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# Semi-quotient mappings and spaces

Moiz ud Din Khan
• Corresponding author
• COMSATS Institute of Information Technology, Chak Shahzad, Islamabad - 45550, Pakistan
• Email:
/ Rafaqat Noreen
• COMSATS Institute of Information Technology, Chak Shahzad, Islamabad - 45550, Pakistan
• Email:
• Punjab Education Department, Pakistan
• Email:
Published Online: 2016-12-19 | DOI: https://doi.org/10.1515/math-2016-0093

## Abstract

In this paper, we continue the study of s-topological and irresolute-topological groups. We define semi-quotient mappings which are stronger than semi-continuous mappings, and then consider semi-quotient spaces and groups. It is proved that for some classes of irresolute-topological groups (G, *, τ) the semi-quotient space G/H is regular. Semi-isomorphisms of s-topological groups are also discussed.

MSC 2010: 54H11; 22A05; 54C08; 54H99

## 1 Introduction

The basic aim of this article is to study properties of topological spaces and mappings between them by weakening the continuity and openness conditions. Semi-continuity [1] and irresolute mappings [2] were a consequence of the study of semi-open sets in topological spaces. In [3] Bohn and Lee defined and investigated the notion of s-topological groups and in [4] Siddique et. al. defined the notion of 5-topological groups. In [5] Siab et. al. defined and studied the notion of irresolute-topological groups by using irresolute mappings. Study of s-paratopological groups and irresolute-paratopological groups is a consequence of the study of paratopological groups (see [6]). For the study of semi-topological groups with respect to semi-continuity and irresoluteness we refer the reader to Oner’s papers [7-9].

In this paper we continue the study of properties of s-topological and irresolute-topological groups. Keeping in mind the existing concepts, semi-quotient topology on a set is defined as a generalization of the quotient topology for spaces and groups. Various results on semi-quotients of topologized groups are proved. A counter example is given to show that the quotient topology is properly contained in the semi-quotient structure. We define also semi-isomorphisms and S-isomorpisms between topologized groups and prove that if certain irresolute-topological groups G and H are semi-isomorphic or S-isomorphic, then their semi-quotients are semi-isomorphic. Investigation of s-openness and s-closedness of mappings on s-topological groups is also presented.

## 2 Definitions and preliminaries

Throughout this paper X and Y are always topological spaces on which no separation axioms are assumed. If ƒ : X → Y is a mapping between topological spaces X and Y and B is a subset of F, then f(B) denotes the pre-image of B. By C1(A) and Int(A) we denote the closure and interior of a set A in a space X. Our other topological notation and terminology are standard as in [10]. If (G, *) is a group, then e or eG denotes its identity element, and for a given x ∈ G, x : GG, y ↦ x ∘ y, and rx : G→ G, yy ∘ x, denote the left and the right translation by x, respectively. The operation * we call the multiplication mapping m : G x GG, and the inverse operation x ↦ x–1 is denoted by i.

In 1963, N. Levine [1] defined semi-open sets in topological spaces. Since then, many mathematicians have explored different concepts and generalized them by using semi-open sets (see [2,11-14]). A subset A of a topological space X is said to be semi-open if there exists an open set U in X such that UA ⊂ Cl(U), or equivalently if A ⊂ Cl(Int(A)). SO(X) denotes the collection of all semi-open sets in X, and SO(X, x) is the collection of semi-open sets in X containing the point x ∈ X. The complement of a semi-open set is said to be semi-closed; the semi-closure of AX, denoted by sCl(A), is the intersection of all semi-closed subsets of X containing A [15, 16]. x ∈ sCl(A) if and only if any U ∈ SO(X, x) meets A.

Clearly, every open (resp. closed) set is semi-open (resp. semi-closed). It is known that a union of any collection of semi-open sets is again a semi-open set. The intersection of two semi-open sets need not be semi-open whereas the intersection of an open set and a semi-open set is semi-open. Basic properties of semi-open sets and semi-closed sets are given in [1], and [15, 16].

