Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 580 https://internationalpubls.com More on Contra Β-Open Mappings in a Quadripartitioned Neutrosophic Topological Spaces 1Mohanarao Navuluri, 2V Sathishkumar 1Department of Mathematics, Annamalai University, Annamalainagar, Tamilnadu, India. (Deputed to Government college of Engineering , Theni, Tamilnadu, India.) 2Department of Mathematics, Rajalakshmi institute of technology (Autonomous), Chennai; Department of Mathematics, Annamalai University, Annamalainagar, Tamilnadu, India. mohanaraonavuluri@gmail.com1, vsathishkumar2020@gmail.com2 Article History: Received: 01-10-2024 Revised: 29-11-2024 Accepted: 07-12-2024 Abstract In this article, we introduce the concept of a quadripartitioned neutrosophic contra β- continuous, quadriparti- tioned neutrosophic contra β-open and a quadripartitioned neutrosophic contra β-closed mappings in a quadri- partitioned neutrosophic topological spaces and studied some of their related properties. Further the work is extended to a quadripartitioned neutrosophic contra β-homeomorphism and a quadripartitioned neutrosophic contra β-Completely homeomorphism in a quadripartitioned neutrosophic topological spaces and establishes some of their related properties. Keywords: Quadripartitioned neutrosophic β-open set, Quadripartitioned neutrosophic contra β-continuous map, Quadripartitioned neutrosophic contra β-open map, Quadripartitioned neutrosophic contra β-closed map, Quadripartitioned neutrosophic contra β-homeomorphism, Quadripartitioned neutrosophic contra β- completely homeomorphism. 1 Introduction In mathematics, Zadeh25 was first presented a idea of fuzzy set between the intervals in order of logic and set hypothesis. The fuzzy set was attempted in general topology by Chang2 as fuzzy topological space. The intu- itionistic fuzzy set which contains a membership and non-membership values was introduced by Atanassov1 in 1983. Coker4 made intuitionistic fuzzy set in a topology entitled as intuitionistic fuzzy topological spaces. The ideas of neutrosophy and neutrosophic set was presented by Smarandache16,17 toward the start of 20th century. Salama and Alblowi14,15 in 2012, originated neutrosophic set and neutrosophic crisp set in a neutrosophic topological space. In the year 2016, Chatterjee et al.3 grounded the idea of quadripartitioned neutrosophic set and defined several similarity measures between two quadripartitioned neutrosophic sets. Iswaraya and Bageerathi9 studied the concept of neutrosophic semi-open sets and neutrosophic semi-closed sets. Push- palatha and Nandhini12grounded the idea of neutrosophic generalized closed sets in NTS’s. The notion of neutrosophic b-open sets in NTS’s was presented by Ebenanjar et al.8 Rao and Srinivasa13 grounded the concept of pre open set and pre closed set via neutrosophic topological spaces. Thereafter, Maheswari et al.10 studied the neutrosophic generalized b-closed sets in NTS’s. In the year 2019, Mohammed Ali Jaffer and Ramesh11 studied the concept of neutrosophic generalized pre-regular closed sets. The generalized neutro- sophic b-open sets in NTS’s was introduced by Das and mailto:mohanaraonavuluri@gmail.com1 mailto:vsathishkumar2020@gmail.com2 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 581 https://internationalpubls.com Pramanik.6 Das and Pramanik7 also defined the neutrosophic Φ-open sets and neutrosophic Φ- continuous mappings via NTS’s. Vadivel and Sundar defined γ open sets,18 γ continuous maps,20,21 β- open sets19 and β continuous maps22–24 in N -neutrosophic crisp topological spaces. In this paper, we develop the concept of quadripartitioned neutrosophic contra β-continuous maps, quadri- partitioned neutrosophic contra β-open maps and quadripartitioned neutrosophic contra β- closed maps in a quadripartitioned neutrosophic topological spaces and also specialized some of their basic properties with examples. Also, we discuss about quadripartitioned neutrosophic contra β- homeomorphism and quadripar- titioned neutrosophic contra β-completely homeomorphism in a quadripartitioned neutrosophic topological spaces and also specialized some of their basic properties with examples. 2 Preliminaries The needful basic definitions & properties are discussed in this section. Definition 2.1. 3 Let Z be a fixed set. Then, a quadripartitioned neutrosophic set (in-short, Q-Nss) U over Z is defined by U = {(u, TU (u), CU (u), IU (u), FU (u)) : u ∈ Z} where TU , CU , IU and FU (∈ [0, 1]) are the truth, contradiction, ignorance, and falsity membership values of u ∈ Z. So, 0 ≤ TU (u) + CU (u) + IU (u) +FU (u) ≤ 4. Definition 2.2. 