Proyecciones Vol. 28, No 1, pp. 1—19, May 2009. Universidad Cat´olica del Norte Antofagasta - Chile

ON g-HOMEOMORPHISMS IN TOPOLOGICAL SPACES e M. CALDAS UNIVERSIDADE FEDERAL FLUMINENSE, BRASIL S. JAFARI COLLEGE OF VESTSJAELLAND SOUTH, DENMARK N. RAJESH PONNAIYAH RAMAJAYAM COLLEGE, INDIA and M. L. THIVAGAR ARUL ANANDHAR COLLEGE Received : August 2006. Accepted : November 2008

Abstract In this paper, we first introduce a new class of closed map called g-closed map. Moreover, we introduce a new class of homeomor- phism called g-homeomorphism, which are weaker than homeomor- phism.e We prove that gc-homeomorphism and g-homeomorphism are independent.e We also introduce g*-homeomorphisms and prove that the set of all g*-homeomorphisms forms a groupe under the operation of composition of maps. e e 2000 Math. Subject Classification : 54A05, 54C08.

Keywords and phrases : g-closed set, g-open set, g-continuous function, g-irresolute map. e e e e 2 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

1. Introduction

The notion homeomorphism plays a very important role in topology. By definition, a homeomorphism between two topological spaces X and Y is a 1 bijective map f : X Y when both f and f − are continuous. It is well known that as J¨anich→ [[5], p.13] says correctly: homeomorphisms play the same role in topology that linear isomorphisms play in linear algebra, or that biholomorphic maps play in function theory, or group isomorphisms in group theory, or isometries in Riemannian geometry. In the course of generalizations of the notion of homeomorphism, Maki et al. [9] intro- duced g-homeomorphisms and gc-homeomorphisms in topological spaces. Recently, Devi et al. [2] studied semi-generalized homeomorphisms and generalized semi-homeomorphisms. In this paper, we first introduce g-closed maps in topological spaces and then we introduce and study g-homeomorphisms, which are weaker than homeomorphisms. We prove that gc-homeomorphisme and g-homeomorphism are independent. We also introducee g -homeomorphisms. It turns out that ∗ the set of all g*-homeomorphisms forms a group under thee operation com- position of functions. e e 2. Preliminaries

Throughout this paper (X, τ)and(Y,σ) represent topological spaces on which no separation axioms are assumed unless otherwise mentioned. For a subset A of a space (X, τ), cl(A), int(A)andAc denote the closure of A, the interior of A and the complement of A in X, respectively.

We recall the following definitions and some results, which are used in the sequel.

Definition 1. A subset A of a space (X, τ) is called: (i) semi-open [6] if A cl(int(A)), (ii) α-open [11] if A ⊂int(cl(int(A))), (iii) regular open [17]⊂ if A = int(cl(A)), (iv) pre-closed [10] if cl(int(A)) A. ⊂ Thesemi-closure[1](resp. α-closure [11]) of a subset A of X,denoted by sclX(A)(resp. αclX (A)) briefly scl(A), (resp. αcl(A)) is defined to be the intersection of all semi-closed (resp. α-closed) sets containing A. On g-homeomorphisms in topological spaces 3

e Definition 2. A subset A of a space (X, τ) is called a: (i) generalized closed (briefly g-closed) set [7] if cl(A) U whenever A U and U is open in (X, τ). ⊂ ⊂ (ii) generalized α-closed (briefly gα-closed) set [8] if αcl(A) U whenever A U and U is α-open in (X, τ). ⊂ (iii)⊂ g-closed set [19] if cl(A) U whenever A U and U is semi-open in (X, τ). ⊂ ⊂ (iv) bg-closed set [20] if cl(A) U whenever A U and U is g-open in (X, τ∗). ⊂ ⊂ (v) #g-semi-closed (briefly #gs-closed) set [21] if scl(A) U bwhenever A U and U is g-open in (X, τ). ⊂ (vi)⊂ g-closed set∗ [4] if cl(A) U whenever A U and U is #gs-open in (X, τ). ⊂ ⊂ e Where g-open (resp. g-open, gα-open, rg-open, g-open, #gs-open, g- open), are defined as the complement of g-closed (resp.∗ g-closed, gα-closed, rg-closed, bg-closed, #gs-closed, g-closed). e ∗ b Definition 3. A function f :(X,e τ) (Y,σ) is called: 1 → (i) g-continuous [18] if f − (V ) is g-closed in (X, τ) for every closed set V in (Y,σ). 1 (ii) g-continuous [19] if f − (V ) is g-closed in (X, τ) for every closed set V in (Y,σ). 1 (iii)bg-continuous [15] if f − (V ) is bg-closed in (X, τ) for every closed set V in (Y,σ). 1 (iv) ge-irresolute [13] if f − (V ) is g-closede in (X, τ) for every g-closed set V in (Y,σ). 1 (v) gce -irresolute [18] if f − (V ) iseg-closed in (X, τ) for each ge-closed set V of (Y,σ). 1 (vi) #gs-irresolute [21] if f − (V ) is #gs-closed in (X, τ) for each #gs- closed set V of (Y,σ). (vii) strongly g-continuous [13] if the inverse image of every g-open set in (Y,σ) is open in (X, τ). e e Definition 4. A function f :(X, τ) (Y,σ) is called: (i) g-open [18] if f(V ) is g-open in (Y,σ→ ) for every g-open set V in (X, τ). (ii) g-open [19] if f(V ) is g-open in (Y,σ) for every g-open set V in (X, τ).