Recall that a set U ⊂ X is a semi-neighbourhood of a point x ∈ X if there exists A ∈ SO(X, x) such that A ⊂ U. If a semi-neighbourhood U of a point x is a semi-open set, we say that U is a semi-open neighbourhood of x. A set A ⊂ X is semi-open in X if and only if A is a semi-open neighbourhood of each of its points. Let X be a topological space and A ⊂ X. Then x ∈ X is called a semi-interiorpoint of A if there exists a semi-open set U such that x UA The set of all semi-interior points of A is called a semi-interior of A and is denoted by sInt(A). A nonempty set A is pre-open (or locally dense) [17] if A ⊂ Int.(Cl(A)). A space X is s-compact [18], if every semi-open cover of X has a finite subcover. Every s-compact space is compact but converse is not always true. For some applications of semi-open sets see [13].

A mapping f : X → Y between topological spaces X and Y is called:

• - semi-continuous [1] (resp. irresolute [2]) if for each open (resp. semi-open) set V ⊂ Y the set f(V). is semi-open in X. Equivalently, the mapping f is semi-continuous (irresolute) if for each x ∈ X and for each open (semi-open) neighbourhood V of f(x), there exists a semi-open neighbourhood U of x such that f(U) V;

• - pre-semi-open [2] if for every semi-open set A of X, the set f(A) is semi-open in Y;

• - s-open (s-closed) if for every semi-open (semi-closed) set A of X, the set f(A) is open (closed) in Y;

• - s-perfect if it is semi-continuous, s-closed, surjective, and f(y) is s-compact relative to X, for each y in Y.

• - semi-homeomorphism [2, 19] if f is bijective, irresolute and pre-semi-open;

• - S-homeomorphism [4] if f is bijective, semi-continuous and pre-semi-open.

We need also some basic information on (topological) groups; for more details see the excellent monograph [20]. If G is a group and H its normal subgroup, then the canonical projection of G onto the quotient group G/H (sending each g ∈ G to the coset in G/H containing g) will be denoted by p. A mapping f : GH between two topological groups is called a topological isomorphism if f is an algebraic isomorphism and a topological homeomorphism.

Let f : X Y be a surjection; a subset C of X is called saturated with respect to f (or f-saturated) if f(f(C) = C [21].

## 3 Semi-quotient mappings

Definition 3.1: A mapping f : X → Y from a space X onto a space Y is said to be semi-quotient provided a subset V of Y is open in Y if and only if f (V) is semi-open in X.Evidently, every semi-quotient mapping is semi-continuous and every quotient mapping is semi-quotient. The following simple examples show that semi-quotient mappings are different from semi-continuous mappings and quotient mappings.

Example 3.2: Let X = Y = {1, 2, 3} and let τX = {, X, {1}, {2} {1, 2}, {1, 3}} and τY = {∅, X, {1}, {2}, {1 2}} be topologies on X and Y. Let f : X → Y be defined by f(x) = x, x ∈ X. Since τY τχ, the mapping ƒ is continuous, hence semi-continuous. On the other hand, this mapping is not semi-quotient because f ({3}) is semi-open in X although {1, 3} is not open in Y.

Example 3.3: Let X = {1, 2, 3, 4}, Y = {a, b}, τχ = {∅, X, {1}, {3}, {1, 3}}, τγ = {∅, Y, {a}}. Define f : X → Y by; f (1) = f(3) = f (4) = a; f (2) = b. The mapping f is not a quotient mapping because it is not continuous. On the other hand, f is semi-quotient: the only proper subset of Y whose preimage is semi-open in X is the set fag which is open in Y.The following proposition is obvious.

Proposition 3.4: (a) Every surjective semi-continuous mapping f : X → Y which is either s-open or s-closed is a semi-quotient mapping.(b) If f : X → Y is a semi-quotient mapping and g : Y → Z a quotient mapping, then g ∘ f : X → Z is semi-quotient.