3 Let Z be a non-empty set & the Q-Nss’s U & U0 in the form U = {(u, TU (u), CU (u),IU (u), FU (u)) : u ∈ Z}, U = {(u, TU (u), CU (u), IU (u), FU ) : u ∈ Z}, then (i) 0QNs = (u, 0, 0, 1, 1) and 1QNs = (u, 1, 1, 0, 0), (ii) U ⊆ U o iff TU (u) ≤ TUo(u), CU (u) ≤ CUo(u), IU (u) ≥ IUo(u) & FU (u) ≥ FUo(u) : u ∈ Z, (iii) 1QNs − U = {(u, FU (u), IU (u), CU (u), TU (u)) : u ∈ Z} = Uc , (iv) U ∪U0 = {(u, max(TU (u), TU0 (u)), max(CU (u), CU0(u)), min(IU (u), IU (u)), min(FU (u), FU0(u))) : u ∈ Z}, (v) U ∩U = {(u, min(TU (u), TU0u)), min(CU (u), CU0(u)), max(IU (u), IU0(u)), max(FU (u), FU0(u))) : u ∈ Z}. Definition 2.3. 5 Let Z be a fixed set. A collection ΓQ of some Q-Nss’s over Z is called a quadripartitioned neutrosophic topology (in-short, Q-Nst) on Z, if the following conditions holds: (i) 0N , 1N ∈ ΓQ. (ii) Gϕ ∩ Gφ ∈ ΓQ for any Gϕ, Gφ ∈ ΓQ. (iii) ∪Gϕ ∈ ΓQ, ∀ {Gϕ : ϕ ∈ Z} ⊆ ΓQ. Then (Z, ΓQ) is called a quadripartitioned neutrosophic topological space (in-short, Q-Nsts) in Z. Every element of ΓQ are called a quadripartitioned neutrosophic open sets (in-short, Q-Nso set). If C∈ΓQ, then Cc is called a quadripartitioned neutrosophic closed sets (in-short, Q-Nsc set). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 582 https://internationalpubls.com Definition 2.4. 5 Let (Z, ΓQ) be Q-Nsts on Z and U be an Q-Nss on Z, then a quadripartitioned neutrosophic interior (resp. closure) of U (in-short, Q-Nsint(U ) (resp. Q-Nscl(U ))) are defined as Q-Nsint(U ) =∪{U o : U o ⊆ U & U o is a Q-Nso in Z}, Q-Nscl(U ) = ∩{U o : U ⊆ U o & U o is a Q-Nsc in Z}, Definition 2.5. 5 Let (Z, ΓQ) be Q-Nsts on Z and U be an Q-Nss on Z. Then U is said to be a quadripartitioned neutrosophic pre (resp. semi, α & b) open set (in-short, Q-Ns ƿo set (resp. Q-Ns α o set, Q-Nsαo set & Q-Nsbo set)) if U⊆Q-Nsint(Q-Nscl(U )) (resp. U⊆Q-Nscl(Q-Nsint(U )), U⊆Q-Nsint(Q-Nscl(Q-Nsint(U ))) & U⊆Q-Nscl(Q-Nsint(U )) ∪ Q-Nsint(Q-Nscl(U ))). The complement of an Q-Ns o set (resp. Q-Ns o set, Q-Nsαo set & Q-Nsbo set) is called a quadripar- titioned neutrosophic pre (resp. semi, α & b) closed set (in-short, Q-Ns c set (resp. Q-Ns ƿ c set, Q- Nsαc set & Q-Nsbc set)) in Z. The family of all Q-NsPo set (resp. Q-NsPc set, Q-NsSo set, Q-NsSc set, Q-Nsαo set, Q-Nsαc set, Q-Nsbo set & Q-Nsbc set) of Z is denoted by Q-NsPOS(Z) (resp. Q-NsPCS(Z), Q- NsSOS(Z), Q- NsSCS(Z), Q-NsαOS(Z), Q-NsαCS(Z), Q-NsbOS(Z) & Q-NsbCS(Z)). Definition 2.6. Let (Z, ΓQ) be Q-Nsts on Z and U be an Q-Nss on Z. Then U is said to be a quadriparti- tioned neutrosophic β open set (in-short, Q-Nsβo) set if U⊆Q-Nscl(Q-Nsint(Q-Nscl(U ))). The complement of an Q-Nsβo set is called a quadripartitioned neutrosophic β closed set (in-short, Q- Nsβc set in Z. The family of all Q-Nsβo set (resp. Q-Nsβc set) of Z is denoted by Q-NsβOS(Z) (resp. Q- NsβCS(Z)). Definition 2.7.The Q-Nsβ interior of U (briefly, Q-Nsβint(U )) and Q-Nsβ closure of U (briefly, Q-Nsβcl(U )) are defined as (i) Q-Nsβint(U ) = ∪{Uo : Uo ⊆ U & Uo is a Q-Nsβo set in Z}. (ii) Q-Nsβcl(U ) = ∩{Uo : U ⊆ Uo & Uo is a Q-Nsβc set in Z}. Definition 2.8. Let (Z1, ΓQ) and (Z2, σQ) be any two Q-Nsts’s. A map K : (Z1, ΓQ) → (Z2, σQ) is said to be quadripartitioned neutrosophic (resp. semi, pre, b & β) continuous (briefly, Q-NsCts (resp. Q-NsSCts, Q-NsPCts, Q-NsbCts & Q-NsβCts)) if the inverse image of every Q-Nso set in (Z2, σQ) is a Q-Nso set (resp. Q-NsSo set, Q-NsPo set, Q-Nsbo set & Q-Nsβo set) in (Z1, ΓQ). Definition 2.9. A map K : (Z1, ΓQ)→ (Z2, σQ) is called a quadripartitioned neutrosophic β-irresolute (briefly, Q-NsβIrr) map if K−1(λ) is a Q-Nsβo set in (Z1, ΓQ) for every Q-Nsβo set λ of (Z2, σQ). Definition 2.10. A Q-Nsts (Z, ΓQ) is said to be an quadripartitioned neutrosophic β U 1/2 (in short Q-NsβU 1/2 )-space, if every Q-Nsβo set in Z is a Q-Nso set in Z. Definition 2.11. Let (Z1, ΓQ) and (Z2, σQ) be any two Q-Nsts’s. A map K : (Z1, ΓQ) → (Z2, σQ) is said to be quadripartitioned neutrosophic (resp. semi, pre, b & β) open map (briefly, Q-NsO Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 583 https://internationalpubls.com (resp. Q-NsSO, Q-NsPO, Q-NsbO & Q-NsβO)) if the inverse image of every Q-Nso set in (Z1, ΓQ) is a Q-Nso set (resp. Q-NsSo set, Q-NsPo set, Q-Nsbo set & Q-Nsβo set) in (Z2, σQ). Definition 2.12. Let (Z1, ΓQ) and (Z2, σQ) be any two Q-Nsts’s. A map K : (Z1, ΓQ) → (Z2, σQ) is said to be quadripartitioned neutrosophic (resp. semi, pre, b & β) closed map (briefly, Q-NsC (resp. Q-NsSC, Q-NsPC, Q-NsbC & Q-NsβC)) if the inverse image of every Q-Nsc set in (Z1, ΓQ) is a Q-Nsc set (resp. Q-NsSc set, Q-NsPc set, Q-Nsbc set & Q-Nsβc set) in (Z2, σQ). Definition 2.13. A bijection K : (Z1, ΓQ) → (Z2, σQ) is called a (i) quadripartitioned neutrosophic homeomorphism (briefly Q-NsHom) if K and K−1 are Q-NsCts. (ii) quadripartitioned neutrosophic β-homeomorphism (briefly Q-NsβHom) if K and K−1 are Q- NsβCts. Definition 2.14. A bijection K : (Z1, ΓQ) (Z2, σQ) is called a quadripartitioned neutrosophic β- Completely homeomorphism (briefly, Q-NsβCHom) if K and K−1 are Q-NsβIrr mappings. Definition 2.15. A Q-Nsts (Z, ΓQ) is said to be a quadripartitioned neutrosophic β ½ space ( briefly, Q -NsβT 1/2 )-space if every Q-Nsβcs is Q-Nsc in (Z, ΓQ). 