A characterization of inducible mappings between hyperspaces

1. Introduction

A continuum is a nonempty, compact, connected metric space. Given a continuum X and a positive integer n, consider the following hyperspaces of X:

All these hyperspaces are considered with the Hausdorff metric (see [4, p. 11]). In this paper, the phrase H(X) is a hyperspace of a continuum X means that H(X) is one of the hyperspaces defined above for X. Readers specially interested in the structure of these hyperspaces are referred to [4].

The word mapping stands for a continuous function. Given a mapping between continua f : X → Y and a hyperspaces H(X) and H(Y ) of X and of Y , respectively, the H-induced function by f, H(f) : H(X) → H(Y ), is defined by

By [5, 5.10.1 of Theorem 5.10, p. 170], the continuity of f implies that of the induced function to the hyperspace 2 ^X , 2 ^f , and since each H(f) = 2 ^f | _H(X) , we deduce that H(f) is also a mapping.

A mapping g between hyperspaces H(X) and H(Y ) of continua X and Y , respectively, is called inducible provided that there exists a mapping f : X → Y such that g = H(f).

In [2, Theorem 2.2, p. 7], the authors characterize the inducible mappings for the case that H(X) ∈ {2 ^X , C ₁(X)}. After, this characterization is extended for the case that H(X) ∈ {C_n (X) : n ∈ } in [3, Theorem 49, p. 802]. Finally, in [1, Theorem 5.2, p. 256], the characterization of this class of mapping is generalized for every hyperspace H(X) listed above. However, the authors do not show that those three necessary and sufficient conditions are independent in the sense that each two of them do not imply the other one when H(X) ∈ {C_n (X), F_n (X), F_∞ (X) : n ≥ 2}. On the other hand, an important tool in all these characterizations is an order in the set of all mappings between hyperspaces defined in [2, p. 6], so they are based in an extrinsic property of mappings between hyperspaces. In the current paper, a new characterization of the inducible mappings, using only intrinsic properties, will be given, and we present examples to prove the independence of our conditions and the previous ones to do complete the study of this class of mappings.

The study of mappings between hyperspace has been focused to induced mappings. The characterizations of inducible mappings provide information on mappings between hyper[1]space that could be used to produce a suitable mapping between hyperspace satisfying a specific property that has not been possible to find between continua, due to the relation between the induced mapping and the mapping between the ground spaces that induces it.

2. Preliminaries and auxiliary results

The symbol denotes the set of all positive integers. A relation from a set X to a set Y is a subset of X × Y . The domain of a relation φ, represented by dom(φ), is the set {x ∈ X : ({x} × Y ) ∩ φ ∅}. For a subset A of X, let φ[A] = {y ∈ Y : (A × {y}) ∩ φ ∅}. Finally, recall that a function f from X to Y can be related with the relation {(x, f(x)) : x ∈ X}, and a relation φ is a function provided that the condition (a, b),(a, c) ∈ φ implies that b = c.

Let X be a continuum. Define

Let x ∈ X and H(X) ∈ (X). The set {A ∈ H(X) : x ∈ A} is denoted by H(x, X).

Let X and Y be continua, and let H(X) ∈ (X) and H(Y ) ∈ (Y ). Given a mapping g : H(X) → H(Y ), define the relation φ_g from X to Y as follows:

Lemma 2.1. Let X and Y be continua and let H(X) ∈ (X) and H(Y ) ∈ (Y ). If g : H(X) → H(Y ) is a mapping, then φ_g [A] ⊆ g(A), for every A ∈ H(X).

Proof. Let A ∈ H(X) and let y ∈ φ_g [A]. Then, there exists x ∈ A such that (x, y) ∈ φ_g . So, by the definition of φ_g , we obtain that y ∈ g(A).

3. Main result

The following result allows us to characterize all mapping between hyperspace that are induced by a mapping between the ground spaces.

Theorem 3.1. Let X and Y be continua, and let H(X) ∈ (X) and H(Y ) ∈ (Y ). A mapping g : H(X) → H(Y ) is inducible if, and only if, g satisfies each one of the following conditions:

(1) _g φ_g is a function such that dom(φ_g ) = X; and

(2) _g g ⁻¹ [H(y, Y )] ⊆ {A ∈ H(X) : y ∈ φ_g [A]}, for every y ∈ Y .