Proof: We prove only (b). A subset V ⊂ Z is open in Z if and only if g(V) is open in Y (because g is a quotient mapping), while the latter set is open in Y if and only if f(g(V)) is semi-open in X (because f is semi-quotient). So, V is open in Z if and only (g ∘ f)(V) is semi-open in X, i.e. g ∘ f is a semi-quotient mapping.The restriction of a semi-quotient mapping to a subspace is not necessarily semi-quotient. Let X and Y be the spaces from Example 3.3, and A = {2, 4}. Then τA = {∅, A}. The restriction fA : A → Y of f to A is not a semi-quotient mapping because fA{(a)} = {4} is not semi-open in A.To see when the restriction of a semi-quotient mapping is also semi-quotient we will need the following simple but useful lemmas.

Lemma 3.5: ([22, Theorem 1]). Let X be a topological space, X0 SO(X) and A ⊂ X0. Then A ∈ SO(X0 ) if and only if A ∈ SO(X).

Lemma 3.6: ([23, Lemma 2.1]). Let X be a topological space, X0 a subspace of X. If A ∈ SO(X0), then A = B ⋂ X0, for some B ∈ SO(X).

Lemma 3.7: ([21]). Let f : X → Y be a mapping, A a subspace of X saturated with respect to f, B a subset of X. If g : A → f(A) is the restriction of f to A, then:(1) g(C) = f(C) for any C ⊂ f(A);(2) f(A B) = f(A) f(B).Now we have this result.

Theorem 3.8: Let f : X → Y be a semi-quotient mapping and let A be a subspace of X saturated with respect to f, and let g : A → f(A) be the restriction of f to A. Then:(a) If A is open in X, then g is a semi-quotient mapping;(b) Iff is an s-open mapping, then g is semi-quotient.

Proof: (a) Let V be an open subset of f(A). Then V = W ⋂ f(A) for some open subset W of X, so that g(V) = f (W f(A)) = f(W) A is a semi-open set in A.Let now V be a subset of f(A) such that g(V) is semi-open in A. We have to prove that V is open in f(A). Since g(V) is semi-open in A and A is open in X we have that g(V) is semi-open in X. By Lemma 3.7, g(V)= f(V); the set f(V) is semi-open in X since f is semi-quotient, hence g(V) is semi-open in f(A). This means that V is open in Y and thus in f(A). This completes the proof that g is a semi-quotient mapping.(b) Let now f be s-open and V a subset of f(A) such that g (V) is semi-open in A. Again we must prove that V is open in f(A). Since g(V)= f(V) and g(V) is semi-open in A, by Lemma 3.6 we have f(V) = UA, for some U semi-open in X. As f is surjective, it holds f(f(V) = V. By Lemma 3.7, then V = f(f(V)) = f(UA) = f(U) ∩ f(A). The set f(U) is open in Y because f is s-open, so that V is open in f(A). Other part is the same as in (a).As a complement to Proposition 3.4 we have the following two theorems.

Theorem 3.9: Let X, Y and Z be topological spaces, f : X → Y a semi-quotient mapping, g : Y → Z a mapping. Then the mapping g o f : X → Z is semi-quotient if and only if g is a quotient mapping.

Proof: If g is a quotient mapping, then g ∘ f is semi-quotient as the composition of a semi-quotient and a quotient mapping (Proposition 3.4).Conversely, let g o f be semi-quotient. We have to prove that a subset V of Z is open in Z if and only ifg(V) is open in Y. For, the set (g ∘ f)(V) is semi-open in X and since f is a semi-quotient mapping we conclude that it will be if and only if g(V) is open in Y.

Theorem 3.10: Let f : X → Y be a mapping and g : X → Z a mapping which is constant on each set f({y}), y ∈ Y. Then g induces a mapping h : Y → Z such that g = h ∘ f. Then:(1) If f is pre-semi-open and irresolute, then h is a semi-continuous mapping if and only if g is semi-continuous;(2) If f is semi-quotient, then h is continuous if and only if g is semi-continuous.

