Class of Crosscap Two Graphs Arising from Lattices–I

Abstract

Let $\mathcal{L}$ be a lattice. The annihilating-ideal graph of $\mathcal{L}$ is a simple graph whose vertex set is the set of all nontrivial ideals of $\mathcal{L}$ and whose two distinct vertices I and J are adjacent if and only if $I\wedge J=0$. In this paper, crosscap two annihilating-ideal graphs of lattices with at most four atoms are characterized. These characterizations provide the classes of multipartite graphs, which are embedded in the Klein bottle.

1. Introduction

According to the well-known theorem of Kuratowski and Wagner, a graph is planar if and only if it does not contain either of the two forbidden graphs $K_5$ and $K_{3,3}$. The Graph Minor Theorem of Robertson and Seymour [1] can be considered a powerful generalization of Kuratowski’s Theorem. In particular, their theorem, which is the “deepest” and “most important” result in the arena of graph theory [2], implies that each graph property, no matter what, is characterized by a corresponding finite list of graphs. Thus, for surfaces (both orientable and non-orientable) in general, it is known that the set of forbidden minors is finite [3]. An analogous characterization for the embedding of graphs on surfaces is known for the crosscap one surface (Möbius strip) where 103 forbidden subgraphs (equivalently 35 forbidden minors) are characterized [4,5]. So, an open problem is to determine the several forbidden subgraphs for crosscap two surfaces (the Klein bottle). In this sequel, finding a family of graphs that has a crosscap two is an interesting one. Note that most of the 103 graphs contain a subgraph that is homeomorphic to $K_{3,3}$, and multipartite graphs play a vital role in finding these 103 forbidden subgraphs for the projective plane. It is worth mentioning that the crosscap value of bipartite and tripartite graphs are well known (refer to Proposition 1). The main goal of this paper is to identify a large class of crosscap two r-partite graphs where $r\ge 4$.

Let us introduce the concept of the annihilating-ideal graph of a lattice, a type of multipartite graph. Note that the annihilating-ideal graph is an extension of the concept of the zero-divisor graph. The idea of the zero-divisor graph of a ring structure is due to Beck [6]. In 2009, Halaš et al. [7] introduced the zero-divisor graph for a partially ordered set, and, in 2012, Estaji et al. [8] extended the concept of the zero-divisor graph to an arbitrary finite bounded lattice. For a clear exposition of the work completed in the area of zero-divisor graphs and their related areas, the reader is referred to the book by Anderson et al. [9]. In 2011, Behboodi et al. [10] defined and investigated the ideal theoretic version of the zero-divisor graph, called the annihilating-ideal graph of a ring, and, thereafter, many facts about zero-divisors were expressed in the language of ideals. The concept of an annihilating-ideal graph of a ring was extended to an arbitrary lattice by Afkhami et al. [11] in 2015. The annihilating-ideal graph of a lattice $\mathcal{L}$, denoted by $\mathbb{AG}\left(\mathcal{L}\right)$, is defined to be a simple graph whose vertex set is the set of all non-trivial ideals of $\mathcal{L}$, and whose two distinct vertices I and J are adjacent if and only if $I\wedge J=0$. The hope when studying the annihilating-ideal graph of a lattice is that the graph theoretic properties of the graph from the lattice will help us to better understand the lattice theoretic properties of the lattice.

One of the most important topological properties of a graph is its genus, which can be orientable or non-orientable (crosscap). The genus of graphs associated with algebraic structures has been studied by many authors (see [12,13,14,15,16,17]). The planar zero-divisor graph was first explicitly characterized by Smith [18], and the characterization of commutative rings with projective zero-divisor graphs was obtained by Chiang-Hsieh [15]. In 2019, Asir et al. [12] enumerated all commutative rings whose zero-divisor graph has a crosscap two. The planar and crosscap one annihilating-ideal graph of lattices were characterized by Shahsavar [19] and Parsapour et al. [20], respectively. Additionally, whether the line graph associated with the annihilating-ideal graph of a lattice is planar or projective was characterized by Parsapour et al. [21]. Moreover, the authors of [22] characterized all lattices $\mathcal{L}$ whose line graph of $\mathbb{AG}\left(\mathcal{L}\right)$ is toroidal.

Now, this paper aims to classify lattices with a number of atoms less than or equal to four whose annihilating-ideal graph can be embedded in the non-orientable surfaces of crosscap two. The main results of this paper are Theorems 2, 3, and 5, in which we have obtained our classifications. As a result, this classification provides a large class of r-partite graphs that can be embedded in the Klein bottle. Further, in the proof of the main theorems, we have shown several minimal r-partite graphs that cannot be embedded in the Klein bottle. Possibly, these graphs may be realized as forbidden subgraphs for crosscap two surfaces (refer to Example 1). Further, in order to cover the missing cases in the proof of Theorem 2.6 [20], which affects the statement of the corresponding theorem, the modified version is included as Theorem 4.

2. Preliminaries

In this section, we present the definitions and results needed to prove the main results in the subsequent sections. First, we recall some definitions and notations on lattices. A lattice is an algebra $\mathcal{L}=(\mathcal{L},\wedge ,\vee )$, where ∧ and ∨ are the binary operations, satisfying the following conditions: for all $a,b,c\in \mathcal{L}$

1.

$a\wedge a=a$, $a\vee a=a;$

2.

$a\wedge b=b\wedge a$, $a\vee b=b\vee a;$

3.

$(a\wedge b)\wedge c=a\wedge (b\wedge c)$; $a\vee (b\vee c)=(a\vee b)\vee c$;

4.

$a\vee (a\wedge b)=a\wedge (a\vee b)=a$.

According to [23] (Theorem 2.1), we can define an order ≤ on $\mathcal{L}$ as follows: for any $a,b\in \mathcal{L}$, we set $a\le b$ if and only if $a\wedge b=a$. Then $(\mathcal{L},\le )$ is an ordered set in which every pair of elements has the greatest lower bound ($glb$) and the least upper bound ($lub$). Conversely, let P be an ordered set such that, for every pair $a,b\in P$, $glb(a,b)$ and $lub(a,b)$ belong to P. For each a and b in P, we define $a\wedge b=glb(a,b)$ and $a\vee b=lub(a,b)$. Then $(P,\wedge ,\vee )$ is a lattice. A lattice $\mathcal{L}$ is said to be bounded if there are the elements 0 and 1 in $\mathcal{L}$ such that $0\wedge a=0$ and $a\vee 1=1$, for all $a\in \mathcal{L}.$ Clearly, every finite lattice is bounded. Let $(\mathcal{L},\wedge ,\vee )$ be a lattice with a least element 0 and I be a non-empty subset of $\mathcal{L}$. Then I is said to be the ideal of $\mathcal{L}$, denoted by $I⊴\mathcal{L}$,

1.

For all $a,b\in I,a\vee b\in I.$

2.

If $0\le a\le b$ and $b\in I,$ then $a\in I.$

In a lattice $(\mathcal{L},\wedge ,\vee )$ with a least element 0, an element a is called an atom if $a\ne 0$, and, for an element $x\in \mathcal{L}$, the relation $0\le x\le a$ implies that either $x=0$ or $x=a.$ We denote the set of all atoms of $\mathcal{L}$ by $A\left(\mathcal{L}\right)$. For basic facts about lattices, we refer the reader to [24].

Next, we recall the following terms regarding graph embedding. For the non-negative integers ℓ and k, let $S_{\ell }$ denote the sphere with ℓ handles, and $N_k$ denote a sphere with k crosscaps attached to it. Note that every connected compact surface is homeomorphic to $S_{\ell }$ or $N_k$ for some non-negative integers ℓ and k. The genus $\gamma \left(G\right)$ of a simple graph G is the minimum ℓ such that G can be embedded in $S_{\ell }$. Similarly, crosscap number (non-orientable genus) $\tilde{\gamma }\left(G\right)$ is the minimum k such that G can be embedded in $N_k$. Note that the projective space is of crosscap one and the Klein bottle is of crosscap two. If $e=xy\in E\left(G\right)$, then the contraction of e in G, denoted as $[x,y]$ is the graph obtained from $G-xy$ by identifying vertices x and y to create a new vertex z incident with all edges of G that were incident with either x or y. We say H is a minor of G, if H can be obtained from G by deleting vertices, edges, and/or contracting edges. For a graph G, we denote $\tilde{G}$ for the subgraph $G-V^{\prime }$ where $V^{\prime }=\{v\in V|deg\left(v\right)=1\}$, and we call this graph the reduction of G. For details on the notion of the embedding of graphs in a surface, we recommend reading [25].

The following three results on the non-orientable embedding of graphs are used frequently in this paper. In what follows, we denote the complete graph with p vertices by $K_p$, the complete bipartite graph with parts of sizes p and q by $K_{p,q}$, the complete tripartite graph with parts of sizes $p,q$, and r by $K_{p,q,r}$, and the complete four-partite graph with parts of sizes $p,q,r$, and s by $K_{p,q,r,s}$.

Proposition 1 

([25,26]). Let $p,q,r$, and s be positive integers greater than or equal to two. Then

(a)

$\tilde{\gamma }\left(K_p\right)=\left\{\begin{array}{ccc}⌈\frac{(p-3)(p-4)}{6}⌉& if& p\ge 3\\ 3& if& p=7.\end{array}\right.$

(b)

$\tilde{\gamma }\left(K_{p,q}\right)=⌈\frac{(p-2)(q-2)}{2}⌉$.

(c)

$\tilde{\gamma }\left(K_{p,q,r}\right)=⌈\frac{(p-2)(q+r-2)}{2}⌉$except for $K_{3,3,3},K_{4,4,1}$and $K_{4,4,3}$. Further, $\tilde{\gamma }\left(K_{3,3,3}\right)=3,$$\tilde{\gamma }\left(K_{4,4,1}\right)=4$ and $\tilde{\gamma }\left(K_{4,4,3}\right)=6.$

(d).

If $p\ge q+r$, then $\tilde{\gamma }\left(K_{p,q,r,s}\right)\ge ⌈\frac{(p-2)(q+r+s-2)}{2}⌉.$

If $p\le q+r$, then $\tilde{\gamma }\left(K_{p,q,r,s}\right)\ge ⌈\frac{(p+s-2)(q+r-2)}{2}⌉$.

Proposition 2 

(([16] Theorem 1.3) (Euler formula)). Let $\varphi :G\to N_k$ be a two-cell embedding of a connected graph G to the non-orientable surface $N_k$. Then $|V|-|E|+|F|=2-k$, where $|V|,|E|$, and $|F|$ are the number of vertices, edges, and faces that $\varphi \left(G\right)$ has, respectively, and k is the crosscap of $N_k$.

The following is an easy observation that will be used in the proof of the main theorem.

Observation 1. 

Let G be a simple graph with $|E|$ edges embedded with $|F|$ faces. Then $\frac{2|E|}{|F|}\ge gr\left(G\right)$ where $gr\left(G\right)$ denotes the length of the shortest cycle in G.

3. Basic Results and Notations

Before going into the classifications, we need to be familiar with the following notations and observations given by Parsapour and Javaheri in [20].

Notation: ([20]) Let $\mathcal{L}$ be a lattice and $A\left(\mathcal{L}\right)=\{a_1,a_2,\dots ,a_n\}$ be the set of all atoms. Let $i_1,i_2,\dots ,i_k$ be integers with $1\le i_1<i_2<\dots <i_k\le n$. The notation $U_{i_1{i}_2\dots i_k}$ stands for the following set:

$$\left\{I⊴\mathcal{L}:\{a_{i_1},a_{i_2},\dots ,a_{i_k}\}\subseteq I\mathrm{and}{a}_{i_j}\notin I\mathrm{for}{i}_j\in \{1,2,\dots ,n\}\setminus \{i_1,i_2,\dots ,i_k\}\right\}.$$

The next result provides the structure of $\mathbb{AG}\left(\mathcal{L}\right)$.