Defibnition 5. A bijectiveb function f :(X, τ) (Y,σb ) is called a: (i) generalized homeomorphism (briefly g-homeomorphism)→ [9] if f is both 4 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar g-continuous and g-open, 1 (ii) gc-homeomorphism [9] if both f and f − are gc-irresolute maps, (iii) g-homeomorphism [19] if f is both g-continuous and g-open.

Defibnition 6. [14] Let (X, τ) be a topologicalb space and Eb X.Wedefine the g-closure of E (briefly g-cl(E)) to be the intersection of⊂ all g-closed sets containing E.i.e., g-cl(eE)= A : E A ande A GC(X, τ) . e ∩{ ⊂ ∈ } Propositione 2.1. [13] If f :(X,e τ) (Y,σ) is g-irresolute, then it is g- continuous. → e e Proposition 2.2. [14] Let (X, τ) be a topological space and E X.The following properties are satisfied: ⊂ (i) g-cl(E)) is the smallest g-closed set containing E and (ii) E is g-closed if and only if g-cl(E)) = E. e e Propositione 2.3. [14] For anye two subsets A and B of (X, τ), (i) If A B,theng-cl(A)) g-cl(B)), (ii) g-cl(⊂A B)) g-cl(A))⊂ g-cl(B)). ∩ ⊂ ∩ e e Propositione 2.4. e[4] Every ge-closed set is gα-closed and hence pre-closed.

Proof. Let A be g-closedin(e X, τ)andU be any α-open set containing A. Since every α-open set is #gs-open [21] and since αcl(A) cl(A)for ⊂ every subset A of X,e we have by hypothesis, αcl(A) cl(A) U and so A is gα-closed in (X, τ). ⊂ ⊂

In [8], it has been proved that every gα-closed set is pre-closed. There- fore, every g-closed set is pre-closed.

Theorem 2.5.e [4] Suppose that B A X, B is a g-closed set relative to A and that A is open and g-closed in⊂ (X,⊂ τ).ThenB is g-closed in (X, τ). e Corollary 2.6. [4] If A ise a g-closed set and F is a closede set, then A F is a g-closed set. ∩ e Theoreme 2.7. [4] A set A is g-open in (X, τ) if and only if F int(A) whenever F is #gs-closed in (X, τ) and F A. ⊂ ⊂ e Definition 7. [14] Let (X, τ) be a topological space and E X.We define the g-interior of E (briefly g-int(E)) to be the union of⊂ all g-open sets contained in E. e e e On g-homeomorphisms in topological spaces 5

e Lemma 2.8. [14] For any E X, int(E) g-int(E) E. ⊂ ⊂ ⊂ Proof. Since every open set is g-open, thee proof follows immediately.

Definition 8. A topological spacee (X, τ) is a, (i) T1/2 space [7] if every g-closed subset of (X, τ) is closed in (X, τ), (ii) Tg space [16] if every g-closed subset of (X, τ) is closed in (X, τ), (iii) semi-T1/2 space [19] if every g-closed subset of (X, τ) is closed in (X, τ). e e 3. g-closed maps b

Malghane [12] introduced the concept of generalized closed maps in topo- logical spaces. In this section, we introduce g-closed maps, g-open maps, g*-closed maps, g*-open maps and obtain certain characterizations of these maps. e e e e Definition 9. Amapf :(X, τ) (Y,σ) is said to be g-closed if the image of every closed set in (X, τ) is g-closed→ in (Y,σ). e Example 3.1. (a) Let X = eY = a, b, c , τ = ,X, a and σ = ,Y, b .Define a map f :(X, τ){ (}Y,σ) by {f∅(a)={ }}b, f(b)=a {and∅ f{(c)=}} c.Thenf is a g-closed map.→ (b) Let X = Y = a, b, c , τ = ,X, a , b , a, b and σ = ,Y, a, b . { } {∅ { } { } { }} {∅ { }} Let f :(X, τ) (Y,σ) be thee identity map. Then f is not a g-closed map. → Proposition 3.2. A mapping f :(X, τ) (Y,σ) is g-closede if and only if g-cl(f(A)) f(cl(A)) for every subset A →of (X, τ). ⊂ e e Proof. Suppose that f is g-closed and A X.Thenf(cl(A)) is g- closed in (Y,σ). We have f(A) f(cl(A)) and by⊂ Propositions 2.2 and 2.3, ⊂ g-cl(f(A)) g-cl(f(cl(A))) = fe(cl(A)). e Conversely,⊂ let A be any closed set in (X, τ). By hypothesis and Proposition 2.2,e we have Ae = cl(A)andsof(A)=f(cl(A)) g-cl(f(A)). Therefore, f(A)=g-cl(f(A)). i.e., f(A)isg-closed and hence⊃ f is g-closed. e Propositione 3.3. If f :(X, τ) e (Y,σ) is a g-closed mapping,e then for each subset A of (X, τ), cl(int(f(→A))) f(cl(A)). ⊂ e Proof. Let f be a g-closed map and A X.Thensincecl(A) is a closed set in (X, τ), we have f(cl(A)) is g-closed⊂ and hence pre-closed by Propo- sition 2.4. Therefore,e cl(int(f(cl(A)))) f(cl(A)). i.e., cl(int(f(A))) ⊂ ⊂ e 6 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar f(cl(A)).

