1. Introduction and Preliminaries
The concept of quasihomeomorphism was first introduced by the Grothendieck school [
1,
2]. This notion is mainly used in algebraic geometry, and it was shown that this concept arises naturally in the theory of some foliations associated to closed connected manifolds [
1].
We recall that subset L of topological space X is called locally closed if it is an intersection of an open set and a closed set of X. We take , , and as the families of all open, closed, and locally closed subsets of X, respectively; we call a continuous map a quasihomeomorphism if represents a bijection from (resp., , ) to (resp., , ).
Topological space
X is called Whyburn [
3] if for every non-closed subset
A of
X and for every
there exists
such that
or, equivalently, there exists
such that
. It is called weakly Whyburn [
4] if for every non-closed subset
A of
X there exists
such that
or, equivalently, there exists
such that
. It was illustrated that Whyburn space is weakly Whyburn, whereas the converse side is not always true; see [
5], Theorem 3.8.
A door space is a topological space in which every subset is either open or closed. By a submaximal space, we mean a topological space in which every subset is locally closed or, equivalently, every dense subset is open.
A principal space, which is also recognized as Alexandroff space, is a topological space in which any intersection of open sets is open. The most fundamental property of Alexandroff spaces is that the category of Alexandroff spaces is isomorphic to the category of qosets. We let Alx denote the category of Alexandroff spaces and Qos the category of quasi-ordered sets. We let be the map defined by where is the specialization topology defined by choosing, as a basis of , where is called the upset determined by x. In this context, the closure is exactly the downset . Therefore, it is clear that is an Alexandroff space. Similarly, we define by where , called corresponding specialization quasi-order, is defined by for any if and only if . Clearly, and are inverse maps one of the other, which means that considering an Alexandroff space is equivalent to considering a qoset.
In this paper, we detect topological properties preserved by quasihomeomorphisms. We show that the Whyburn (resp., weakly Whyburn, Submaximal, door) property is preserved by quasihomeomorphisms. Furthermore, we lay bare the necessary conditions on continuous map where Y is Whyburn (resp., weakly Whyburn) to ensure that X is Whyburn (resp., weakly Whyburn).
This paper is organized as follows. In
Section 2, we show that quasihomeomorphisms preserve the properties of being Whyburn, weakly Whyburn, submaximal, and door spaces. We devote
Section 3 to revealing the necessary conditions on continuous map
where
Y is Whyburn (resp., weakly Whyburn, submaximal, door) in order to assure that
X is Whyburn (resp., weakly Whyburn, submaximal, door). Finally, in
Section 4, we prove that given a quasihomeomorphism,
between two Alexandroff spaces. Then, if
q is an onto (resp., one-to-one) and
X (resp.,
Y) is strongly primal, so is
Y (resp.,
X).
2. Quasihomeomorphisms and Some Topological Properties
In this section, we detect topological properties preserved by quasihomeomorphisms.
Firstly, let us start by recalling the following results.
Lemma 1 ([
6]).
We let be a continuous onto map. Then, the following statements are equivalent:- 1.
q is a quasihomeomorphism;
- 2.
q is open and equality holds for any open set U in X;
- 3.
q is closed and equality holds for any closed set F in X.
Lemma 2 ([
7]).
If is a quasihomeomorphism, then the following statements are equivalent:- 1.
q is onto;
- 2.
for any set A in Y.
Theorem 1. We let be an onto quasihomeomorphism. If X is Whyburn, then Y is Whyburn.
Proof. We let A be a non-closed subset of Y and . Since q is an onto, there exists such that . It can be seen easily that . Using the fact that X is Whyburn, we let B be a subset of that satisfies . Let us show that . Since q is continuous and closed, we obtain and thus . Now, we suppose that ; then, is a contradiction, so that .
Conversely, let , then . Now , so . Finally and consequently . □
In the following example, we show that the surjectivity of quasihomeomorphism q is necessary to conclude that Y is Whyburn.
Example 1. We let be a finite space whose open sets are and and let equipped with topology . Then, X is a Whyburn space. In contrast, is not a closed subset of Y and . We notice that subset B of satisfying does not exist, so Y is not Whyburn. Now, we let be the canonical injection. Clearly, q is a quasihomeomorphism that is not an onto.
The following example shows that if in Theorem 1, q is not a quasihomeomorphism, then Y need not be a Whyburn space.
Example 2. We let , equipped with the topology whose basis of open sets is We let be a space whose basis of open sets is Then, X is a Whyburn space and Y is not Whyburn. We let be defined by and . Hence, q is an onto continuous map and whenever O is an open set in Y. Thus, q is not an quasihomeomorphism.