Proposition 3. 

Let $\mathcal{L}$ be a lattice with n atoms. Then $\mathbb{AG}\left(\mathcal{L}\right)$ is a $2^n-2$-partite graph.

Proof. 

Let $|A(\mathcal{L})|=n$. For $1\le i_1<i_2<\dots <i_k\le n$ and $1\le j_1<j_2<\dots <j_{k^{\prime }}\le n$, if the index sets $\{i_1,i_2,\dots ,i_k\}$ and $\{j_1,j_2,\dots ,j_{k^{\prime }}\}$ of $U_{i_1{i}_2\dots i_k}$ and $U_{j_1{j}_2\dots j_{k^{\prime }}}$ respectively, are distinct, then $U_{i_1{i}_2\dots i_k}\cap U_{j_1{j}_2\dots j_{k^{\prime }}}=\varnothing$. Clearly, $V\left(\mathbb{AG}\left(\mathcal{L}\right)\right)={\displaystyle \bigcup _{1\le i_1<i_2<\dots <i_k\le n}}{U}_{i_1{i}_2\dots i_k}$. Therefore, for $1\le i_1<i_2<\dots <i_k\le n$, the set $U_{i_1{i}_2\dots i_k}$ forms a partition of $V\left(\mathbb{AG}\right(\mathcal{L}\left)\right).$ Since $0\ne a_{i_1}$ belongs to every ideal in $U_{i_1{i}_2\dots i_k}$, no pair of distinct vertices in $U_{i_1{i}_2\dots i_k}$ are adjacent in $\mathbb{AG}\left(\mathcal{L}\right)$. Note that the number of distinct $U_{i_1{i}_2\dots i_k}$s is $2^n-1$. This, together with the fact that every vertex in $U_{12\dots n}$ is isolated in $\mathbb{AG}\left(\mathcal{L}\right)$, implies that $\mathbb{AG}\left(\mathcal{L}\right)$ is a $2^n-2$-partite graph. □

According to the abovementioned result regarding the structure of $\mathbb{AG}\left(\mathcal{L}\right)$, in order to identify the crosscap two r-partite graph or to classify the forbidden r-partite graphs of a non-orientable surface of order two for some $3\le r\in \mathbb{N}$, one may be interested in finding all crosscap two annihilating-ideal graphs. This is the main objective of this paper.

We shall also need the following notations:

Notations: Before proving our main results, the following points are assumed for convenience in notations and clarity in proofs. Let us take $|A(\mathcal{L})|=n$.

• To avoid repetition, we assume $|U_1|\ge |U_2|\ge \dots \ge |U_n|$.
• We denote the vertices of the set $U_{i_1{i}_2\dots i_k}$ by $\{I_{i_1{i}_2\dots i_k},I_{i_1{i}_2\dots i_k}^{\prime },I_{i_1{i}_2\dots i_k}^{″},\dots \}$.
• For an integer $p,$ an integer different from p will be denoted by $p^{\prime }$.
• For the sake of convenience, we shall denote $U_{{(i_1{i}_2\dots i_k)}^c}=U_{j_1{j}_2\dots j_{\ell }}$ where $j_1,j_2,\dots ,j_{\ell }=\{1,2,\dots ,n\}\setminus \{i_1,i_2,\dots ,i_k\}$ and the notation $U_{{(i_1{i}_2\dots i_k)}^c}$ exists only when $U_{i_1{i}_2\dots i_k}\ne \varnothing$.
• The edge between the two vertices I and J is denoted by $(I,J)$.
• The notations $|F|$ and $f_i$ denote the number of faces and number of i-gons in an embedding of G in $N_k$, respectively.
• There may be sets $U_{i_1{i}_2\dots i_k}$ such that each vertex of $U_{i_1{i}_2\dots i_k}$ is isolated, ends, or is adjacent to exactly two ends of an edge in $\mathbb{AG}\left(\mathcal{L}\right)$. In such places, the vertices of $U_{i_1{i}_2\dots i_k}$ do not affect the crosscap number of $\mathbb{AG}\left(\mathcal{L}\right)$, which leads to ignoring the set $U_{i_1{i}_2\dots i_k}$ from the corresponding embedding. This fact is used throughout the article and is sometimes not explicitly pointed out.
• For convenience in any drawing, we provide a particular type of $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$. This means that instead of drawing graphs for the case $U_{ij}$ with $1\le i\le j\le 3$, we assume $i=1$ and $j=2$ in figures. Additionally, the notation ⋯ is used to denote the possibility of embedding any number of vertices.

We show a few simple, but useful, properties of a crosscap on $\mathbb{AG}\left(\mathcal{L}\right).$ We now state and prove the following lemma, which provides a subgraph and super-graph structure of $\mathbb{AG}\left(\mathcal{L}\right)$.

Lemma 1. 

Let $\mathcal{L}$ be a lattice, $|A(\mathcal{L})|=n$, and $n\ge k\in \mathbb{N}$. Let ${\alpha }_{i_1{i}_2\dots i_k}=|U_{i_1{i}_2\dots i_k}|$, $\lambda =max\left\{{\alpha }_{i_1{i}_2\dots i_k}\right\}$ for all $1\le i_1<i_2<\dots <i_k\le n$. Then

(a).

$K_{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n}$  is a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$.

(b).

$K_{(2^n-2)\left(\lambda \right)}$ is a super-graph of $\mathbb{AG}\left(\mathcal{L}\right)$.

Proof. 

Let H be the induced subgraph of $\mathbb{AG}\left(\mathcal{L}\right),$ induced by the vertex subset ${\bigcup }_{i=1}^n{U}_i$. It is clear that no two distinct vertices in $U_i$ are adjacent, and every vertex in $U_i$ is adjacent to all of the vertices of $U_j$ for $i\ne j$ in $\mathbb{AG}\left(\mathcal{L}\right)$. Thus $H=K_{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n}$.

The second part follows from the facts that $V\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=\bigcup U_{i_1{i}_2\dots i_k}$; the number of vertex subsets $U_{i_1{i}_2\dots i_k}$, except $U_{12\dots n}$, in $V\left(\mathbb{AG}\right(\mathcal{L}\left)\right)$ is $\left(\genfrac{}{}{0pt}{}{n}{1}\right)+\left(\genfrac{}{}{0pt}{}{n}{2}\right)+\dots +\left(\genfrac{}{}{0pt}{}{n}{n-1}\right)=2^n-2$; and $\lambda =max\left\{{\alpha }_{i_1{i}_2\dots i_k}\right\}$. □

We are now in the position to provide a lower bound for the crosscap of $\mathbb{AG}\left(\mathcal{L}\right)$. Applying Proposition 1c,d in the first part of the above lemma, we obtain the following result.

Theorem 1. 

Let $\mathcal{L}$ be a lattice, $|A(\mathcal{L})|=n\ge 3$, and $|U_1|\ge |U_2|\ge \dots \ge |U_n|$.

(a).

If $n=3$, then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge ⌈\frac{(|U_1|-2)(|U_2|+|U_3|-2)}{2}⌉.$ Moreover, the equality holds whenever $U_{ij}=\varnothing$ for all $1\le i\le j\le 3.$

(b).

If $n\ge 4$, then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge \left\{\begin{array}{cc}⌈\frac{(|U_1|-2)(|U_2|+|U_3|+|U_4|-2)}{2}⌉\hfill & if|U_1|\ge |U_2|+|U_3|\hfill \\ ⌈\frac{(|U_1|+|U_4|-2)(|U_2|+|U_3|-2)}{2}⌉\hfill & if|U_1|<|U_2|+|U_3|.\hfill \end{array}\right.$

We now enter into the core part of the paper. We first observe that $\mathbb{AG}\left(\mathcal{L}\right)$ is totally disconnected when $|A(\mathcal{L})|=1$, and $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_7$ as a subgraph when $|A(\mathcal{L})|\ge 7$. Further, according to Proposition 1a, the crosscap of $K_7$ is three. Thus, one obtains the following result, which provides a bound for the number of atoms in lattice $\mathcal{L}$ with $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Proposition 4. 

Let $\mathcal{L}$ be a lattice. If the crosscap of the annihilating-ideal graph $\mathbb{AG}\left(\mathcal{L}\right)$ is two, then $2\le |A(\mathcal{L})|\le 6$.

We start the characterization by analyzing the simple case that $|A(\mathcal{L})|=2.$ If $|A(\mathcal{L})|=2$, then Theorem 2.6 [20] implies that $\mathbb{AG}\left(\mathcal{L}\right)\cong K_{|U_1|,|U_2|}$, and so

$$\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=⌈\frac{(|U_1|-2)(|U_2|-2)}{2}⌉$$

whenever $|U_1|,|U_2|\ge 2$. Now, a simple calculation has yielded the following result, which characterized lattice $\mathcal{L}$ with a crosscap two $\mathbb{AG}\left(\mathcal{L}\right)$ in the case of $|A(\mathcal{L})|=2$.

Theorem 2. 

Let $\mathcal{L}$ be a lattice and $|A(\mathcal{L})|=2.$ Then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ if and only if $|U_1|=|U_2|=4$ or $|U_i|=3$ and $|U_j|\in \{5,6\}$ where $i,j\in \{1,2\}$ with $i\ne j$.

To finish this section we show two results that will be used to prove the main results. The graphs given in Figure 1 and Figure 2 play a vital role in characterizing a lattice with crosscap two annihilating-ideal graphs, and, therefore, we draw the graph with its embedding in the first result.

Figure 1. The graph $H_1$ and its $N_2$-embedding.

Figure 2. The graph $H_2$ and its $N_2$-embedding.

Lemma 2. 

For the graphs $H_1$ and $H_2$, as shown in Figure 1 and Figure 2, we have $\tilde{\gamma }\left(H_1\right)=\tilde{\gamma }\left(H_2\right)=2.$

The graphs $H_3$ and $H_4$ given in Figure 3 play a vital role in our main theorems.

Figure 3. The graphs $H_3$ and $H_4$.

Lemma 3. 

For the graphs $H_3$ and $H_4$, as shown in Figure 3, we have $\tilde{\gamma }\left(H_3\right)\ge 3$ and $\tilde{\gamma }\left(H_4\right)\ge 3$.

Proof. 

(a). Consider the subgraph $H_3^{\prime }=H_3-\left\{u_1\right\}$. Clearly $H_3^{\prime }\cong K_7-e$ where $e=(u_2,u_3)$, and there are 13 faces in any $N_2$-embedding of $H_3^{\prime }$ of which 12 are triangular, and 1 is rectangular. Now, we try to recover an $N_2$-embedding of $H_3$ by inserting $u_1$ with its edges. Since $u_1$ is adjacent to four vertices of $H_3^{\prime }$, $u_1$ should be inserted into the rectangular face of $H_3^{\prime }$. However, all vertices of $H_3^{\prime }$ are adjacent to each other, except for $u_2$ and $u_3$, so the rectangular face of $H_3^{\prime }$ must contain either $u_2$ or $u_3$, which is in contradiction to $u_2$ and $u_3$ not belonging to the neighborhood set of $u_1$. Therefore, $\tilde{\gamma }\left(H_3\right)\ge 3$.