Proof. Assume that there exists a mapping f : X → Y such that g = H(f). In order to see that (1) _H(f) is satisfied, we are going to verify that f = φ_{H (f)} . The definition of H(f) implies that f(x) ∈ H(f)(A) for every x ∈ X and for every A ∈ H(x, X). So, {(x, f(x)) : x ∈ X} ⊆ φ_{H (f)} . Next, let (w, z) ∈ φ_{H (f)} . From the definition of φ_{H (f)} , we get that z ∈ H(f)({w}) = {f(w)}. Thus, z = f(w). This proves that f = _φH(f) . Now, let y ∈ Y and B ∈ H(f)⁻¹ [H(y, Y )]. Hence, y ∈ H(f)(B) = {f(b) : b ∈ B} = φ_{H (f)} [B]. Thus, (2) _H(f) holds.

In order to show the second part, let A ∈ H(X) and let y ∈ g(A) be arbitrary. Then, A ∈ g ⁻¹ [H(y, Y )]. By (2) _g , we get that y ∈ φ_g [A]. So, g(A) ⊆ φ_g [A]. This and Lemma 2.1 together imply that g(A) = φ_g [A] for every A ∈ H(X) and, since (1) _g is true, we have that g({x}) = φ_g [{x}] = {φ_g (x)} for every x ∈ X. Thus, dom(φ_g ) = X and, from the fact that X and F ₁(X) are isometric, it follows that the continuity of g implies that of φ_g . Therefore, φ_g : X → Y is a mapping such that g = H(φ_g ).

4. Comparation with the first characterization

In this section all the possible interrelations between the conditions of Theorem 3.1 and the conditions of [2, Theorem 2.2, p. 7] will be verified, obtaining one more characteri[1]zation of inducible mappings more.

Define the order ≺ on the set of all mapping between hyperspaces H(X) and H(Y ) as follows: h ≺ g if, and only if, h(A) ⊆ g(A), for every A ∈ H(X) (see [2, p. 6]).

The result below is the general version of [2, Theorem 2.2, p. 7] presented in [1, Theorem 5.2, p. 256].

Theorem 4.1. Let X and Y be continua, and let H(X) ∈ (X) and H(Y ) ∈ (Y ). A mapping g : H(X) → H(Y ) is inducible if, and only if, each one of the following conditions is satisfied:

(I) _g g[F ₁(X)] ⊆ F ₁(Y );

(II) _g A ⊆ B implies g(A) ⊆ g(B), for every A, B ∈ H(X);

(III) _g g is minimal with respect to the order ≺, i. e., if a mapping h : H(X) → H(Y ) satisfies (II) _h and h ≺ g, then h = g.

Theorem 4.2. Let X and Y be continua, let H(X) ∈ (X) and H(Y ) ∈ (Y ), and let g : H(X) → H(Y ) be a mapping. If (I) _g and (II) _g are satisfied, then (1) _g holds.

Proof. Let us show that φ_g = {(x, y) ∈ X × Y : g({x}) = {y}}.

First, let (x, y) ∈ φ_g be arbitrary. Then y ∈ g({x}). This and condition (I) _g together imply that g({x}) = {y}. So, the inclusion φ_g ⊆ {(x, y) ∈ X × Y : g({x}) = {y}} holds.

Next, let (x, y) ∈ X × Y be such that g({x}) = {y} and let A ∈ H(x, X). A consequence of condition (II) _g is that g({x}) ⊆ g(A). This implies that y ∈ g(A). So, we conclude that (x, y) ∈ φ_g .

Therefore, φ_g is a function.

Example 5.1 proves that the converse of Theorem 4.2 is false.

Theorem 4.3. Let X and Y be continua, let H(X) ∈ (X) and H(Y ) ∈ (Y ), and let g : H(X) → H(Y ) be a mapping. If (2) _g is satisfied, then (II) _g holds.

Proof. Let A, B ∈ H(X) be such that A ⊆ B and let y ∈ g(A). Condition (2) _g guarantees that there exists x ∈ A such that (x, y) ∈ φ_g . This and our assumption A ⊆ B imply that y ∈ φ_g [B]. Finally, Lemma 2.1 ensures that φ_g [B] ⊆ g(B). Then y ∈ g(B).

