Special Issue - 2020 International Journal of Engineering Research & Technology (IJERT) ISSN: 2278-0181 RTICCT - 2020 Conference Proceedings

Weakly Quotient Map and Space

Chidanand Badiger T Venkatesh Department of Mathematics, Department of Mathematics, Rani Channamma University, Belagavi-591 156, Rani Channamma University, Belagavi-591 156, Karnataka, INDIA. Karnataka, INDIA.

Abstract- In this paper, we introduce the new weakly quotient Definition 2.1 [9] A subset F of a space M is said to be map (briefly, 푤-quotient map), strongly 푤-quotient map and Weakly closed (briefly, w-closed) if cl(F) ⊆ U whenever 푤*-quotient map. Here will discuss composition of few such F ⊆ U and U is semi-open in M. Respectively called w-open quotient maps. Also, we investigate some important properties set, if M ∖ F is w-closed set in M. We denote the set of all w- and consequences associated with usual quotient and bi, tri- quotient maps. open sets in M by wO(M). Definition 2.2 A map h: M → N is said to be 푤-continuous −1 Key Words: 푤-quotient map, strongly 푤-quotient map, 푤*-quotient map, if h (F) is 푤-closed set of M, for every closed set F of map, 푤-quotient space. N. AMS Classification: 54A05, 54C10, 54H99. Definition 2.3 A map h: M → N is said to be 푤-irresolute map, if h−1(F) is 푤-closed set of M, for every 푤-closed set F I. INTRODUCTION of N. Construction is needed in any branch of science, Definition 2.4 A bijection h: M → N is called 푤- which can be sufficiently achieved by the theory of quotient , if both h and h−1 are 푤-continuous. notion in Mathematics. We can see how identifications Definition 2.5 (Consider) A M is called explain or yield simple objects like circle, sphere, suspension, ηw-space, if every 푤-closed set is closed set. and torus. In fact not only special objects, most of the objects Theorem 2.6 Every 푤-irresolute maps are 푤-continuous. are actually quotient creations. Allen Hetcher [1] has III. 푤-QUOTIENT MAP discussed mapping cylinder, mapping cone etc. This takes We introduce here some class of maps 푤-quotient map, frontiers part of abstract quotient construction. Mobius strips, strongly 푤-quotient map, 푤*-quotient map in topological Klein bottle are quotient spaces, which have classical space. Investigate some relations between them with usual importance having special topological properties. quotient maps. Also we characterize few notions. Visualization is new aspect in the theory of gluing, but many Definition 3.1 Let M and N be two topological spaces then a time we unable to catch the object. surjective map h: M → N is said to be 푤-quotient map if h is There are many situations in where we build a 푤-continuous and h−1(V) is open in M implies V is 푤-open topological space by starting with some set (space) and doing set in N. some kind of “gluing” or “identifications” that makes suitable Example 3.2 Let h: ( ℝ, η ) → ( ℝ, η ) and η be usual object for our claim. In the literature of Mathematics, u u u topology on ℝ , the map h is defined by h(x) = 5x is quotient maps are generally called strong continuous maps or obviously (See 3.3) 푤-quotient map. identification maps, because of strong conditions of Theorem 3.3 Every quotient map is 푤-quotient map. continuity i.e. h−1(V) is open in M if and only if V is open in Proof: Let h: M → N be any quotient map, by definition h is N of a surjective fiunction h: M → N and their importance for surjective. It is known that every continuous map is 푤- the philosophy of gluing. [9] In 2000, M.Sheik John, introduced and studied w-closed sets respective properties continuous map, hence h is 푤-continuous. Since h is a −1( ) map in topological space. Lellis Thivagar [7], Chidanand quotient map, h V is open in M, implies V is open in N. Badiger et al. [2] and Balamani et al. [10] discussed As every open set is 푤-open set, implies V is 푤-open set in N. generalized quotient maps, rw-quotient map and ψ∗α - Therefore h is 푤-quotient map. quotient maps in topological space respectively. Bi-quotient, Theorem 3.4 [5; § 22] If h: M → N is 푤-continuous and Tri-quotient maps studied by E. Michal [4,3]. k: N → M is continuous, such that h ∘ k: N → N is identity We introduce such class of maps called 푤-quotient map, then h is 푤-quotient map. strongly 푤-quotient map, 푤*-quotient map in topological Proof: Since h ∘ k = Id implies h is bijective. h is 푤- −1 space. Which have many rich consequences concern to usual continuous by hypothesis. For any V ⊂ N with h (V) be −1 −1 quotient and bi, tri-quotient maps. Here we have given open in M , continuity of k gives the k (h (V)) = example and counter examples for respective results. (h ∘ k)−1(V) = Id−1(V) = V is open in N, implies V is 푤- open set in N. Therefore h is 푤-quotient map. II. PRELIMINARIES Theorem 3.5 h become 푤-quotient map, whenever h: M → N Throughout this paper L, M, N, R and S represent the is surjective, continuous and open map. topological spaces on which no separation axioms are Proof: By theorem 3.3, h is 푤-quotient map, because every assumed unless otherwise mentioned. Map here mean h: M → N is surjective, continuous and open maps are function and for a subset F of topological space M, the M\F quotient map. denotes the complement of F in M. We recall the following Theorem 3.6 h become 푤-quotient map, whenever h: M → N definitions. is surjective, continuous and closed map.