(b). Apply a similar argument as in (a) for the subgraph $H_4^{\prime }=H_4-\left\{u_1\right\}\cong K_7-2e$. Here, notice that the largest face in any $N_2$-embedding of $H_4^{\prime }$ is a unique pentagon, and $u_1$ is adjacent to the five vertices $v_1,v_2,v_3,v_4$, and $u_4$. □

4. The Case When $|\mathit{A}(\mathcal{L})|=\mathbf{3}$

Let us start the classification result with a lattice containing exactly three atoms. Note that the following theorem provides a class of multipartite graphs, which are embedded in the Klein bottle (refer to Example 1 for an illustration).

Theorem 3. 

Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})|=3$, and let $1\le i\ne j\ne k\le 3.$ Then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ if and only if one of the following conditions hold:

(i).

$|{\bigcup }_{n=1}^3{U}_n|=9$; there is $U_i$ with $|U_i|=6$ and $U_{jk}=\varnothing$.

(ii).

$|{\bigcup }_{n=1}^3{U}_n|=8$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|=6$ and $|U_{jk}|=1$.

[b]

There exist $U_i$ and $U_j$ such that $|U_i|\in \{5,4\}$ and $|U_j|=2$ with $U_{jk}=\varnothing .$

[c]

There exist $U_i$ and $U_j$ such that $|U_i|=4$ and $|U_j|=3$ with $U_{ik}=U_{jk}=\varnothing .$

[d]

There exist $U_i$ and $U_j$ such that $|U_i|=|U_j|=3$ with $U_{ij}=U_{ik}=U_{jk}=\varnothing .$

(iii).

$|{\bigcup }_{n=1}^3{U}_n|=7$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|\in \{5,4\}$ and $|U_{jk}|=1.$

[b]

There exist $U_i$ and $U_j$ such that $|U_i|=|U_j|=3$ with either $|U_{ik}|\in \{1,2\}$ and $U_{jk}=\varnothing$ or $U_{ik}=\varnothing$ and $|U_{jk}|\in \{1,2\}.$

[c]

There exist $U_i$ and $U_j$ such that $|U_i|=3$, $|U_j|=2$ with $|U_{jk}|\in \{1,2\}$. Further, if $|U_{jk}|=1$, then either $U_{ij}=\varnothing$ or $U_{ik}=\varnothing$ and, if $|U_{jk}|=2$, then $U_{ij}=U_{ik}=\varnothing$.

(iv).

$|{\bigcup }_{n=1}^3{U}_n|=6$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|=4$ and $|U_{jk}|=2.$

[b]

There is $U_i$ with $|U_i|=3$ and $|U_{jk}|\in \{2,3\}.$

(v).

$|{\bigcup }_{n=1}^3{U}_n|=5$; there is $U_i$ with $|U_i|=3$ and $|U_{jk}|\in \{3,4\}.$

Proof. 

Assume that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$ First of all, if $|{\bigcup }_{n=1}^3{U}_n|\le 4$, then $\mathbb{AG}\left(\mathcal{L}\right)$ is planar (see [19]). Suppose $|{\bigcup }_{n=1}^3{U}_n|\ge 10$. If $|U_2|\ge 2$, then by Theorem 1 we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge ⌈\frac{(|U_1|-2)(|U_2|+|U_3|-2)}{2}⌉\ge 3$, which is a contradiction. Suppose $|U_2|=1$. Then $|U_3|=1$. Note that every vertex in $U_{12}$, $U_{13}$, and $U_{23}$ is adjacent to all of the vertices of $U_3$, $U_2$, and $U_1$, respectively. So, if $U_{23}=\varnothing$, then clearly $\mathbb{AG}\left(\mathcal{L}\right)$ is planar. If not, the vertices in $U_1$ are adjacent to all of the vertices of $U_2\cup U_3\cup U_{23}$. Since $|U_1|\ge 8$, $K_{8,3}$ is a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ that has a crosscap of more than three, refer to Proposition 1a. Thus, $5\le |{\bigcup }_{n=1}^3{U}_n|\le 9.$

Case 1 Let $|{\bigcup }_{n=1}^3{U}_n|=9.$ Then, clearly, $|U_1|\le 7$. If $|U_1|=7$, then a slight modification to the discussion made in the above paragraph would show that $\mathbb{AG}\left(\mathcal{L}\right)$ is planar whenever $U_{23}=\varnothing$ and the graph $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{7,3}$ as a subgraph when $U_{23}\ne \varnothing$. If $|U_1|=6$, then $|U_2|=2$ and $|U_3|=1$. Now, if $U_{23}\ne \varnothing$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{6,4}$ as a subgraph, which is a contradiction. So, $U_{23}=\varnothing$. Here, all of the vertices in $U_{12}$ are adjacent to a single vertex of $U_3$, and, therefore, the vertices in $U_{12}$ do not affect the crosscap. In Figure 4a, we provide the canonical representation of the embedding of the resulting graph in $N_2$ so that, in this case, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Next, if $|U_1|=5$ or 4, then $|U_2|+|U_3|\ge 4$, and so, by Theorem 1a, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Thus, $|U_1|=3$, and, therefore, $|U_2|=|U_3|=3$. Here, $K_{3,3,3}$ is a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$, and, therefore, according to Proposition 1c, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$.

Figure 4. $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$.

Case 2 Let $|{\bigcup }_{n=1}^3{U}_n|=8.$

If $|U_1|=6$, then $|U_2|=|U_3|=1.$ Clearly, by [19], $\mathbb{AG}\left(\mathcal{L}\right)$ is planar in the case that $U_{23}$ is empty. If $|U_{23}|\ge 2$, then the partite sets $X=U_1$ and $Y=U_2\cup U_3\cup U_{23}$ form $K_{6,4}$ as a subgraph in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction. Therefore, $|U_{23}|=1$. In this case, the vertices in $U_{13}\cup U_{12}$ are all end vertices, and, therefore, it does not affect the crosscap. Thus, the resulting graph is $K_{6,3}\cup \left\{(I_2,I_3)\right\}$, which is a subgraph of a graph given in Figure 2a, and, therefore, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$

Suppose $|U_1|\in \{5,4\}$. Then, according to Theorem 1a, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2$. If $U_{23}\ne \varnothing$, then the sets $X=U_1$ and $Y=U_2\cup U_3\cup U_{23}$ form $K_{5,4}$ as a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$, and so $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $U_{23}=\varnothing$. Let $|U_2|=2,|U_{12}|\ge 0$, and $|U_{13}|\ge 0$. For the embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ in $N_2$, in the case of $|U_1|=5$, we can obtain help from Figure 4a because the number of vertices and edges of $\mathbb{AG}\left(\mathcal{L}\right)$ is less than that of in Figure 4a. Further, Figure 4b provides an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ in the case of $|U_1|=4.$ Here, notice that the open neighborhood of each vertex in $U_{13}$ is $\{I_2,I_2^{\prime }\}$, and, in Figure 4a,b, there is a face in an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ that contains both $I_2$ and $I_2^{\prime }$ so that every vertex of $U_{13}$ can be embedded in $N_2$ no matter what its cardinality may be. Let $|U_2|=3$. This implies that $|U_1|=4$. If $U_{13}=\varnothing$ (recall that $U_{23}=\varnothing$), then $\mathbb{AG}\left(\mathcal{L}\right)$ is a subgraph of the graph $H_1$ in Figure 1, and, therefore, according to Lemma 2, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. If not, consider that the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{(I_3,I_1),(I_3,I_1^{\prime }),(I_3,I_1^{″}),(I_3,I_1^{‴})\}$ contains $K_{3,6}$. By Euler’s formula, any embedding of $K_{3,6}$ in $N_2$ has nine faces. Further, by solving the equations $2|E|=4f_4+6f_6$ and $|F|=f_4+f_6$, we have all the faces as rectangular faces in any $N_2$-embedding of $K_{3,6}$. Now we try to recover the embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ by inserting all edges $(I_3,I_1),(I_3,I_1^{\prime }),(I_3,I_1^{″}),(I_3,I_1^{‴})$ into the embedding of $K_{3,6}$. Since $de{g}_{K_{3,6}}\left(I_3\right)=3$, the vertex $I_3$ is in the boundary of three rectangular faces of any $N_2$-embedding of $K_{3,6}$. In addition, note that, at the maximum, each rectangular face can adopt one edge incident with $I_3$. So, we cannot insert all four edges of $I_3$ into $N_2$ without crossing, which is a contradiction. Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3.$

Suppose $|U_1|=3$. If $U_{ij}=\varnothing$ for all $1\le i<j\le 3$, then, by Proposition 1c, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Next, our claim is that $U_{ij}=\varnothing$ for all $1\le i<j\le 3.$

Assume that $U_{12}\ne \varnothing .$ Then the minor subgraph is

$$\mathbb{AG}\left(\mathcal{L}\right)-\{(I_1,I_3^{\prime }),(I_1^{\prime },I_3^{\prime }),(I_1^{\prime \prime },I_3^{\prime }),(I_2,[I_3,I_{12}]),(I_2^{\prime },[I_3,I_{12}]),(I_2^{\prime \prime },[I_3,I_{12}])\}\cong K_{4,4}$$

with the partite sets $X=U_2\cup \left\{[I_3,I_{12}]\right\}$ and $Y=U_1\cup \left\{I_3^{\prime }\right\}$. By Euler’s formula, any $N_2$-embedding of $K_{4,4}$ has eight rectangular faces. Next, we attempt to obtain an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ from any $N_2$-embedding of $K_{4,4}$. For this, we try to embed the six omitted edges of $\mathbb{AG}\left(\mathcal{L}\right)$ into an arbitrary $N_2$-embedding of $K_{4,4}$. First, to embed the three edges $(I_1,I_3^{\prime }),(I_1^{\prime },I_3^{\prime })$, and $(I_1^{″},I_3^{\prime })$, three rectangular faces are required, denoted as $F_1,F_2$, and $F_3$, all of which contains $I_3^{\prime }$ (refer to Figure 5a). Since $de{g}_{K_{4,4}}\left(I_3^{\prime }\right)=4,$ exactly one more face should have $I_3^{\prime }$; it is denoted as $F_4.$ Intentionally, we label the diagonals of $F_4$ as the vertices $I_2$ and $[I_3,I_{12}]$ because $F_4$ can adopt one diagonal edge that can be used to embed the fourth edge $(I_2,[I_3,I_{12}])$. Finally, to embed the rest of the two edges $(I_2^{\prime },[I_3,I_{12}])$ and $(I_2^{″},[I_3,I_{12}])$, two distinct faces are required, denoted by $F_5$ and $F_6$, which should have the vertex $[I_3,I_{12}]$. Note that, in any $N_k$-embedding, every edge of a graph is in exactly two faces. Since the edge $(I_1,[I_3,I_{12}])$ is in $F_2$ and the edge $(I_1^{\prime },[I_3,I_{12}])$ is in $F_4$, the common edge between $F_5$ and $F_6$ must be $(I_1^{″},[I_3,I_{12}]).$ Now, the choice for the unlabelled vertex of $F_5$ and $F_6$ is either $I_1$ or $I_1^{\prime }$. Without a loss of generality, we label $I_1$ for $F_5$ and $I_1^{\prime }$ for $F_6$ (refer to Figure 5b). Since any $N_2$-embedding of $K_{4,4}$ has eight faces, there are two more faces, lets say $F_7$ and $F_8$, that have to be formed using all of the remaining vertices and edges of $K_{4,4}$. Notice that, in any $N_2$-embedding of $K_{4,4}$, each vertex is present in exactly four faces, and each edge is present in exactly two faces. Since the vertices $I_2\in X$ and $I_1^{\prime }\in Y$ are used twice in the faces $F_1,\dots ,F_6$, the faces $F_7$ and $F_8$ must share the edge $(I_2,I_1^{\prime })$ (refer to Figure 5c). Now, the choices for the third and fourth vertices of $F_7$ and $F_8$ are $I_2^{\prime },I_2^{″}\in X$ and $I_1,I_1^{″}\in Y$, respectively. Clearly, we have to select distinct vertices for $F_7$ and $F_8$, in which one is from $\{I_2^{\prime },I_2^{″}\}$ and the other is from $\{I_1,I_1^{″}\}$. A contradiction to this fact is that the edges $(I_2^{\prime },I_1)$ and $(I_2^{″},I_1^{″})$ are used twice in the faces $F_1,\dots ,F_6$.