The converse of this Proposition need not be true in general as seen from the following example. Example 3.4. Let X = Y = a, b, c , τ = ,X, a , b , a, b and σ = ,Y, a, b .Letf :(X, τ{) (}Y,σ) be{∅ the identity{ } { } map.{ }} Then for each{∅ subset{ }}A X,wehavecl(→int(f(A))) f(cl(A)), but f is not a g-closed map. ⊂ ⊂ Theorem 3.5. Amapf :(X, τ) (Y,σ) is g-closed if and only if for e → 1 each subset S of (Y,σ) and for each open set U containing f − (S) there is a g-open set V of (Y,σ) such that S V and f e1(V ) U. ⊂ − ⊂ Proof. Suppose that f is a g-closed map. Let S Y and U be an open e 1 ⊂ c c subset of (X, τ) such that f − (S) U.ThenV =(f(U )) is a g-open set 1 ⊂ containing S such that f − (V )e U. ⊂ 1 c c For the converse, let S be a closed set of (X, τ). Then f − ((f(Se)) ) S and Sc is open. By assumption, there exists a g-open set V of (⊂Y,σ) c 1 c 1 c such that (f(S)) V and f − (V ) S and so S (f − (V )) . Hence c ⊂ 1 c c ⊂ ⊂ c c V f(S) f((f − (V )) ) V which implies fe(S)=V . Since V is g-closed,⊂ f(S⊂)isg-closed and⊂ therefore f is g-closed. Proposition 3.6. If f :(X, τ) (Y,σ) is #gs-irresolute g-closed and A e e e is a g-closed subset of (X, τ),then→ f(A) is g-closed. e Proof. Let U be a #gs-open set in (Y,σ) such that f(A) U.Sincef e 1 e ⊂ is #gs-irresolute, f − (U)isa#gs-open set containing A.Hencecl(A) 1 ⊂ f − (U)asA is g-closedin(X, τ). Since f is g-closed, f(cl(A)) is a g-closed set contained in the #gs-open set U, which implies that cl(f(cl(A)) U ⊂ and hence cl(f(eA)) U. Therefore, f(A)isae g-closed set. e ⊂ The following example shows that the compositione of two g-closed maps need not be g-closed. e Example 3.7. Let X = Y = Z = a, b, c , τ = ,X, a , b, c , σ = e ,Y, a, c and γ = ,Z, b , a, c{ .De}fine a map{∅ f :({ X,} { τ) }}(Y,σ) {by∅ f(a{)=}}f(b)=b and{∅f(c)={ }a {and}} a map g :(Y,σ) (Z, γ) by g→(a)=c, g(b)=b and g(c)=a.Thenbothf and g are g-closed→ maps but their composition g f :(X, τ) (Z, γ) is not a g-closed map, since for the ◦ → closed set b, c in (X, τ), (g f)( b, c )= a, b ,e which is not a g-closed { } ◦ { } { } set in (Z, γ). e e On g-homeomorphisms in topological spaces 7

e Corollary 3.8. Let f :(X, τ) (Y,σ) be a g-closed map and g :(Y,σ) (Z, γ) be g-closed and #gs-irresolute→ map, then their composition g f→: ◦ (X, τ) (Z, γ) is g-closed. e → e Proof. Let A bee a closed set of (X, τ). Then by hypothesis f(A)is a g-closed set in (Y,σ). Since g is both g-closed and #gs-irresolute by Proposition 3.6, g(f(A)) = (g f)(A)isg-closedin(Z, γ) and therefore ◦ g ef is g-closed. e ◦ e Propositione 3.9. If f :(X, τ) (Y,σ) and g :(Y,σ) (Z, γ) are g- closed maps and (Y,σ) is a T space,→ then their composition→g f :(X, τ) g ◦ → (Z, γ) is a g-closed map. e e Proof. eLet A be a closed set of (X, τ). Then by assumption f(A)is g-closed in (Y,σ). Since (Y,σ)isaTg space, f(A)isclosedin(Y,σ)and again by assumption g(f(A)) is g-closed in (Z, γ). i.e., (g f)(A)isg-closed ◦ ine (Z, γ)andsog f is g-closed. ◦ e e e Proposition 3.10. If fe:(X, τ) (Y,σ) is g-closed, g :(Y,σ) (Z, γ) is g-closed and (Y,σ) is a T space,→ then their composition g f →:(X, τ) g ◦ → (Z, γ) is g-closed. e e Proof. Similar to Proposition 3.9.

Proposition 3.11. Let f :(X, τ) (Y,σ) be a closed map and g : (Y,σ) (Z, γ) be a g-closed map, then→ their composition g f :(X, τ) (Z, γ) →is g-closed. ◦ → e Proof.e Similar to Proposition 3.9.