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Theorem 3.7 h become 푤-quotient map, whenever h: M → N Remark 5.4 Composition of two 푤-quotient maps need not is surjective, 푤-continuous and 푤-open map. be 푤-quotient map. Theorem 3.8 h become 푤-quotient map, whenever h: M → N Theorem 5.5 If N is ηw topological space, h: M → N is 푤- is surjective, 푤-continuous and 푤-closed map. quotient map and k: N → R is quotient map then k ∘ h is 푤- Proof: Two conditions are obvious by the hypothesis. For last quotient map. condition, any V ⊂ N with h−1(V) be open in M, implies M ∖ Proof: Obviously k ∘ h is surjective and for every open set h−1(V) is closed set in M. Since h is 푤-closed map implies U ⊂ R, the (k ∘ h)−1(U) = h−1(k−1(U)) which is 푤-open set h(M ∖ h−1(V)) = N ∖ V is 푤-closed in N. Therefore V is 푤- in M, implies k ∘ h is 푤-continuous. For every U ⊂ R with open set in N, h is 푤-quotient map. (k ∘ h)−1(U) = h−1(k−1(U)) is open in M, implies k−1(U) is IV. STRONGLY 푤-QUOTIENT AND 푤∗-QUOTIENT 푤-open set in N, by the hypothesis k is quotient map. Hence MAP U is open set as well 푤-open set in R. Definition 4.1 Let M and N be two topological spaces, a Theorem 5.6 If N is ηw topological space, h: M → N is surjective map h: M → N is said to be strongly 푤-quotient quotient map and k: N → R is 푤-quotient map then k ∘ h is 푤- map provided V ⊂ N is open in N if and only if h−1(V) is 푤- quotient map. open in M. Proof: Similar arguments with essential changes in 5.5 can Definition 4.2 Let M and N be two topological spaces, a works. surjective map h: M → N is said to be 푤*-quotient map if h is Corollary 5.7 If N is ηw topological space, h: M → N is 푤- 푤-irresolute and h−1(V) is 푤-open in M implies V is open set quotient map and k: N → R is 푤-quotient map then k ∘ h is 푤- in N. quotient map. Theorem 4.3 Every injective 푤*-quotient map is 푤*-open Proof: It fallows by combining the theorems 5.5 and 5.6 and map. (Converse need not hold). Proof: Let h: M → N be any injective 푤*-quotient map, For Theorem 5.8 If h: M → N is a strong 푤–quotient map and every 푤-open set V in M, injective gives h−1(h(V)) = V is k: N → R is a quotient map then koh is strong 푤-quotient 푤-open set in M. Since h is 푤*-quotient map, implies h(V) is map. open set as well 푤-open in N. Proof: Here surjective of k and h gives koh is surjective and Theorem 4.4 Every injective 푤*-quotient map is 푤*-closed koh is 푤–continuous map due to composition of continuous map. and 푤- continuous. Lastly if h−1(k−1(V)) is open in M. Proof: Let h: M → N be injective 푤*-quotient map, For every Since h is strong 푤-quotient map and quotient of k implies V 푤-closed set F in M , implies M ∖ F is 푤-open set in M . is open in R. Therefore koh is strong 푤-quotient