Figure 5. Representation of faces of $N_2$-embedding of $K_{4,4}$.

Assume that $U_{i3}\ne \varnothing$ for some $i\in \{1,2\}$. Then, the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{i3},(I_i,I_3),(I_i^{\prime },I_3),(I_i^{″},I_3)\}$ contains $K_{4,4}-e$ with the partite sets $X=U_i\cup \left\{I_3\right\}$ and $Y=U_{i^{\prime }}\cup \left\{I_3^{\prime }\right\}$ where $i^{\prime }\in \{1,2\}\setminus \left\{i\right\}$ and $e=(I_3,I_3^{\prime })$. By Proposition 2, any $N_2$-embedding of $K_{4,4}-e$ has one hexagonal and six rectangular faces. Note that the hexagonal face should have either $I_3$ or $I_3^{\prime }$, and the vertex $I_{i3}$ is adjacent to $\{I_{i^{\prime }},I_{i^{\prime }}^{\prime },I_{i^{\prime }}^{″}\}\subset Y$. So, $I_{i3}$ with its edges must be inserted into the hexagonal face, which implies that $I_3$ is in the hexagonal face. Since $de{g}_{K_{4,4}-e}\left(I_3\right)=3,$ exactly two rectangular faces contain $I_3$ in which it is not possible to embed all of the three edges $(I_i,I_3),(I_i^{\prime },I_3)$, and $(I_i^{″},I_3)$, which is a contradiction. Thus, $U_{ij}=\varnothing$ for all $i,j\in \{1,2,3\}$.

Case 3 Let $|{\bigcup }_{n=1}^3{U}_n|=7.$

Suppose $|U_1|\in \{5,4\}$. Clearly, $\mathbb{AG}\left(\mathcal{L}\right)$ is either planar or projective when $U_{23}=\varnothing$ (refer to [19,20]), and $K_{5,4}$ is a subgraph of the contraction of $\mathbb{AG}\left(\mathcal{L}\right)$ when $|U_{23}|\ge 2$. Therefore, $|U_{23}|$ will be one. Then, $\mathbb{AG}\left(\mathcal{L}\right)$ is a subgraph of the graph given in Figure 4a when $|U_1|=5$, and $\mathbb{AG}\left(\mathcal{L}\right)$ is a subgraph of the graph given in Figure 4b when $|U_1|=4$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Assume that $|U_1|=|U_2|=3$. Then, $\mathbb{AG}\left(\mathcal{L}\right)$ is projective when $U_{i3}=\varnothing$ for all $i=1,2$, and the graph $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,7}$ as a subgraph when $|U_{i3}|\ge 3$ for some $i=1,2$. Suppose $U_{13}\ne \varnothing$ and $U_{23}\ne \varnothing$. Now, the graph $\mathbb{AG}\left(\mathcal{L}\right)-\left\{I_3\right\}$ is isomorphic to $K_{4,4}-\left\{e\right\}$ with the bipartite sets $\{I_1,I_1^{\prime },I_1^{″},I_{13}\}$ and $\{I_2,I_2^{\prime },I_2^{″},I_{23}\}$ where $e=(I_{13},I_{23})$. Note that $\tilde{\gamma }(K_{4,4}-\left\{e\right\})=2$, and there are seven faces in any $N_2$-embedding of $K_{4,4}-\left\{e\right\}$, of which six are rectangular, and one is hexagonal. Since $\tilde{\gamma }\left(K_{4,4}\right)=2$ and every face in any $N_2$-embedding of $K_{4,4}$ is rectangular, the hexagonal face of any $N_2$-embedding of $K_{4,4}-\left\{e\right\}$ must have the vertices $I_{13}$ and $I_{23}.$ Now, we try to recover an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ from an $N_2$-embedding of $K_{4,4}-\left\{e\right\}$ by inserting $I_3$ with its edges. Here, $I_3$ is adjacent to the six vertices $I_1,I_1^{\prime },I_1^{″},I_2,I_2^{\prime }$, and $I_2^{\prime \prime }$. However, the hexagonal face of $K_{4,4}-\left\{e\right\}$ does not contain two of them so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, either $U_{13}=\varnothing$ or $U_{23}=\varnothing$. Now, with the help of Figure 6, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ when $1\le |U_{i3}|\le 2$ for a unique $i\in \{1,2\}$.

Figure 6. $|{\bigcup }_{n=1}^3{U}_n|=7$ with $|U_1|=|U_2|=3,U_{13}=\varnothing$ and $|U_{23}|=2$.

Assume that $|U_1|=3$ and $|U_2|=2$. If $|U_{23}|\ge 3$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,7}$ as a subgraph, and, if $U_{23}=\varnothing ,$ then, by Theorem 2.4iii [20], $\mathbb{AG}\left(\mathcal{L}\right)$ is projective. Suppose $|U_{23}|=2$. If $U_{1j}\ne \varnothing$ for $j=2$ or 3, then consider a subgraph $G_1=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{1j},I_{23}^{\prime },e_1,e_2,e_3,e_4\}$ where $e_1=(I_2,I_3),e_2=(I_2,I_3^{\prime }),e_3=(I_2^{\prime },I_3)$, and $e_4=(I_2^{\prime },I_3^{\prime })$. Clearly, $G_1$ contains $K_{3,5}$ with the partite sets $X=\{I_1,I_1^{\prime },I_1^{″}\}$ and $Y=\{I_2,I_2^{\prime },I_3,I_3^{\prime },I_{23}\}$. Note that any $N_2$-embedding of $K_{3,5}$ has one hexagonal and six rectangular faces. Now, we try to recover an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ from any $N_2$-embedding of $K_{3,5}$. Since $I_{23}^{\prime }$ is adjacent to all three vertices of X, the embedding of $I_{23}^{\prime }$ requires the hexagonal face of $K_{3,5}$ to have $I_1,I_1^{\prime }$, and $I_1^{\prime \prime }$. Notice that each rectangular face may adopt at most one edge into it. So, to insert $e_f$s, for $1\le f\le 4$, into any $N_2$-embedding of $K_{3,5}$, four rectangular faces with diagonals as the end vertices of each $e_f$ are required. At last, to insert $I_{1j}$, a rectangular face with the diagonals $I_{j^{\prime }}$ and $I_{j^{\prime }}^{\prime }$ for $j^{\prime }\in \{2,3\}\setminus \left\{j\right\}$ is required. Therefore, it requires one hexagonal face with five rectangular faces containing the vertices $I_2,I_2^{\prime },I_3$, and $I_3^{\prime }$ in at least three different faces. Since the degree of $I_2,I_2^{\prime },I_3$, and $I_3^{\prime }$ in $K_{3,5}$ is three, all four vertices are placed in exactly three faces of any $N_2$-embedding of $K_{3,5}$. So, the sixth rectangular face of $K_{3,5}$ could not be formed using the only left-out vertex in X (namely $I_{23}$), which is a contradiction. Thus, $U_{12}=U_{13}=\varnothing$, and an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ for this case is provided in Figure 7a.

Figure 7. $|{\bigcup }_{n=1}^3{U}_n|=7$ with $|U_1|=3$ and $|U_2|=2$.

Suppose $|U_{23}|=1$. If $U_{1j}\ne \varnothing$ for $j=2$ and 3, then the minor subgraph is

$$\begin{array}{c}{G}_2=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{13},e_1,e_2,e_3,e_4,e_5\}\cong K_{4,4}-\left\{e\right\},\end{array}$$

(1)

with the bipartite sets $\{I_1,I_1^{\prime },I_1^{″},I_3\}$ and $\{I_2,I_2^{\prime },[I_3^{\prime },I_{12}],I_{23}\}$ where $e_1=(I_1,I_3),e_2=(I_1^{\prime },I_3)$, $e_3=(I_1^{″},I_3),e_4=(I_2,[I_3^{\prime },I_{12}])$, $e_5=(I_2^{\prime },[I_3^{\prime },I_{12}])$, and $e=(I_3,I_{23})$. Note that any $N_2$-embedding of $K_{4,4}-\left\{e\right\}$ has six rectangular faces and a hexagonal face, and the hexagonal face must have the vertices $I_3$ and $I_{23}$. Let us denote the six rectangular faces by $F_1,\dots ,F_6$ and the hexagonal face by $F_7$. Now, let us try to recover an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ by inserting the vertex $I_{13}$ and the edges $e_i$ for all $i=1,\dots ,5$. If we embed the edge $e_4$, the edge $e_5$, or the vertex $I_{13}$ together with its edges into $F_7$, then we cannot insert the edges $e_1$, $e_2$, or $e_3$ into $F_7$. Since $de{g}_{G_2}\left(I_3\right)=3,$ the vertex $I_3$ is in exactly three faces of an $N_2$-embedding of $G_2$. So, in such cases, the edges $e_1,e_2$ and $e_3$ cannot be embedded in two rectangular faces which contains $I_3$. Therefore we have to add at least one of the edges $e_1,e_2$ or $e_3$ into $F_7$. For the best possibility, say $e_1$ and $e_2$ are embedded in $F_7$. Then, $e_3$ has to be embedded into one of the two rectangular faces that contains $I_3$, for example, $F_1$. Notice that there are two rectangular faces, say $F_2$ and $F_3$, that contain $I_{23}$, in which one should not embed any of $e_4$, $e_5$, or $I_{13}$ with its edges. So, the edges $e_4$ and $e_5$ have to be embedded into different rectangular faces, say $F_4$ and $F_5$, respectively. Therefore, after embedding the edges from $e_1$ to $e_5$ nicely, we are left with the single rectangular face $F_6$ that could not be formed using the diagonal vertices $I_2$ and $I_2^{\prime }$. Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Hence, either $U_{12}=\varnothing$ or $U_{13}=\varnothing$. In this case, with the help of Figure 7b, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Case 4 Let $|{\bigcup }_{n=1}^3{U}_n|=6.$ Suppose $|U_1|=4.$ If $|U_{23}|\ge 3$, then $K_{4,5}$ is contained in $\mathbb{AG}\left(\mathcal{L}\right)$, and if $|U_{23}|=1$, then $\mathbb{AG}\left(\mathcal{L}\right)$ is projective. Therefore $|U_{23}|=2$. Clearly, $\mathbb{AG}\left(\mathcal{L}\right)$(except for the end vertices) is a subgraph of the graph $H_1$ given in Figure 1a, and so Lemma 2 implies $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$

Suppose $|U_1|=3.$ Then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,7}$ when $|U_{23}|\ge 4$, and $\mathbb{AG}\left(\mathcal{L}\right)$ is projective when $|U_{23}|\le 1$. Thus, $2\le |U_{23}|\le 3$. Then, $\mathbb{AG}\left(\mathcal{L}\right)-\left\{U_{13}\right\}$ is a subgraph of the graph $H_2$ (see Figure 2a), so that $\tilde{\gamma }(\mathbb{AG}\left(\mathcal{L}\right)-\left\{U_{13}\right\})=2.$ Note that every vertex in $U_{13}$ is adjacent to exactly two vertices of $U_2$ in $\mathbb{AG}\left(\mathcal{L}\right)$. Therefore, replace the labels $u_4$ and $u_5$ with $I_2$ and $I_2^{\prime }$, respectively, in the $N_2$-embedding of $H_2$ provided in Figure 2b, and then label all of the other vertices accordingly. Now, we can insert any number of vertices of $U_{13}$ into a face that contains both $I_2$ and $I_2^{\prime }$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Moreover, if $|U_1|=2$, then $\mathbb{AG}\left(\mathcal{L}\right)$ is either planar or projective (refer to [19,20]).

