EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6124 ISSN 1307-5543 – ejpam.com Published by New York Business Global Total Exact Domination in Graphs Ritchie G. Aguinod1, Edward M. Kiunisala2, Cris L. Armada3,4 1 College of Computing, Artificial Intelligence and Sciences, Cebu Normal University, 6000 Cebu City, Philippines 2 Mathematics Department, College of Computing, Artificial Intelligence and Sciences, Cebu Normal University, 6000 Cebu City, Philippines 3 Vietnam National University Ho Chi Minh City, Linh Trung Ward, Thu Duc City, Ho Chi Minh City, Vietnam 4 Department of Applied Mathematics, Faculty of Applied Science, Ho Chi Minh City Uni- versity of Technology (HCMUT), 268 Ly Thuong Kiet, District 10, Ward 14, Ho Chi Minh City, Vietnam Abstract. Let G = (V (G), E(G)) be a simple nontrivial undirected graph. A set T ⊆ V (G) is said to be a total exact dominating set if T is both a total dominating set and an exact dominating set of G. The cardinality of a minimum total exact dominating set is called the total exact domination number and is denoted by γte(G). The total exact domination numbers of various types of special graphs, such as path, cycle, star, complete bipartite, and graphs resulting from binary operations such as join, corona, and lexicographic product are obtained in this study. Furthermore, if G does not have a total exact dominating set, then G is called a non− γte-graph. Examples include complete graphs, fan graphs, and wheel graphs with more than two vertices. We also consider some disconnected graphs in the corona and lexicographic product, making the study more interesting. In defining total exact domination, a condition of exact domination was modified because it contradicted the definition of total domination. 2020 Mathematics Subject Classifications: 05C69,05C38, 05C76 Key Words and Phrases: Total domination, exact domination, total exact domination 1. Introduction Graph domination is a thoroughly researched area in graph theory, which is essential for optimizing networks, managing resources, and modeling social networks. The basic concept of a dominating set and its associated domination number was initially presented by O. Ore in 1962 in his foundational paper on Graph Theory [1]. Since then, various forms of domination have been studied, each offering unique insights into the structural DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6124 Email addresses: aguinodr@cnu.edu.ph (R. G. Aguinod), kiunisalae@cnu.edu.ph (E. M. Kiunisala), cris.armada@hcmut.edu.vn (C. L. Armada) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 2 of 26 properties of graphs. Graph domination is a well-explored topic within graph theory, playing a crucial role in fields such as network optimization, resource management, and social network modeling. Cockayne, Dawes, and Hedetniemi [2] were among the first to formally define total domination in graphs in 1980. In their study, a set T ⊆ V (G), where V (G) is the vertex set of G, was defined as a total dominating set if every vertex in V (G) is adjacent to at least one vertex in T . Building upon this definition, we explore its intersection with exact domination in this paper. The idea of exact domination in graphs was formally put forth by Kinsley and Joeshi in 2020, building upon earlier work in domination theory [3]. Where N(v) denotes the set of all vertices adjacent to v, a set T ⊆ V (G) is called an exact dominating set if it satisfies the following conditions: i |N(v) ∩ T | = 1 for all v ∈ V (G) \ T and ii |N(u) ∩ T | ≤ 1 for all u ∈ T . In this paper, we introduce an integration of total and exact domination. Combining the two domination conditions will modify the second condition for an exact dominating set, changing |N(u) ∩ T | ≤ 1 to |N(u) ∩ T | = 1, since |N(u) ∩ T | < 1 for all u ∈ T contradicts the definition of a total dominating set. Furthermore, not all graphs have a total exact dominating set; we denote them as non-γte-graphs. 2. Terminology and Notation Let G = (V (G), E(G)) be a simple nontrivial undirected graph where V (G) is the vertex set and E(G) is the edge set of G. The set of neighbors of a vertex u in G is called the open neighborhood of u in G and is denoted by NG(u) = N(u) = {v ∈ V (G) : uv ∈ E(G)} and the closed neighborhood of u is the set N [u] = N(u)∪{u}. The open neighborhood of a subset T of V (G) is the set NG(T ) = N(T ) = ∪v∈TNG(v) and its closed neighborhood is the set NG[T ] = N [T ] = N(T ) ∪ T [4]. A set T ⊆ V (G) is a dominating set (resp. total dominating set) of G if N [T ] = V (G) (resp. N(T ) = V (G)). The domination number γ(G) (resp. total domination number γt(G)) of G is the minimum cardinality of a dominating set (resp. total dominating set) [4]. A set that has a cardinality of γt(G) is a γt-set. A set T ⊆ V (G) is said to be an exact dominating set if (i) |N(v) ∩ T | = 1 ∀ v ∈ V (G) \ T and (ii) |N(u) ∩ T | ≤ 1 ∀ u ∈ T . The minimum cardinality of an exact dominating set is the exact domination number of the graph denoted by γe(G). A set that has a cardinality of γe(G) is a γe-set [3]. A set T ⊆ V (G) is said to be the total exact dominating set if it satisfies (i) |N(v) ∩ T | = 1 ∀ v ∈ V (G) \ T and (ii) |N(u) ∩ T | = 1 ∀ u ∈ T . Combining the two conditions, we can express it concisely as: |N(u)∩T | = 1 ∀ u ∈ V (G). The cardinality of a minimum total exact dominating set is the total exact domination number, denoted by γte(G). A set that has a cardinality of γte(G) is a γte-set. Graph G is considered a non−γte−graph if it does not contain a total exact dominating