# On Quasi-ω-Confluent Mappings

Authors: Abdo Qahis; Mohd. Salmi Md. Noorani
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We introduce a new class of mappings called quasi-ω-confluent maps, and we study the relation between these mappings, and some other forms of confluent maps. Moreover, we prove several results about some operations on quasi-ω-confluent mappings such as: composition, factorization, pullbacks, and products.


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## Body

## 1. Introduction

A generalization of the notion of the classical open sets which has received much attention lately is the so-calledω-open sets. These sets are characterized as follows  [1]: a subset W of a topological space (X,τ) is an ω-open set if and only if for each x∈W, there exists U∈τ such that x∈U and U-W is countable. One can then show that the family of all ω-open subsets of a space (X,τ), denoted by τw, forms a topology on X finer than τ. Using this notion of ω-open sets, one can then define notions such as ω-compact and ω-connected sets whose definitions follow closely the definitions of the related classical notions. For example, a space X is called ω-connected provided that X is not the union of two disjoint nonempty ω-open sets. And X is said to be ω-compact if every ω-open cover of X has a finite subcover. For more information regarding these notions and some recent related results, see [2–4].Recall that a subsetK of a space X is said to be a continuum if K is connected and compact. Using this idea of a continuum, Charatonik introduced and studied the idea of a confluent mapping between topological spaces  [5] as follow: A mapping f:X→Y is said to be confluent provided that for each continuum K of Y and for each component C of f-1(K), we have f(C)=K.In  [6], motivated by Charatonik's work, we have introduced the notion of ω-confluent mappings and studied its basic properties. In particular, we say a space X is an ω-continuum if it is ω-connected and ω-compact at the same time, and a subset K of a space X is said to be ω-continuum if K is ω-connected and ω-compact as a subspace of X. Moreover, a mapping f:X→Y is said to be ω-confluent provided that for each ω-continuum K of Y and for each component C of f-1(K), we have f(C)=K.In this paper, we are interested in the further generalizations of the work of Charatonik in the context ofω-open sets and the idea of quasicomponents. Recall that a quasicomponent of space X containing a point p∈X is the intersection of all nonempty clopen sets in X containing p  [7]. In particular, we will introduce the notion of quasi-ω-confluent maps and study its relation with the classical mappings such as confluent, ω-confluent, and quasiconfluent maps. We also study operations on such mappings like compositions, pullback of quasi-ω-confluent, factorizations, and products.