Case 5 Let $|{\bigcup }_{n=1}^3{U}_n|=5.$ Then $\mathbb{AG}\left(\mathcal{L}\right)$ is planar or projective when $|U_1|=2$. This implies that $|U_1|=3$. If $|U_{23}|\ge 5$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,7}$, and, if $|U_{23}|\le 2$, then $\mathbb{AG}\left(\mathcal{L}\right)$ is projective. Thus, $|U_{23}|=3$ or 4. Then, clearly, $\mathbb{AG}\left(\mathcal{L}\right)$ is a subgraph of the graph $H_1$, as in Figure 2a, so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$ □

All of the results proved in this paper have a similar structure to that of those given in the statement of Theorem 3. To familiarize readers with the connection between the multipartite graph and the statement of Theorem 3, we illustrate two four-partite graphs, G and H, with $\tilde{\gamma }\left(G\right)=2$ and $\tilde{\gamma }\left(H\right)\ne 2$, respectively, in the following example.

Example 1. 

Consider Case (iii)[c] in Theorem 3. Let $|U_1|=3,|U_2|=2,|U_3|=2$, and $|U_{23}|=1$. If $|U_{12}|=k\in {\mathbb{Z}}^{+}$ and $U_{13}=\varnothing$, then the corresponding four-partite graph G is a crosscap two, which is given in Figure 8a. Additionally, if $|U_{12}|=1$ and $|U_{13}|=1$, then the crosscap of the corresponding four-partite graph H, given in Figure 8b, is not equal to two. It is worth mentioning that the four-partite graph H in Figure 8b is minimal with respect to $\tilde{\gamma }\left(H\right)\ne 2;$ that is, there exists an edge e in H such that $\tilde{\gamma }(H-e)=2$. Further, the graph H may be realized as one of the forbidden subgraphs for a crosscap two surface.

Figure 8. Four-partite graphs.

By using the proof of Theorem 3, we establish the following points, which will be used in the subsequent results.

Remark 1. 

If a graph G is isomorphic to $K_{6,3}\cup (K_4-e)$ or $K_{4,5}-e$ where e is an edge, then $\tilde{\gamma }\left(G\right)\ge 3$.

5. The Case When $|\mathit{A}(\mathcal{L})|=\mathbf{4}$

Next, we fix the number of atoms as four. As mentioned in the introduction, for $1\le i\ne j\le 4$, we denote $U_{{\left(ij\right)}^c}=U_{k\ell }$ where $k,\ell \in \{1,2,3,4\}\setminus \{i,j\}$, and the notation $U_{{\left(ij\right)}^c}$ exists only when $U_{ij}\ne \varnothing$. Before going into the characterization of the crosscap two $\mathbb{AG}\left(\mathcal{L}\right)$ with $|A(\mathcal{L})|=4$, we provide modifications for Theorem 2.6 [20]. To be precise, the missing cases and the corresponding conditions for the projectiveness of $\mathbb{AG}\left(\mathcal{L}\right)$ are given below.

(i) First of all, consider the missing case $|{\bigcup }_{n=1}^4{U}_n|=4.$ Then, $|U_i|=1$ for all $1\le i\le 4$. Clearly, $\mathbb{AG}\left(\mathcal{L}\right)$ is planar whenever ${\displaystyle \bigcup _{U_{ij}\ne \varnothing }}{U}_{{\left(ij\right)}^c}=\varnothing .$ Therefore, ${\displaystyle \bigcup _{U_{ij}\ne \varnothing }}{U}_{{\left(ij\right)}^c}\ne \varnothing .$ If $|U_{ij}\cup U_{{\left(ij\right)}^c}|\ge 4$ with $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$, then the subgraph induced by the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y={\displaystyle \bigcup _{k\ne i,j}}{U}_k\cup U_{{\left(ij\right)}^c}$ contains $K_{4,4}$ or $K_{3,5}$ as a subgraph. This implies $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2$. Therefore, $2\le |U_{ij}\cup U_{{\left(ij\right)}^c}|\le 3$ if $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$ for $1\le i\ne j\le 4$.

Suppose $|U_{ij}\cup U_{{\left(ij\right)}^c}|=3$ for some $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$ with $1\le i\ne j\le 4$. If $U_{k\ell },U_{{(k\ell )}^c}\ne \varnothing$ for $k\ell \ne ij$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{U_{ij}\cup U_{{\left(ij\right)}^c}\}$ contains $K_{3,3}$ with the partite sets $X=U_k\cup U_{\ell }\cup U_{k\ell }$ and $Y={\displaystyle \bigcup _{m\ne k,\ell }}{U}_m\cup U_{{(k\ell )}^c}$. Note that $\tilde{\gamma }\left(K_{3,3}\right)=1.$ Now, we try to embed all of the vertices of $U_{ij}\cup U_{{\left(ij\right)}^c}$ with their edges in any $N_1$-embedding of $K_{3,3}.$ Since $|U_{ij}\cup U_{{\left(ij\right)}^c}|=3$, either $|U_{ij}|=2$ or $|U_{{\left(ij\right)}^c}|=2$. Without a loss of generality, let $|U_{ij}|=2$. Since the vertex $I_{{\left(ij\right)}^c}\in U_{{\left(ij\right)}^c}$ is adjacent to $I_{ij},I_{ij}^{\prime }\in U_{ij}$, all of the three vertices $I_{ij},I_{ij}^{\prime }$, and $I_{{\left(ij\right)}^c}$ must be embedded into a single face of the $N_1$-embedding of $K_{3,3},$ denoted as $F_1$. Now, draw the path $I_{ij}-I_{{\left(ij\right)}^c}-I_{ij}^{\prime }$ into $F_1$ and then draw the edges $(I_{ij},I_m),(I_{ij},I_n),(I_{ij}^{\prime },I_m)$, and $(I_{ij}^{\prime },I_n)$ where $m,n\notin \{i,j\}$. Now, the edges $(I_{{\left(ij\right)}^c},I_i)$ and $(I_{{\left(ij\right)}^c},I_j)$ cannot be embedded into $F_1$. Therefore, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2$. Thus, ${\displaystyle \bigcup _{k\ell \ne ij,{\left(ij\right)}^c;U_{k\ell }\ne \varnothing }}{U}_{{(k\ell )}^c}=\varnothing .$

Suppose $|U_{ij}\cup U_{{\left(ij\right)}^c}|=2$ for all $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$ with $1\le i\ne j\le 4$. Then, Figure 9 guarantees that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=1.$

Figure 9. $|{\bigcup }_{n=1}^4{U}_n|=4$ with $|U_{ij}\cup U_{{\left(ij\right)}^c}|\le 2$ for all $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$.

(ii) Let $|{\bigcup }_{n=1}^4{U}_n|=5.$ Then, $|U_i|=2$ for some $1\le i\le 4$, and the condition for the projectiveness of $\mathbb{AG}\left(\mathcal{L}\right)$ given in Theorem 2.6i [20] is that $|U_{jk}|=1$ or 2, in which at most one of the $U_{jk}$s has exactly two elements for $1\le i\ne j\ne k\le 4$. However, if $|U_{jk}|=2$ with $U_{{\left(jk\right)}^c}\ne \varnothing$, then the sets $X=U_i\cup U_{\ell }\cup U_{{\left(jk\right)}^c}$ and $Y=U_j\cup U_k\cup U_{jk}$, where $\ell \notin \{i,j,k\}$, contain $K_{4,4}$ in $\mathbb{AG}\left(\mathcal{L}\right)$ so that we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2.$ In fact, if $|U_{jk}|=2$ for some $j,k\ne i$, then ${\displaystyle \bigcup _{p,q\ne i;U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing .$ Otherwise, the sets $X=U_j\cup U_k\cup U_{jk}\cup [I_{pq},I_{{\left(pq\right)}^c}]$ and $Y=U_1\cup U_{\ell }$, where $\ell \notin \{i,j,k\}$, form $K_{5,3}$, so we can conclude that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2$. Further, if $|U_{jk}|\le 1$ for all $j,k\ne i$, then $|{\displaystyle \bigcup _{p,q\ne i;U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}|\le 1.$ For if $|U_{{\left(pq\right)}^c}|\ge 2$, then the sets $X=U_p\cup U_q\cup U_{pq}$ and $Y=U_i\cup U_r\cup U_{{\left(pq\right)}^c}$, where $r\notin \{i,p,q\}$, form $K_{3,5}$, and, if $|U_{{\left(pq\right)}^c}|,|U_{{\left(p_1{q}_1\right)}^c}|=1$ for some $1\le p_1\ne q_1\le 4$ with $p_1{q}_1\ne pq$, then the sets $X=U_p\cup U_q\cup U_{pq}\cup \left\{[I_{p_1{q}_1},I_{{\left(p_1{q}_1\right)}^c}]\right\}$ and $Y=U_i\cup U_r\cup U_{{\left(pq\right)}^c}$ form $K_{4,4}-\left\{e\right\}$ in $\mathbb{AG}\left(\mathcal{L}\right)$ where $r\notin \{i,p,q\}$.

(iii) Let $|{\bigcup }_{n=1}^4{U}_n|=6.$ If there exists $|U_i|=3$ for some $1\le i\le 4$, then the statement of ([20] Theorem 2.6(ii)(a)) says that if $U_{jk\ell }=\varnothing$ for $1\le i\ne j\ne k\ne \ell \le 4$, $|U_{jk}|\le 1$, and at most one of the $U_{jk}$s has exactly one element, then $\mathbb{AG}\left(\mathcal{L}\right)$ is projective. However, for instance, if $|U_{jk}|=1$ with $U_{{\left(jk\right)}^c}=U_{i\ell }\ne \varnothing$, then the partite sets $X=U_i\cup U_{\ell }\cup U_{i\ell }$ and $Y=U_j\cup U_k\cup U_{jk}$ contain $K_{5,3}$ as a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 2.$ Therefore, the condition $U_{{\left(jk\right)}^c}=\varnothing$ has to be added to the statement of ([20] Theorem 2.6iia).

As a result of the above remarks (i), (ii), and (iii), we modify the statement of ([20] Theorem 2.6) as follows.

Theorem 4. 

Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})|=4$. Let $1\le i\ne j\ne k\ne \ell \le 4$ and $1\le p\ne q\le 4$. Then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=1$ if and only if one of the following conditions hold:

(i).

$|{\bigcup }_{n=1}^4{U}_n|=4$; there exist two non-empty sets $U_{ij}$ and $U_{{\left(ij\right)}^c}$ such that $2\le |U_{ij}\cup U_{{\left(ij\right)}^c}|\le 3$. Moreover, if $|U_{ij}\cup U_{{\left(ij\right)}^c}|=3$, then ${\displaystyle \bigcup _{pq\ne ij,{\left(ij\right)}^c;U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing .$

(ii).

$|{\bigcup }_{n=1}^4{U}_n|=5$; there is $U_i$ with $|U_i|=2$, $|{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}|\le 4$ in which at most one of the $U_{pq}$s has a maximum of two elements, and $|{\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}|\le 1$. Moreover, if $|U_{pq}|=2$, then ${\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing$, and, if ${\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}=\varnothing$, then $U_{jk\ell }\ne \varnothing$.

(iii).