Next, we show that the property weakly Whyburn is preserved by onto quasihomeomorphisms.
Theorem 2. We let be an onto quasihomeomorphism. If X is weakly Whyburn, then Y is weakly Whyburn.
Proof. We let
A be a non-closed subset of
Y. Since
q is an onto quasihomeomorphism, then
is a non-closed subset of
X. Regarding
X being weakly Whyburn, we consider set
satisfying
. Hence,
Therefore, Y is weakly Whyburn. □
Remark 1. - (1)
The fact that q is a quasihimeomorphism in Theorem 2 is necessary as shown by Example 2. Of course, X is weakly Whyburn. However, Y is not. Indeed,
- (2)
The condition of suyjectivity in Theorem 2 is necessary as shown by Example 1.
Theorem 3. We let be an onto quasihomeomorphism. If X is submaximal, then Y is submaximal.
Proof. We let B be a dense subset of Y. Then, . Since q is an onto quasihomeomorphism, we obtain . Hence, is an open subset of X and is open in Y. Therefore, Y is a submaximal space. □
Example 3. We let be equipped with topology . Then, the dense subset is open in X and thus X is submaximal. Now, we let be a finite space whose open sets are and . Then, we have as a dense subset of Y and it is not open. We let be the canonical injection. It can be seen easily that q is a quasihomeomorphism that is not an onto.
Example 4. We let be a space whose open sets are and . Then, every dense subset of X is open and thus X is submaximal. We let be equipped with topology . It can easily be seen that Y is not a submaximal space. Now, we let be the identity map. Thus, q is an onto that is not quasihomeomorphism.
Theorem 4. We let be an onto quasihomeomorphism. If X is door, then Y is door.
Proof. We let A be a subset of Y. Since X is door, we obtain that is either open or closed. Hence is either open or closed in Y. □
Example 5. We let be a space whose open sets are and let be quipped with topology . It can be seen easily that X is door and Y is not a door space. We let be the canonical injection. Hence, q is a quasihomeomorphism that is not an onto.
Example 6. We let be a space whose open sets are and let be equipped with indiscrete topology. Then, X is door and Y is not. We let be defined by and . Thus, the continuous map q is an onto but not a quasihomeomorphism.
Remark 2. We remark that in Theorem 4, it is enough to suppose that q is an onto closed continuous map.
3. Inverse Preserved Topological Properties
We let be a continuous map. In this section, we are interested in additional conditions to X in order to satisfy properties satisfied by Y. Hence, we introduce the following definition:
Definition 1. Continuous map is said to be satisfying the complement closed property if and only if is not closed in X whenever A is not closed in Y.
We recall that Alexandroff spaces are topological spaces in which arbitrary intersections of open sets are open. It is clear that if space X is Alexandroff, then , for every subset A of X.
Theorem 5. We let be a one-to-one continuous map from an arbitrary topological space X to a Whyburn space Y. Then, the following properties hold:
- (1)
If X is Alexandrov, then it is Whyburn.
- (2)
If q satisfies the complement closed property, then it is Whyburn.
Proof. - (1)
We let A be a non-closed subset of X and . Then, there exists such that and . We suppose that there exists such that , and . Then, and . Since Y is Whyburn, we have and so a contradiction. Hence, .Therefore, X is a Whyburn space.
- (2)
We let A be a non-closed subset of X and . Then, . Since Y is a Whyburn space, there exists such that . Hence, ; and since is not a closed subset of X, we obtain . Thus, . Therefore, X is Whyburn. □
Example 7. We let and both be equipped with indiscrete topology. Then, X is not Whyburn and Y is a Whyburn space. We let be defined by and . It may be checked easily that q is continuous and q is not one-to-one.
Now, we study the weakly Whyburn property.
Theorem 6. We let be a one-to-one continuous map from an arbitrary topological space X to a weakly Whyburn space Y. Then, the following properties hold:
- (1)
If X is Alexandrov, then it is weakly Whyburn.
- (2)
If q satisfies the complement closed property, then it is weakly Whyburn.
Proof. - (1)
We let A be a non-closed subset of X, then there exists . Then, , for some such that . We suppose that there exists and such that . Thus, . Since Y is weakly Whyburn, we obtain and so a contradiction.
- (2)
We let A be a non-closed subset of X; then, there exists and so . Since Y is weakly Whyburn, there exists such that for some . Hence, . Now, we suppose that , then , which is a contradiction. Thus, . □
Theorem 7. We let be a one-to-one continuous map. Then, the following properties hold:
- (1)
If Y is submaximal, then X is submaximal.
- (2)
If Y is door, then X is door.