Proof: Since g is constant on the set f({y}), y ∈ Y, then for each y ∈ Y, the set g(f({y})) is a one-point set in Z, say h(y). Define now h : Y Z by the ruleh(y) = g(f(y)) (y ∈ Y).Then for each x ∈ X we haveg(x) = g(f(f(x))) = h(f(x)), i.e. g = h º f.(1) Suppose g is a semi-continuous mapping. If V is an open set in Z, then h(V) = f(g(V)) ∈ SO(Y) because f is pre-semi-open and g(V) is semi-open in X. Thus h is a semi-continuous mapping.Conversely, suppose h is a semi-continuous. Let V be an open set in Z. The set g(V) = fh(V)) is semi-open in X because f is irresolute and h(V) SO(Y). So, g is semi-continuous.(2) If g is semi-continuous, then for any open set V in Z we have g(V) is a semi-open set in X. But, g(V) = f(h(V)). Since f is semi-quotient it follows that h(V) is open in Y. So, h is continuous.Conversely, suppose h is continuous. For a given open set V in Z, h(V) is an open set in Y. We have then g(V) = f(h(V)) is semi-open in X because f is semi-quotient. Hence g is semi-continuous.At the end of this section we describe now a typical construction which shows how the notion of semi-quotient mappings may be used to get a topology or a topology-like structure on a set.Construction: Let X be a topological space and Y a set. Let f : X→ Y be a mapping. DefineQ := {V ⊂ Y : f(V) ∈ SO(X)},It is easy to see that the family Q ia a generalized topology on Y (i.e. ∅ ∈ Q and union of any collection of sets in Q is again in Q) generated by f; we call it the semi-quotient generalized topology. But Q need not be a topology on Y. It happens if X is an extremally disconnected space, because in this case the intersection of two semi-open sets in X is semi-open [24]. It is trivial fact that in the latter case sτQ is the finest topology ς on Y such that f : X → (Y, σ) is semi-continuous. In fact, f : X → (Y, sτQ) is a quotient mapping in this case.In particular, let p be an equivalence relation on X. Let p : X → X/ρ be the natural (or canonical) projection from X onto the quotient set X/ρ: for each x in X, p sends x to the equivalence class ρ(x). Then the family Q generated by p is a generalized topology on the quotient set Y/ρ, and a topology when X is extremally disconnected. This topology will be called the semi-quotient topology on X/ρ. Observe, that we forced the mapping p to be semi-continuous, that is semi-quotient.This kind of construction will be applied here to topologized groups: to s-topological groups and irresolute-topological groups.The following example shows that a quotient topology on a set generated by a mapping and the semi-quotient (generalized) topology generated by the same mapping are different.

Example 3.11: Let the set X = {1, 2, 3, 4} be endowed with the topologyτ = (∅,X, {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.Then the set SO(X) is(∅, X, {1},{2},{3},{1,2},{1,3},{2,3},{1,4},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4}}.Define the relation R on X by xRy if and only if x + y is even. Therefore,R = {(1,1),(1,3),(2,2),(2,4),(3,1),(3,3),(4,2),(4,4),}is an equivalence relation, and X/R = {R(1),R(2)} = {{1,3},{2,4}}. Let p : X → X/R be the canonical projection. Then, p(R(1)) = {1, 3} ∈ SO(X), and p(R(2)) = f2; 4} ∈ SO(X) , so thatQ = {∅, X/R, {R(1)},{R(2)}}is the semi-quotient topology on X/R. On the other hand, p({R(1)}) = {1,3} ∈ τ, but p({R(2)}) = {2,4} ∉ τ. Therefore, the quotient topology on X/R isτQ = {∅ X/R, {R(1)}}.

## 4 Topologized groups

In this section we give some information on s-topological groups and irresolute-topological groups introduced and studied first in [4] and [5], respectively.