$|{\bigcup }_{n=1}^4{U}_n|=6$, and one of the following is satisfied:

[a] There is $U_i$ with $|U_i|=3$. If $|U_{jk\ell }|=1$, then $U_{jk}=U_{j\ell }=U_{k\ell }=\varnothing$ and if $U_{jk\ell }=\varnothing$, then $|U_{jk}\cup U_{j\ell }\cup U_{k\ell }|\le 1$. Moreover, $U_{{\left(pq\right)}^c}=\varnothing$ whenever $U_{pq}\ne \varnothing$.

[b] There exist $U_i$ and $U_j$ such that $|U_i|=|U_j|=2$ with $|U_{k\ell }|\le 1$. Additionally, $U_{{\left(pq\right)}^c}=\varnothing$ whenever $U_{pq}\ne \varnothing$. Moreover, if $|U_{ik}|,|U_{i\ell }|\le 1$ or $|U_{jk}|,|U_{j\ell }|\le 1$, then $|U_{k\ell }|\le 1$. Furthermore, if $|U_{ik}|=|U_{jk}|=1$ or $|U_{i\ell }|=|U_{j\ell }|=1$, then $U_{k\ell }=\varnothing$.

(iv).

$|{\bigcup }_{n=1}^4{U}_n|=7$ and one of the following is satisfied:

[a] There is $U_i$ with $|U_i|=4$ and $U_{jk\ell }=U_{jk}=\varnothing .$

[b] There exist $U_i$ and $U_j$ such that $|U_i|=3$ and $|U_j|=2$. Additionally, $U_{k\ell }=\varnothing$, and $U_{jk\ell }=\varnothing$ whenever $U_{ik}=U_{i\ell }=U_{jk}=U_{j\ell }=\varnothing$.

We are now in the position to state and prove the second result which classifies all lattices $\mathcal{L}$ with four atoms whose $\mathbb{AG}\left(\mathcal{L}\right)$ has a crosscap two.

Theorem 5. 

Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})|=4$. Let $1\le i\ne j\ne k\ne \ell \le 4$ and $1\le p,q,r,s,t\le 4$. Then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ if and only if one of the following conditions hold:

(i).

$|{\bigcup }_{n=1}^4{U}_n|=9$; there is $U_i$ with $|U_i|=6$ and $U_{jk}=U_{j\ell }=U_{k\ell }=U_{jk\ell }=\varnothing .$

(ii).

$|{\bigcup }_{n=1}^4{U}_n|=8$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|=5$ and $U_{jk}=U_{j\ell }=U_{k\ell }=U_{jk\ell }=\varnothing .$

[b]

There exist $U_i$ and $U_j$ such that $|U_i|=4,$$|U_j|=2$ and ${\displaystyle \bigcup _{pq\ne ij}}{U}_{pq}=U_{jkl}=\varnothing$.

[c]

There exist $U_i$ and $U_j$ such that $|U_i|=|U_j|=3$ and ${\displaystyle \bigcup _{pq\ne ij}}{U}_{pq}=U_{ik\ell }=U_{jk\ell }=\varnothing$.

[d]

There exist $U_i,U_j$, and $U_k$ such that $|U_i|=3,|U_j|=|U_k|=2$, and $\bigcup U_{pq}={\displaystyle \bigcup _{pqr\ne ijk}}{U}_{pqr}=\varnothing$ for $1\le p\ne q\ne r\le 4.$

(iii).

$|{\bigcup }_{n=1}^4{U}_n|=7$, and one of the following cases is satisfied:

[a] There is $U_i$ with $|U_i|=4$ and $|{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}\cup U_{jk\ell }|=1$. Moreover, $U_{{\left(pq\right)}^c}=\varnothing$ whenever $|U_{pq}|=1$ for $p,q\ne i.$

[b] There exist $U_i$ and $U_j$ such that $|U_i|=3,|U_j|=2$ and $|{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}\cup U_{jk\ell }|\le 1$. Moreover, if $|{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}\cup U_{jk\ell }|=1,$ then $U_{{\left(pq\right)}^c}=\varnothing$ and $U_{ik}=U_{i\ell }=U_{ik\ell }=\varnothing$, and if ${\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}\cup U_{jk\ell }=\varnothing$, then $|U_{ik}\cup U_{i\ell }|\in \{1,2\}.$

[c] There exist $U_i,U_j$, and $U_k$ such that $|U_i|=|U_j|=|U_k|=2$ with $|\bigcup U_{pq}|\le 2$, in which at most one of the $U_{p\ell }$s has exactly one element, and, also, at most two distinct sets’ $U_{rst}$s are non-empty for all $rst\ne ijk$. Moreover, if $|U_{pq}|=2$ or $|U_{p\ell }|=1$ for $p,q\ne \ell$, then at most one of the $U_{rst}$s is non-empty.

(iv).

$|{\bigcup }_{n=1}^4{U}_n|=6$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|=3$, $|{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}\cup U_{jk\ell }|\in \{2,3\}$ in which $|U_{pq}|\le 2$, and $|{\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}|\le 1$. Moreover, if $|U_{pq}|\in \{1,2\}$ with $|U_{jk\ell }|=2,$ then ${\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing .$

[b]

There exist $U_i$ and $U_j$ such that $|U_i|=|U_j|=2$ and $|U_{ij}\cup U_{k\ell }|\le 3$ with $|U_{ij}|,|U_{k\ell }|\le 2$. Additionally, if $|U_{ij}|=2$, then $|U_{k\ell }|\le 1$ and ${\displaystyle \bigcup _{pq\ne ij,k\ell }}{U}_{pq}=U_{ik\ell }=U_{jk\ell }=\varnothing$, and, if $|U_{ij}|=1$, then $|U_{k\ell }|\le 1$ and $|{\displaystyle \bigcup _{pq\ne ij,k\ell }}{U}_{pq}|\le 1$. Moreover, in the case of $U_{ij}=\varnothing$, one of the following hold:

[b1]

If $|U_{k\ell }|=2,$ then $|{\displaystyle \bigcup _{pq\ne ij,k\ell }}{U}_{pq}|\le 2$ in which $|U_{pq}|\le 1$ and ${\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing$.

[b2]

If $|U_{k\ell }|=1,$ then $|U_{rs}|\le 3$ with $U_{{\left(rs\right)}^c}=\varnothing$ where $|U_{rs}|={\displaystyle \underset{pq\ne ij,k\ell }{max}}|U_{pq}|$ and $|{\displaystyle \bigcup _{mn\ne ij,k\ell ,rs,{\left(rs\right)}^c}}{U}_{mn}|\le 1.$

[b3]

If $U_{k\ell }=\varnothing ,$ then $|{\displaystyle \bigcup _{pq\ne ij,k\ell }}{U}_{pq}|\le 4$ in which at most three $U_{pq}$s are non-empty. Furthermore, if $|U_{pq}|\in \{2,3\}$, then $U_{{\left(pq\right)}^c}=\varnothing$.

(v).

$|{\bigcup }_{n=1}^4{U}_n|=5$; there exists $U_i$ such that $|U_i|=2$ and $1\le |{\displaystyle \bigcup _{p,q\ne i}}{U}_{pq}|\le 6$ in which  $|U_{pq}|\le 4$. Moreover,

[a]

If$|U_{pq}|=4,$then$U_{{\left(pq\right)}^c}=\varnothing$, $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|\le 1$, and ${\displaystyle \bigcup _{U_{rs}\ne \varnothing }}{U}_{{\left(rs\right)}^c}=\varnothing$.

[b]

If $|U_{pq}|=3,$ then $U_{{\left(pq\right)}^c}=\varnothing$, $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|\le 2$ and $U_{{\left(rs\right)}^c}=\varnothing$ whenever $|U_{rs}|=2$.

[c]

In the case of $|U_{pq}|=2,$ one of the following holds

[c1]

If $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|=4$, then ${\displaystyle \bigcup _{U_{rs}\ne \varnothing }}{U}_{{\left(rs\right)}^c}=\varnothing$.

[c2]

If $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|\in \{2,3\}$, then $|{\displaystyle \bigcup _{U_{rs}\ne \varnothing }}{U}_{{\left(rs\right)}^c}|\le 1$. In addition, $|{\displaystyle \bigcup _{U_{rs}\ne \varnothing }}{U}_{{\left(rs\right)}^c}|=1$ whenever $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|=2$ in which exactly two $U_{rs}$s are non-empty.

[c3]

If $|{\displaystyle \bigcup _{r,s\ne i;rs\ne pq}}{U}_{rs}|\le 1$, then either $U_{{\left(pq\right)}^c}=\varnothing$ with $1\le |{\displaystyle \bigcup _{U_{rs}\ne \varnothing }}{U}_{{\left(rs\right)}^c}|\le 2$ or $U_{{\left(rs\right)}^c}=\varnothing$ with $|U_{{\left(pq\right)}^c}|\le 1$.

[d]

If $|U_{pq}|\le 1$ for all $1\le p\ne q\ne i\le 4$, then $2\le |{\displaystyle \bigcup _{U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}|\le 3$ in which at most two distinct $U_{{\left(pq\right)}^c}$s are non-empty.

(vi).

$|{\bigcup }_{n=1}^4{U}_n|=4$; there exist two non-empty sets $U_{ij}$ and $U_{{\left(ij\right)}^c}$ such that $2\le |U_{ij}\cup U_{{\left(ij\right)}^c}|\le 5$, and one of the following cases is satisfied:

[a]

There is $U_i$ with $|U_i|=3$, $|\bigcup _{p,q\ne i}{U}_{pq}\cup U_{jk\ell }|\in \{2,3\}$ in which $|U_{pq}|\le 2$, and $|\bigcup _{U_{pq}\ne \varnothing }{U}_{{\left(pq\right)}^c}|\le 1$. Moreover, if $|U_{pq}|\in \{1,2\}$ with $|U_{jk\ell }|=2,$ then $\bigcup _{U_{pq}\ne \varnothing }{U}_{{\left(pq\right)}^c}=\varnothing .$

[b]

If$|U_{ij}\cup U_{{\left(ij\right)}^c}|=5$, then either$|U_{ij}|=4$or$|U_{{\left(ij\right)}^c}|=4$. Further,${\displaystyle \bigcup _{pq\ne ij,{\left(ij\right)}^c;U_{pq}\ne \varnothing }}{U}_{{\left(pq\right)}^c}=\varnothing$.

[c]

If$|U_{ij}\cup U_{{\left(ij\right)}^c}|=4$, then$|U_{pq}\cup U_{{\left(pq\right)}^c}|=2$ whenever $U_{pq},U_{{\left(pq\right)}^c}\ne \varnothing$ for $pq\ne ij$. Further, if $|U_{ij}|=|U_{{\left(ij\right)}^c}|=2$, then at most one pair of $U_{pq},U_{{\left(pq\right)}^c}$ is nonempty for all $pq\ne ij$. [c] If $|U_{ij}\cup U_{{\left(ij\right)}^c}|=3$, then $|U_{pq}\cup U_{{\left(pq\right)}^c}|\in \{2,3\}$ whenever $U_{pq},U_{{\left(pq\right)}^c}\ne \varnothing$ for $pq\ne ij$. Further, if $U_{{\left(rs\right)}^c}\ne \varnothing$ for $1\le r\ne s\le 4$ and $rs\ne pq,ij$, then $|U_{rs}\cup U_{{\left(rs\right)}^c}|\in \{2,3\}$ with $|(U_{pq}\cup U_{{\left(pq\right)}^c})\cup (U_{rs}\cup U_{{\left(rs\right)}^c})|\in \{4,5\}$.

Proof. 