set, following R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 3 of 26 a definition analogous to that of a non-γp0-graph as presented in [5]. The join of two graphs, G and H, denoted by G + H, is the graph with vertex set V (G+H) = V (G) ∪ V (H) and edge set E(G+H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)} [6]. The corona G ◦H of two graphs G and H is the graph obtained by taking one copy of G and |V (G)| copies of H and then forming the join ⟨{v}⟩+Hv = v +Hv, where Hv is a copy of H, for each v ∈ V (G) [6]. The lexicographic product or composition of two graphs G and H is the graph G[H] with vertex set V (G[H]) = V (G) × V (H) and edge set E(G[H]) = {(x, u)(y, v) | xy ∈ E(G) or x = y and uv ∈ E(H)}. Any subset (C) of (V (G[H])) can be expressed as C = ⋃ x∈S({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S. The set S is called the G-projection of C, and ⋃ x∈S Tx is called the H-projection of C [7]. The path graph Pn is a tree with two nodes of vertex degree 1, and the other n − 2 nodes of vertex degree 2 [8]. A cycle graph Cn are defined as simple, connected, undirected graphs consisting of a single cycle passing through all n vertices, with each vertex having degree 2 and the graph forming a closed loop [9]. The complete graph Km has every pair of its m points adjacent [7]. The complement of complete graphs Km are totally disconnected graphs and are regular of degree 0 [7]. A complete bipartite graph Km,n is a graph equivalent to Km +Kn such that n,m ≥ 1 [6]. The star graph Sn, of order n, is a tree on n vertices with one vertex of degree n− 1 and the remaining n− 1 vertices of degree 1 [7]. A wheel graph Wn, of order n, is formed by adding a vertex adjacent to all vertices of a cycle [7]. A fan graph Fn is formed by joining a path to a single vertex [7]. The generalized fan graph Fm,n is defined as the graph join Km + Pn, where Km is the empty graph on m vertices and Pn is the path graph on n vertices [10]. The generalized wheel graph Wm,n is a graph obtained by joining the vertices of Km to every vertex of a cycle Cn. That is, Wm,n = Cn+Km [11]. A windmill graph Wm n is a graph formed by connecting m copies of a complete graph Kn−1 to a single common vertex [9]. The friendship graph Fn is a graph consisting of n triangles that share a common central vertex. It is also called the Dutch windmill graph [12]. Example 1. Let G be the graph in Figure 1 and let T = {t4, u1, u4, u5, u8, u9, r3, r4, w3, w4, z1, x1}. The vertices in T are colored red, and the vertices in V (G) \ T are colored blue. Clearly, each blue vertex is adjacent to exactly one red vertex in T , that is, each vertex in V (G) \ T is adjacent to exactly one vertex in T or |N(v) ∩ T | = 1 ∀v ∈ V (G) \ T . Observe further that each red vertex is adjacent to exactly one red vertex in T , that is, each vertex in T are adjacent to exactly one vertex in T or |N(u) ∩ T | = 1 ∀u ∈ T . R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 4 of 26 Thus, T is a total exact dominating set of G and γte(G) = |T | = 12. In Figure 2, clearly, S = {u1, u4, u5, u8, u9, r3, w3, z1, x1} is a γe-set of G since |N(u) ∩ S| = 1 ∀v ∈ V (G) \ S and |N(v) ∩ S| ≤ 1 ∀v ∈ S with |N(u1) ∩ S| = |N(r3) ∩ S| = |N(w3) ∩ S| = 0. Thus, γe(G) = |S| = 9. Figure 1: Graph G with γte(G) = 12. Figure 2: Graph G with γe(G) = 9. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 5 of 26 3. Known Results The following results will be used to prove the main results. Theorem 1. [2]. For any connected graph G with p ≥ 3 vertices, then γt(G) ≤ 2p 3 . Proposition 1. [6] The total domination number of a cycle Cn or a path Pn on n ≥ 3 vertices are given by γt(Cn) = γt(Pn) =  n 2 , if n ≡ 0 (mod 4), n+2 2 , if n ≡ 2 (mod 4), n+1 2 , otherwise. Theorem 2. [13] Let G and H be connected graphs. Then C ⊆ V (G + H) is a total dominating set of G+H if and only if it satisfies at least one of the following: (i) C ∩ V (G) is a total dominating set of G. (ii) C ∩ V (H) is a total dominating set of H. (iii) C ∩ V (G) ̸= ∅ and C ∩ V (H) ̸= ∅. Corollary 1. [13] Let G and H be connected graphs. Then C = {x, y}, where x ∈ V (G) and y ∈ V (H), is a minimum total dominating set of G+H and γt(G+H) = 2. Corollary 2. [14] Let G be a connected graph of order m and let H be any graph of order n. Then γt(G ◦H) = m. 4. Main Results This section contains results corresponding to the total exact domination number of paths, cycles, complete bipartite graphs and graphs resulting from some binary operations. Furthermore, some non− γte − graphs are shown. Clearly, every total exact dominating set of G is an exact dominating set and a total dominating set. Thus, the next statement is immediate from this observation, definition of total exact dominating set and Theorem 1. Remark 1. Let G be any graph such that G has a total exact dominating set with p ≥ 3 vertices. Then γe(G) ≤ γte(G) and 2 ≤ γte(G) ≤ 2p 3 . R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 6 of 26 Theorem 3. For any positive integers p and q with 1 ≤ p ≤ q, there exists a simple graph G such that γe(G) = p and γte(G) = q. Proof. Consider the following cases: Case 1: p = q Let G be the graph shown in Figure 3. Clearly, the set T = {u1, u2, u3, u4, u5, u6, . . . , up−1, up} is both a γe-set and a γte-set of G. Therefore, γe(G) = p = q = γte(G). Figure 3: Graph G with γte(G) = γe(G) = p. Case 2: p < q Let G be the graph shown in Figure 4 or Figure 5. Let m = q − p and m ∈ Z+. Observe that S1 = {u1, u2, u3, . . . , up−m} ∪ {vi : i = 1, 2, . . . ,m} is a γe-set of G, and S2 = {u1, u2, u3, . . . , up−m} ∪ {vi : i = 1, 2, . . . ,m} ∪ {wi : i = 1, 2, . . . ,m} is a γte-set of G. It follows that γe(G) = |S1| = p−m+m = p and γte(G) = |S2| = (p−m) +m+m = p+m = p+ (q − p) = q. Therefore, γe(G) = p < q = γte(G). R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 7 of 26 Figure 4: Graph G with γe(G) = p. Figure 5: Graph G with γte(G) = q. This confirms the claim. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 8 of 26 The next result follows from Theorem 3. Corollary 3. The difference γte − γe can be made arbitrarily large. Theorem 4. If T is a total exact dominating set of a graph G, then |T | is even. Proof. Let T be a total exact dominating set with n vertices. By definition, |N(v) ∩ T | = 1 ∀ v ∈ T . Thus, for every vertex v ∈ T , there exists exactly one other vertex x ∈ T such that N(v) ∩ T = {x} and N(x) ∩ T = {v}.Thus the vertices of T can be paired uniquely into disjoint sets of two. Therefore, |T | must be even. Theorem 5. If T is a γt-set of a graph G and |T | is odd, then T is not a γte-set of G. Proof. Let T be a γt-set of a graph G and |T | is odd. Then there exists three vertices, say ui, uj , uk such that the vertex uj is adjacent to ui and uk by definition of total dominating set. Hence, |N(uj) ∩ T | = 2, a contradiction to the definition of exact dominating set. Therefore, T is not a γte-set of G. Theorem 6. Let n be a positive integer such that n ≥ 5. If n ≡ 1 (mod 4), then the path graph Pn is non− γte − graph. Proof. Let n = 5. By Proposition 1, γt(P5) = 5+1 2 = 3. Let T1 = {u2, u5} and T2 = {u2, u4}. Clearly, T1 and T2 are not total dominating sets since γt(P5) = 3. Therefore, they are not γte-sets. It is also clear that T3 = {u2, u3, u4} is the only γt-set of P5 and |T3| is odd. By Theorem 5, T3 is not a γte − set of P5. Let T4 = {u1, u2, u4, u5}. Clearly, |N(ui) ∩ T4| = 1 for ui ∈ V (P5) \ {u3} and |N(u3) ∩ T4| = 2. Thus, T4 is not a γte − set of P5. Since there is no possible way to create a set that is both total and exact dominating set, P5 is a non− γte − graph. Now, suppose that n > 5. Let p = n−1 4 and j = 1, 2, . . . , p. Group the vertices of Pn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−8, un−7, un−6, un−5}, and Rp = {un−4, un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 5. Case 1: Suppose that we pick u2 and u3 first to form T . Let T = {u2, u3, u6, u7, u10, u11, . . . , un−7, un−6, un−3, un−2, un−1}, R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 9 of 26 where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 3 vertices in Rp. Thus, |T | = 2(p− 1) + 3 = 2p+ 1 = 2 ( n−1 4 ) + 1 = n+1 2 , and clearly, N(T ) = V (Pn). By Proposition 1, T is a γt-set of Pn. Also, since |T | is odd, by Theorem 5, T is not a γte-set of Pn, and un−1 must not be in T . Removing un−1 from T or replacing un−1 by un in T means that un is not adjacent to a vertex in T , that is, |N(un) ∩ T | = 0 and |N(ui) ∩ T | = 1 for all ui ∈ V (Pn) \ {un}. Thus, T is not a total exact dominating set. Hence, it is not possible to form a γte-set T of Pn. Case 2: Suppose that we pick u1 and u2 first to form T . Let T = {u1, u2, u5, u6, u9, u10, . . . , un−8, un−7, un−4, un−3, un−1, un}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p − 1 and 4 vertices in Rp. Clearly, |T | is even and N(T ) = V (Pn). Thus, T is a total dominating set of Pn. It is also clear that |N(ui) ∩ T | = 1 for all ui ∈ V (Pn) \ {un−2} since un−2 is adjacent to un−3 and un−1 in T , that is, |N(un−2) ∩ T | = 2 and so, T is not an exact dominating set. Therefore, T is not a γte-set and un−1 must not be in T . Removing un−1 from T means that un is not adjacent to a vertex in T , that is, |N(un) ∩ T | = 0 and |N(ui) ∩ T | = 1 for all ui ∈ V (Pn) \ {un}. Thus, T is not a total exact dominating set. Hence, it is not possible to form a γte-set T of Pn. In both cases, it is not possible to create a set that is both a total and exact dominating set of Pn. Therefore, Pn is a non-γte-graph if n ≡ 1 (mod 4). Theorem 7. Let n be a positive integer such that n ≥ 2. Then the total exact domination number of a path Pn of order n, where n ̸≡ 1 (mod 4), is given by γte(Pn) =  n 2 , if n ≡ 0 (mod 4) n+2 2 , if n ≡ 2 (mod 4) n+1 2 , if n ≡ 3 (mod 4). Proof. Let V (Pn) = {u1, u2, u3, . . . , un−1, un}. Consider the following cases: Case 1: Suppose that n ≡ 0 (mod 4). Let n = 4. By Proposition 1, γt(P4) = 4 2 = 2. Let T = {u2, u3}. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (P4). Thus, T is a γte-set of P4, and γte(P4) = |T | = 2. Now, suppose that n > 4. Let p = n 4 and j = 1, 2, . . . , p − 1, p. Group the vertices of Pn into p disjoint subsets Rj , such that R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 10 of 26 R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−7, un−6, un−5, un−4}, and Rp = {un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1, p. Let T = {u2, u3, u6, u7, u10, u11, . . . , un−6, un−5, un−2, un−1}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1, p. It follows that |T | = 2p = 2 (n 4 ) = n 2 . Also, note that N(T ) = V (Pn). Thus, T is a γt-set of Pn by Proposition 1. It is also clear that |N(ui) ∩ T | = 1 for all ui ∈ V (Pn). Therefore, T is a γte-set. This implies that γte(Pn) = |T | = n 2 . Case 2: Suppose that n ≡ 2 (mod 4). When n = 2, S = {u1, u2} is a γte-set of P2 since |N(u1)∩S| = 1 and |N(u2)∩S| = 1. Thus, γte(P2) = |S| = 2. Let n = 6. By Proposition 1, γt(P6) = 6+2 2 = 2. Let T = {u1, u2, u5, u6}. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (P6). Thus, T is a γte-set of P6, and γte(P6) = |T | = 4. Now, suppose that n > 6, let p = n−2 4 and j = 1, 2, . . . , p− 1, p. Group the vertices of Pn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−9, un−8, un−7, un−6}, and Rp = {un−5, un−4, un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 6. Let T = {u1, u2, u5, u6, u9, u10, . . . , un−9, un−8, un−5, un−4, un−1, un}, R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 11 of 26 where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 4 vertices in Rp. It follows that |T | = 2(p− 1) + 4 = 2p+ 2 = 2 ( n− 2 4 ) + 2 = n+ 2 2 . Also, note that N(T ) = V (Pn). Thus, T is a γt-set of Pn by Proposition 1. It is also clear that |N(ui) ∩ T | = 1 for all ui ∈ V (Pn). Therefore, T is a γte-set and γte(Pn) = |T | = n+ 2 2 . Case 3: Suppose that n ≡ 3 (mod 4). When n = 3, S = {u1, u2} is a γte-set of P3 since |N(u1) ∩ S| = 1, |N(u2) ∩ S| = 1 and |N(u3)∩S| = 1. Thus, γte(P3) = |S| = 2. Let n = 7. By Proposition 1, γt(P7) = 7+1 2 = 4. Let T = {u1, u2, u5, u6}. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (P7). Thus, T is a γte-set of P7, and γte(P7) = |T | = 4. Also, R = {u2, u3, u6, u7} is another γte-set of P7 since |N(ui) ∩R| = 1 for all ui ∈ V (P7). Now, suppose that n > 7, let p = n−3 4 and j = 1, 2, . . . , p− 1, p. Group the vertices of Pn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−10, un−9, un−8, un−7}, and Rp = {un−6, un−5, un−4, un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 7. Let T = {u1, u2, u5, u6, u9, u10, . . . , un−10, un−9, un−6, un−5, un−2, un−1}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 4 vertices in Rp. It follows that |T | = 2(p− 1) + 4 = 2p+ 2 = 2 ( n− 3 4 ) + 2 = n+ 1 2 . Also, note that N(T ) = V (Pn). Thus, T is a γt-set of Pn by Proposition 1. It is also clear that |N(ui) ∩ T | = 1 for all ui ∈ V (Pn). Therefore, T is a γte-set. This implies that γte(Pn) = |T | = n+ 1 2 . R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 12 of 26 Theorem 8. Let n be a positive integer such that n ≥ 3. If n ̸≡ 0 (mod 4), then the cycle graph Cn is a non− γte − graph. Proof. Let V (Cn) = {u1, u2, u3, . . . , un−1, un} with deg(ui) = 2 for all ui ∈ V (Cn). Consider the following cases: Case 1: n ≡ 1 (mod 4) Let n = 5. By Proposition 1, γt(C5) = 3. Clearly, T1 = {u1, u2, u3}, T2 = {u2, u3, u4}, T3 = {u3, u4, u5}, T4 = {u4, u5, u1}, and T5 = {u5, u1, u2} are the only γt-sets of C5. For i = 1, 2, 3, ..., 5, |Ti| is odd and so, by Theorem 5, Ti is not a γte − set. Suppose that T has 4 vertices. Then there exists one vertex, say ui, such that ui is adjacent to 2 vertices of C5, that is, |N(ui) ∩ T | = 2. Thus, T is not a γte − set. Since there is no possible way to create a set that is both total and exact dominating set, C5 is a non− γte − graph. Now, suppose that n > 5. Let p = n−1 4 and j = 1, 2, . . . , p− 1, p. Group the vertices of Cn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−8, un−7, un−6, un−5}, and Rp = {un−4, un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 5. Let T = {u2, u3, u6, u7, u10, u11, . . . , un−7, un−6, un−3, un−2, un−1}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 3 vertices in Rp. It follows that |T | = 2(p− 1) + 3 = 2p+ 1 = 2 ( n− 1 4 ) + 1 = n+ 1 2 . Also, it is clear that N(T ) = V (Cn) and so, T is a γt-set by Proposition 1. Note that |T | is odd. By Theorem 5, T is not a γte-set. Clearly, |N(ui)∩T | = 1 for all ui ∈ V (Cn)\{un−2} and |N(un−2)∩T | = 2 and so, un−1 must not be in T . Note that u1 must not be in T also since otherwise, |N(u2)∩T | would become 2. Thus, the only option left is to replace un−1 with un; however, |T | remains odd and so T is not a γte-set. Since T is arbitrarily chosen, there is no possible way to create a set that is both a total and exact dominating set of Cn. Thus, Cn is a non-γte-graph if n ≡ 1 (mod 4). Case 2: n ≡ 2 (mod 4) Let n = 6. By Proposition 1, γt(C6) = 6+2 2 = 4. Clearly, T1 = {u1, u2, u3, u4}, T2 = {u1, u2, u4, u5}, T3 = {u1, u2, u5, u6}, T4 = {u2, u3, u4, u5}, T5 = {u2, u3, u5, u6}, R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 13 of 26 T6 = {u2, u3, u6, u1}, T7 = {u3, u4, u5, u6}, T8 = {u3, u4, u6, u1}, and T9 = {u4, u5, u6, u1} are the only γt-sets of C6. It is also clear that for i = 1, 2, 3, ..., 9, there always exists a vertex uk in V (C6) such that |N(uk) ∩ Ti| = 2; that is, u2 is associated with T1, u3 with T2, u1 with T3, u3 with T4, u4 with T5, u2 with T6, u4 with T7, u2 with T8, and u5 with T9. It follows that for i = 1, 2, 3, ..., 9, Ti is not a γte-set of C6. Hence, there is no possible way to create a set that is both a total and exact dominating set of C6 and so, C6 is a non-γte-graph. Now, suppose that n > 6. Let p = n−2 4 and j = 1, 2, . . . , p. Group the vertices of Cn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−9, un−8, un−7, un−6}, and Rp = {un−5, un−4, un−3, un−2, un−1, un} where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 6. Let T = {u2, u3, u6, u7, u10, u11, . . . , un−8, un−7, un−4, un−3, un−1, un}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 4 vertices in Rp. It follows that |T | = 2(p− 1) + 4 = 2p+ 2 = 2 ( n− 2 4 ) + 2 = n+ 2 2 . Also, it is clear that N(T ) = V (Cn) and so, T is a γt-set by Proposition 1. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (Cn) \ {un−2, u1}, |N(un−2) ∩ T | = 2 and |N(u1) ∩ T | = 2. Therefore, T is not a γte-set, and both un−1 and un must not be in T so that |N(un−2) ∩ T | = 1 and |N(u1) ∩ T | = 1. Note also that un−2 and u1 must not be in T since otherwise, |N(un−3) ∩ T | and |N(u2) ∩ T | would become 2. Thus, no two vertices can replace both un−1 and un in T . Since T is arbitrarily chosen, there is no possible way to create a set that is both a total and exact dominating set of Cn. Thus, Cn is a non-γte-graph if n ≡ 2 (mod 4). Case 3: n ≡ 3 (mod 4) Let n = 7. By Proposition 1, γt(C7) = 4. Clearly, T1 = {u1, u2, u4, u5}, T2 = {u1, u2, u5, u6}, T3 = {u2, u3, u5, u6}, T4 = {u2, u3, u6, u7}, T5 = {u3, u4, u6, u7}, T6 = {u3, u4, u7, u1}, and T7 = {u4, u5, u7, u1} are the only γt-sets of C7. It is also clear that for i = 1, 2, 3, ..., 7, there always exists a vertex uk in V (C7) such that |N(uk)∩Ti| = 2; that is, u3 is associated with T1, u7 with T2, u4 with T3, u1 with T4, u5 with T5, u2 with R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 14 of 26 T6, and u6 with T7. It follows that for i = 1, 2, 3, ..., 7, Ti is not a γte-set of C7. Hence, there is no possible way to create a set that is both a total and exact dominating set of C7 and so, C7 is a non-γte-graph. Now, suppose that n > 7. Let p = n−3 4 and j = 1, 2, . . . , p. Group the vertices of Cn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−10, un−9, un−8, un−7}, and Rp = {un−6, un−5, un−4, un−3, un−2, un−1, un}, where |Rj | = 4 for j = 1, 2, . . . , p− 1 and |Rp| = 7. Let T = {u2, u3, u6, u7, u10, u11, . . . un−9, un−8, un−5, un−4, un−1, un}, where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1 and 4 vertices in Rp. It follows that |T | = 2(p− 1) + 4 = 2p+ 2 = 2 ( n− 3 4 ) + 2 = n+ 1 2 . Also, it is clear that N(T ) = V (Cn) and so, T is a γt-set by Proposition 1. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (Cn) \ {u1} and |N(u1) ∩ T | = 2. Therefore, T is not a γte-set. Note that |N(u1) ∩ T | = 1 if either u2 or un, but not both, is not in T . If u2 /∈ T and un ∈ T , |N(u3) ∩ T | = 0. If un /∈ T and u2 ∈ T , |N(un−1) ∩ T | = 0. In either case, T is not a γte-set. Since T is arbitrarily chosen, there is no possible way to create a set that is both a total and exact dominating set of Cn. Thus, Cn is a non-γte-graph if n ≡ 3 (mod 4). Therefore, in any case, Cn is a non-γte-graph if n ̸≡ 0 (mod 4). Theorem 9. Let n be a positive integer such that n ≥ 4. Then the total exact domination number of a cycle Cn of order n, where n ≡ 0 (mod 4), is given by γte(Cn) = n 2 . Proof. Suppose that n ≡ 0 (mod 4). Let n = 4. By Proposition 1, γt(C4) = 4 2 = 2. Clearly, T1 = {u1, u2}, T2 = {u2, u3}, T3 = {u3, u4}, and T4 = {u4, u1} are the only γt-sets of C4. It is also clear that for i = 1, 2, 3, 4, |N(uk) ∩ Ti| = 1 for all vertices uk in V (C4). Thus, for i = 1, 2, 3, 4, Ti is a γte-set of C4 and so, γte(C4) = |Ti| = 2. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 15 of 26 Now, suppose that n > 4. Let p = n/4 and j = 1, 2, . . . , p. Group the vertices of Cn into p disjoint subsets Rj , such that R1 = {u1, u2, u3, u4}, R2 = {u5, u6, u7, u8}, R3 = {u9, u10, u11, u12}, ... Rp−1 = {un−7, un−6, un−5, un−4}, and Rp = {un−3, un−2, un−1, un}. where |Rj | = 4 for j = 1, 2, . . . , p− 1, p. Let T = {u2, u3, u6, u7, u10, u11, . . . , un−6, un−5, un−2, un−1} where T is formed by getting 2 vertices in each Rj for j = 1, 2, . . . , p− 1, p. It follows that |T | = 2p = 2 (n 4 ) = n 2 . Also, it is clear that N(T ) = V (Cn) and so, T is a γt-set by Proposition 1. Clearly, |N(ui) ∩ T | = 1 for all ui ∈ V (Cn). Therefore, T is a γte-set. This implies that γte(Cn) = |T | = n 2 . Theorem 10. The total exact domination number of a complete bipartite graph Km,n is given by γte(Km,n) = 2 . Proof. The vertex set of Km,n can be partitioned into two disjoint sets, U and V , where each vertex in U is adjacent to every vertex in V and and there are no edges within U and within V . Let U = {u1, u2, . . . , um} and V = {v1, v2, . . . , vn}. Consider T ⊆ V (Km,n) containing exactly one vertex up ∈ U and one vertex vq ∈ V , that is, T = {up, vq}. Clearly, |N(ui)∩T | = |{vq}| = 1 for all ui ∈ U and |N(vj)∩T | = |{up}| = 1 for all vj ∈ V . Hence, T is a γte-set of Km,n and so, γte(Km,n) = |T | = 2. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 16 of 26 4.1. Total Exact Dominating Sets in the Join of Graphs This section contains results when the join G + H has either a γte − set or has no γte − set and its total exact domination number. Theorem 11. Let G and H be any graphs. Then G+H is a non-γte-graph if and only if it satisfies one of the following: (i) both G and H have no isolated vertices; or (ii) exactly one of the graphs G and H has at least one isolated vertex, while the other has none. Proof. Suppose that both G and H have at least one isolated vertex. Let u and v be isolated vertices in G and H, respectively. Clearly, T = {u, v} is a total dominating set of G + H by Theorem 2(iii). Clearly, |N(u) ∩ T | = |{v}| = 1 and |N(v) ∩ T | = |{u}| = 1. Since u is an isolated vertex in G, any vertex p ̸= u in G is adjacent to the vertex v in G +H. That is, in G +H, |N(p) ∩ T | = |{v}| = 1 for all p ∈ V (G). Also, since v is an isolated vertex in H, any vertex q ̸= v in H is adjacent to the vertex u in G +H. That is, in G+H, |N(q) ∩ T | = |{u}| = 1 for all q ∈ V (H). Therefore, T is a γte − set and so, G+H is not a non-γte-graph. Suppose that (i) holds, that is, both G and H have no isolated vertices. By Corollary 1, γt(G+H) = 2. Consider the following cases: Case 1: T is a total dominating set that contains two vertices in V (G) or V (H). Without loss of generality, let T = {p, q} where p, q ∈ V (G). Note that for any vertex r ∈ V (H), r is adjacent to both p and q in G+H, that is, |N(r)∩T | = 2 for all r ∈ V (H). Thus, T is not a total exact dominating set. Similarly, if T is a total dominating set that contains two vertices in V (H), then T is not a total exact