Remark 3.12. If f :(X, τ) (Y,σ) is g-closed and g :(Y,σ) (Z, γ) is closed, then their composition→ need not be a g-closed map as seen→ from the following example. e e Example 3.13. Let X = Y = Z = a, b, c , τ = ,X, a , b, c , σ = ,Y, a, c and γ = ,Z, b , a, c { .Defi}ne a map{∅ f :({ X,} { τ) }}(Y,σ) {by∅ f(a{)=}}f(b)=b and{∅ f(c{)=} {a and}} g :(Y,σ) (Z, γ) be the→ identity map. Then f is a g-closed map and g is a closed map.→ But their composition g f :(X, τ) (Z, γ) is not a g-closed map, since for the closed set b, c ◦ → { } in (X, τ), (g f)(eb, c )= a, b , which is not a g-closed set in (Z, γ). ◦ { } { } e e 8 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

Theorem 3.14. Let f :(X, τ) (Y,σ) and g :(Y,σ) (Z, γ) be two mappings such that their composition→ g f :(X, τ) (Z,→ γ) be a g-closed mapping. Then the following statements◦ are true if:→ (i) f is continuous and surjective, then g is g-closed. e (ii) g is g-irresolute and injective, then f is g-closed. (iii) f is g-continuous, surjective and (X, τ)e is a T1/2 space, then g is g- closed. e e (iv) f is g-continuous, surjective and (X, τ) is a semi-T1/2 space, then g eis g-closed. (v) g is stronglye g-continuous and injective, then f is closed. e e 1 Proof. (i) Let A be a closed set of (Y,σ). Since f is continuous, f − (A) 1 is closed in (X, τ) and since g f is g-closed, (g f)(f − (A)) is g-closed in (Z, γ). i.e., g(A)isg-closedin(◦ Z, γ), since f is◦ surjective. Therefore, g is a g-closed map. e e (ii) Let B be a closede set of (X, τ). Since g f is g-closed, (g f)(B)isg- ◦ ◦ closede in (Z, γ). Since g is g-irresolute, g 1((g f)(B)) is g-closedin(Y,σ). − ◦ i.e., f(B)isg-closedin(Y,σ), since g is injective.e Thus, f is a g-closede map. e e 1 (iii) Let A bee a closed set of (Y,σ). Since f is g-continuous, f −e (A)is 1 g-closedin(X, τ). Since (X, τ)isaT1/2 space, f − (A)isclosedin(X, τ) andsoasin(i),g is a g-closed map. 1 (iv) Let A be a closed set of (Y,σ). Since f is g-continuous, f − (A)is 1 g-closed in (X, τ). Sincee (X, τ)isasemi-T1/2 space, f − (A)isclosedin (X, τ) and so as in (i), g is a g-closed map. e (v)e Let D be a closed set of (X, τ). Since g f is g-closed, (g f)(D)is ◦ ◦ g-closedin(Z, γ). Since g is stronglye g-continuous, g 1((g f)(D)) is closed − ◦ in (Y,σ). i.e., f(D)isclosedin(Y,σ), since g is injective.e Therefore, f is aclosedmap.e e

As for the restriction fA of a map f :(X, τ) (Y,σ) to a subset A of (X, τ), we have the following: →

Theorem 3.15. Let (X, τ) and (Y,σ) be any topological spaces. Then if: (i) f :(X, τ) (Y,σ) is g-closed and A is a closed subset of (X, τ),then → fA :(A, τA) (Y,σ) is g-closed. → (ii) f :(X, τ) (Y,σ) ise #gs-irresolute and g-closed and A is an open → subset of (X, τ),thenfAe :(A, τA) (Y,σ) is g-closed. → (iii) f :(X, τ) (Y,σ) is g-closed (resp. closed)e and A = f 1(B) for → − e e On g-homeomorphisms in topological spaces 9

e some closed (resp. g-closed) set B of (Y,σ),thenfA :(A, τA) (Y,σ) is g-closed. → e Proof. (i) Let B be a closed set of A.ThenB = A F for some closed e set F of (X, τ)andsoB is closed in (X, τ). By hypothesis,∩ f(B)isg-closed in (Y,σ). But f(B)=fA(B) and therefore fA is a g-closed map. (ii) Let C be a closed set of A.ThenC is g-closed in A.SinceAeis both open and g-closed, C is g-closed, by Theorem 2.5.e Since f is both #gs- irresolute and g-closed, f(C)isg-closed in (Y,σ), by Proposition 3.6. Since f(C)=fAe(C), fA is a g-closede map. (iii) Let D bee a closed set of Ae.ThenD = A H for some closed set H ∩ 1 of (X, τ). Now fA(D)=e f(D)=f(A H)=f(f (B) H)=B f(H). Since f is g-closed, f(H)isg-closed and∩ so B f(H)is∩g-closedin(∩ Y,σ) ∩ by Corollary 2.6. Therefore, fA is a g-closed map. e e e The next theorem shows that normalitye is preserved under continuous g-closed maps. Theorem 3.16. If f :(X, τ) (Y,σ) is a continuous, g-closed map from e anormalspace(X, τ) onto a space→ (Y,σ) then (Y,σ) is normal. e Proof. It follows from Theorem 1.12 of [12] and the fact that every g-closed map is g-closed. e Analogous to a g-closed map, we define a g-open map as follows: Definition 10. Amapf :(X, τ) (Y,σ) is said to a g-open map if the e e image f(A) is g-open in (Y,σ) foreachopenset→ A in (X, τ). e Proposition 3.17. For any bijection f :(X, τ) (Y,σ),thefollowing e statements are equivalent: → 1 (i) f − :(Y,σ) (X, τ) is g-continuous, (ii) f is a g-open→ map and (iii) f is a g-closed map. e e Proof. (i) (ii): Let U be an open set of (X, τ). By assumption 1 1 e ⇒ (f − )− (U)=f(U)isg-openin(Y,σ)andsof is g-open. (ii) (iii): Let F be a closed set of (X, τ). Then F c is open in (X, τ). ⇒ By assumption, f(F c)ise g-openin(Y,σ). i.e., f(Fec)=(f(F ))c is g-open in (Y,σ) and therefore f(F )isg-closed in (Y,σ). Hence f is g-closed. (iii) (i): Let F be a closede set in (X, τ). By assumption f(F )isg-closede ⇒ 1 1 1 in (Y,σ). But f(F )=(f − )− (Fe ) and therefore f − is g-continuouse on Y . e e 10 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