## 2. Quasi-ω-Confluent Mappings

In this section, we introduce and study a new form ofω-confluent mapping, which is a quasi-ω-confluent mapping. Throughout this paper, all mappings are assumed to be continuous.Now, we introduce the following notion.Definition 2.1.
A mappingf:X→Y is said to be quasi-ω-confluent (resp., quasiconfluent) if for each ω-continuum (resp., continuum) K in Y and for each quasicomponent QC of f-1(K), we have f(QC)=K.First, we need the following theorem.Theorem 2.2 (see [6]).
LetX be a topological space. Then,   (1)
everyω-connected subset K of X is connected,(2)
everyω-compact subset K of X is compact,(3)
everyω-continuum subset K of X is a continuum.Proposition 2.3.
(1) Everyω-confluent mapping is quasi- ω-confluent.
(2) Every quasiconfluent mapping is quasi-ω-confluent.Proof.
(1) Suppose thatf:X→Y be an ω-confluent mapping, let K be any ω-continuum in Y, and let x be any point in f-1(K) and QCx be the quasicomponent of x in f-1(K). Then, any component Cx of x in f-1(K) contained in the quasicomponents QCx, or Cx⊂QCx. Thus, f(Cx)⊂f(QCx). Since f is an ω-confluent, then f(Cx)=K. This implies, K⊂f(QCx). But we have QCx⊂f-1(K). So, f(QCx)⊂K. Thus, f(QCx)=K. Therefore, f is quasi-ω-confluent mapping.
(2) LetK be any ω-continuum in Y and QC be any quasicomponent of f-1(K). Then, K is a continuum in Y by the Theorem 2.2(3). Since, f is quasiconfluent. So that, f(QC)=K. Thus, f is quasi-ω-confluent mapping.Remark 2.4.
It is clear that everyω-confluent (confluent or quasiconfluent) mapping is quasi-ω-confluent, but the converses are not necessarily true, as shown by the following examples.Example 2.5.
LetK={1/n:nisapositiveinteger}, D=K×[0,1].
(a) LetX=D∪{(0,0),(0,1)} subspaces of ℝ2 under the usual topology τu, and Y={0,1}, with the topology τY={ϕ,Y}. Let f:X→Y be the mapping defined by
(2.1)f(x,y)={0,for(x,y)∈{(0,0),(0,1)},1,for(x,y)∈{k}×[0,1],foreachk∈K.
Then, f is quasi-ω-confluent but not quasiconfluent. Since, if we take the continuum K={0,1} in Y, then the quasicomponents of f-1(K) are {(0,0),(0,1)} and D. So, f({(0,0),(0,1)})≠K, and f(D)≠K.
(b) LetX=D∪{(0,0),(0,1)}∪([0,1]×{0}) subspaces of ℝ2 under the usual topology τu, and Y={0,1}, with the topology τY={ϕ,Y}. Let f:X→Y be the mapping defined by
(2.2)f(x,y)={0,for(x,y)=(0,1),1,otherwise.
Then, f is quasi-ω-confluent, but not confluent. Since if we take the continuum K={0,1} in Y, then the components of f-1(K) are {(0,1)} and X∖{(0,1)}. So, f({(0,1)})≠K, and f(X∖{(0,1)}≠K.Example 2.6.
LetX={p,q,r} and Y={a,b,c} with topologies τ={ϕ,X,{p},{q},{p,q}} and σ={ϕ,Y,{a}} defined on X and Y, respectively. Let f:X→Y be a mapping defined by f(p)=a, f(q)=b, f(r)=c. Then, f is quasi-ω- confluent, but it is not confluent.Remark 2.7.
Quasi-ω-confluent does not imply ω-confluent in general, since the quasicomponent containing p, QC(X,p) may be different from the component containing p, C(X,p), as the following example shows.Example 2.8.
LetX=K×[0,1]∪{(0,0),(0,1)}∪([0,1]×{0}) be a subspaces of ℝ2 under the usual topology τu, where K be as in Example 2.5, and let Y=[0,1] with the topology τind={ϕ,Y,}.Let f:X→Y be the mapping defined by
(2.3)f(x,y)=x,∀(x,y)∈X.
Then, f is quasi-ω-confluent, but f is not ω-confluent. Note that if we take the ω-continuum K=[0,1], then the components of f-1(K) are C1={(0,1)} and C2=X-{(0,1)}. Thus, f(C1)≠K and f(C2)=K.The following diagram summarizes the relations between confluent mapping, quasiconfluent mapping, andω-confluent mapping with quasi-ω-confluent mapping.The following theorem shows that under certain conditions, quasi-ω-confluent mapping will be ω-confluent.Theorem 2.9.