Definition 4.1: ([3]). An s-topological group is a group (G, *) with a topology τ such that for each x, y ∈ G and each neighbourhood W of x * y1 there are semi-open neighbourhoods U of x and V of y such that U * V—1 W.

Definition 4.2: ([5]). A triple (G, *, τ) is an irresolute-topological group if (G, *) is a group, and τ a topology on G such that for each x, y ∈ G and each semi-open neighbourhood W of x * y—1 there are semi-open neighbourhoods U of x and V of y such that U * V—1 W.

Lemma 4.3: ([25]). If (G, *, x) is an s-topological group, y ∈ G, and K an s-compact subset of G, then y * K–1 is s-compactin G. Inparticular, K–1 is s-compact.

Theorem 4.4: If a mapping f : X → Y between topological spaces X and Y is s-perfect, then for any compact subset K of Y, the pre-image f(K) is an s-compact subset of X.

Proof: Let {Ui : i ∈ } be a semi-open cover of f(K). Then for each x ∈ K the set f(x) can be covered by finitely many Ui; let U(x) denote their union. Then O(x) = Y \ f [X \ ⊂ U(x)] is an open neighbourhood of x in Y because f is an s-closed map. So, $K\subset {\bigcup }_{x\in K}O\left(x\right),$ and because K is assumed to be compact, there are finitely many points x1, x2, . . . , xn in K such that $K\subset {\bigcup }_{i=1}^{n}O\left({x}_{i}\right).$ It follows that ${f}^{←}\left(K\right)\subset {\bigcup }_{i=1}^{n}{f}^{←}\left(O\left({x}_{i}\right)\right)\subset {\bigcup }_{i=1}^{n}U\left({x}_{i}\right)$ hence f(K) is s-compact in X.The following results are related to s-topological groups, and they are generalizations of some results for topological groups.

Theorem 4.5: Let G, H and K be s-topological groups, φ : G → H a semi-continuous homomorphism, ψ G → K an irresolute endomorphisms, such that ker ψ ⊂ ker φ. Assume also that for each open neighbourhood U of eH there is a semi-open neighbourhood V of eK with ψ(V) ⊂ ψ(U). Then there is a semi-continuous homomorphism f : K → H such that φ = fψ.

Proof: The existence of a homomorphism f such that φ = f ∘ ψ is well-known fact in group theory. We verify the semi-continuity of f. Suppose U is an open neighbourhood of eH in H. By our assumption, there is a semi-open neighbourhood V of eK in K such that ψ(V) φ(U). Then φ = fψ implies f(V) = φ(ψ(V)) ⊂ φ(φU) U, which means that f is semi-continuous at the identity element eK of K. By [4, Theorem 3.5], f is semi-continuous on K.

Theorem 4.6: Suppose that G, H and K are s-topological groups. Let φ : G → H be a semi-continuous homomorphism, ψ : G → K an irresolute endomorphism such that ker ψ ⊂ ker φ. If ψ is pre-semi-open, then there is a semi-continuous homomorphism f : K → H such that φ = fψ .

Proof: By Theorem 4.5, there exists a homomorphism f : K → H satisfying φ = fψ. We prove that f is semi-continuous. Let V be an open set in H. From φ = fψ it follows f(V) = ψ(φ(V)). Since, φ is semi-continuous, the set φ(V) is semi-open in G, and pre-semi-openness of ψ implies that ψ(ψ(V)) is semi-open, i.e. f(V) is semi-open in K. This means that f is semi-continuous.

Theorem 4.7: Let (G, *, τG) and (H, ·, τH) be s-topological groups, and f : G → H a homomorphism of G onto H such that for some non-empty open set U → G, the set f(U) is semi-open in H and the restriction f |U : U → f(U) is a pre-semi-open mapping. Then f is pre-semi-open.