dominating set. Case 2: T is a total dominating set that contains one vertex in V (G) and one vertex in V (H). Let T = {s, t} where s ∈ V (G) and t ∈ V (H). Note that s and t are adjacent in G+H. Since G has no isolated vertices, there exists at least one vertex w ∈ V (G) that is adjacent to the vertex s also. Note that w is also adjacent to the vertex t in G+H. Thus, |N(w) ∩ T | = 2. Hence, T is also not a total exact dominating set. Suppose that (ii) holds. Without loss of generality, suppose that G has no isolated vertices, and H has at least one isolated vertex. Let u be an isolated vertex in H, and let v ∈ V (G). Pick T = {u, v}. Then there exists at least one vertex z ∈ V (G) such that z is adjacent to v since G has no isolated vertices. Note that z is also adjacent to u in G+H , that is , |N(z) ∩ T | = 2. Hence, T is also not a total exact dominating set. Thus, a total exact dominating set cannot exist. Therefore, under conditions (i) and (ii), G+H is a non-γte-graph. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 17 of 26 Theorem 12. Let G and H be any graphs. Then a subset T of V (G+H) is a γte-set of G+H if and only if T = {u, v}, where u is an isolated vertex in V (G) and v is an isolated vertex in V (H). Proof. Let T ⊆ V (G+H) be a γte-set of G+H. By the proof of Theorem 11, G+H has a total exact dominating set if both G and H have at least one isolated vertex, say u and v, respectively. Clearly, the vertices u and v are adjacent in G+H and set {u, v} is a total dominating set of G +H by Theorem 2 (iii). Note that in the graph G +H, all vertices in V (G) \ {u} is adjacent to the vertex v but not in u while all vertices in V (H) \ {v} is adjacent to the vertex u but not in v. Thus, |N(r) ∩ {u, v}| = 1 for all r ∈ V (G + H). Therefore, T = {u, v}. The converse is clear. The next results follow directly from Theorems 11 and 12. Corollary 4. If both graphs G and H have at least one isolated vertex, then γte(G+H) = 2. Example 2. The following example illustrates the validity of Corollary 4. Figure 6: Join of graphs G and H with γte(G+H) = 2. Corollary 5. Let G and H be nontrivial connected graphs, and let u and v be isolated vertices, then ((G∪H)+{u, v}) is a non−γte−graph and γte((G∪{u})+(H ∪{v})) = 2. Corollary 6. If either G or H or both have no isolated vertex, then G + H is a non-γte graph. Corollary 7. The following are graphs having γte(G) = 2. (i) Star graph Sn = K1 +Kn, n ≥ 1 (ii) Complete bipartite graph Km,n = Km +Kn m ≥ 2 and n ≥ 1. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 18 of 26 Example 3. The following examples verify the results of Corollary 7. Figure 7: Graph S6 with γte(S6) = 2 and graph K3,4 with γte(K3,4) = 2. Corollary 8. The following are non-γte-graphs. (i) Fan graph Fn = K1 + Pn, n ≥ 2 (ii) Wheel graph Wn = K1 + Cn, n ≥ 3 (iii) Friendship graph Fn = K1 + nP2, n ≥ 2 (iv) Windmill graph Wm n = K1 +mKn−1, n ≥ 3 and m ≥ 2. (v) Generalized fan graph Fm,n = Km + Pn, m ≥ 2 and n ≥ 2. (vi) Generalized wheel graph Wm,n = Km + Cn, m ≥ 2 and n ≥ 3. 4.2. Total Exact Dominating Set in the Corona of Graphs This section contains results when the corona G ◦ H has either a γte − set or has no γte − set and its total exact domination number. Theorem 13. Let G be a connected graph, and H be any graph. Then G ◦ H is a non-γte-graph if and only if either of the following is satisfied: (i) |V (G)| ≥ 3, (ii) |V (G)| = 1 and H has no isolated vertices. Proof. Suppose that |V (G)| = 2. By Corollary 2, γt(G ◦H) = 2. Thus, V (G) = {u, v} is a γt-set of G ◦ H. Take T = V (G). Clearly, in G ◦ H, |N(u) ∩ T | = |N(v) ∩ T | = 1, |N(au) ∩ T | = |{u}| = 1 for all vertices au ∈ V (Hu) and |N(av) ∩ T | = |{v}| = 1 for all vertices av ∈ V (Hv). Hence, T is a γte-set of G◦H. Therefore, G◦H is not a non-γte-graph. Suppose that |V (G)| = 1 and H has at least one isolated vertex, say r. Let V (G) = {s} and so, rs is an isolated vertex in Hs. Take T = {s, rs}. Since rs and s are adjacent in G◦H, |N(s)∩T | = |N(rs)∩T | = 1. Also, in G ◦H, it is clear that |N(ts)∩T | = |{s}| = 1 for all R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 19 of 26 vertices ts ∈ V (Hs). Hence, T is a γte-set of G◦H. Therefore, G◦H is not a non-γte-graph. Suppose that (i) holds, that is, G is of order m ≥ 3. By Corollary 2, γt(G ◦H) = m. Note that V (G) is a γt-set of G ◦H since N(V (G)) = V (G ◦H) and |V (G)| = m. Let T be a total exact dominating set of G ◦H. Suppose that T ̸= V (G), that is, not all vertices in V (G) are in T , say z ∈ V (G) \ T . Then there exist at least two adjacent vertices in V (Hz), say az and bz such that the vertices az and bz must be in T and |N(az)∩T | = |N(bz)∩T | = 1. Note that z is adjacent to both az and bz in G ◦H, that is, |N(z) ∩ T | = |{az, bz}| = 2. This is a contradiction to the definition of a total exact dominating set. Thus, for all z ∈ V (G), no two vertices in V (Hz) can be chosen to form a total exact dominating set of G ◦H and T = V (G). Now, suppose that T = V (G), that is, all vertices in G are in T . Since G is connected and |V (G)| ≥ 3, there exists at least one vertex u ∈ V (G) that is adjacent to two vertices in G, and so, |N(u) ∩ T | = 2 for some u ∈ V (G), a contradiction. Hence, T ̸= V (G) and T is not a total exact dominating set of G ◦H. Suppose that (ii) holds, that is, |V (G)| = 1 and H has no isolated vertices. Let V (G) = {p}. By the previous argument, no two vertices in V (Hp) can be chosen to form a total exact dominating set of G ◦ H. Take T = {p, qp} where qp is a vertex in V (Hp). Clearly, T is a total dominating set of G ◦ H since N(T ) = V (G ◦ H). Since H has no isolated vertices, there exists at least one vertex cp that is adjacent to qp. Note that cp is also adjacent to p in G ◦H, that is, |N(cp)∩ T | = |{p, qp}| = 2, a contradiction. Hence, T is not a total exact dominating set of G ◦H. Thus, it is not possible to create a total exact dominating set of G ◦ H. Therefore, G ◦H is a non-γte-graph. Theorem 14. Let G be a connected graph, and H be any graph. Then the set T ⊆ V (G◦H) is a γte-set of G ◦H if and only if either of the following conditions is satisfied: (i) G ∼= P1 and T = V (P1) ∪ {u}, where u is an isolated vertex in H or (ii) G ∼= P2 and T = V (P2). Proof. Suppose that T ⊆ V (G ◦ H) is a γte-set of G ◦ H. Then by using the same argument from Theorem 13, the graph G ◦ H will have a γte-set if and only if either (i) |V (G)| = 1 and H has at least one isolated vertex, or (ii) |V (G)| = 2. Suppose that |V (G)| = 1 and H has at least one isolated vertex, say u. Clearly, G ∼= P1. Since both G and H have isolated vertices and P1◦H = P1+H, by Theorem 12, T = {x, u} where x ∈ V (P1) and u is an isolated vertex in V (H). Therefore, T = V (P1) ∪ {u} where u is an isolated vertex in V (H). R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 20 of 26 Suppose that |V (G)| = 2. Then G ∼= P2. By Corollary 2, γt(G ◦ H) = 2. Clearly, V (P2) is a γte-set of G ◦H since |N(v) ∩ V (P2)| = 1 for all v ∈ G ◦H. Thus, T = V (P2). The converse is easy. The next results follow directly from Theorems 13 and 14: Corollary 9. Let G ∼= P1 graph and H be any graph with an isolated vertex. Then γte(G ◦H) = 2. Corollary 10. Let G ∼= P2 graph and H be any graph. Then γte(G ◦H) = 2. Example 4. Consider the graphs G1 ◦ H1 and G2 ◦ H2 in Figure 8. Let T1 = {a, ua} and T2 = {a, b}. It is clear that |N(ra) ∩ T1| = 1 for all vertices ra ∈ V (G1 ◦ H1) and |N(sa) ∩ T1| = |N(sb) ∩ T1| = 1 for all vertices sa, sb ∈ V (G2 ◦ H2). Hence, T1 and T2 are γte-sets of G1 ◦ H1 and G2 ◦ H2, respectively. It follows that γte(G1 ◦ H1) = 2 and γte(G2 ◦H2) = 2. Figure 8: Graphs G1 ◦H1 and G2 ◦H2 with γte(G1 ◦H1) = γte(G2 ◦H2) = 2. Corollary 11. If G ≇ P1 or G ≇ P2, then for any graph H, G ◦H is a non-γte-graph. Corollary 12. Let G be a disconnected graph with k components such that each component is isomorphic to either P1 or P2 and H be any graph with an isolated vertex. Then γte(G ◦H) = 2k. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 21 of 26 Example 5. The following example verifies the result of Corollary 12. Figure 9: Graph G ◦H with γte(G ◦H) = 2(5) = 10. Corollary 13. Let G be a disconnected graph with k components such that each component is isomorphic to P2 and H be any graph. Then γte(G ◦H) = 2k. 4.3. Total Exact Dominating Set in the Lexicographic Product of Graphs This section contains results when the lexicographic product G[H] has either a γte−set or has no γte − set and its total exact domination number. Theorem 15. Let G and H be graphs such that at least one of them is connected, and at most one is an empty graph. Then G[H] is a non-γte-graph if and only if one of the following is satisfied: (i) both G and H are nontrivial connected graphs, (ii) G is an empty graph and H is a connected non- γte-graph, or (iii) H is an empty graph and G is a connected non- γte-graph. Proof. Suppose that G is an empty graph and H is a connected graph, and let T be a γte-set of H. Since G is an empty graph, G[H] is a disjoint union of |V (G)| copies of H. Clearly, |C| = |V (G)||T | and C is a γte-set of G[H]. Hence, G[H] is not a non-γte-graph. Suppose that H is an empty graph and G is a connected graph, and S is a γte-set of G. Let C = ⋃ x∈S({x} × y) where y ∈ V (H). Since H is an empty graph, all vertices of the |V (H)| copies of G is adjacent to the exactly one vertex in C. Thus, |N(u, v)∩C| = 1 for all vertices (u, v) ∈ V (G[H]). Therefore, C is a γte-set of G[H]. Hence, G[H] is not a R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 22 of 26 non-γte-graph. Suppose that (i) holds, that is, both G and H are nontrivial connected graphs. Let C be a γte-set of G[H]. Note that for any two adjacent vertices u, v ∈ V (G) and any two adjacent vertices p, q ∈ V (H), the vertices formed in G[H] are (u, p), (v, p), (u, q), and (v, q). These vertices are all adjacent to each other, forming a complete subgraph K4. This holds for every pair of adjacent vertices in V (G) and V (H) when constructing the vertices of G[H]. Moreover, note that two vertices in C must be adjacent. Thus, there exists a vertex in V (G[H]) \ C such that |N(s, t) ∩ C| = 2. This contradicts the definition of C. Hence, G[H] is non-γte-graph. Suppose that (ii) holds, that is, G is an empty graph and H is a connected non- γte-graph. Let C be a γte-set of G[H]. Since G is an empty graph, G[H] is a disjoint union of |V (G)| copies of H and so, the set C must be the union of γte-sets from each copy of H. However, since H is a non-γte-graph, no such γte-set can exist in any copy of H, making it impossible to form C. Hence, G[H] is non-γte-graph. Suppose that (iii) holds, that is, H is an empty graph and G is a connected non- γte-graph. Let C be a γte-set of G[H]. Since H is an empty graph and G is a connected graph, G[H] is a connected graph such that the neighbors of (u, p) ∈ V (G[H]) come only from adjacent copies of H in G[H]. That is, N(u, p) contains all vertices from independent sets corresponding to neighbors of u in G. Since the independent sets are fully connected through the edges of G and G is a non- γte-graph, it is impossible to pick a set C such that every vertex has exactly one