Every quasi-ω-confluent mapping f:X→Y of a compact Hausdorff space X into a Hausdorff space Y is ω-confluent.Proof.
LetK be any ω-continuum in Y and C any component of f-1(K). Then, by the Theorem 2.2, K is continuum subset of Y. Since Y is Hausdorff, then K is closed in Y and since f is continuous, then f-1(K) is closed in X, since X is compact Hausdorff space, so that f-1(K) is compact Hausdorff space. Thus, the quasicomponents are connected and coincide with components of f-1(K). Thus, f(C)=K. Therefore, f is ω-confluent.Proposition 2.10.
IfX is hereditarily locally connected, then any quasi-ω-confluent mapping f:X→Y is  ω-confluent.Proof.
It follows that from the fact that in locally connected space, the components and quasicomponents are the same.Definition 2.11 (see [2]).
A space(X,τ) is said to be ω-space if every ω-open set is open in X.
It is easy to see that in anω-space that the continuum and ω-continuum sets coincide.Proposition 2.12.
IfY is an ω-space and if f:X→Y is a mapping of a compact Hausdorff space X into a Hausdorff space Y, then the following are equivalent:  (1)
f is confluent,(2)
f is ω-confluent,(3)
f is quasiconfluent,(4)
f is quasi-ω-confluent.Proof.
(1)⇒ (2). Obvious.
(2)⇒ (3). Let f be an ω-confluent mapping, K any continuum in Y, and QC any quasicomponent of f-1(K). Since Y is an ω-space, then K is an ω-continuum, since Y is Hausdorff and X is compact Hausdorff, so that the components and quasicomponents of f-1(K) are the same. Hence, f(QC)=K by assumption. Thus, f is quasiconfluent mapping.
(3)⇒ (4). It follows from Proposition 2.3(2).
(4)⇒ (1). Let f is quasi-ω-confluent mapping, K⊆Y any continuum, and C be an arbitrary component of f-1(K), since Y is an ω-space, then K is an ω-continuum in Y, since X is a compact Hausdorff and Y is a Hausdorff. Then, C is a quasicomponent of f-1(K). Thus, f(C)=K. Therefore, f is confluent mapping.Theorem 2.13.
Letf:X→Y be a mapping of zero-dimensional space X into space Y.  Then, the following are equivalent:  (1)
f is an ω-confluent,(2)
f is quasi-ω-confluent.Proof.
(1)⇒ (2). Obvious.
(2)⇒ (1). Let f be quasi-ω-confluent mapping, K⊆Y any ω-continuum, and C any component of f-1(K). Since X is a zero-dimensional space, then it is totally disconnected. Then the components of f-1(K) are coincide with quasicomponents. Thus, C is a quasicomponent of f-1(K). Then, f(C)=K, by the assumption. Therefore, f is an ω-confluent.Proposition 2.14.
Letf:X→Y be a mapping of space X into zero-dimensional space Y.  Then, the following are equivalent:  (1)
f is quasiconfluent,(2)
f is quasi-ω-confluent.Proof.
(1)⇒ (2). It follows immediately from the Proposition 2.3(2).
(2)⇒ (1). Let K be any ω-continuum, and let QC be any quasicomponent of f-1(K). Since Y is zero-dimensional space. Then, the connected subsets of Y are precisely the singleton sets. Thus, the ω-continuum are coincide with continuum sets in Y, therefore, K is a continuum in Y, so that f(QC)=K. Hence, f is quasiconfluent mapping.Proposition 2.15.
Letf:X→Y be any mapping. If X is a hereditarily locally connected space, then the following conditions (1) and (2) are equivalent, and the conditions (3) and (4) are equivalent:  (1)
f is ω-confluent mapping,(2)
f quasi-ω-confluent mapping,(3)
f is confluent mapping,(4)
f quasiconfluent mapping.Proof.
Similar to the proof of Proposition2.10.