Definition 11. Let x be a point of (X, τ) and V be a subset of X.Then V is called a g-neighbourhood of x in (X, τ) if there exists a g-open set U of (X, τ) such that x U V . ∈ ⊂ e e In the next two theorems, we obtain various characterizations of g-open maps. e Theorem 3.18. Let f :(X, τ) (Y,σ) be a mapping. Then the following statements are equivalent: → (i) f is a g-open mapping. (ii) For a subset A of (X, τ), f(int(A)) g-int(f(A). ⊂ (iii) For eache x X and for each neighborhood U of x in (X, τ),there ∈ exists a g-neighbourhood W of f(x) in (Y,σe ) such that W f(U). ⊂ Proof.e (i) (ii): Suppose f is g-open. Let A X.Sinceint(A)is open in (X, τ),⇒f(int(A)) is g-openin(Y,σ). Hence f⊂(int(A)) f(A)and ⊂ we have, f(int(A)) g-int(f(A)). e ⊂ (ii) (iii): Suppose (ii)e holds. Let x X and U be an arbitrary ⇒ ∈ neighborhood of x ine (X, τ). Then there exists an open set G such that x G U. By assumption, f(G)=f(int(G)) g-int(f(G)). This im- plies∈ f(⊂G)=g-int(f(G)). Therefore, f(G)isg-open⊂ in (Y,σ). Further, f(x) f(G) f(U)andso(iii)holds,bytakingW e= f(G). ∈ ⊂ (iii) (i): Supposee (iii) holds. Let U be anye open set in (X, τ), x U and f⇒(x)=y.Thenx U and for each y f(U), by assumption there∈ ∈ ∈ exists a g-neighbourhood Wy of y in (Y,σ) such that Wy f(U). Since Wy ⊂ is a g-neighbourhood of y, there exists a g-open set Vy in (Y,σ) such that y Vy e Wy.Therefore,f(U)= Vy : y f(U) . Since the arbitrary ∈ ⊂ ∪{ ∈ } unione of g-open sets is g-open, f(U)isage-open set of (Y,σ). Thus, f is a g-open mapping. e e e Theoreme 3.19. A function f :(X, τ) (Y,σ) is g-open if and only if for → 1 any subset B of (Y,σ) and for any closed set S containing f − (B),there exists a g-closed set A of (Y,σ) containing B suche that f 1(A) S. − ⊂ Proof.e Similar to Theorem 3.5.

Corollary 3.20. A function f :(X, τ) (Y,σ) is g-open if and only if f 1(g-cl(B)) cl(f 1(B)) for every subset→ B of (Y,σ). − ⊂ − e 1 Proof.e Suppose that f is g-open. Then for any B Y , f − (B) 1 ⊂ ⊂ cl(f − (B)). By Theorem 3.19, there exists a g-closed set A of (Y,σ) e e On g-homeomorphisms in topological spaces 11

e 1 1 1 such that B A and f − (A) cl(f − (B)). Therefore, f − (g-cl(B)) f 1(A) cl(f⊂ 1(B)), since A is⊂ a g-closed set in (Y,σ). ⊂ − ⊂ − e Conversely, let S be any subsete of (Y,σ)andF be any closed set con- 1 taining f − (S). Put A = g-cl(S). Then A is a g-closed set and S A. By assumption, f 1(A)=f 1(g-cl(S)) cl(f 1(S)) A and therefore⊂ by − − ⊂ − ⊂ Theorem 3.19, f is g-open. e e e Finally in this section,e we define another new class of maps called g*- closed maps which are stronger than g-closed maps. e Definition 12. Amapf :(X, τ) (eY,σ) is said to be a g*-closed map if the image f(A) is g-closed in (Y,σ→) for every g-closed set A in (X, τ). e For instance thee map f in Example 3.1 ise a g*-closed map

Remark 3.21. Since every closed set is a g-closede set we have every g*- closed map is a g-closed map. The converse is not true in general as seen from the following example. e e e Example 3.22. Let X = Y = a, b, c , τ = ,X, a, b , σ = ,Y, a , a, b and f :(X, τ) (Y,σ) be the identity{ } map.{ Then∅ {f is}} a g-closed{∅ map{ but} { }} not a g*-closed→ map, since a, c is a g-closed set in (X, τ), but its image { } under f is a, c , which is not g-closed in (Y,σ). e { } e e Proposition 3.23. A mappinge f :(X, τ) (Y,σ) is g*-closed if and only if g-cl(f(A)) f(g-cl(A)) for every subset→A of (X, τ). ⊂ e eProof. Similare to Proposition 3.2.

Analogous to g*-closed map we can also define g*-open map.

Proposition 3.24.e For any bijection f :(X, τ) (eY,σ),thefollowingare equivalent: → 1 (i) f − :(Y,σ) (X, τ) is g-irresolute, (ii) f is a g*-open→ and (iii) f is a g*-closed map. e e Proof. Similare to Proposition 3.17. Proposition 3.25. If f :(X, τ) (Y,σ) is #gs-irresolute and g-closed, then it is a g*-closed map. → e e 12 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

Proof. Follows from Proposition 3.6.