map. Obvious h(M ∖ F) = N ∖ h(F) is 푤-open set in N. Hence Theorem 5.9 If h: M → N is open surjective, 푤-irresolute and h(F) is 푤-closed set in N. k: N → R is 푤-quotient map then k ∘ h is 푤-quotient map. Theorem 4.5 Every strongly 푤-quotient map is 푤-quotient Proof: Obviously k ∘ h surjective. For every open set U ⊂ R, map. the k is 푤-continuous and h is 푤-irresolute gives (k ∘ Proof: Let h: M → N be strongly 푤-quotient map, obviously h)−1(U) = h−1(k−1(U)) is 푤-open set in M. Now for every −1 the first two conditions hold. For V ⊂ N with h (V) be an W ⊂ R with (k ∘ h)−1(W) = h−1(k−1(W)) be open in M. open set in M, also that become 푤-open in M. Since h is −1 −1 −1 strongly 푤-quotient map, implies V is open set in N, therefore Since h is open map h (h (k (W))) = k (W) is open in V is 푤-open set in N. N, given k is 푤-quotient map implies W is 푤-open set in R. Theorem 4.6 Every 푤*-quotient map is strongly 푤-quotient Hence k ∘ h is 푤-quotient map. map. Theorem 5.10 If h: M → N is 푤*-open surjective and 푤- Proof: Let h: M → N be strongly 푤*-quotient map, two irresolute and k: N → R is strongly 푤-quotient map then k ∘ h conditions are obvious. Because V is any open set in N is also is strongly 푤-quotient map. 푤-open set in N. Since h is 푤-irresolute implies h−1(V) is 푤- Proof: Obviously k ∘ h surjective and 푤-continuous. For open set in M. For W ⊂ N with h−1(W) be an open set in M, every U ⊂ R with (k ∘ h)−1(U) = h−1(k−1(U)) be 푤-open in implies h−1(W) is 푤-open in M. Since h is 푤*-quotient map, M , since h is 푤*-open map implies h(h−1(k−1(U)) ) = implies W is open set in N. k−1(U) is 푤-open in N. Since k is strongly 푤-quotient map Theorem 4.7 Every 푤*-quotient map is 푤-quotient map. implies U is open set in R. Hence k ∘ h is strongly 푤-quotient Proof: It followed by theorem 4.5 and 4.6 map. V. CONSEQUENCES ON COMPOSITIONS Theorem 5.11 If h: M → N is 푤*-open, surjective and 푤- Theorem 5.1 Composition of two quotient maps is 푤- irresolute and k: N → R is 푤*-quotient map then k ∘ h is 푤*- quotient map. quotient map. Proof: We know composition of two quotient maps is Proof: Arguments in 5.9 and 5.10 yield the proof. quotient map [5; 3.29]. Hence by theorem 3.3 which is 푤- Theorem 5.12 If h: M → N and k: N → R are 푤*-quotient quotient map. maps then k ∘ h is 푤*-quotient map. Remark 5.2 Composition of quotient map with 푤-quotient Proof: Obviously k ∘ h surjective and 푤 irresolute. For every map need not be 푤-quotient map. U ⊂ R with (k ∘ h)−1(U) = h−1(k−1(U)) be 푤-open in M , Remark 5.3 Composition of 푤-quotient map with quotient since h is 푤*-quotient map implies k−1(U) is open and 푤- map need not be 푤-quotient map. open in N. Since k is 푤*-quotient map implies U is open set in R. Hence k ∘ h is 푤*-quotient map.