3 Quadripartitioned neutrosophic contra β-continuous maps In this section, quadripartitioned neutrosophic contra β-continuous maps are introduced and some of its properties are discussed. Definition 3.1. A mapping K : (Z1, ΓQ) → (Z2, σQ) is said to be a quadripartitioned neutrosophic contra (resp. semi, pre, b & β) continuous (in short, Q-NsɕCts (resp. Q-Ns ɕSCts, Q-NsɕƿCts, Q-NsɕbCts & Q-NsɕβCts)) if the inverse image of each Q-Nso set of (Z2, σQ) is Q-Nsc (resp. Q- Ns Sc, Q-Ns ρc, Q-Nsbc & Q-Nsβc) set in (Z1, ΓQ). Example 3.2. Let V = {Va, Vb, Vc} = W and define Q-Nss’s V1, V2 & V3 in V and W1 in W are V1 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.3, 0.5, 0.5, 0.7), (Vc, 0.4, 0.5, 0.5, 0.6)}, V2 = {(Va, 0.1, 0.5, 0.5, 0.9), (Vb, 0.1, 0.5, 0.5, 0.9), (Vc, 0.4, 0.5, 0.5, 0.6)}, V3 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.4, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.4, 0.5, 0.5, 0.6)}. Then we have ΓQ ={ 0QNs , V1, V2, 1QNs } and σQ = {0QNs , W1, 1QNs}. Let K : (V, ΓQ)(W, σQ) be an identity mapping, then K is Q-NsɕβCts function. Proposition 3.3. A map K : (Z1, ΓQ) (Z2, σQ), then the statements are hold but the converse does not true. Every (i) Q-NsɕCts is a Q-NsɕSCts. (ii) Q-NsɕCts is a Q-NsɕPCts. (iii) Q-NsɕSCts is a Q-NsɕbCts. (iv) Q-NsɕPCts is a Q-NsɕbCts. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 584 https://internationalpubls.com (v) Q-NsɕbCts is a Q-NsɕβCts. Proof. (i) Let η be a Q-Nso set in Z2. Since K is Q-NsɕCts, K−1(η) is a Q-Nsc set in Z1. Since every Q-Nsc set is a Q-NsSc set, K−1(η) is a Q-NsSc set in Z1. Hence K is a Q-NsɕSCts. (ii) Let η be a Q-Nso set in Z2. Since K is Q-NsɕCts, K−1(η) is a Q-Nsc set in Z1. Since every Q-Nsc set is a Q-NsPc set, K−1(η) is a Q-NsPc set in Z1. Hence K is a Q-NsCPCts. (iii) Let η be a Q-Nso set in Z2. Since K is Q-NsɕSCts, K−1(η) is a Q-NsSc set in Z1. Since every Q-NsSc set is a Q-Nsbc set, K−1(η) is a Q-Nsbc set in Z1. Hence K is a Q-NsCbCts. (iv) Let η be a Q-Nso set in Z2. Since K is Q-NsɕPCts, K−1(η) is a Q-NsPc set in Z1. Since every Q-NsPc set is a Q-Nsbc set, K−1(η) is a Q-Nsbc set in Z1. Hence K is a Q-NsCbCts. (v) Let η be a Q-Nso set in Z2. Since K is Q-NsɕbCts, K−1(η) is a Q-Nsbc set in Z1. Since every Q-Nsbc set is a Q-Nsβc set, K−1(η) is a Q-Nsβc set in Z1. Hence K is a Q-NsɕβCts. Figure 1: Q-NsɕβCts maps in Q-Nsts Example 3.4. In Example 3.2, K is Q-NsɕbCts but not Q-NsCPCts, the set K−1(W1) =V3 c is a Q-Nsbc set but not Q-NsPc set. Example 3.5. Let V = {Va, Vb, Vc} = W and define Q-Nss’s V1, V2, V3 & V4 in V and W1 in W are V1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.5, 0.5, 0.5, 0.5)}, V2 = {(Va, 0.4, 0.5, 0.5, 0.6), (Vb, 0.2, 0.5, 0.5, 0.8), (Vc, 0.6, 0.5, 0.5, 0.4)}, V3 = {(Va, 0.4, 0.5, 0.5, 0.6), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.6, 0.5, 0.5, 0.4)}, V4 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.4, 0.5, 0.5, 0.6)} W1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.4, 0.5, 0.5, 0.6)}. Then we have ΓQ={0QNs , V1, V2, V3, V1 ∩ V2, 1QNs } and σQ ={0QNs , W1, 1QNs }. Let K : (Z1, ΓQ) →(Z2, σQ) be an identity mapping, then K is Q-NsCbCts but not Q-NsɕSCts, the set K−1(W1) = V c is a Q-Nsbc set but not Q-NsSc set. Example 3.6. Let V = {Va, Vb} = W and define Q-Nss’s V1 & V2 in V and W1 in W are V1 = {(Va, 0.3, 0.5, 0.5, 0.5), (Vb, 0.2, 0.5, 0.5, 0.5)}, V2 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.6)}. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 585 https://internationalpubls.com Then we have ΓQ = {0QNs , V1, 1QNs} and σQ = {0QNs , W1, 1QNs} . Let K : (Z1, ΓQ)(Z2, σQ) be an identity mapping, then K is Q-NsɕβCts but not Q-NsɕbCts, the set K−1(W1) = V2 c is a Q- Nsβc set but not Q-Nsbc set. Theorem 3.7. A map K : (Z1, ΓQ) →(Z2, σQ) is Q-NsɕβCts iff the inverse image of every Q- Nscs in Z2 is Q-Nsβos in Z1. Proof. Consider a Q-Nscs Ψ̃ in Z2. Then Ψ̃c is Q-Nsos in Z2. As K is Q-NsɕβCts, K−1(Ψ̃c) is Q-Nsβcs in Z1. As K−1(Ψ̃c) = (K−1(Ψ̃))c, K−1(Ψ̃) is a Q-Nsβos in Z1. Conversely, consider a Q-Nscs Ψ̃ in Z2. So Ψ̃c is a Q-Nsos in Z2. By presumption, K−1(Ψ̃c) is Q- Nsβcs in Z1. As K−1(Ψ̃c) = (K−1(Ψ̃))c, (K−1(Ψ̃))c is a Q-Nsβcs in Z1. Hence K−1(Ψ̃) is a Q- Nsβos in Z1. Thus K is Q-NsɕβCts. Theorem 3.8. Let K : (Z1, ΓQ)→(Z2, σQ) be Q-NsCβCts. If Z1 is a Q-NsβU1/2 -space, then K is a Q-NsɕCts. Proof. Consider a Q-Nsos Ψ̃ in Z2. So K−1(Ψ̃) is a Q-Nsβcs in Z1, by presumption. As Z1 is a Q- NsβU1/2 -space, K−1(Ψ̃) is a Q-Nscs in Z1. Thus K is a Q-NsɕCts. Theorem 3.9. Let K : (Z1, ΓQ) → (Z2, σQ) be a Q-NsɕβCts map and G : (Z2, σQ) → (Z3, ρQ) be a Q-NsCts, then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is a Q-NsɕβCts. Proof. Let à be a Q-Nsos in Z3. By presumption, G−1(Ã) is a Q-Nsos in Z2. As K is a Q-NsɕβCts map, K−1(G−1(Ã)) is a Q-Nsβcs in Z1. Thus G ◦ K is a Q-NsCβCts map. Theorem 3.10. Let K : (Z1, ΓQ) (Z2, σQ) be a Q-NsɕβCts map. Then, the succeeding conditions are true. (i) K(Q-Nsβcl(Ψ̃)) ⊇ Q-Nsint(K(Ψ̃)), ∀ (Ψ̃) in Z1. (ii) Q-Nsβcl(K−1(Φ̃)) ⊇ K−1(Q-Nsint(Φ̃)), ∀ (Φ̃) in Z2. Proof. (i) As Q-Nsβcl(K(Ψ̃)) is a Q-Nsβcs in Z2 and K is Q-NsɕβCts, K−1(Q-Nsβcl(K(Ψ̃))) is Q-Nsβo in Z1. Now, as (Ψ̃)⊇K−1(Q-Nsint(K(Ψ̃))), Q-Nsβcl(Ψ̃)⊇K−1(Q-Nsint(K(Ψ̃))). Therefore, K(Q-Nsβcl(Ψ̃)) Q-Nsint(K(Ψ̃)). (ii) By replacing (Ψ̃) with (Φ̃) in (i), we get K(Q-Nsβcl(K−1(Φ̃)))⊇Q- Nsint(K(K−1(Φ̃)))⊇Q- Nsint(Φ̃). Hence, Q-Nsβcl(K−1(Φ̃)) ⊇ K−1(Q-Nsint(Φ̃)). 