Lemma 3.26. Let A be a subset of X.Thenp g-cl(A) if and only if for any g-neighborhood N of p in X, A N = . ∈ ∩ 6 ∅ e Definition 13. Let A be a subset of X. A mapping r : X A is called e → a g-continuous retraction if r is g-continuous and the restriction rA is the identity mapping on A. e e Definition 14. A topological space (X, τ) is called a g-Hausdorff if for each pair x, y of distinct points of X,thereexistsg-neighborhoods U1 and U2 of x and y, respectively, that are disjoint. e e Theorem 3.27. Let A be a subset of X and r : X A be a g-continuous retraction. If X is g-Hausdorff,thenA is a g-closed→ set of X. e Proof. Supposee that A is not g-closed.e Then there exists a point x in X such that x g-cl(A) but x/A.Itfollowsthatr(x) = x because ∈ ∈ 6 r is g-continuous retraction. Since eX is g-Hausdorff, there exists disjoint g-open sets U and Vein X such that x U and r(x) V .NowletW be ∈ ∈ an arbitrarye g-neighborhood of x.TheneW U is a g-neighborhoodof x. ∩ Sincee x g-cl(A), by Lemma 3.26, we have (W U) A = .Therefore ∈ ∩ ∩ 6 ∅ there exists ae point y in W U A.Sincey A,wehavee r(y)=y U ∩ ∩ ∈ ∈ and hence er(y) / V . This implies that r(W ) V because y W .Thisis contrary to the ∈g-continuity of r. Consequently,6⊂ A is a g-closed∈ set of X.

Theorem 3.28. Let Xi : i I be any family of topological spaces. If e { ∈ } e f : X Xi is a g-continuous mapping, then Pri f : X Xi is → ◦ → g-continuous for each i I,wherePri is the projection of Xj onto Xi. Q ∈ e Q e Proof. We shall consider a fixed i I. Suppose Ui is an arbitrary open 1 ∈ set in X .ThenPri− (Ui)isopenin Xi.Sincef is g-continuous, we 1 1 1 have, f (Pr− (Ui)) = (Pri f) (Ui) g-open in X. Therefore Pri f is − i ◦ − Q ◦ g-continuous. e e 4.e g-Homeomorphisms

In thise section we introduce and study two new homeomorphisms namely g- homeomorphism and g*-homeomorphism. We prove that gc-homeomorphism and g-homeomorphism are independent and g*-homeomorphism is an equiv-e alence relation betweene topological spaces. e e On g-homeomorphisms in topological spaces 13

e Definition 15. A bijection f :(X, τ) (Y,σ) is called g-homeomorphism if f is both g-continuous and g-open. → e Propositione 4.1. Every homeomorphisme is a g-homeomorphism but not conversely. e Proof. It follows from definitions.

TheconverseoftheaboveProposition4.1neednotbetrueasseenfrom the following example.

Example 4.2. Let X = Y = a, b, c , τ = ,X, a, b and σ = ,Y, a , a, b . The identity map f on X is g-homeomorphism{ } {∅ but{ not}} a homeomorphism,{∅ { } { }} because it is not continuous. e Thus, the class of g-homeomorphisms properly contains the class of homeomorphisms. Next we show that the class of g-homeomorphisms prop- erly contains the class ofe g-homeomorphisms.

Proposition 4.3. Everyeg-homeomorphism is a g-homeomorphism (resp. g-homeomorphism) but not conversely. e b Proof. Since every g-continuous map is g-continuous (resp. g-continuous) and every g-open map is g-open (resp. g-open), the proof follows. e b TheconversesoftheaboveProposition4.3neednotbetrueasseenb b from the following examples.

Example 4.4. Let X = Y = a, b, c , τ = ,X, a and σ = ,Y, b . Define a map f :(X, τ) (Y,σ{) by f}(a)={c∅, f(b)={ }}a and f(c)={∅ b.Then{ }} f is a g-homeomorphism.→ However, f is not a g-homeomorphism.

Example 4.5. Let X = Y = a, b, c , τ = e ,X, a , b, c and σ = ,Y, a , b , a, b , b, c .Let{f :(X,} τ) {(Y,σ∅ ) {be} the{ identity}} map. {Then∅ {f is} {g-homeomorphism.} { } { }} However, f is→ not a g-homeomorphism.

Propositionb 4.6. Let f :(X, τ) (Y,σ) be a bijectione g-continuous map. Then the following statements are→ equivalent: (i) f is a g-open map. b (ii) f is a g-homeomorphism. (iii) f is aeg-closed map. e e 14 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

Proof. It follows from Proposition 3.17.