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VI. STANDARD COMPARISONS AND denote this point, then which is clear that l is well–defined on APPLICATIONS N and can see as each x ∈ M , l(h(x)) = k(x). Theorem 6.1 If h: M → N is any map, where M and N are ηw ii) If l is continuous and h is 푤-continuous implies loh = k topological spaces then following are equivalent. is 푤 continuous. On other hand let U be any open set in R i) h is 푤*-quotient map then k−1(U) is an 푤-open set due to k is 푤-continuous. But ii) h is strongly 푤-quotient map k−1(U) = h−1(l−1(U)) is 푤-open in M. Since h is a strong iii) h is 푤-quotient map 푤–quotient map, l−1(U) is a open set. Hence l is continuous. Proof: (i) ⇒ (ii) Surjective is obvious and since every 푤- iii) If l is quotient map and h is 푤–quotient map by theorem irresolute map is 푤-continuous map. Third condition is trivial 5.11, loh = k is strong 푤-quotient map. On other hand, by the definition of 푤*-quotient map. since loh = k surjective implies l is surjective and l is (ii) ⇒ (iii) Obvious by the similar arguments in above. continuous by above result (ii). If l−1(U) is open in N and 푤- (iii) ⇒ (i) Surjective is obvious and h is 푤-irresolute, continuity of h implies h−1(l−1(U)) = k−1(U) is 푤-open in because every continuous map is 푤-irresolute because N is M. Since k is strong 푤-quotient map implies U is open in R. −1 −1 ηw. Let h (V) be 푤-open set in M, implies h (V) is open Hence l is quotient map. set in M. The 푤-quotient map of h implies V is 푤-open set The following theorems 6.5 to 6.13 followed obviously from and open set in N. the results of [4] i.e. every bi-quotient map is quotient map Theorem 6.2 If h: (M, η) → (N, ξ) is surjective, 푤-continuous and from [4,9] implies that every tri quotient maps are map then following are equivalent. quotient maps Hence by theorem 3.3 both class of maps i) h is strongly 푤-quotient map. become h as 푤-quotient maps. ii) For any k: (N, ξ) → (R, τ) then k is continous if and Theorem 6.5 Arbitrary product map of bi quotient maps are only if k ∘ h is 푤-continuous. 푤-quotient. iii) For fixed topology η on M then ξ is the maximal Theorem 6.6 If N is Hausdorff space and any map h: M → N topology for h to be 푤-continuous. is surjective continuous then h and h × IdR are 푤-quotient Proof: (i) ⇒ (ii) For every open set U ⊂ R , the (k ∘ maps for every space R. h)−1(U) = h−1(k−1(U)) is 푤-open set in (M, η) because of Theorem 6.7 If N is , map h: M → N is k is continuous and h is 푤-continuous. Therefore k ∘ h is 푤 surjective continuous map and h × k is quotient maps for continuous. Conversely, for every open set U of (R, τ), the every quotient maps k then h 푤-quotient maps. (k ∘ h)−1(U) = h−1(k−1(U)) is 푤-open set in (M, η) since h Theorem 6.8 If maps h: M → N and k: R → S are quotient is strongly 푤-quotient map gives k−1(U) is open in (N, ξ). maps and M and N × S are Hausdorff k-spaces then h × k is Therefore k is continuous. 푤-quotient map. (ii) ⇒ (iii) On contrary, with fixed toplogy η on M, there Theorem 6.9 Any surjective contiguous map h from M to a exists another topology ξ′ ⊋ ξ such that h: (M, η) → (N, ξ′) is hausdorff space N is 푤-quotient map whenever h is compact 푤-continuous. Obviously we see identity map id: (N, ξ) → covering, and N is locally compact. (N, ξ′) become not continuous, because ξ′ ⊋ ξ. But id ∘ h = Theorem 6.10 The h: M → N is continous surjective map h: (M, η) → (N, ξ′) become 푤-continuous. Hence id ∘ h is 푤- become 푤-quotient map whenever either h is perfect or h is continuous which contradicts to the hypothesis in (ii). compact-covering and Y is locally compact and Hausdorff. Therefore ξ is the maximal topology for h is 푤-continuous. Theorem 6.11 If M regular sieve-complete space, N is (iii) ⇒ (i) Surjective and 푤-continuous of h are obvious. For paracompact space then every inductively perfect map −1 last condition, on contrary, there exist a V0 ⊂ N with h (V0) h: M → N is 푤-quotient. is 푤-open set in (M, η), but V0 is not an open set in (N, ξ). Let Theorem 6.12 If M completely metrizable, N is paracompact ∗ ℬ = ξ ∪ {V0} , which induces topology ξ on N , which space then following gives h: M → N is 푤-quotient. ∗ ∗ contains V0 , also ξ ⊋ ξ . But h: (M, η) → (N, ξ ) also 푤- (i) Either N is metrizable or N is completely metrizable. continuous. Which contradicts to (iii), therefore h is strongly (ii) Y is a countably bi-k-space. 푤- quotient map. Theorem 6.13 Product map h × k: M × R → N × S is 푤- Theorem 6.3 [5; § 22] If h: M → N is 푤-continuous and quotient map whenever h: M → N and K: R → S are tri- k: N → M is continuous, such that h ∘ k: N → N is identity quotient. Theorem 6.14 Suppose h: M → N and k: N → R are with N is ηw- topological space then h is 푤*-quotient map. continuous maps then Proof: Arguments in theorem 3.6 and N is ηw-space can give the proof. (i) If h and k are tri-quotient, or bi quotient, or quotient then Theorem 6.4 [5; §22] If h: M → N is a strong 푤–quotient is koh is 푤-quotient map. map and k: M → R is a map that is constant on each set (ii) If koh is tri-quotient, or bi quotient, or quotient, or 푤 h−1({y}) for y ∈ N, then quotient, or strong 푤 quotient map then is k is 푤-quotient i) k induces a map l: N → R such that loh = k. map. ii) The induced map l is continuous iff k is 푤-continuous. Proof: i) Composition of two tri-quotients, or bi quotients, or iii) The induced map l is quotient iff k strong 푤-quotient quotients are respectively tri-quotients, or bi quotients, or map. quotients. Hence by Theorem 3.3 koh is 푤-quotient map. Proof: i) Since k is constant on h−1({y}) for y ∈ N, the set ii) Refer [3;Theorem 7.1 and Theorem 3.6] or Arguments are k(h−1({y})) is a one point set in R . By considering l(y) easy.