The converse of Theorem 4.3 fails (see examples 5.5, 5.7 and 5.8).

Corollary 4.4. Let X and Y be continua, and let H(X) ∈ (X) and H(Y ) ∈ (Y ). A mapping g : H(X) → H(Y ) is inducible if, and only if, each one of the following conditions is satisfied:

(I) _g g(F ₁(X)) ⊆ F ₁(Y );

(2) _g g ⁻¹ [H(y, Y )] ⊆ {A ∈ H(X) : y ∈ φ_g [A]}, for every y ∈ Y .

Proof. First, by theorems 3.1 and 4.1, if g is inducible, then (I) _g and (2) _g holds.

Now, suppose that g satisfies the conditions (I) _g and (2) _g . Theorem 4.3 implies that (II) _g holds. Hence, by Theorem 4.2, we have that (1) _g is satisfied. Finally, from Theorem 3.1, it follows that g is inducible.

5. Examples

The aim of this section is to show that conditions (1) _g , (2) _g , (I) _g and (III) _g are independent in the sense that none of them is implied by the other. Also, it will be proved that condition (II)g does not imply any condition of Theorem 3.1 and (II) _g is not implied by (1) _g .

Our first example proves that the conditions (I) _g , (II) _g , (III) _g and (2) _g are not implied by (1) _g .

Example 5.1. Let H([0, 1]) ∈ ([0, 1]) − {C ₁([0, 1])} and let g : H([0, 1]) → H([0, 1]) be defined by

for each A ∈ H([0, 1]). Notice that g is a mapping. First, to see that (1) _g is satisfied, we shall prove that φ_g = [0, 1] × {0}. From the definition of φ_g , it follows that the inclusion [0, 1]× {0} ⊆ φ_g holds. On the other hand, if (a, b) ∈ φ_g , then b ∈ φ_g ({0, a}) = {0, min{0, a}} = {0}. Hence, we conclude that φ_g = [0, 1] × {0}.

Next, we are going to argue that (I)g, (II) _g , (III) _g and (2) _g fail. Set B = {1} and C = {0, 1}. Observe that g(B) = {0, 1} and g(C) = {0}. Since B ∈ F ₁([0, 1]) and g(B) F ₁([0, 1]), we have that (I) _g fails. Notice that B is a subset of C but g(B) is not contained in g(C). Then (II) _g is not satisfied. From this and Theorem 4.3, we infer that (2) _g does not hold. Observe that the constant mapping h : H([0, 1]) → H([0, 1]) defined by h(A) = {0} for each A ∈ H([0, 1]) fulfils (II) _h , h ≺ g and h g. Then (III) _g fails.

Finally, in the case that H([0, 1]) = C₁ ([0, 1]), we define g : C ₁([0, 1]) → C ₁([0, 1]) by

for each A ∈ C ₁([0, 1]). Similar arguments above prove that (1) _g holds and (2) _g , (I) _g , (II) _g and (III) _g fail.

The second example shows that each one of the following statements holds for each H(X) ∈ (X):

(a) (2) _g does not imply (1) _g .

(b) (2) _g does not imply neither (I) _g nor (III) _g .

(c) (II) _g does not imply (1) _g .

(d) (II) _g does not imply neither (I) _g nor (III) _g .

Example 5.2. Let X and Y be continua and let H(X) ∈ (X). Fix B ∈ H(Y ) − F ₁(Y ). Let g : H(X) → H(Y ) be the mapping defined by

for each A ∈ H(X). Notice that g is continuous. We are going to prove that (2) _g and (II) _g are satisfied and the conditions (1) _g , (I) _g and (III) _g are not.

Notice that φ_g = X × B. Hence, φ_g is not a function and so (1) _g is not satified. A consequence of the fact that g[F ₁(X)] ∩ F ₁(Y ) = ∅ is that (I) _g does not hold. Now, let y ∈ Y . Notice that {A ∈ H(X) : y ∈ φ_g [A]} = H(X) if y ∈ B, and g ⁻¹ [H(y, Y )] = ∅ otherwise. Using the two last equalities, we get that g satisfies (2) ^g . Theorem 4.3 guarantees that (II) _g holds. Finally, fix b ∈ B. The mapping h : H(X) → H(Y ) defined by h(A) = {b} fulfils that h ≺ g, (II) _h holds and h g. In conclusion, (III) _g is not satisfied.