Proof: We have to prove that if x ∈ G and W ∈ SO.(G, x), then f(W) SO.(H, f(x). Pick a fixed point y ∈ U and consider the mapping ${\ell }_{y\ast {x}^{-1}}:G\to G$. Evidently, ${\ell }_{y\ast {x}^{-1}}\left(x\right)=y,$ and by [4, Theorem 3.1], ${\ell }_{y\ast {x}^{-1}}$ is an S-homeomorphismof G onto itself. Thus the set $V=U\cap {\ell }_{y\ast {x}^{-1}}\left(W\right)$ is semi-open as the intersection of an open set U and a semi-open set ${\ell }_{y\ast {x}^{-1}}\left(w\right)$ i.e. V is a semi-open neighbourhood of y in U. By assumption on f, the set f(V) is semi-open in f(U) and also in H by Lemma 3.5. Set z = f (x * y–1) and consider the mapping z HH. We have z(f(y)) = z · f(y) = f(x). Clearly, ${\ell }_{Z}\circ f\circ {\ell }_{y\ast {x}^{-1}}=f$ hence ${\ell }_{z}\circ f\circ {\ell }_{y\ast {x}^{-1}}\left(w\right)=f\left(w\right).$ However, z is an S-homeomorphism of H onto itself. As f(V) is semi-open in H, the set z(f(V)) is also semi-open in H. Therefore, the set f(W) contains a semi-open neighbourhood z.(f(V)) of f(x) in H, so that f(W) SO(H, f(x)) as required.

## 5 Semi-quotients of topologized groups

In this section we apply the construction of sτQ described in Section 3 to topologized groups and establish some properties of their semi-quotients.

If G is a topological group and H a subgroup of G, we can look at the collection G/H of left cosets of H in G (or the collection H\G of right cosets of H in G), and endow G/H (or G\H) with the semi-quotient structure induced by the natural projection p : G → G/H. Recall that G/H is not a group under coset multiplication unless H is a normal subgroup of G.

The following simple lemmas may be quite useful in what follows.

Lemma 5.1: ([20]). Let p : G → G/H be a canonicalprojection map. Then for any subset U of G, p(p(U)) = U * H.

Lemma 5.2: ([25]). Let (G, *, τ) be an s-topological group, K an s-compact subset of G, and F a semi-closed subset of G. Then F * K and K * F are semi-closed subsets of G.

Lemma 5.3: ([4]). Let (G, *, τ) be an s-topological group. Then each left (right) translation in G is an S-homeomorphism. Moreover they and symmetry mappings are actually semi-homeomorphisim (see [1, Remark 1]).

Lemma 5.4: ([26]). If ƒ : XY is a semi-continuous mapping and X0 is an open set in X, then the restriction $f{|}_{{X}_{0}}:{X}_{0}\to Y$ is semi-continuous.

Theorem 5.5: Let (G, *, τ) be an extremally disconnected irresolute-topological group and H its invariant subgroup. Then ρ : $p:\left(G,\ast ,\tau \right)\to \left(G/H,\overline{\ast },s{\tau }_{Q}\right)$ is pre-semi-open.

Proof: Let VG be semi-open. By the definition of semi-quotient topology, p(V)⊂ G/H is open if and only if p(p(V))⊂ G is open. By Lemma 5.1 p(p(V)) = V * H. Since V is semi-open, V * H is semi-open and so p(V)is semi-open. Hence p is pre-semi-open.

Theorem 5.6: Let (G, *, τ) be an extremally disconnected irresolute-topological group, Η its invariant subgroup. Then $\left(G/H,\overline{\ast },s{\tau }_{Q}\right)$ is an irresolute-topological group.

Proof: First, we observe that Q is a topology on G/H. Let x * H, y * H ∈ G/H and let W ⊂ G/H be a semi-open neighbourhood of $\left(x\ast !H\right)\overline{\ast }\left(y\ast {H}^{-1}\right)$. By the definition of Q (induced by p), the set p(W)is a semi-open neighbourhooud of x * y-1in G, and since G is an irresolute-topological group, there are semi-open sets U ⊂ SO(G, x)and V ⊂ SO(G, y)such that U * V-1p(W). By Theorem 5.5, the sets p(U) = U * H and p(V) = V * H are semi-open in G/H, contain x * H and y * H, respectively, and satisfy $U∗H∗¯V∗H−1=U∗V−1∗H=pU∗V−1⊂pp←w⊂W.$This just means that $\left(G/H,\overline{\ast },s{\tau }_{Q}\right)$ is an irresolute-topological group.

