21 (1997) 239–250. [25] H. Hosokawa, Induced mappings on hyperspaces II, Tsukuba J. Math. 21 (1997) 773–783. [26] A. Illanes, The openness of induced mappings on hyperspaces, Colloq. Math. 74 (1997) 219–224. [27] K. Kunen, Set Theory. An Introduction to Independence Proofs, Studies in Logic and the Foundations of Mathematics, vol. 102, North-Holland Publishing Co., Amsterdam, 1980. [28] A.Y.W. Lau, A note on monotone maps and hyperspaces, Bull. Pol. Acad. Sci. Ser. Sci. Math. Astron. Phys. 24 (1976) 121–123. [29] S. Lin, On sequence-covering s-mappings, Adv. Math. (China) 25 (1996) 548–551. [30] S. Lin, K. Li, Y. Ge, Convergent-sequence spaces and sequence-covering mappings, Houst. J. Math. 39 (4) (2013) 1367–1383. [31] M. de, J. López, S. Macías, Induced mappings on n-fold hyperspaces, Houst. J. Math. 33 (2007) 1047–1057. [32] D. Maya, P. Pellicer-Covarrubias, R. Pichardo-Mendoza, Cardinal functions of the hyperspace of convergent sequences, Math. Slovaca (2017), in press. [33] D. Maya, P. Pellicer-Covarrubias, R. Pichardo-Mendoza, General properties of the hyperspace of convergent sequences, in: Topology Proceedings, submitted for publication.

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