Assume that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$ Then, by Theorem 1b, we have $|{\bigcup }_{n=1}^4{U}_n|\le 9$. So, $4\le |{\bigcup }_{n=1}^4{U}_n|\le 9.$

Case 1 Let $|{\bigcup }_{n=1}^4{U}_n|=9.$ Then, by Theorem 1b, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ implies $|U_1|=6.$ If $U_{ij}\ne \varnothing$ or $U_{ijk}\ne \varnothing$ for some $i\ne 1$, then the sets $X=U_1$ and $Y=V\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\setminus U_1$ contain $K_{6,4}$, which has a crosscap four. So, $U_{ij},U_{ijk}=\varnothing$ for all $i\ne 1$. Here, remember that every vertex in $U_{1jk}$ is an end vertex, and every vertex in $U_{1j}$ is of degree two. Let $G_{12}$ be the induced subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ induced by the vertex subset ${\bigcup }_{n=1}^4{U}_n$. It is clear that $G_{12}\cong K_{6,1,1,1}$, and $G_{12}$ is a subgraph of the graph $H_2$ given in Figure 2a with the labels $u_{\ell }\in U_1$ (for $\ell =1,\dots ,6$), $I_2=v_1,I_3=v_2$, and $I_4=v_3$. By Figure 2b, the $N_2$-embedding of $G_{12}$ contains three different faces with vertices $I_2,I_3$; $I_3,I_4;$, and $I_2,I_4$, respectively. So, any number of vertices in $U_{1j}$ can be embedded into the $N_2$-embedding of $G_{12}$ without edge-crossing. Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Case 2 Let $|{\bigcup }_{n=1}^4{U}_n|=8.$

Case 2.1 Suppose $|U_1|\in \{5,4\}.$ If $U_{ij}\ne \varnothing$ or $U_{ijk}\ne \varnothing$ for some $i\ne 1$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{5,4}$ as a subgraph, which is a contradiction. Therefore, $U_{ij}=\varnothing$ and $U_{ijk}=\varnothing$ for all $i\ne 1$. Now, if $|U_1|=5$, then $\mathbb{AG}\left(\mathcal{L}\right)$ is a subgraph of the annihilating-ideal graph in Case 1 with $|U_1|=6$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Suppose $|U_1|=4.$ Here, $|U_2|=2$. If $I\in {\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{234}$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains a copy of $K_{4,5}$ where the partite sets are $U_1$ and $U_2\cup U_3\cup U_4\cup \left\{I\right\}$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. If $U_{1j}\ne \varnothing$ for some $j\in \{3,4\}$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{5,4}-e$ as a subgraph with the partition sets $U_1\cup U_{1j}$ and $U_2\cup U_3\cup U_4$ so that, by Remark 1, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, ${\displaystyle \bigcup _{ij\ne 12}}{U}_{ij}=\varnothing$ and $U_{234}=\varnothing$. In this case, one can retrieve an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ from Figure 4b by changing the label $I_3^{\prime }$ to $I_4$ and its related edges such that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Case 2.2 Suppose $|U_1|=3$. Let $|U_2|=3$. If $U_{ij}\ne \varnothing$ or $U_{ijk}\ne \varnothing$ for $ij\ne 12$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{4,5}-e$, which is a contradiction. Therefore, $U_{ij}=\varnothing$ and $U_{ijk}=\varnothing$ for all $ij\ne 12$. In this case, the crosscap of $\mathbb{AG}\left(\mathcal{L}\right)$ is same as the crosscap of $K_{3,3,1,1}$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Let $|U_2|=2$ and $I\in {\displaystyle \bigcup _{ijk\ne 123}}{U}_{ij}\cup U_{ijk}$.

• In the case that $I\in U_{ij}$ for $ij\in \{12,13\}$, the contraction of $\mathbb{AG}\left(\mathcal{L}\right)$ induced by the partite sets $X=U_i\cup U_4$ and $Y=U_j\cup \{I_k,[I_k^{\prime },I_{ij}]\}$, where $k\notin \{i,j,4\}$, forms a copy of $H_4$.
• In the case that $I\in U_{ij}$ for $ij\in \{14,23,24,34\}$, the graph $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{5,4}$ with the partite sets $U_i\cup U_j\cup U_{ij}$ and $U_k\cup U_{\ell }$ where $k,\ell \notin \{i,j\}$.
• In the case that $I\in {\displaystyle \bigcup _{ijk\ne 123}}{U}_{ijk}$, the contraction of $\mathbb{AG}\left(\mathcal{L}\right)$ induced by $({\bigcup }_{n=1}^4{U}_n\setminus \left\{I_{\ell }\right\})\cup \left\{[I_{\ell },I]\right\}$ forms $H_4$ where ℓ is the least integer in $\{1,2,3,4\}\setminus \{i,j,k\}$.

Thus, ${\displaystyle \bigcup _{ijk\ne 123}}{U}_{ij}\cup U_{ijk}=\varnothing$, and, so, the crosscap of $\mathbb{AG}\left(\mathcal{L}\right)$ is the crosscap of $K_{3,2,2,1}$, which is two.

Case 2.3 Suppose $|U_1|=2$. Then, $K_{2,2,2,2}$ is a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$. Suppose $\tilde{\gamma }\left(K_{2,2,2,2}\right)=2$. Then, by Euler’s formula, the number of faces in an $N_2$ embedding of $K_{2,2,2,2}$ is 16 so that all the faces are triangular, which contradicts the fact that $K_{2,2,2,2}$ has no triangular embedding (see [27]). Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$.

Case 3 Let $|{\bigcup }_{n=1}^4{U}_n|=7.$

Case 3.1 Suppose $|U_1|=4.$ If $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\ge 2$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{4,5}$ with one partite set $X=U_1$, and, so, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Further, by Theorem 4iv, $\mathbb{AG}\left(\mathcal{L}\right)$ is projective whenever $U_{ij}=U_{ijk}=\varnothing$ for all $i\ne 1$. Therefore, $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|=1$, and let $I\in {\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}$. Now, if $U_{1j}=\varnothing$ for all $2\le j\le 4$, then it is easy to verify that $\mathbb{AG}\left(\mathcal{L}\right)$ is isomorphic to a subgraph of the graph $H_1$ (see Figure 1a). Therefore, by Lemma 2, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. So, let $U_{1j}\ne \varnothing$ for some $2\le j\le 4$. Suppose $U_{k\ell }=\varnothing$ for $2\le j\ne k\ne \ell \le 4$. Here, the open neighbor of each vertex in $U_{1j}$ is $I_k$ and $I_{\ell }$ in $\mathbb{AG}\left(\mathcal{L}\right)$. Let $G_{13}$ be the induced subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ induced by the vertex subset ${\bigcup }_{n=1}^4{U}_n\cup \left\{I\right\}$. Clearly, $G_{13}$ is a subgraph of the graph $H_1$ given in Figure 1a with the labels $u_{\ell }\in U_1$ (for $\ell =1,\dots ,4$), $v_1=I_2,v_2=I_3,v_3=I_4$, and $v_4=I$. Since $(I_3,I_4),(I_2,I_4),(I_2,I_3)\in E\left(\mathbb{AG}\left(\mathcal{L}\right)\right)$, any number of vertices in $U_{1j}$ (for $2\le j\le 4$) can be embedded in the $N_2$-embedding of $G_{13}$ without edge-crossing, and, therefore, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Now, take $U_{k\ell }\ne \varnothing$ for $2\le j\ne k\ne \ell \le 4$. Note that the set $U_{k\ell }$ is nothing but the singleton set $\left\{I\right\}$. Now, consider the subgraph $G_{14}=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{1j},(I_j,I_k),(I_k,I_{\ell }),(I_j,I_{\ell }),(I,I_j)\}$, which is isomorphic to $K_{4,4}$ with the partition sets $X=U_1$ and $Y=\{I_j,I_k,I_{\ell },I\}$. Note that any $N_2$-embedding of $G_{14}$ has eight rectangular faces so that each face shares exactly two vertices from X and Y. In $\mathbb{AG}\left(\mathcal{L}\right)$, the vertex $I_{1j}$ is adjacent to three vertices of Y, namely $I_k,I_{\ell }$, and I. Therefore, one cannot insert $I_{1j}$ with its edges into $N_2$ without crossing, which is a contradiction.

Case 3.2 Suppose $|U_1|=3.$ Then, $|U_2|=2$. If $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\ge 2$, then it is easy to check that the contraction of $\mathbb{AG}\left(\mathcal{L}\right)$ contains either $K_{4,5}-e$ or $K_{3,6}\cup (K_4-e)$ as a subgraph, and, so, by Remark 1, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\le 1$.

Assume $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|=1.$ If $U_{ij}\ne \varnothing$, then $U_{{\left(ij\right)}^c}=\varnothing$; otherwise, the graph induced by the partition sets $X=U_1\cup U_3$ and $Y=U_2\cup U_4\cup [I_{ij},I_{{\left(ij\right)}^c}]$ form $H_4$ in $\mathbb{AG}\left(\mathcal{L}\right)$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3.$ Further, if $I\in U_{13}\cup U_{14}\cup U_{134}$, then consider the graph $\mathbb{AG}\left(\mathcal{L}\right)-\{I,e_1,e_2,e_3,e_4,e_5\}\cong K_{4,4}-e$ with the bipartite sets $\{I_1,I_1^{\prime },I_1^{″},I_j\}$ and $\{I_i,I_i^{\prime },I_k,I_{ijk}\}$ where $e_1=(I_1,I_j),e_2=(I_1^{\prime },I_j),e_3=(I_1^{″},I_j),e_4=(I_i,I_k),e_5=(I_i^{\prime },I_k)\}$, and $e=(I_j,I_{ijk})$. Now, a similar argument given for $G_2$ (refer to Equation 1) leads to $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|=1$ with $U_{13}=U_{14}=U_{134}=\varnothing$. In this case, with the help of Figure 10a, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Notice that in Figure 10a, we take $|U_{34}|=1.$

Figure 10. $|{\bigcup }_{n=1}^4{U}_n|=7$ with $|U_1|=3$.

Assume ${\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}=\varnothing$. If $|U_{1j}|\ge 3$ for some $j\in \{3,4\}$, then the sets $X=U_2\cup U_{j\prime }$ and $Y=U_1\cup U_j\cup U_{1j}$, where $j^{\prime }\in \{3,4\}\setminus \left\{j\right\},$ form $K_{3,7}$. So, $|U_{1j}|\le 2$ for $j=3,4$. Suppose $|U_{13}\cup U_{14}|\ge 3.$ Let $|U_{1j}|\ge 2$ and $|U_{1k}|\ge 1$ for $j,k\in \{3,4\}.$ Then, the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{1k},(I_1,I_j),(I_1^{\prime },I_j),(I_1^{″},I_j)\}$ contains $K_{3,6}$ with the partite sets $X=U_2\cup U_k$ and $Y=U_1\cup U_j\cup U_{1j}$. Since $de{g}_{K_{3,6}}\left(I_j\right)=3$, $I_j$ is contained in exactly three rectangular faces in any $N_2$-embedding of $K_{3,6}.$ Since $\{I_1,I_1^{\prime },I_1^{″},I_j\}\subset Y$, to embed the edges $(I_1,I_j),(I_1^{\prime },I_j)$, and $(I_1^{″},I_j)$, the vertices $I_1,I_1^{\prime }$, and $I_1^{\prime \prime }$ on the diagonals of the three rectangular faces that contain $I_j$, respectively, are required. Now, after embedding the three edges, $I_j$ is in exactly six triangular faces, all of which were formed by using two vertices from Y and one vertex from $X.$ Therefore, the vertex $I_{1k}$ cannot be embedded because it is adjacent to $I_j$ as well as two vertices from $X.$ So, $|U_{13}\cup U_{14}|\le 2$. However, $\mathbb{AG}\left(\mathcal{L}\right)$ is projective if $U_{13}\cup U_{14}=\varnothing .$ Thus, $1\le |U_{13}\cup U_{14}|\le 2.$ Now, one can obtain help from Figure 10b to say that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$

Case 3.3 Suppose $|U_1|=2$.