## 3. Composition and Factorization of Quasi-ω-Confluent Mappings

In this section, we study the composition and factorization of quasi-ω-confluent mapping. So, we need to recall the following theorem.Theorem 3.1 (see [6]).
Letf:X→Y and g:Y→Z be two ω-confluent mappings, where f is a surjective. Then, h=g∘f is an ω-confluent mapping.Theorem 3.2.
Letf:X→Y be a surjective quasi-ω-confluent of compact Hausdorff space X into space Y and g:Y→Z a quasi-ω-confluent of space Y into Hausdorff space Z. Then, h=g∘f is quasi-ω-confluent mapping.Proof.
SinceX and Y are two compact Hausdorff spaces and since f and g are two quasi-ω-confluent mappings, then f and g are ω-confluent mappings by Theorem 2.9. Therefore, h=g∘f is an ω-confluent mapping by Theorem 3.1. Then, from Proposition 2.3, h=g∘f is quasi-ω-confluent mapping.Proposition 3.3.
IfX is hereditarily locally connected space and if f:X→Y and g:Y→Z are two quasi-ω-confluent mapping such that f is onto closed or open map, then h=g∘f is quasi-ω-confluent mapping.Proof.
The proof follows immediately from Proposition2.10 and Theorem 3.1.Theorem 3.4.
Letf:X→Y be a mapping of strongly connected space X into Hausdorff space Y, and let f be a canonical decomposition (f=inc∘f′∘pRf) of the following mappings:
(3.1)f′:XRf⟶f(X),inc:f(X)⟶Y,andpRf:X⟶XRf,
where pRf is the quotient surjection map, inc is the inclusion map, and f′ is the bijection mapping, where X/Rf denote to quotients space over the kernel relation Rf={(x,x′):f(x)=f(x′)}. Then, f is a canonical decomposition of ω-confluent mappings.Proof.
We have to prove that these mappingspRf, i, and f′ are ω-confluent mappings. Let K be any arbitrary ω-continuum in the quotients space X/Rf and C any component of pRf-1(K). Since pRf is continuous mapping, then X/Rf is a Hausdorff, so that K is closed in X/Rf. Then, by the continuity of pRf, we have pRf-1(K) is closed in X. But X is strongly connected. Therefore, pRf-1(K) is connected. This means pRf-1(K)=C. So, pRf(C)=K. Thus, pRf is an ω-confluent mapping.
It is clearly thatf′ and inc are ω-confluent mappings, since Y is a Hausdorff, then the subspace f(X) is Hausdorff, and since X is strongly connected, then X/Rf is strongly connected and also Hausdorff. Thus, f′ and the inclusion map inc are ω-confluent. Hence, f is canonical decomposition of ω-confluent mappings.Remark 3.5.
In the above theorem, ifX is strongly connected compact Hausdorff space, then the mapping f is the canonical decomposition of quasi-ω-confluent mappings.Corollary 3.6.
IfX,  Y, and Z are Hausdorff spaces, X is a compact space, and if f:X→Y is a surjective ω-confluent mapping and g:Y→Z is a quasi-ω-confluent mapping, then h=g∘f is ω-confluent mapping.Corollary 3.7.
IfX,Y, and Z are Hausdorff spaces, X is a compact space, and if f:X→Y is a surjective quasi-ω-confluent mapping and g:Y→Z is ω-confluent mapping, then h=g∘f is a quasi-ω- confluent mapping.Now, we study Whyburn's factorization theorem for quasi-ω-confluent mappings. Thus, we recall the definition of a factorable mapping.Definition 3.8 (see [8]).
Iff:X→Y be a mapping, any representation of f in the form f=f2∘f1, where f1:X→Z and f2:Z→Y are two mappings and Z is a certain space, will said to be factorization of f, and f is said be a factorable mapping and Z a middle space.Before we study the factorization property, we state the following theorem.Theorem 3.9 (see [6]).
Iff:X→Y is an ω-confluent of strongly connected compact space X into Hausdorff space Y, then there exists a unique factorization for f into two ω-confluent mappings
(3.2)f(x)=f2∘f1(x),∀x∈X,
such that f1 is confluent mapping.Now, we can get the factorization of a quasi-ω-confluent mapping in the following proposition.Proposition 3.10.
Iff:X→Y be a quasi-ω-confluent of strongly connected compact Hausdorff space X into Hausdorff space Y, then there exists a unique factorization for f into two quasi-ω-confluent mappings in the form f=f2∘f1.Proof.
Sincef and g are two quasi-ω-confluent mappings and since X is strongly connected compact Hausdorff space and Y is a Hausdorff space, then from Theorem 2.9f and g are ω-confluent mappings. Thus, f has unique factorization in the form f=f2∘f1 by Theorem 3.9.Next, we study the product property of quasiconfluent mappings.Let{Xi}i∈I and {Yi}i∈I be any two families of topological spaces. The product space of {Xi}i∈I and {Yi}i∈I is denoted by ∏i∈IXi and ∏i∈IYi, respectively. Let fi:Xi→Yi be a mapping for each i∈I. Let f:∏i∈IXi→∏i∈IYi be the product mappings as follows: f((xi))=(fi(xi)) for each (xi)∈∏i∈IXi. The projection of ∏i∈IXi and ∏i∈IYi onto Xi and Yi, respectively, is denoted by pi and qi.Before we get the following result, we need to state the following theorem.Theorem 3.11 (see [6]).
Letfi:Xi→Yi be an ω-confluent mapping, for each i∈I of space Xi into Hausdorff space Yi.  Then,
(3.3)f:∏i∈IXi⟶∏i∈IYi
is an ω-confluent mapping if the following equality holds:
(3.4)(∏i∈Iτi)ω=∏i∈I(τi)ω,∀i∈I.As immediate consequence of the above theorem, we get the following corollary.Corollary 3.12.
Letfi:Xi→Yi be a quasi-ω-confluent mapping of compact Hausdorff space X into Hausdorff space Y for each i∈I,  then
(3.5)f:∏i∈IXi⟶∏i∈IYi
is quasi-ω-confluent mapping if the following equality holds:
(3.6)(∏i∈Iτi)ω=∏i∈I(τi)ω,∀i∈I.Proof.
Since,Xi is compact Hausdorff, then the product space ∏i∈IXi is compact Hausdorff, and since Yi is Hausdorff, then the product space ∏i∈IYi is also Hausdorff. From Theorem 2.9, we infer that fi:Xi→Yi is an ω-confluent for each i∈I. Then, by Theorem 3.11, f:∏i∈IXi→∏i∈IYi is an ω-confluent mapping. Therefore, f is quasi-ω-confluent by Proposition 2.3.