neighbor in C. Thus, G[H] is non-γte-graph. Theorem 16. Let G be a connected graph, and H be any graph of order m. Then C = ⋃ x∈S ({x} × Tx) ⊆ V (G[H]) is a γte-set of G[H] if and only if either one of the following is satisfied: (i) S is a γte-set of G and H is an empty graph or (ii) Tx is a γte-set of H for every x ∈ V (G) and G is an empty graph. Proof. Let C be a γte-set of G[H]. By Theorem 15, either (i) H is an empty graph and G is a connected graph and not a non-γte-graph or (ii) G is an empty graph and H is a connected graph and not a non-γte-graph. Consider the following cases: (i) Suppose that H is an empty graph and G is a connected graph and not a non-γte-graph. Let u ∈ V (G). Pick any v ∈ V (H). Since C is a γte-set , there exists (y, z) ∈ C such that NG[H](u, v) ∩ C = {(y, z)}. This implies that NG(u)∩ S = {y}. Hence, every u ∈ V (G) is dominated by exactly one vertex in S, and so, S is a γte-set of G. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 23 of 26 (ii) Suppose that G is an empty graph and H is a connected graph and not a non-γte-graph. Suppose that Tx is not a γte-set of H for every x ∈ V (G). Then, since G is an empty graph, G[H] is a disjoint union of |V (G)| copies of H. Thus, C = ⋃ x∈S ({x} × Tx) is not a γte-set of G[H] since Tx is not a γte-set of H for every x ∈ V (G). This is a contradiction. Therefore, Tx is a γte-set of H for every x ∈ V (G). The converse is easy. Corollary 14. Let G be a connected graph such that G has a total exact dominating set and H be an empty graph. Then γte(G[H]) = γte(G). Proof. Since H is an empty graph, by Theorem 16 (i), γte(G[H]) = γte(G). Corollary 15. Let H be a connected graph such that H has a total exact dominating set and G be an empty graph of order n. Then γte(G[H]) = nγte(H). Proof. Since G is an empty graph of order n and H has a total exact dominating set, by Theorem 16 (ii), γte(G[H]) = nγte(H). Example 6. The following examples verify the results of Corollaries 14 and 15. Figure 10: Graph G1[H1] with γte(G1[H1]) = γte(G1) = 4. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 24 of 26 Figure 11: Graph G2[H2] with γte(G2[H2]) = 3(γte(H2)) = 3(4) = 12. The next results follow directly from Corollaries 14 and 15. Corollary 16. Let m and n be positive integers where m,n ≥ 2. Then γte(Km[Pn]) =  mn 2 , if n ≡ 0 (mod 4) m ( n+2 2 ) , if n ≡ 2 (mod 4) m ( n+1 2 ) , if n ≡ 3 (mod 4) and γte(Pn[Km]) = γte(Pn) =  n 2 , if n ≡ 0 (mod 4) n+2 2 , if n ≡ 2 (mod 4) n+1 2 , if n ≡ 3 (mod 4). Corollary 17. The total exact domination number of the Lexicographic product of Km and Cn is given by γte(Km[Cn]) = mn 2 , ∀ n ≡ 0 (mod 4) and γte(Cn[Km]) = n 2 , ∀ n ≡ 0 (mod 4). The next result follows directly from Theorems 6, 8 and 15. Corollary 18. The following lexicographic products are non-γte-graphs: i. Pn[Km] ∀ n ≡ 1 (mod 4), ii. Km[Pn] ∀ n ≡ 1 (mod 4), iii. Cn[Km] ∀ n ̸≡ 0 (mod 4), iv. Km[Cn] ∀ n ̸≡ 0 (mod 4). R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 25 of 26 5. Conclusion In this study, we explored and introduced the concept of total exact domination in graphs. We computed γte(G) for several families of special graphs such as paths, cycles, stars, and complete bipartite graphs, as well as for graphs resulting from some binary operations including the join, corona, and lexicographic product. Graphs that do not have total exact dominating sets were identified as non-γte-graphs, examples of which include complete graphs, fan graphs, and wheel graphs with more than two vertices. In addition, we considered some disconnected graphs in the binary operations, enriching the scope of the investigation. The total exact domination has applications in communication and sensor networks, where nodes must be fully covered (total domination) with minimal overlapping influence (exact domination), similar to applications discussed in network design scenarios [15]. The total exact domination model ensures efficient placement of resources or control centers with distinct coverage, reducing redundancy. For future research, we recommend studying the total exact domination number under other binary operations not mentioned in this study, such as the strong product, Cartesian product, and tensor product of graphs. Additionally, extending the study to weighted graphs, directed graphs, or random graphs could broaden its computational relevance and applicability. Acknowledgements The authors would like to thank the anonymous referees for their notable and valuable suggestions and comments that helped shape the quality of this study. They also express their heartfelt gratitude to the Department of Science and Technology – Science and Technology Regional Alliance of Universities for National Development (DOST- STRAND) for the financial support, which is vital in the actualization of this study, Cebu Normal University - College of Computing, Artificial Intelligence and Sciences, which also provided guidance and made the publication of this work possible. We acknowledge Ho Chi Minh City University of Technology (HCMUT), VNU-HCM for supporting this study. References [1] O. Ore. Theory of Graphs. American Mathematical Society, 1962. [2] E.J. Cockayne, R.M. Dawes, S.T. Hedetniemi. Total Domination in Graphs. Networks, 10(3):211–219, 1980. [3] A. Kinsley, A. Vetha. Exact Domination in Graphs. Malaya Journal of Matematik, S(1):236–242, 2020. R. G. Aguinod, E. M. Kiunisala, C. L. Armada / Eur. J. Pure Appl. Math, 18 (2) (2025), 6124 26 of 26 [4] C. Armada. Forcing Subsets for γ∗tpw-Sets in Graphs. 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