4 Quadripartitioned neutrosophic contra β-irresolute maps The quadripartitioned neutrosophic contra β-irresolute maps are introduced and some of its properties are discussed in this section. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 586 https://internationalpubls.com Definition 4.1. A map K : (Z1, ΓQ) → (Z2, σQ) is known as a quadripartitioned neutrosophic contra β- irresolute (in short, Q-NsCβIrr) map if K−1(Ψ̃) is a Q-Nsβcs in (Z1, ΓQ) for each Q- Nsβos Ψ̃ of (Z2, σQ). Theorem 4.2. Let K : (Z1, ΓQ)→ (Z2, σQ) be a Q-NsCβIrr map. Then K is Q-NsCβCts. But the converse need not be true. Proof. Assume K is a Q-NsCβIrr map. Consider a Q-Nsos Ψ̃ in Z2. As each Q-Nsos is a Q-Nsβos, Ψ̃ is a Q-Nsβos in Z2. By presumption, K−1(Ψ̃) is a Q-Nsβcs in Z1. Thus K is a Q-NsɕβCts map. Example 4.3. Let V = {Va, Vb, Vc} = W and define Q-Nss’s V1, V2 & V3 in V and W1 & W2 in W are V1 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.3, 0.5, 0.5, 0.7), (Vc, 0.4, 0.5, 0.5, 0.6)}, V2 = {(Va, 0.1, 0.5, 0.5, 0.9), (Vb, 0.1, 0.5, 0.5, 0.9), (Vc, 0.4, 0.5, 0.5, 0.6)}, V3 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.4, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.1, 0.5, 0.5, 0.9), (Vb, 0.1, 0.5, 0.5, 0.9), (Vc, 0.4, 0.5, 0.5, 0.6)}, W2 = {(Va, 0.1, 0.5, 0.5, 0.9), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.5, 0.5, 0.5, 0.5)}. Then we have ΓQ ={0QNs , V1, V2, 1QNs} and σQ ={0QNs , W1, 1QNs}. Let K : (Z1, ΓQ)→(Z2, σQ) be an identity mapping, then K is Q-NsɕβCts but not Q-NsɕβIrr, the set W2 is a Q- Nsβc set in W but K−1(W2) is not Q-Nsβc set in V . Theorem 4.4. Let K : (Z1, ΓQ)→(Z2, σQ) be a Q-NsɕβIrr. If Z1 is a Q-NsβU1/2 -space, then K is a Q-NsɕCts map. Proof. Consider a Q-Nsos Ψ̃ in Z2. Then Ψ̃ is a Q-Nsβos in Z2. Hence K−1(Ψ̃) is a Q-Nsβcs in Z1. As Z1 is a Q-NsβU1/2 -space, K−1(Ψ̃) is a Q-Nscs in Z1. Thus K is a Q-NsɕCts map. Theorem 4.5. Let K : (Z1, ΓQ) → (Z2, σQ) be a Q-NsCβIrr map and G :(Z2, σQ) → (Z3, ρQ) be Q-NsβCts map. Then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is a Q-NsɕβCts map. Proof. Consider a Q-Nsos à in Z3. Then G−1(Ã) is a Q-Nsβos in Z2. As K is a Q-NsɕβIrr, K−1(G−1(Ã)) is a Q-Nsβcs in Z1. Thus G ◦ K is a Q-NsɕβCts map. Theorem 4.6. Let K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) be mappings. Then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is: (i) Q-NsɕβCts if K is Q-NsβIrr and G is Q-NsɕβCts. (ii) Q-NsɕβIrr if K is Q-NsɕβIrr (resp. Q-NsβIrr) and G is Q-NsβIrr (resp.Q-NsɕβIrr). Proof. (i) Let à be a Q-Nsos in Z3. Then G−1(Ã) is a Q-Nsβcs in Z2. As K is a Q-NsβIrr, K−1(G−1(Ã))is a Q-Nsβcs in Z1. Thus G K is a Q-NsɕβCts map. The other cases are similar. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 587 https://internationalpubls.com Theorem 4.7. Let K : (Z1, ΓQ) → (Z2, σQ) be a mapping. (i) If (Z1, ΓQ) is Q-NsβU1/2 -space, then the concepts of Q-NsɕCts and Q-NsɕβCts are equivalent. (ii) If (Z2, σQ) is Q-NsβU1/2 -space, then the concepts of Q-NsɕβCts and Q-NsɕβIrr are equivalent. (iii) If (Z1, ΓQ) and (Z2, σQ) are Q-NsβU1/2 -spaces, then the concepts of Q-NsɕCts, Q- NsɕβCts andQ-NsɕβIrr are equivalent. Proof. (i) Let Ψ̃ be a Q-Nscs in Z2. Then G−1(Ψ̃) is a Q-Nsβos in Z1 if K is Q-NsɕβCts. As (Z1, ΓQ) is a Q-NsβU 1/2 -space, G−1(Ψ̃) is a Q-Nsos in Z1. Hence K is also Q-NsɕCts map. The other cases are similar. Theorem 4.8. Let K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) be Q-NsɕβCts mappings and (Z2, σQ) be a Q-NsβU1/2 -space. Then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is Q-NsβCts. Proof. Let à be a Q-Nscs in Z3. Then G−1(Ã) is a Q-Nsβos in Z2, since G is Q-NsɕβCts. As (Z2, σQ) is a Q-NsβU1/2 -space, G−1(Ã) is a Q-Nsos in Z2. Then, K(G−1(Ã)) is Q-Nsβcs in Z1 because K is Q-NsɕβCts. Hence, G ◦ K is a Q-NsβCts map. Theorem 4.9. Let K : (Z1, ΓQ) → (Z2, σQ) be a map from a Q-Nst Z1 into a Q-Nst Z2. If Z1 and Z2 are Q-NsβU 1/2 -spaces, then the following are equivalent: (i) K is a Q-NsɕβIrr map. (ii) K−1(Ψ̃) is a Q-Nsβos in Z1 for every Q-Nsβcs Ψ̃ in Z2. (iii) Q-Nscl(K−1(Ψ̃)) ⊇ K−1(Q-Nsint(Ψ̃)) for each Q-Nss Ψ̃ of Z2. Proof. (i) → (ii): Consider a Q-Nsβcs Ψ̃ in Z2. So Ψ̃c is a Q-Nsβos in Z2. As K is Q-NsɕβIrr, K−1(Ψ̃c)is a Q-Nsβcs in Z1. We know that, K−1(Ψ̃c) = (K−1(Ψ̃))c. Thus K−1(Ψ̃) is a Q-Nsβos in