## 4. Pullback of Quasi-ω-Confluent Mappings

In this section, we study the pullback of quasi-ω-confluent mappings. So, we recall the following definitions.Definition 4.1 (see [9]).
A fiber structure is a triple(X,f,Y) consisting of two spaces X and Y and a mapping f:X→Y. The space X is said to be the fibered (or, total) space, f is termed the projection, and Y is the base space. Next, we recall the definition of the pullback.Definition 4.2 (see [9]).
Let(X,f,Y) be a fiber structure. Let Z be any space, and let g:Z→Y be any mapping into the base Y. Let Ef be a subspace of the cartesian product X×Z, where Ef={(x,z):f(x)=g(z)}, and let p:Ef→Z be the projection of Ef onto Z such that p(x,z)=z,∀(x,z)∈Ef. The fiber structure (Ef,p,Z) is said to be the fiber structure over Z induced by the mapping g, and the projection p is said to be the pullback of f by g.Now, letγ:Ef→X be the projection such that γ(x,z)=x, ∀(x,z)∈Ef.We observe that the following diagram is commutative.Before we prove the main results in this section, we state the following lemma.Lemma 4.3 (see [6]).
Letf:X→Y be a mapping, let Z be any space, and let g:Z→Y be any mapping, and if K⊆Z, then p-1(K)=f-1(g(K))×K, where p is the pullback of f by g.Theorem 4.4.
The pullback of a quasiconfluent mapping is quasi-ω-confluent.Proof.
Letf:X→Y be a quasiconfluent mapping, let Z be any space, and let g:Z→Y be any mapping. Let K⊆Z be any ω-continuum and QC any quasicomponent of p-1(K). Then, QC is a quasicomponent of f-1(g(K))×K by Lemma 4.3. Since every ω-continuum is continuum, then K is a continuum by Theorem 2.2. Thus, g(K) is continuum in Y. Since f is quasiconfluent mapping, then f(QC′)=g(K) for each quasicomponent QC′ of f-1(g(K)). Since p-1(K)=f-1(g(K))×K, so K=p(f-1(g(K))×K)=p(QC′×K) such that QC=QC′×K for some quasicomponent QC′ of f-1(g(K)). Thus, p(QC)=P(QC′×K)=K. Therefore, p is quasi-ω-confluent.The pullback of quasi-ω-confluent mapping is not necessarily quasi-ω-confluent as shown by the following example.Example 4.5.
LetX=ℝ be the real number with upper limit topology, Y={a,b} with the topology τY={ϕ,{a},Y}, and Z=ℝ with topology τZ={ϕ,ℝ,ℝ-{1},ℝ-{2},ℝ-{1,2}}.
Letf:X→Y be a mapping defined by
(4.1)f(x)={a,ifx>0,b,ifx≤0,
and let g:Z→Y be a mapping defined by
(4.2)g(z)={a,ifz∈R-{1,2},b,ifz∈{1,2}.
Let Ef be a subspace of the cartesian product X×Z, where
(4.3)Ef={(x,z):f(x)=g(z)}.
Then, the pullback of f by g is the projection p:Ef→Z which is defined by
(4.4)p(x,z)=z,∀(x,z)∈Ef.We note thatf is quasi-ω-confluent mapping, but p is not quasi-ω-confluent mapping.Since if we take theω-continuum K=[0,∞)⊂Z, then by Lemma 4.3, we get p-1(K)=f-1(g(K))×K. But g(K)={a,b} is not ω-continuum in Y.Under certain condition, the pullbackp of quasi-ω-confluent mapping f will be quasi-ω-confluent as shown by the following theorem.Theorem 4.6.
IfY is a zero -dimensional space and if f:X→Y is a quasi-ω-confluent mapping, then the pullback p of f is quasi-ω-confluent.Proof.
Letf:X→Y be a quasi-ω-confluent mapping, let Z be any space, and let g:Z→Y be an mapping. Let K be any ω-continuum in Z, and let QC be any quasicomponent of p-1(K), where p is the pullback of f by g. Then QC is the quasicomponent of f-1(g(K))×K by Lemma 4.3. By Theorem 2.2, K is continuum. Thus, g(K) is continuum in Y by the continuity of g. Since Y is zero-dimensional space, then the quasiconfluent mapping equivalent to the quasi-ω-confluent by Proposition 2.14. This implies the continuum and ω-continuum sets coincide in Y. Thus, g(K) is an ω-continuum in Y. Since f is a quasi-ω-confluent, then f(QC′)=g(K) for each quasicomponents QC′ of f-1(g(K)), and since p-1(K)=f-1(g(K))×K. K=p(f-1(g(K))×K)=p(QC′×K) such that QC=QC′×K for some quasicomponent of f-1(g(K)). Thus, p(QC)=P(QC′×K)=K. Therefore, p is a quasi-ω-confluent.Corollary 4.7.
Iff:X→Y is a quasi- ω-confluent mapping of space  X into ω-space Y, then the pullback of f is quasi-ω-confluent mapping.



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