The example below proves that, for each H([0, 1]) ∈ ([0, 1]), there exists a mapping g : H([0, 1]) → H([0, 1]) such that g satisfies (I) _g and (III) _g , but not (1) _g , (2) _g and (II) _g .

Example 5.3. Let H([0, 1]) ∈ ([0, 1]). Define g : H([0, 1]) → H([0, 1]) as follows: for each A ∈ H([0, 1]), let

In order to prove that the function g is a mapping, define l : 2^[0,1] → 2^[0,2] by l(A) = {t + max A − min A : t ∈ A} to get a mapping. Then g(A) = {min{1, s} : s ∈ l(A)}, for every A ∈ H([0, 1]). Hence, g is a mapping. Let us argue that (I) _g and (III) _g are satisfied but (1) _g , (2) _g and (II) _g are not.

For each x ∈ [0, 1], we have that g({x}) = {x}. Thus, the inclusion g[F ₁([0, 1])] ⊆ F ₁([0, 1]) holds. Hence (I) _g is satisfied. Now, in order to see that (III) _g holds and (1) _g , (2) _g and (II) _g do not hold, set B = {0} and let C ∈ H([0, 1]) be such that min C = 0 and max C = 1. Observe that g(B) = B and g(C) = {1}. Then g(B) is not contained in g(C). This shows that (II) _g fails. Hence, from Theorem 4.3, it follows that (2) _g is not true. Since there is no mapping h : H([0, 1]) → H([0, 1]) satisfying h ≺ g and (II) _h , we have that (III) _g is vacuously true. Finally, to prove that (1) _g fails, let b ∈ [0, 1] be such that (0, b) ∈ φ_g. By the definition of φ_g , we deduce that b ∈ g(B) and b ∈ g(C). So, b = 0 and b = 1. Hence, φ_g is not a function and so (1) _g does not hold.

The condition (III) _g does not imply any of the conditions (1) _g , (2) _g and (II) _g for each H(X) ∈ (X) (compare [2, Example 3.2]).

Example 5.4. Let H([0, 1]) ∈ ([0, 1]). Define g : H([0, 1]) → H([0, 1]) by

In [2, Example 3.2], the authors prove that (I) _g and (III) _g are satisfied, while the condition (II) _g does not hold. Hence, from Theorem 4.3, it follows that g does not fulfil (2) _g .

In order to verify that (1) _g does not hold, suppose to the contrary that φ_g is function. Let b ∈ [0, 1] be such that ( , b) ∈ φ_g. The definition of φ_g ensures that b ∈ g({ }) = { }, and so b = . Now, take A ∈ H( , [0, 1]) ∩ H(0, [0, 1]). Then b ∈ g(A) = {0}. This is a contradiction. In conclusion, (1) _g does not hold.

The last examples are not valid for all hyperspaces in (X). In each one of them, we indicate for which hyperspaces in (X) the example works.

In the case that H(X) ∈ (X) − {F_n (X) : n ∈ ∪ {∞}}, the example below shows that (II) _g does not imply (2) _g and (I) _g does not imply (III) _g (compare [2, Example 3.1, p. 8]).

Example 5.5. Let H([0, 1]) ∈ {2^[0,1] , C _∞([0, 1])} ∪ {C_n ([0, 1]) : n ∈ }. Define g : H([0, 1]) → H([0, 1]) by

Notice that g is a mapping. Now, arguments in [2, Example 3.1, p. 8] prove that (I) _g and (II) _g hold and (III) _g is not satisfied. Finally, by Theorem 4.1, g is not inducible. Therefore, by Corollary 4.4, the condition (2) _g must fails.

The mapping g in the example below proves that (I) _g does not implies (III) _g when H(X) ∈ {F_n (X) : n ∈ (− {2}) ∪ {∞}}.

Example 5.6. Let H([0, 1]) ∈ {F_n ([0, 1]) : n ∈ (− {2}) ∪ {∞}} and let g : H([0, 1]) → H([0, 1]) be defined by g(A) = {0, max A − min A}. Then g is a mapping.