Theorem 5.7: Let (G, *, τ) be an s-topological group and H a subgroup of G. Then for every semi-open set U ⊂ G, the set p(U) belongs to sτQ. In particular, if G is extremally disconnected, then p is an s-open mapping from G to (G/H, sτQ).

Proof: Let V ⊂ G be semi-open. By definition of Q, p(V) ∈ sτQ if and only if p(p(V)) G is semi-open, i.e. V * H is semi-open in G. But V * H is semi-open in G because V ∈ SO(G) and (G, *, τ)is an s-topological group. Clearly, if Q is a topology, the last condition actually says that p is an s-open mapping.The following theorem is similar to Theorem 5.7.

Theorem 5.8: If Η is an s-compact subgroup of an s-topological group (G, *, τ), then for every semi-closed set F ⊂ G, the set p(G \ F) belongs to sτQ. If sτQ is a topology, then p is an s-perfect mapping.

Proof: Let F ⊂ G be semi-closed. By Lemma 5.2 the set p(p(F)) = F * H ⊂ G is semi-closed. By definition of Q, G/H \ (F * H) ∈Q.Let now Q be a topology on G/H. Take any semi-closed subset F of G. The set F * H is semi-closed in G and F * H = p(p(F)). This implies, p(F)is closed in the semi-quotient space G/H. Thus p is an s-closed mapping. On the other hand, if z * Η ∈ G/H and p(x) = z * Η for some x ∈ G, then p(z * H) = p(p(x)) = x * Η, and by Lemmas 4.3 and 5.3 this set is s-compact in G. Therefore, p is s-perfect.

Corollary 5.9: Let (G, *,τ) be an extremally disconnected s-topological group and Η its s-compact subgroup. If the semi-quotient space (G/H, sτQ) is compact, then G is s-compact.

Proof: By Theorem 5.8, the projection p : G → G/H is s-perfect. Then by Theorem 4.4 we obtain that p(p(G)) = G * H = G is s-compact.

Theorem 5.10: Suppose that (G,*, τ) is an extremally disconnected s -topological group, H an invariant subgroup of G, p:G → (G/H, sτQ) the canonical projection. Let U and V be semi-open neighbourhoods of e in G such that V-1* V ⊂ U. Then C1(p(V)) p(U).

Proof: Let p(x) C1(p(V)). Since V * x is a semi-open neighbourhood of xG and, by Theorem 5.7, p is s-open, we have that p(V * x)is an open neighbourhood of p(x). Therefore, p(V * x)⋂ p(V)≠ø. It follows that for some a, b V we have p(a * x) = p(b), that is a * x * h1 = b * h2for some h1, h2Η. Hence, $x=a−1∗b∗h2∗h1−1=a−1∗b∗h2∗h1−1∈U∗H$since a-1 * b V-1* VU and H is a subgroup of G. Therefore, p (x)∈ p(U * H) = U * H * H = U *H = p(U).

Theorem 5.11: Let (G,*, τ) be an extremally disconnected irresolute-topological group and H an invariant subgroup of G. Then the semi-quotient space (G/H, sτQ) is regular.

Proof: Let W be an open neighbourhood of p(eG) = Η in G/H. By semi-continuity of p, we can find a semi-open neighbourhood U of eG such that p(U)⊂ W. As G is extremally disconnected and irresolute-topological group, it follows from ${e}_{G}\ast {e}_{G}^{-1}={e}_{G}$ that there is a semi-open neighbourhood V of eG such that V * V-1U. By Theorem 5.10 we have C1(p(V))⊂ p(U)⊂ W. By Theorem 5.7, p(V)is an open neighbourhood of p(eG). This proves that (G/H, sτQ) is a regular space.If (G, *) is a group, H its subgroup, and a G, then we define the mapping λa : G/H G/H by λa(x * Η) = a * (x * Η). This mapping is called a left translation of G/H by a [20].