Z1. (ii) → (iii): Consider a Q-Nss Ψ̃ in Z2 and Q-Nsint(Ψ̃)⊆ (Ψ̃). Then K−1(Q-Nsint(Ψ̃))⊆ K−1(Ψ̃). As Q-Nsint(Ψ̃) is a Q-Nsos in Z2, Q-Nsint(Ψ̃) is a Q-Nsβos in Z2. Therefore (Q- Nsint(Ψ̃))c is a Q-Nsβcs in Z2. By presumption, K−1((Q-Nsint (Ψ̃))c) is a Q-Nsβos in Z1. As K−1((Q-Nsint(Ψ̃))c) = (K−1(Q- Nsint(Ψ̃)))c, K−1(Q-Nsint(Ψ̃)) is a Q-Nsβos in Z1. As Z1 is Q-NsβU1/2 -space, K−1(Q-Nsint(Ψ̃)) is a Q-Nsos in Z1. Thus, Q-Nscl(K−1(Ψ̃))⊇ Q-Nscl(K−1(Q-Nsint(Ψ̃))) = K−1(Q-Nsint(Ψ̃)). That is Q- Nscl(K−1(Ψ̃)) ⊇K−1(Q-Nsint(Ψ̃)). (iii) → (i): Consider a Q- in Z2. As Z2 is Q-NsβU1/2 - space, Ψ̃ is a Q-Nscs in Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 588 https://internationalpubls.com Z2 and Q -Nscl(Ψ̃) = (Ψ̃). Hence K−1(Ψ̃) = K−1(Q-Nscl(Ψ̃)) Q-Nsint(K−1(Ψ̃)). But clearly K−1(Ψ̃) Q- Nsint(K−1(Ψ̃)). Therefore Q-Nsint(K−1(Ψ̃)) = K−1(Ψ̃). So, K−1(Ψ̃) is a Q-Nsos and hence it is a Q- Nsβos in Z1. Thus K is a Q-NsɕβIrr map. 5 Quadripartitioned Neutrosophic contra β-open mapping The quadripartitioned neutrosophic contra β-open maps are introduced in this section and some of their characteristics are analyzed. Definition 5.1. A mapping K : (Z1, ΓQ) → (Z2, σQ) is quadripartitioned neutrosophic contra (resp. semi, pre, b & β) open (in short, Q-NsɕO (resp. Q-NsɕSO, Q-NsɕPO, Q-NsɕbO & Q-NsɕβO)) if the image of each Q-Nso set of (Z1, ΓQ) is Q-Nsc (resp. Q-NsSc, Q-NsPc, Q-Nsbc & Q-Nsβc) set in (Z2, σQ). Proposition 5.2. A map K : (Z1, ΓQ) → (Z2, σQ), then the statements are hold but the converse does not true. Every (i) Q-NsɕO is a Q-NsɕSO. (ii) Q-NsɕO is a Q-NsɕPO. (iii) Q-NsɕSO is a Q-NsɕbO. (iv) Q-NsɕPO is a Q-NsɕbO. (v) Q-NsɕbO is a Q-NsɕβO. Proof. (i) Let η be a Q-Nso set in Z1. Since K is Q-NsɕO, K(η) is a Q-Nsc set in Z2. Since every Q-Nsc set is a Q-NsSc set, K(η) is a Q-NsSc set in Z2. Hence K is a Q-NsɕSO. (ii) Let η be a Q-Nso set in Z1. Since K is Q-NsɕO, K(η) is a Q-Nsc set in Z2. Since every Q-Nsc set is a Q-NsPc set, K(η) is a Q-NsPc set in Z2. Hence K is a Q-NsɕPO. (iii) Let η be a Q-Nso set in Z1. Since K is Q-NsɕSO, K(η) is a Q-Ns Sc set in Z2. Since every Q-Ns Sc set is a Q-Nsbc set, K(η) is a Q-Nsbc set in Z2. Hence K is a Q-NsɕbO. (iv) Let η be a Q-Nso set in Z1. Since K is Q-Nsɕ ρO, K(η) is a Q-Ns ρc set in Z2. Since every Q-Ns bc set is a Q-Nsbc set, K(η) is a Q-Nsbc set in Z2. Hence K is a Q-NsɕbO. (v) Let η be a Q-Nso set in Z1. Since K is Q-NsɕbO, K(η) is a Q-Nsbc set in Z2. Since every Q- Nsbc set is a Q-Nsβc set, K(η) is a Q-Nsβc set in Z2. Hence K is a Q-NsɕβO. Example 5.3. Let V = {Va, Vb, Vc} = W and define Q-Nss’s V1 in V and W1, W2 & W3 in W are V1 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.4, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.3, 0.5, 0.5, 0.7), (Vc, 0.4, 0.5, 0.5, 0.6)}, W2 = {(Va, 0.1, 0.5, 0.5, 0.9), (Vb, 0.1, 0.5, 0.5, 0.9), (Vc, 0.4, 0.5, 0.5, 0.6)}, W3 = {(Va, 0.2, 0.5, 0.5, 0.8), (Vb, 0.4, 0.5, 0.5, 0.6), (Vc, 0.4, 0.5, 0.5, 0.6)}. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 589 https://internationalpubls.com 3 Then we have ΓQ = {0QNs , V1, 1QNs } and σQ = {0QNs , W1, W2, 1QNs }. Let K : (Z1, ΓQ) → (Z2, σQ) be an identity mapping, then K is Q-NsCbO but not Q-NsCPO, the set K(V1) = Wc is a Q-Nsbc set but not Q-NsPc set. Example 5.4. Let V = {Va, Vb, Vc} = W and define Q-Nss’s V1 in V and W1, W2, W3 & W4 in W are V1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.4, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.5, 0.5, 0.5, 0.5)}, W2 = {(Va, 0.4, 0.5, 0.5, 0.6), (Vb, 0.2, 0.5, 0.5, 0.8), (Vc, 0.6, 0.5, 0.5, 0.4)}, W3 = {(Va, 0.4, 0.5, 0.5, 0.6), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.6, 0.5, 0.5, 0.4)}, W4 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.5), (Vc, 0.4, 0.5, 0.5, 0.6)}. Then we have Γ ={0QNs , V1, 1QNs } and σQ= 0QNs , W1, W2, W3, W1∩W2, 1QNs }. Let K : (Z1, ΓQ) →(Z2, σQ) be an identity mapping, then K is Q-NsɕbO but not Q-NsɕSO, the set K(V1) = W3c is a Q-Nsbc set but not Q-NsSc set. Example 5.5. Let V = {Va, Vb} = W and define Q-Nss’s V1 in V and W1 & W2 in W are V1 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.6)}, W1 = {(Va, 0.3, 0.5, 0.5, 0.5), (Vb, 0.2, 0.5, 0.5, 0.5)}, W2 = {(Va, 0.3, 0.5, 0.5, 0.7), (Vb, 0.5, 0.5, 0.5, 0.6)}. Then we have ΓQ={0QNs , V1, 1QNs and σQ ={ 0QNs , W1, 1QNs}.Let K : (Z1, ΓQ) → (Z2, σQ) be an identity mapping, then K is Q-NsɕβO but not Q-NsɕbO, the set K(V1) = Wc is a Q- Nsβc set but not Q-Nsbc set. Figure 2: Q-NsCβO maps in Q-Nsts Theorem 5.6. A mapping K : (Z1, ΓQ) → (Z2, σQ) is Q-NsCβO iff for every Q-Nss (Ψ̃) of (Z1, ΓQ), K(Q-Nsint(Ψ̃)) ⊇ Q-Nsβcl(K(Ψ̃)). Proof. Necessity: Assume K is a Q-NsɕβO mapping and (Ψ̃) is a Q-Nsos in (Z1, ΓQ). Now, Q- Nsint(Ψ̃) ⊆ (Ψ̃) implies K(Q-Nsint(Ψ̃)) ⊆ K(Ψ̃). Since K is a Q-NsɕβO mapping, K(Q- Nsint(Ψ̃)) is Q-Nsβcs in (Z2, σQ) such that K(Q-Nsint(Ψ̃)) ⊇ K(Ψ̃). Therefore K(Q-Nsint(Ψ̃)) ⊇ Q-Nsβcl(K(Ψ̃)). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 590 https://internationalpubls.com Sufficiency: Assume (Ψ̃) is a Q-Nsos of (Z1, ΓQ). Then we have K(Ψ̃) = K(Q-Nsint(Ψ̃)) ⊇ Q- Nsβcl(K(Ψ̃)). But Q-Nsβcl(K(Ψ̃)) ⊇K(Ψ̃). So, K(Ψ̃) = Q-Nsβcl(Ψ̃) which implies K(Ψ̃) is a Q-Nsβcs of (Z2, σQ) and hence K is a Q-NsCβO. Theorem 5.7. If K : (Z1, ΓQ) → (Z2, σQ) is a Q-NsɕβO mapping, then Q-Nsint(K−1(Ψ̃)) ⊆ K−1(Q-Nsβcl(Ψ̃)) for every Q-Nss (Ψ̃) of (Z2, σQ). Proof. Consider a Q-Nss (Ψ̃) in (Z2, σQ). We know that Q-Nsint(K−1(Ψ̃)) is a Q-Nsos in (Z1, ΓQ). Since K is Q-NsβO, K(Q-Nsint(K−1(Ψ̃))) is Q-Nsβcs in (Z2, σQ) and hence K(Q- Nsint(K−1(Ψ̃)))⊆ Q- Nsβcl(K(K−1(Ψ̃))) ⊆ Q-Nsβcl(Ψ̃). Thus Q-Nsint(K−1(Ψ̃)) ⊆ K−1(Q- Nsβcl(Ψ̃)). Theorem 5.8. A mapping K : (Z1, ΓQ) → (Z2, σQ) is Q-NsCβO iff for each Q-Nss Ψ̃ of (Z2, σQ) and for each Q-Nsos (Ψ̃) of (Z1, ΓQ) containing K−1(Ψ̃), there is a Q-Nsβos à of (Z2, σQ) such that (Ψ̃) ⊆ (Ã) and K−1(Ã) ⊆ (Ψ̃). Proof. Necessity: Let K be a Q-NsɕβO mapping. Consider a Q-Nscs Ψ̃ in (Z2, σQ) and a Q-Nsos (Ψ̃) in (Z1, ΓQ) such that K−1(Ψ̃) ⊆ (Ψ̃). Then (Ã) = (K(Ψ̃)c)c is Q-Nsβos of (Z2, σQ) such that K−1(Ã) ⊆ (Ψ̃). Sufficiency: Assume (Ψ̃) is a Q-Nsos of (Z1, ΓQ). So K−1((K(Ψ̃))c) ⊆ (Ψ̃)c and (Ψ̃)c is Q- Nscs in (Z1, ΓQ). By presumption, there is a Q-Nsβos à of (Z2, σQ) such that (K(Ψ̃))c⊆(Ã) and K−1(Ã)⊆(Ψ̃)c. Therefore (Ψ̃)⊆(K−1(Ã))c .Hence (Ã)c⊆K(Ψ̃)⊆ K((K−1(Ã))c) ⊆(Ã)c which implies K(Ψ̃) = (Ã)c.As (Ã)c is Q-Nsβcs of (Z2, σQ), K(Ψ̃) is Q-Nsβcs in (Z2, σQ) and hence K is Q-NsCβO mapping. Theorem 5.9. A mapping K : (Z1, ΓQ) → (Z2, σQ) is Q-NsCβO iff K−1(Q-Nsβcl(Ψ̃)) ⊇ Q- Nsint(K−1(Ψ̃)) for every Q-Nss Ψ̃ of (Z2, σQ). Proof. Necessity: Let K be a Q-NsCβO mapping. For any Q-Nss Ψ̃ of (Z2, σQ), K−1(Ψ̃) ⊆ Q-Nscl(K−1(Ψ̃)). Therefore by Theorem 5.8, there exists a Q-Nsβos (Ψ̃) in (Z2, σQ) ∋ (Ψ̃) ⊇ (Ψ̃) & K−1(Ψ̃) ⊇Q-Nsint(K−1(Ψ̃)). Hence K−1(Q-Nsβcl(Ψ̃)) ⊇ K−1(Ψ̃) ⊇ Q-Nsint(K−1(Ψ̃)). Sufficiency: Let Ψ̃ be a Q-Nss in (Z2, σQ) and (Ψ̃) be a Q-Nscs of (Z1, ΓQ) containing K−1(Ψ̃). Put (Ã) = Q-Nscl(Ψ̃), then (Ψ̃)⊆(Ã) and à is Q-Nsβc and K−1(Ã) ⊆Q- Nsint(K−1(Ψ̃)) ⊆(Ψ̃). Thus by Theorem 5.8, K is Q-Q-NsɕβO mapping. Theorem 5.10. If K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) be two quadripartitioned neutrosophic mappings and G◦K : (Z1, ΓQ) → (Z3, ρQ) is Q-NsɕβO. If G : (Z2, σQ) → (Z3, ρQ) is Q-NsɕβIrr, then K : (Z1, ΓQ) → (Z2, σQ) is Q-NsɕβO mapping. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 591 https://internationalpubls.com Proof. Let (Ψ̃) be a Q-Nsos in (Z1, ΓQ). Then G◦K(Ψ̃) is Q-Nsβcs of (Z3, ρQ) because G◦K is Q- NsɕβO mapping. As G is Q-NsCβIrr and G K(Ψ̃) is Q-Nsβcs of (Z3, ρQ), G−1(G ∘ K(Ψ̃)) = K(Ψ̃) is Q-Nsβos in (Z2, σQ). Hence K is Q-NsɕβO mapping. Theorem 5.11. If K : (Z1, ΓQ) → (Z2, σQ) is Q-NsO & G : (Z2, σQ) → (Z3, ρQ) is Q-NsɕβO mappings, then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is Q-NsɕβO. Proof. Let (Ψ̃) be a Q-Nsos in (Z1, ΓQ). Then K(Ψ̃) is a Q-Nsos of (Z2, σQ) because K is a Q- NsO map- ping. As G is Q-NsɕβO, G(K(Ψ̃)) = (G∘ K)(Ψ̃) is a Q-Nsβcs of (Z3, ρQ). Thus G K is Q-NsɕβO mapping. 6 Quadripartitioned Neutrosophic contra β-closed mapping In this section, quadripartitioned neutrosophic contra β-closed maps are introduced and some of its properties are discussed. Definition 6.1. A mapping K : (Z1, ΓQ) → (Z2, σQ) is quadripartitioned neutrosophic contra (resp. semi, pre, b & β) closed (in short, Q-NsCC (resp. Q-NsCSC, Q-NsCPC, Q-NsCbC & Q-NsCβC)) if the image of each Q-Nsc set of (Z1, ΓQ) is Q-Nso (resp. Q-NsSo, Q-NsPo, Q-Nsbo & Q-Nsβo) set in (Z2, σQ). Proposition 6.2. A map K : (Z1, ΓQ) → (Z2, σQ), then the statements are hold but the converse does not true. Every (i) Q-NsɕC is a Q-NsɕSC. (ii) Q-NsɕC is a Q-NsɕPC. (iii) Q-NsɕSC is a Q-NsɕbC. (iv) Q-NsɕPC is a Q-NsɕbC. (v) Q-NsɕbC is a Q-NsɕβC. Proof. (i) Let η be a Q-Nsc set in Z1. Since K is Q-NsɕC, K(η) is a Q-Nso set in Z2. Since every Q-Nsoset is a Q-NsSo set, K(η) is a Q-NsSo set in Z2. Hence K is a Q-NsɕSC. (ii) Let η be a Q-Nsc set in Z1. Since K is Q-NsɕC, K(η) is a Q-Nso set in Z2. Since every Q-Ns ρo set is a Q-NsPo set, K(η) is a Q-NsPo set in Z2. Hence K is a Q-NsɕPC. (iii) Let η be a Q-Nsc set in Z1. Since K is Q-NsɕC, K(η) is a Q-NsЅo set in Z2. Since every Q- Ns Ѕo set is a Q-Nsbo set, K(η) is a Q-Nsbo set in Z2. Hence K is a Q-NsɕbC. (iv) Let η be a Q-Nsc set in Z1. Since K is Q-NsɕC, K(η) is a Q-Nsb o set in Z2. Since every Q- Nsρo set is a Q-Nsbo set, K(η) is a Q-Nsbo set in Z2. Hence K is a Q-NsCbC. (v) Let η be a Q-Nsc set in Z1. Since K is Q-NsɕbC, K(η) is a Q-Nsbo set in Z2. Since every Q- Nsbo set is a Q-Nsβo set, K(η) is a Q-Nsβo set in Z2. Hence K is a Q-NsɕβC. Example 6.3. In Example 5.3, K is Q-NsɕbC but not Q-NsɕPC. Example 6.4. In Example 5.4, K is Q-NsɕbC but not Q-NsɕSC. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 592 https://internationalpubls.com Example 6.5. In Example 5.5, K is Q-NsɕβC but not Q-NsɕbC. Figure 3: Q-NsɕβC maps in Q-Nsts Theorem 6.6. A mapping K : (Z1, ΓQ) → (Z2, σQ) is Q-NsɕβC iff for each Q-Nss Ψ̃ of (Z2, σQ) and for each Q-Nscs (Ψ̃) of (Z1, ΓQ) containing K−1(Ψ̃), there is a Q-Nsβcs à of (Z2, σQ) such that (Ψ̃) ⊆ (Ã) and K−1(Ã) ⊆ (Ψ̃). Proof. Necessity: Let K be a Q-NsCβC mapping. Consider a Q-Nsos Ψ̃ in (Z2, σQ) and a Q- Nscs in (Z1, ΓQ) such that K−1(Ψ̃) ⊆ (Ψ̃) .Then (Ψ̃) = 1QN- K−1((Ψ̃)c) is Q-Nsβcs of (Z2, σQ) such that Sufficiency: Assume (Ψ̃) is a Q-Nscs of (Z1, ΓQ). Then (K(Ψ̃))c is a Q-Nss of (Z2, σQ) and (Ψ̃)c isQ-Nsos in (Z1, ΓQ) such that K−1((K(Ψ̃))c)⊆(Ψ̃)c. By presumption, there is a Q-Nsβcs à of (Z2, σQ) such that (K(Ψ̃))c⊆(Ψ̃) and K−1(Ψ̃)⊆(Ψ̃)c. Therefore (Ψ̃)⊆(K−1(Ψ̃))c. Hence (Ψ̃)c⊆K(Ψ̃) K⊆((K−1(Ψ̃))c)⊆(Ψ̃)c which implies K(Ψ̃) = (Ψ̃)c. As (Ψ̃)c is Q-Nsβos of (Z2, σQ), K(Ψ̃) is Q-Nsβo in (Z2, σQ) and hence K is Q-NsɕβC mapping. Theorem 6.7. If K : (Z1, ΓQ) → (Z2, σQ) is Q-NsC and G : (Z2, σQ) → (Z3, ρQ) is Q-NsɕβC. Then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is Q-NsɕβC. Proof. Let (Ψ̃) be a Q-Nscs in (Z1, ΓQ). As K is Q-NsC mapping, K(Ψ̃) is Q-Nscs in (Z2, σQ). As G is Q- NsCβC mapping, (G ◦ K)(Ψ̃) = G(K(Ψ̃)) is Q-Nsβos in (Z3, ρQ). Hence G ◦ K is Q- NsɕβC mapping. Theorem 6.8. If K : (Z1, ΓQ) → (Z2, σQ) is Q-NsCβC map, then Q-Nsβint(K(Ψ̃)) ⊇ K(Q- Nsint(Ψ̃)). Proof. The proof is obvious from Definition 2.7 and Definition 6.1. Theorem 6.9. Let K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) be Q-NsɕβC mappings. If every Q-Nsβos of (Z2, σQ) is Q-Nsos, then G ◦ K : (Z1, ΓQ) → (Z3, ρQ) is Q-NsβC. Proof. Let (Ψ̃) be a Q-Nscs in (Z1, ΓQ). As K is Q-NsɕβC mapping, K(Ψ̃) is Q-Nsβos in (Z2, σQ). By presumption, K(Ψ̃) is Q-Nsos of (Z2, σQ). As G is Q-NsɕβC mapping, G(K(Ψ̃)) = (G◦K)(Ψ̃) is Q-Nsβcs in (Z3, ρQ). Hence G ◦ K is Q-NsβC mapping. Theorem 6.10. Consider a bijective mapping K : (Z1, ΓQ) → (Z2, σQ). Then the following statements are equivalent: Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 593 https://internationalpubls.com (i) K is a Q-NsɕβO mapping. (ii) K is a Q-NsɕβC mapping. (iii) K−1 is Q-NsβCts mapping. Proof. (i) ⇒ (ii): Assume K is a Q-NsɕβO mapping. If Q-Nsos (Ψ̃) in (Z1, ΓQ), by presumption K(Ψ̃) is a Q-Nsβcs in (Z2, σQ). But now, (Ψ̃) is Q-Nscs in (Z1, ΓQ). So, 1QNs − (Ψ̃) is a Q- Nsos in (Z1, ΓQ). By assumption, K(1QNs - (Ψ̃)) is a Q-Nsβcs in (Z2, σQ). Hence, 1QNs - K(1QNs - (Ψ̃)) is a Q-Nsβos in (Z2, σQ). Thus, K is a Q-NsɕβC mapping. (ii) ⇒ (iii): Consider a Q-Nscs (Ψ̃) in (Z1, ΓQ). By assumption, K(Ψ̃) is a Q-Nsβos in (Z2, σQ). Hence, K(Ψ̃) = (K−1)−1(Ψ̃). So K−1 is a Q-Nsβos in (Z2, σQ). Thus, K−1 is Q-NsβCts. (iii) ⇒ (i): Consider a Q-Nsos (Ψ̃) in (Z1, ΓQ). By assumption, (K−1)−1(Ψ̃) = K(Ψ̃) is a Q-NsCβO mapping. 7 Quadripartitioned Neutrosophic contra β-homeomorphism In this section, the concept of quadripartitioned neutrosophic contra β-homeomorphism is introduced and its properties are discussed. Definition 7.1. A bijection K : (Z1, ΓQ) → (Z2, σQ) is called a (i) quadripartitioned neutrosophic contra homeomorphism (briefly Q-NsCHom) if K and K−1 are Q- NsCCts mapping. (ii) quadripartitioned neutrosophic contra β-homeomorphism (briefly Q-NsCβHom) if K and K−1 are Q- NsCβCts mapping. Theorem 7.2. Each Q-NsɕHom is a Q-NsɕβHom. But the converse not true. Proof. Assume K is Q-NsɕHom. Then K and K−1 are Q-NsɕCts. We know that each Q-NsɕCts function is Q-NsɕβCts. So, K and K−1 are Q-NsɕβCts. Thus, K is a Q-NsɕβHom. Example 7.3. Let V = {a, b, c} = W and define Q-Nss’s V1, V2 & V3 in V and W1 in W are V1 = {(a, 0.2, 0.5, 0.5, 0.8), (b, 0.3, 0.5, 0.5, 0.7), (c, 0.4, 0.5, 0.5, 0.6)}, V2 = {(a, 0.1, 0.5, 0.5, 0.9), (b, 0.1, 0.5, 0.5, 0.9), (c, 0.4, 0.5, 0.5, 0.6)}, V3 = {(a, 0.2, 0.5, 0.5, 0.8), (b, 0.4, 0.5, 0.5, 0.6), (c, 0.4, 0.5, 0.5, 0.6)}, W1 = {(a, 0.2, 0.5, 0.5, 0.8), (b, 0.4, 0.5, 0.5, 0.6), (c, 0.4, 0.5, 0.5, 0.6)}. Then we have ΓQ = {0QNs , V1, V2, 1QNs} and σQ = {0QNs , W1, 1QNs} . Let K : (Z1, ΓQ) → (Z2, σQ) be an identity mapping, then K is Q-NsHom but not Q-NsɕHom. Theorem 7.4. Consider a bijective mapping K : (Z1, ΓQ) →(Z2,σQ). The followings statements are equivalent if K is Q-NsɕβCts. (i) K is a Q-NsɕβC mapping. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 594 https://internationalpubls.com (ii) K is a Q-NsɕβO mapping. (iii) K is a Q-NsɕβHom. Proof. (i)⇒(ii) : Let K be a bijective mapping and a Q-NsɕβC mapping. Therefore, K−1 is a Q- NsɕβCts mapping. As each Q-Nsos in (Z1, ΓQ) is a Q-Nsβcs in (Z2, σQ), K is a Q-NsɕβO mapping. (ii) ⇒ (iii) : Assume K is a bijective and Q-NsɕβO mapping. Also, K−1 is a Q-NsCβCts mapping. Therefore, K and K−1 are Q-NsɕβCts. Thus, K is a Q-NsɕβHom. (iii) ⇒ (i): Assume K is a Q-NsɕβHom. So, K and K−1 are Q-NsɕβCts. As every Q- Nscs in (Z1, ΓQ) is a Q-Nsβos in (Z2, σQ), K is a Q-NsɕβC mapping. Theorem 7.5. Let K : (Z1, ΓQ) (Z2, σQ) be a Q-NsɕβHom. If (Z1, ΓQ) and (Z2, σQ) are Q- NsβT 1 - spaces, then K is a Q-NsɕHom. Proof. Consider a Q-Nscs Ψ̃ in (Z2, σQ). So, K−1(Ψ̃) is a Q-Nsβos in (Z1, ΓQ). As (Z1, ΓQ) is a Q-NsβT1/2 -space, K−1(Ψ̃) is a Q-Nsos in (Z1, ΓQ). Therefore, K is Q-NsɕCts. By hypothesis, K−1 is Q-NsɕβCts. Let (Ψ̃) be a Q-Nscs in (Z1, ΓQ). Then, K(Ψ̃) is a Q-Nsβos in (Z2, σQ), by presumption. Since (Z2, σQ) is a Q-NsβT1/2 -space, K(Ψ̃) is a Q-Nsos in (Z2, σQ). Therefore, K−1 is Q-NsɕCts. Thus, K is a Q-NsɕHom. Theorem 7.6. Let K : (Z1, ΓQ)→ (Z2, σQ) be a Q-Nsts. If (Z2, σQ) is a Q-NsβT1/2 -space, then the following are equivalent: (i) K is Q-NsɕβC mapping. (ii) If (Ψ̃) is a Q-Nsos in (Z1, ΓQ), then K(Ψ̃) is Q-Nsβcs in (Z2, σQ). (iii) K(Q-Nsint(Ψ̃)) ⊆ Q-Nscl (Q-Nsint(K(Ψ̃))) for every Q-Nss (Ψ̃) in (Z1, ΓQ). Proof. (i) ⇒ (ii): Obvious. (ii)⇒ (iii): Consider a Q-Nss (Ψ) in (Z1, ΓQ). We know that, Q-Nsint(Ψ) is a Q-Nsos in (Z1, ΓQ). Then, K(Q-Nsint(Ψ̃)) is a Q-Nsβcs in (Z2, σQ). Since (Z2, σQ) is a Q-NsβT1/2 -space, K(Q- Nsint(Ψ̃)) is a Q-Nscs in (Z2, σQ). Therefore, K(Q-Nsint(Ψ̃)) = Q-Nscl(K(Q-Nsint(Ψ̃))) ⊆ Q-Nscl(Q-Nsint(K(Ψ̃))). (iii)⇒ (i): Let (Ψ̃) be a Q-Nscs in (Z1, ΓQ). Then, (Ψ̃)c is a Q-Nsos in (Z1, ΓQ). As K(Q- Nsint(Ψ̃)c) ⊆ Q-Nscl(Q-Nsint(K(Ψ̃)c)), we get K((Ψ̃)c)⊆ Q-Nscl(Q-Nsint(K(Ψ̃)c)). Therefore, K((Ψ̃)c) is Q-Nsβcs in (Z2, σQ). Thus, K(Ψ̃) is a Q-Nsβos in (Z1, ΓQ). Hence, K is a Q-NsCβC mapping. Theorem 7.7. Let K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) be Q-NsɕβC, where (Z1, ΓQ)and (Z3, ρQ) are two Q-Nsts’s and (Z2, σQ) a Q-NsβT1/2-space, then the composition G∘ K is Q-NsβC. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 595 https://internationalpubls.com Proof. Consider a Q-Nscs (Ψ̃) in (Z1, ΓQ). As K is Q-NsɕβC and K(Ψ̃) is a Q-Nsβos in (Z2, σQ), by assumption, K(Ψ̃) is a Q-Nsos in (Z2, σQ). Since G is Q-NsɕβC, then G(K(Ψ̃)) is Q- Nsβcs in (Z3, ρQ) and G(K(Ψ̃)) = (G ◦ K)(Ψ̃). Thus, G ◦ K is Q-NsβC. Theorem 7.8. Let K : (Z1, ΓQ) → (Z2, σQ) and G :(Z2, σQ) → (Z3, ρQ) be two Q-Nsts’s, then the following hold: (i) If G ◦ K is Q-NsɕβO and K is Q-NsCts, then G is Q-NsɕβO. (ii) If G ◦ K is Q-NsO and G is Q-NsɕβCts, then K is Q-NsɕβO. Proof. The proof is obvious from Definition 3.1 and Definition 5.1. 8 Quadripartitioned neutrosophic contra β-C homeomorphism The quadripartitioned neutrosophic contra β-C homeomorphism is introduced in this section and some of its properties are analyzed. Definition 8.1. A bijection K : (Z1, ΓQ) → (Z2, σQ) is called a quadripartitioned neutrosophic contra β- Completely homeomorphism (briefly, Q-NsɕβCHom) if K and K−1 are Q-NsɕβIrr mappings. Theorem 8.2. Each Q-NsɕβCHom is a Q-NsɕβHom. But not conversely. Proof. Consider a Q-Nsos Ψ̃ in (Z2, σQ). Then Ψ̃ is a Q-Nsβos in (Z2, σQ). By presumption, K−1(Ψ̃) is a Q-Nsβcs in (Z1, ΓQ). Therefore, K is a Q-NsɕβCts mapping. So, K and K−1 are Q- NsɕβCts mappings. Thus, K is a Q-NsɕβHom. Example 8.3. Let V = {a, b, c} = W and define Q-Nss’s V1 & V2 in V and W1 in W are V1 = {(a, 0.2, 0.5, 0.5, 0.8), (b, 0.3, 0.5, 0.5, 0.7), (c, 0.4, 0.5, 0.5, 0.6)}, V2 = {(a, 0.1, 0.5, 0.5, 0.9), (b, 0.1, 0.5, 0.5, 0.9), (c, 0.4, 0.5, 0.5, 0.6)}, W1 = {(a, 0.4, 0.5, 0.5, 0.6), (b, 0.3, 0.5, 0.5, 0.7), (c, 0.2, 0.5, 0.5, 0.8)}. Then we have ΓQ = {0QNs , V1, V2, 1QNs} and σQ = {0QNs , W1, 1QNs} . Let K : (Z1, ΓQ) (Z2, σQ) be a mapping, defined as K(a) = c, K(b) = b & K(c) = a, then K is Q-NsɕβHom but not Q-NsɕβCHom. Theorem 8.4. If K : (Z1, ΓQ) → (Z2, σQ) is a Q-NsɕβCHom, then Q-Nsβint(K−1(Ψ̃)) ⊆ K−1(Q-Nscl(Ψ̃)) for every Q-Nss Ψ̃ in (Z2, σQ). Proof. Consider a Q-Nss Ψ̃ in (Z2, σQ). Since, Q-Nscl(Ψ̃) is a Q-Nscs in (Z2, σQ) and every Q-Nscs is a Q-Nsβcs in (Z2, σQ). As K is Q-NsɕβIrr, K−1(Q-Nscl(Ψ̃)) is a Q-Nsβos in (Z1, ΓQ). Then, Q-Nsint(K−1(Q-Nscl(Ψ̃))) = K−1(Q-Nscl(Ψ̃)). Here, Q-Nsβint(K−1(Ψ̃)) ⊆ Q-Nsβint(K−1(Q-Nscl(Ψ̃)))= K−1(Q-Nscl(Ψ̃)). Therefore, Q-Nsβint(K−1(Ψ̃)) ⊆ K−1(Q- Nscl(Ψ̃)) for every Q-Nss Ψ̃ in (Z2, σQ). Theorem 8.5. Let K : (Z1, ΓQ) → (Z2, σQ) be a Q-NsβCHom. Then Q-Nsβint(K−1(Ψ̃)) ⊆ K−1(Q-Nsβcl(Ψ̃)) for every Q-Nss Ψ̃ in (Z2, σQ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 596 https://internationalpubls.com Proof. As K is a Q-NsCβCHom, K is a Q-NsCβIrr mapping. Consider a Q-Nss Ψ̃ in (Z2, σQ). It is obvious that, Q-Nsβcl(Ψ̃) is a Q-Nsβcs in (Z2, σQ). As K−1(Ψ̃) ⊆ K−1(Q-Nsβcl(Ψ̃)), we have Q−Nsβint(K−1(Ψ̃)) ⊆ Q−Nsβint(K−1(Q−Nsβcl(Ψ̃))) ⊆ K−1(Q−Nsβcl(Ψ̃)). ⇒Q-Nsβint(K−1(Ψ̃))⊆K−1(Nsβcl(Ψ̃)). Theorem 8.6. If K : (Z1, ΓQ) → (Z2, σQ) and G : (Z2, σQ) → (Z3, ρQ) are Q-NsɕβCHom’s, then G∘K is a Q-NsβCHom. Proof. Assume that K and G are two Q-NsɕβCHom’s. Let Ψ̃ be a Q-Nsβcs in (Z3, ρQ). Then, G−1(Ψ̃) is a Q-Nsβos in (Z2, σQ). By presumption, K−1(G−1(Ψ̃)) is a Q-Nsβcs in (Z1, ΓQ). Therefore, (G ◦ K)−1 is a Q-NsβIrr mapping. Assume (Ψ̃) is a Q-Nsβcs in (Z1, ΓQ). 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