First, if t ∈ [0, 1], then g({t}) = {0, t − t} = {0}. So, g[F ₁([0, 1])] = {{0}} ⊆ F ₁([0, 1]). We obtain that (I) _g is satisfied. Now, in order to prove that (III) _g does not hold, let h : H([0, 1]) → H([0, 1]) be defined by h(A) = {0}. We have that h is a mapping, (II) _h is satisfied and h ≺ g. Therefore, g is not minimal with respect to ≺, in other words, (III) _g does not hold.

Our next example shows that (2) _g is not implied by (II) _g when H(X) = F_n (X) for some n ≥ 2.

Example 5.7. Let n ≥ 2. Define g : F_n ([0, 1]) → F_n ([0, 1]) by

Notice that g is a mapping.

Now, observe that g(A) ⊆ g(B) for each A, B ∈ F_n ([0, 1]) such that A ⊆ B and g({x}) = {0}, for each x ∈ [0, 1]. This means (I) _g and (II) _g hold. In light of Theorem 4.2, (1) _g is satisfied. We have that φ_g = [0, 1] × {0}.

Notice that { : m ∈ {0, 1, . . . , n − 1}} ∈ g ⁻¹ [F_n ( , [0, 1])] and {A ∈ F_n ([0, 1]) : ∈ φ_g [A]} = ∅. Hence, (2) _g fails.

Our next example exhibit a mapping g from F _∞([0, 1]) into itself satisfying (II) _g , while (2) _g fails.

Example 5.8. Define g : F _∞([0, 1]) → F _∞([0, 1]) by g(A) = {|x − y| : x, y ∈ A} for each A ∈ F _∞([0, 1]). Observe that g is a mapping.

Next, from the definition of g, it follows that (II) _g is satisfied and [0, 1]× {0} ⊆ φ_g . Note that if (x, y) ∈ φ_g , then y ∈ g({x}) = {0}, and so, we conclude that φ_g = [0, 1] × {0}. Then, g fulfils (1) _g .

Finally, since {0, 1} ∈ g ⁻¹ [F _∞(1, [0, 1])] and {A ∈ F _∞([0, 1]) : 1 ∈ φ_g [A]} = ∅, we obtain that (2) _g fails.

The next example shows that there exists a continuum X for which the statement (III) _g does not imply (I) _g holds for every H(X) ∈ (X) − {C ₁(X)}.

Example 5.9. Let S ¹ be the set of all complex numbers having norm equal to 1. Consider the mapping exp : → S ¹ defined by exp(s) = cos(2πs) + isin(2πs). Let H([0, 1]) ∈ ([0, 1]) − {C ₁([0, 1])} and H(S ¹) ∈ (S ¹) − {C ₁(S ¹)}. Define g : H([0, 1]) → H(S ¹) by

Notice that g is a mapping. Now, we will show that g satisfies (III) _g , while the condition (I) _g does not hold.

First, for each x ∈ [0, 1], by the definition of g, we have that g({x}) = {exp(x), exp(x + )}. So, (I) _g does not hold.

Now, let h : H([0, 1]) → H(S ¹) be a mapping such that h ≺ g. Let A ∈ H(X) be such that 0 ∈ A and max A = . Observe that the equality g(A) ∩ g({0}) = {i, −i} ∩ {1, −1} = ∅ implies that h(A) ∩ h({0}) = ∅. This shows that (II) _h fails. So we have that (III) _g is vacuously true.

Example described in [2, Example 3.4] can be used to argue that the condition (III) _g does not imply (I) _g when H(X) = C ₁(X).

Acknowledgments

The authors wish to thank the referees for their valuable comments and suggestions, which helped to improve greatly this paper.

References

Author notes

dmayae@uaemex.mx

Additional information

To cite this article: J.G. Anaya, D. Maya and F. Orozco-Zitli, A characterization of inducible mappings between hyperspaces, Rev. Integr. temas mat. 38 (2020), No. 2, 109-117. doi: 10.18273/revint.v38n2-2020004

Revista Integración
ISSN: 0120-419X
Vol. 38
Num. 2
Año. 2020

A characterization of inducible mappings between hyperspaces

JoséDavidFernando G. AnayMayaOrozco-Zitl
Universidad Autónoma del Estado de MéxicoUniversidad Autónoma del Estado de MéxicoUniversidad Autónoma del Estado de México,MéxicoMéxicoMéxico