Proof. - (1)
We let A be a dense subset of X. Then, is a dense subset of . But it can easily be seen that is a dense subset of Y and since Y is submaximal, is open and thus is open. Therefore, X is submaximal.
- (2)
We let A be a subset of X; then, , which is open or closed. □
Example 8. We let be a space whose open sets are and . We let equipped with topology . Subset is a dense subset of X and it is not open, and thus X is non-submaximal. On the other hand, it can be seen easily that Y is a submaximal space. Now, we let be defined by . Then, q is a continuous map and q is non-one-to-one.
4. Quasihomeomorphisms and -Primal Spaces
Ayatollah Zadeh Shirazi and Golestani [
8] on the one hand and Echi [
9] on the other hand, working independently, explicitly introduced a class of Alexandroff topologies on
X called by Echi primal spaces and by Shirazi and Golestani functional Alexandroff spaces. In this paper, we use the terminology of primal spaces. Given map
, we define the Alexandroff topology on
X by taking the closure of point
in this topology orbit
, where
is the set of all natural numbers including 0. Therefore, subset
A in this topology is closed if and only if it is an
f-invariant set, that is,
. Equivalently, in the language of qosets, we define
by for any
,
if and only if
for
. Such a topology is denoted by
.
Topology
on set
X is primal if it is
for some
. Since their recent introduction, functional Alexandroff topologies, were further investigated in [
10,
11,
12,
13,
14,
15,
16,
17].
The equivalence between Alexandroff spaces and quasi-ordered sets is the motivation to introduce the following definition.
In [
9], Echi introduced the notion of primal spaces as follows:
Definition 2. We let X be a non-empty set and f be a map from X to itself. Subset A of X is said f-invariant if . The family of all f-invariant sets forms closed sets of a topology on X called primal topology and denoted by .
In [
11], the authors showed that when giving a quasihomeomorphism
q from
X to
Y, if
q is onto and
X is primal, then
Y is primal, and if
q is one-to-one and
Y is primal, then so is
X. Some examples showing the importance of conditions of quasihomeomorphism, onto and one-to-one, were given (for more information, see [
11,
18]).
On the other hand, in [
19], the authors introduced the notion of
k-primal spaces as follows:
Definition 3. For any non-zero positive integer k and any family of functions from a given set X to itself, we define the k-primal space as the intersection of primal spaces . That is, for any if and only if for every .
Hence, the following definition is immediate:
Definition 4. An Alexandroff space is called a strongly primal space if and only if it is a k-primal space for some positive integer k.
The characterization of weakly primal spaces, in finite cases, is given by [
19].
Theorem 8. We let be a finite qoset. Then, is a k-primal space for some , if and only if for every cyclic point a and every we have In this section, we study the relation between strongly primal spaces and quasihomeomorphisms in the class of finite Alexandroff spaces.
Before offering the main result of this section, we need the following lemma:
Lemma 3. We let be quasihomeomorphisms between two Alexandroff spaces. Then, for every .
Proof. If , then , which is included in by continuity of q and thus .
Conversely, we let such that . Since q is a quasihomeomorphism, there exists a unique closed subset F of Y satisfying . Now, using and , we obtain and thus which means that . □
Now, we are in a position to cite and prove our main theorem in this section.
Theorem 9. We let be quasihomeomorphisms between two finite Alexandroff spaces. Then, the following properties hold:
- (1)
If q is onto and X is strongly primal, then so is Y.
- (2)
If q is one-to-one and Y is strongly primal, then so is X.
Proof. - (1)
We let b be a cyclic point in Y, and two points such that and . Since q is an onto, there exist with , and . It can easily be seen that a is a cyclic point in X such that and . Now, since X is strongly primal, and consequently, which complete the proof.
- (2)
Now, we consider cyclic point a in X and with and . Since q is one-to-one and a is cyclic, then is a cyclic point in Y such that and . Hence, by the fact that Y is strongly primal, we obtain . Therefore, and X is strongly primal. □
Example 9. - (1)
We consider equipped with the indiscrete topology and Y is an infinite set equipped with the indiscrete topology. We let q from X to a Y be a constant map. Clearly, q is a non-onto quasihomeomorphism. However, X is primal and thus a strongly primal space but not Y (see [19], Proposition 2.2). This example shows that the surjectivity of q in Theorem 9 is necessary. - (2)
We consider X an infinite set equipped with the indiscrete topology and equipped with the indiscrete topology. We let q from X to a Y be a map which associates zero to a fixed point and one to the complement of . Clearly, q is a non-one-to-one quasihomeomorphism. However, Y is primal and thus a strongly primal space, but not X. This example shows that the injectivity of q in Theorem 9 is necessary.