The composition of two g-homeomorphism maps need not be a g-homeomorphism as can be seen from the following example. e Examplee 4.7. Let X = Y = Z = a, b, c , τ = ,X, a , b , a, b , σ = ,Y, a, b and γ = ,Z, a{, a, b} .Let{∅f :({X,} τ{) } { (Y,σ}}) and g{∅:(Y,σ{ ) }} (Z, γ) be{ identity∅ { } maps.{ }} Then both f and →g are g- homeomorphisms→ but their composition g f :(X, τ) (Z, γ) is not a g- ◦ → homeomorphism, because for the open set b in (X, τ), (g f)( b )= be, { } ◦ { } { } which is not a g-open set in (Z, γ).Thereforeg f is not a g-open mape and so g f is not a g-homeomorphism. ◦ ◦ e e We next introducee a new class of maps called g*-homeomorphisms which forms a sub class of g-homeomorphisms. This class of maps is closed under composition of maps. e e Definition 16. A bijection f :(X, τ) (Y,σ) is said to be g*-homeomorphism 1 → if both f and f − are g-irresolute. e We denote the familye of all g-homeomorphisms (resp. g*-homeomorphism and homeomorphisms) of a topological space (X, τ) onto itself by g-h(X, τ) (resp. g*-h(X, τ)andh(X, τ)).e e e Propositione 4.8. Every g*-homeomorphism is a g-homeomorphism but not conversely. i.e., for any space (X, τ), g*-h(X, τ) g-h(X, τ). ⊂ e e Proof. It follows from Proposition 2.1e and the fact that every g*-open map is g-open. e Thee function g in Example 4.7 is a g-homeomorphism but not a g*- homeomorphism, since for the g-closed set a, c in (Y,σ), (g 1) 1( a, c )= { } − − { } g( a, c )= a, c which is not g-closed ine (Z, γ). Therefore, g 1 is noteg- { } { } − irresolute and so g is not a g*-homeomorphism.e e e Proposition 4.9. Every g*-homeomorphisme is a g-homeomorphism but not conversely. e Proof. Follows from Propositions 4.8 and 4.3.

The map f in Example 4.4 is a g-homeomorphism but not a g*-homeomorphism.

e On g-homeomorphisms in topological spaces 15

e Remark 4.10. g-homeomorphism and gc-homeomorphism are indepen- dent as can be seen from the following examples. e Example 4.11. Let X = Y = a, b, c , τ = ,X, a , a, b and σ = ,Y,, b b, c .Define f :(X,{ τ) }(Y,σ) by{∅ f(a{)=} {c, f(}}b)=b and {f∅(c)={a.Then}{ }}f is a gc-homeomorphism→ but not g-homeomorphism, be- cause f is not a g-continuous map. e e The map f in Example 4.4 is a g-homeomorphism but not a gc-homeomorphism, 1 because f − is not a gc-irresolute map. e Proposition 4.12. If f :(X, τ) (Y,σ) and g :(Y,σ) (Z, γ) are g*- homeomorphisms, then their composition→ g f :(X, τ) → (Z, γ) is also ◦ → g*-homeomorphism. e

1 1 1 e Proof. Let U be a g-open set in (Z, γ). Now, (g f)− (U)=f − (g− (U)) = 1 1 ◦ f − (V ), where V = g− (U). By hypothesis, V is g-open in (Y,σ)and so again by hypothesis,e f 1(V )isg-openin(X, τ). Therefore, g f is − ◦ g-irresolute. Also for a g-open set G in (X, τ), wee have (g f)(G)= ◦ g(f(G)) = g(W ), where W = f(G).e By hypothesis f(G)isg-openin(Y,σ) ande so again by hypothesis,e g(f(G)) is g-openin(Z, γ). i.e., (g f)(G)is ◦ g-openin(Z, γ) and therefore (g f) 1 is g-irresolute. Hencee g f is a ◦ − ◦ g*-homeomorphism. e e e Theoreme 4.13. The set g*-h(X, τ) is a group under the composition of maps. e Proof. Define a binary operation *: g*-h(X, τ) g*-h(X, τ) g*- h(X, τ)byf g = g f for all f,g g*-h(X, τ)and is× the usual operation→ ∗ ◦ ∈ ◦ of composition of maps. Then by Propositione 4.12, g e f g*-h(X,e τ). ◦ ∈ We know that the composition of mapse is associative and the identity map I :(X, τ) (X, τ) belonging to g*-h(X, τ) servers as the identitye element. If f g*-h→(X, τ), then f 1 g*-h(X, τ) such that f f 1 = f 1 f = I and ∈ − ∈ ◦ − − ◦ so inverse exists for each elemente of g*-h(X, τ). Therefore, (g*-h(X, τ), ) ◦ is a groupe under the operatione of composition of maps e e Theorem 4.14. Let f :(X, τ) (Y,σ) be a (g*-homeomorphism. Then f induces an isomorphism from→ the group g*-h(X, τ) onto the group g*- h(Y,σ). e e e 16 M. Caldas, S. Jafari, N. Rajesh and M. L. Thivagar

Proof. Using the map f,wedefine a map θf : g*-h(X, τ) g*-h(Y,σ) 1 → by θf (h)=f h f − for every h g*-h(X, τ). Then θf is a bijection. ◦ ◦ ∈ 1 Further, for all h1,h2 g*-h(X, τ), θf (h1 h2)=f e(h1 h2) f − e=(f h1 1 1 ∈ ◦ ◦ ◦ ◦ ◦ ◦ f ) (f h2 f )=θf (h1) θf (h2).e Therefore, θf is a homeomorphism − ◦ ◦ ◦ − ◦ and so it is an isomorphisme induced by f.

Theorem 4.15. g*-homeomorphism is an equivalence relation in the col- lection of all topological spaces. e Proof. Reflexivity and symmetry are immediate and transitivity follows from Proposition 4.12.