Theorem 5.12: If (G, *, τ) is an extremally disconnected irresolute-topological group, Η a subgroup of G, and a ∈ G, then the mapping λa is a semi-homeomorphism and pa = λa p holds.

Proof: Since G is a group, it is easy to see that λa is a (well defined) bijection on G/H. We prove that λap = pa· Indeed, for each xG we have (pa)(x) = p(a * x) = (a*x)* Η = a* (x * Η) = λa(p(x)) = (λap)(x). This is required. It remains to prove that λa is irresolute and pre-semi-open.This follows from the following facts. Let x * Η G/H. For any semi-open neighbourhood U of eG, p(x * U * H)is a semi-open neighbourhood of x * Η in G/H. Similarly, the set p (a * x * U * H)is a semi-open neighbourhood of a * x * Η in G/H. Sinceλa(p(x * U * H)) = p(la(x * U * H)) = p (a * x * U * H),it follows that λa is a semi-homeomorphism.

Definition 5.13: A mapping ƒ : X → Y is:

• -

an S-isomorphism if it is an algebraic isomorphism and (topologically) an S -homeomorphism;

• -

a semi-isomorphism if it is an algebraic isomorphism and a semi-homeomorphism.

Theorem 5.14: Let (G,*,τg) and (Η, ·,τη) be extremally disconnected irresolute-topological groups and ƒ : G → H a semi-isomorphism. If G0 is an invariant subgroup of G and H0 = f (G0), then the semi-quotient irresolute-topological groups (G/G0,sτQ) and (H/H0,sτQ) are semi-isomorphic.

Proof: Let ρ : G → G/G0, x ↦ x * G0, and π : Η → Η/H0, f(x0)↦ f(x0). H0(x0 G0) be the canonical projections. Consider the mapping φ : G/Go → H/H0defined byφ(x * G0) = f(x) · f(G0), xG, y = f(x).Then for x1 * G0, x2 * G0G/G0 we haveφ(x1 * G0* x2 * G0) = φ(x1 * x2 * G0) = f(x1* x2) · f(G0) = y1 · y2 · H0 = φ(x1 * G0) · φ(x2 * G0),i.e. φ is a homomorphism. Let us prove that φ is one-to-one. Let x * G0be an arbitrary element of G/G0. Set y = f(x). If φ(x * G0) = H0, then π(y) = H0, which implies x ∈ G0, y ∈ H0, and ker φ = Go. So, φ is one-to-one.Next, we have φ(x * G0) = y · H0, i.e. φ(p(x)) = π(y) = π(f(x)). This implies φ ∘ p = π ∘ f. Since f is a semi-homeomorphism, and p and π are s-open, semi-continuous homomorphisms, we conclude that φ is open and continuous. Hence φ is semi-homeomorphism and a semi-isomorphism.

Theorem 5.15: Let (G, *, τ) be an extremally disconnected irresolute-topological group, H an invariant subgroup of G, M an open subgroup of G, and p GG/H the canonical projection. Then the semi-quotient group M * H/H is semi-isomorphic to the subgroup p(M) of G/H.

Proof: It is clear that M * H = p(p(M)). As p is s-open and semi-continuous and M is open in τ, the restriction π of p to M * H is an s-open and semi-continuous mapping of M * H onto p(M)by Lemma 5.4. Since M is a subgroup of G and p is a homomorphism it follows that p(M)is a subgroup of G/H, M * H is a subgroup of G, and π: M * H p(M)is a homomorphism. We have π(π(eG)) = p(p(eG)) = H, i.e. ker π = H. It is easy now to conclude that M * H/H and p(M)are semi-isomorphic.

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Accepted: 2016-10-06

Published Online: 2016-12-19

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, Export Citation