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Theorem 6.15 A maph: M → N is harmonious then h is 푤- ACKNOWLEDGMENT quotient map. The research work of the first author is supported by Proof: Arguments in [8; Proposition 3.8] says harmonious the Council of Scientific and Industrial Research, India under implies tri quotient and [3,8] gives the proof. grant award no: 09/1284(0001)2019-EMR-I with JRF exam Construction 6.16 Let M be a topological space and N be a roll no 400528. set. Let h: M → N be any surjective map construct a set REFERENCES −1 Qηw = { V ⊂ N | h (V) is w − open set in M } then obviously the collection Qη is not nessasarly topology on N [1] A. Hetcher, Algebraic Topology, Cambridge University press w (2002). (See basic union intersection properties of two 푤-open sets [2] Chidanand Badiger, T.Venkatesh and Basayya B. Mathad, Regular need obey axioms of topology) but some time (If M with weakly quotient map and space, Malaya Journal of Matematik, topology ηw then clear that Qηw become topology on N) this Vol. S, No. 1, 115-120, 2020. become topology. With this topology on N , h: M → N [3] E. Michel, Complete Spaces And Tri-Quotient Maps, University Of Washington Seattle, Washington, (1976), 716-733. become 푤-quotient map and quotient map. This topology [4] E. Michel, Bi-quotient maps and Cartesian products of quotient Qηw we are calling 푤-quotient topology and (N, Qηw) is maps, Annales de l’ institute Fourier, tome, 18,2 (1968), 287-302. called 푤-quotient topological space. [5] J. R. Munkres, Topology, Prentice Hall India Learn. Priv. Ltd; With which we can see weak sense of gluing or Second edition, (2002). [6] J. M. Lee, Introduction to Topological Manifolds, springer, (2000). identifying or joining in standard constructions like Cone, [7] L. Thivagar, A Note on quotient mappings, Bull, Malaysian cylinder, Sphere, Mobius strip, Torus, Klein bottle and math.Sci.(Second series,, 14(1991), 21-30. suspension etc. [1]. [8] M. Pillot, Tri-quotient maps become inductively perfect with the For standard example If ~ is any equivalence relations on M aid of consonance and continuous selections, Topology and its Applications, 104(2000), 237-253. with ηw then cannanical projection ρ: M → M⁄~ by, ρ(x) = [9] M.Sheik John, On w-closed sets in topology, ActaCienciaIndica,, [x] is clear 푤-quotient map, M⁄~ is 푤-quotient topological 4(2000), 389-392. [10] N. Balamani and A.Parvati, On topological 휓∗훼 -quotient space under 푤-quotient topology Qηw . mappings, Int. J. of Math. Trends and Tech, 48(3)(2017), 209-213. Theorem 6.17 If M with topology ηw , h: M → N is 푤 quotient map and M∗ = {h−1({y}) ∶ y ∈ N} then N is 푤- homeomorphic to M∗. ∗ Proof. M is Qηw topological space From theorem 3.58. define the map k: M∗ → N , by k([x]) = h(x) became 푤- homeomorphism. Application 6.18 The 푤-quotient map and 푤-quotient space are finding applications in gluing in some weak sense. Discussed map and space are weak sense of gluing or identifications and weakly glued space. As seen every gluing is 푤-gluing but not converse. Anyone can see the characterization notion that, If M is ηw given any space N and map from M to N then “the map is 푤-quotient map if and only if it is quotient map”. Hence gluing and weak gluing are same when M (involved space) is a ηw space.

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