Theorem 4.16. If f :(X, τ) (Y,σ) is a g*-homeomorphism, then g- cl(f 1(B)) = f 1(g-cl(B)) for all→ B Y . − − ⊂ e e Proof. Since f ise a g*-homeomorphism, f is g-irresolute. Since g-cl(f(B)) 1 is a g-closed set in (Y,σ), f − (g-cl(f(B))) is g-closedin(X, τ). Now, 1 1 1 1 f − (B) f − (g-cl(B))e and so by Proposition 2.3,e g-cl(f − (B)) ef − (g-cl(B)). ⊂ 1 ⊂ 1 Againe since f is a g*-homeomorphism,e f − is g-irresolute.e Since g-cl(f − (B)) 1 1 1 1 is g-closedin(X,e τ), (f − )− (g-cl(f − (B))) = f(ge-cl(f − (B))) is g-closede in (Y,σ). Now, Be (f 1) 1(f 1(B))) (f e1) 1(g-cl(f 1(B)))e = ⊂ − − − ⊂ − − − f(ge-cl(f 1(B))) and so g-cl(Be) f(g-cl(f 1(B))).e Therefore, e − ⊂ − f 1(g-cl(B)) f 1(f(g-cl(f 1(B)))) g-cl(f 1(Be)) and hence the equal- − ⊂ − − ⊂ − itye holds. e e e e e Corollary 4.17. If f :(X, τ) (Y,σ) is a g*-homeomorphism, then g- cl(f(B)) = f(g-cl(B)) for all B → X. ⊂ e e 1 Proof. Sincee f :(X, τ) (Y,σ)isag*-homeomorphism, f − :(Y,σ) (X, τ)isalsoag*-homeomorphism.→ Therefore, by Theorem 4.16, → g-cl((f 1) 1(B)) = (f 1) 1(g-cl(B)) fore all B X.i.e.,g-cl(f(B)) = f(g- − − − − ⊂ cl(B)). e e e e e Corollary 4.18. If f :(X, τ) (Y,σ) is a g*-homeomorphism, then f(g- int(B)) = g-int(f(B)) for all B→ X. ⊂ e e c c Proof. eFor any set B X, g-int(B)=(g-cl(B )) . Thus, by utilizing Corollary 4.17, we obtain ⊂f(g-int(B)) = f((g-cl(Bc))c)=(f(g-cl(Bc)))c = (g-cl(f(Bc)))c =(g-cl((f(B))c))ec = g-int(f(Be)). e e e Corollarye 4.19. eIf f :(X, τ) e(Y,σ) is a g*-homeomorphism, then f 1(g-int(B)) = g-int(f 1(B)) for→ all B Y . − − ⊂ e e e On g-homeomorphisms in topological spaces 17

e 1 Proof. Since f − :(Y,σ) (X, τ)isalsoag*-homeomorphism, the proof follows from Corollary 4.18.→ e Theorem 4.20. Let f :(X, τ) (Y,σ) is a g-continuous function and → Gg(f)= (x, y) X Y : y = f(x) ,whereX Y is the product topology { ∈ × } × and Gg(f) is called the g-graph of f. Then thee following properties are satisfied: (1) Gg(f), as a subspacee of X Y ,isg-homeomorphic to X. × (2) If Y is g-Hausdorff space, then Gg(f) is g-closed in X Y . × e Proof. (1).e The function g : X Gg(f)isdee fined by g(x)=(x, f(x)) → 1 for each x X is g-continuous and g− is also g-continuous. It is obvious that g is an∈ injective function. Suppose D (resp. E) is an arbitrary g- neighborhood of xe X (resp. g(x)=(x, f(x)) ine Gg(f)). So there exist g- ∈ open sets U and V in X and Y containing x and f(x) respectively for whiche (U V ) Gg(f) E and U D and f(U) V .LetM =(U V ) Gg(fe). Then× (x,∩ f(x)) ⊂M and x ⊂ g 1(M) U⊂ D. It implies that× g∩ 1 is g- ∈ ∈ − ⊂ ⊂ − continuous. Moreover, g(U) U f(U) U V1 and g(U) Gg(f). ⊂ × ⊂ × ⊂ Therefore, g(U) (U V ) Gg(f) E.Henceg is g-continuous whiche means that g is a⊂g-homeomorphism.× ∩ ⊂ (2). Suppose that (x, y) / Gg(f). Then y1 = f(x) =ey. By hypothesis, ∈ 6 there exist disjointe g-open sets V1 and V in Y such that y1 V1, y V . Since f is g-continuous, there exists an open set U in X containing∈ x∈such that f(U) V 1. Thene g(U) U V1. It follows from this and the fact ⊂ ⊂ × that V1 Ve = that (U V ) Gg(f)= . ∩ ∅ × ∩ ∅

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M. Caldas Departamento de Matem´atica Aplicada, Universidade Federal Fluminense, Rua M´ario Santos Braga, s/no 24020-140, Niter´oi, RJ BRAZIL e-mail: [email protected]ff.br

S. Jafari College of Vestsjaelland South, Herrestraede 11 4200 Slagelse, DENMARK e-mail: [email protected]

N. Rajesh Department of Mathematics, Ponnaiyah Ramajayam College, Thanjavur, TamilNadu, INDIA e-mail: [email protected]

and

M. L. Thivagar Department of Mathematics, Arul Anandhar College, Karumathur, Madurai, TamilNadu, INDIA e-mail: [email protected]