Claim A: At most two distinct $U_{ij}$s are non-empty in which at most one $U_{i4}$ is non-empty for $1\le i\ne j\le 4$. Additionally, at most two distinct $U_{\ell mn}$s are non-empty for $\ell mn\ne 123.$

Assume on the contrary that at least three $U_{ij}$s are non-empty for $1\le i,j\le 4$; say, $U_{i_1{i}_2},U_{i_3{i}_4}$ and $U_{i_5{i}_6}$ are non-empty. Let $p\in \{1,2,3\}\setminus \{i_1,i_2\},q\in \{1,2,3\}\setminus \{p,i_3,i_4\}$ and $r\in \{1,2,3\}\setminus \{p,q,i_5,i_6\}.$ If r exists, then the minor subgraph induced by the vertices $[I_p,I_{i_1{i}_2}],I_p^{\prime },[I_q,I_{i_3{i}_4}],I_q^{\prime },[I_r,I_{i_5{i}_6}],I_r^{\prime }$, and $I_4$ forms $K_7$ in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction. If r does not exist, then take r as $\{1,2,3\}\setminus \{p,q\}$ and form a minor of $\mathbb{AG}\left(\mathcal{L}\right)$ with the partite sets $X=\{I_r,I_r^{\prime },I_4,I_{r4}\}$ and $Y=\{[I_p,I_{i_1{i}_2}],I_p^{\prime },[I_q,I_{i_3{i}_4}],I_q^{\prime }\}$, which is isomorphic to either $H_3$ or $H_4$, as in Figure 3. So, by Lemma 3, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, only at most two distinct $U_{ij}$s can be non-empty for $1\le i\ne j\le 4.$ Further, if $U_{m4},U_{n4}\ne \varnothing$ for some $1\le m\ne n\le 4$, then the subgraph induced by the sets $X=U_m\cup U_{m4}\cup \left\{I_k\right\}$ and $Y=U_n\cup U_4\cup \left\{[I_k^{\prime },I_{n4}]\right\}$, where $k\ne m$ or n, form $H_4$ which has a crosscap of at least three.

Note that all the vertices in $U_{123}$ are end vertices in $\mathbb{AG}\left(\mathcal{L}\right)$. If $U_{ijk},U_{\ell mn}$, and $U_{pqr}$ are non-empty for $ijk,\ell mn,pqr\ne 123$, then the minor subgraph induced by $\{[I_{{\left(ijk\right)}^c},I_{ijk}],I_{{\left(ijk\right)}^c}^{\prime },[I_{{(\ell mn)}^c},I_{\ell mn}],I_{{(\ell mn)}^c}^{\prime },[I_{{\left(prq\right)}^c},I_{pqr}],I_{{\left(pqr\right)}^c}^{\prime },I_4\}$ is $K_7$, which is a contradiction. Therefore, at most two distinct $U_{\ell mn}$s are non-empty for $\ell mn\ne 123.$

Claim B: $|U_{ij}|\le 2$ and $|U_{i4}|\le 1$ for all $1\le i<j\le 3.$

If $|U_{ij}|\ge 3$ for some $1\le i,j\le 3$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{7,3}$ as a subgraph with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_k\cup U_4$ where $k\in \{1,2,3\}\setminus \{i,j\}$. Additionally, if $|U_{i4}|\ge 2$ for some $1\le i\le 3$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{5,4}$ as a subgraph with the partite sets $X=U_i\cup U_4\cup U_{i4}$ and $Y=U_j\cup U_k$ where $j,k\in \{1,2,3\}\setminus \left\{i\right\}$. Thus, $|U_{ij}|\le 2$ and $|U_{i4}|\le 1$ for all $1\le i<j\le 3.$

Assume $|U_{ij}|=2$ for some $1\le i,j\le 3$. Suppose $U_{k\ell }\ne \varnothing$ for some $1\le k<\ell \le 4$ and $k\ell \ne ij$. Let us take $j\notin \{k,\ell \}\cap \{i,j\}$. Then, $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{6,3}\cup (K_4-e)$ with the partite sets $X=\{I_i,I_i^{\prime },I_j,[I_j^{\prime },I_{k\ell }],I_{ij},I_{ij}^{\prime }\}$ and $Y=U_m\cup U_4$ where $m\in \{1,2,3\}\setminus \{i,j\}$. So, by Remark 1, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $U_{k\ell }=\varnothing$. In this case, the number of $U_{ijk}$ cannot be more than one because here $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{6,3}\cup (K_4-e)$. For the remaining cases, by Figure 11a, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Figure 11. $|{\bigcup }_{n=1}^4{U}_n|=7$ with $|U_1|=2$.

Assume $|U_{ij}|\le 1$ for all $1\le i,j\le 3$. Suppose $|U_{k4}|=1$ for some $1\le k\le 3$. If there are two $U_{\ell mn}$s that are non-empty for $\ell mn\ne 123$, then it is not hard to verify that $\mathbb{AG}\left(\mathcal{L}\right)$ contains a subgraph similar to the structure of $H_3$, which has a crosscap of at least three. For all the remaining cases, that is $|U_{ij}|=|U_{k4}|=1$ with unique $U_{\ell mn}\ne \varnothing$ or $|U_{ij}|\le 1$ and $|U_{pq}|\le 1$ with at most two $U_{\ell mn}$s that are non-empty for $1\le i,j,k,p,q\le 3$ and $\ell mn\ne 123$, one can use Figure 11b to obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Case 4 Let $|{\bigcup }_{n=1}^4{U}_n|=6.$

Case 4.1 Suppose $|U_1|=3$. Note that each vertex of $U_{ij}$ for $i=1$ is adjacent to exactly two vertices $I_{i^{\prime }}$ and $I_{j^{\prime }}$ for $i^{\prime },j^{\prime }\notin \{i,j\}$ and $(I_{i^{\prime }},I_{j^{\prime }})\in E\left(\mathbb{AG}\left(\mathcal{L}\right)\right)$, so we do not want to bother about $U_{1j}$ and $U_{1jk}$ for all $2\le j<k\le 4$. If $|U_{ij}|\ge 3$ for some $i\ne 1$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{4,5}$ as a subgraph with the partite sets $X=U_1\cup U_k$ and $Y=U_i\cup U_j\cup U_{ij}$ where $k\in \{2,3,4\}\setminus \{i,j\}$, which is a contradiction. So, $|U_{ij}|\le 2$ for all $i\ne 1.$

((i). Assume $|U_{ij}|=2$ for some $i\ne 1$. If $U_{{\left(ij\right)}^c}\ne \varnothing ,$ then the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_k\cup U_{{\left(ij\right)}^c}$ form $K_{4,5}$ in $\mathbb{AG}\left(\mathcal{L}\right)$, and, if $U_{k\ell }\ne \varnothing$ for some $k\ne 1$ with $k\ell \ne ij$ or $U_{234}\ne \varnothing$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{4,5}-e$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. If not, that is $U_{{\left(ij\right)}^c},U_{k\ell },U_{234}=\varnothing$ for all $k\ne 1$ with $k\ell \ne ij$, then by Figure 12a, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Figure 12. $|{\bigcup }_{n=1}^4{U}_n|=6$ with $|U_1|=3$.

(ii). Assume $|U_{ij}|\le 1$ for all $i\ne 1$. If $U_{{\left(i_1{j}_1\right)}^c}\ne \varnothing$ and $U_{{\left(i_2{j}_2\right)}^c}\ne \varnothing$ for some $U_{i_1{j}_1}\ne \varnothing$ and $U_{i_2{j}_2}\ne \varnothing ,$ then the sets $X=U_{i_1}\cup U_{j_1}\cup U_{i_1{j}_1}\cup \left\{[I_{i_2{j}_2},I_{{\left(i_2{j}_2\right)}^c}]\right\}$ and $Y=U_1\cup U_m\cup U_{{\left(i_1{j}_1\right)}^c}$, where $m\ne i_1,j_1$, contains $K_{4,5}-e$ in $\mathbb{AG}\left(\mathcal{L}\right)$. Additionally, if $|U_{{\left(ij\right)}^c}|\ge 3$, then the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_m\cup U_{{\left(ij\right)}^c}$, where $m\ne i,j$, form $K_{3,7}$ in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction. So, at most one of the sets $U_{{\left(ij\right)}^c}$ is non-empty with $|U_{{\left(ij\right)}^c}|\le 2.$

Let $|U_{{\left(ij\right)}^c}|=2$. If $I\in {\displaystyle \bigcup _{k\ell \ne ij}}{U}_{k\ell }\cup U_{234}$, then the sets $X=\{I_i,I_j,I_{ij}\}$ and $Y=\{I_1,I_1^{\prime },[I_1^{″},I],I_m,I_{{\left(ij\right)}^c},I_{{\left(ij\right)}^c}^{\prime }\}$, where $m\ne i,j$, form $K_{3,6}\cup (K_4-e)$ so that, by Remark 1, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, ${\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }\cup U_{234}=\varnothing$. For this case, readers can verify the $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right).$

Let $|U_{{\left(ij\right)}^c}|=1$. If $I,J\in {\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }\cup U_{234}$ with $|U_{k\ell }|\le 1$, then the sets $\{I_i,I_j,I_m,I_1,[I_1^{\prime },I],[I_1^{″},J],[I_{ij},I_{{\left(ij\right)}^c}]\}$ form $K_7$. Therefore, $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }\cup U_{234}|=1.$

Let ${\displaystyle \bigcup _{i\ne 1}}{U}_{{\left(ij\right)}^c}=\varnothing$. Then, by Theorem 4iii[a], $\mathbb{AG}\left(\mathcal{L}\right)$ is projective if $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\le 1.$ If $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\ge 4,$ then $K_{3,7}$ is a subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ with the partite sets $X=U_1$ and $Y=V\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\setminus U_1$. So, in the case of ${\displaystyle \bigcup _{i\ne 1}}{U}_{{\left(ij\right)}^c}=\varnothing$, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $2\le |{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{ijk}|\le 3$ with $|U_{ij}|\le 1$ (refer to Figure 12b).

Case 4.2 Suppose $|U_1|=2$. Then, $|U_2|=2$ and $|U_3|=|U_4|=1$. If $|U_{34}|\ge 3$, then the partite sets $X=U_1\cup U_2$ and $Y=U_3\cup U_4\cup U_{34}$ form $K_{4,5}$ as a subgraph in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction.

Case 4.2.1 Assume $|U_{34}|=2$. Then, $U_{{\left(pq\right)}^c}=\varnothing$ for all $U_{pq}\ne \varnothing$; otherwise, the sets $X=U_1\cup U_2$ and $Y=U_3\cup U_4\cup U_{34}\cup \left\{[I_{pq},I_{{\left(pq\right)}^c}]\right\}$ form $K_{4,5}$ in $\mathbb{AG}\left(\mathcal{L}\right)$. In particular, $U_{12}=\varnothing .$

If $|U_{ij}|\ge 2$ for some $ij\ne 12,34$ and $i<j$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{34},I_{34}^{\prime },(I_i,I_j),(I_i^{\prime },I_j)\}$ contains $K_{3,5}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime }\in \{1,2\}\setminus \left\{i\right\}$ and $j^{\prime }\in \{3,4\}\setminus \left\{j\right\}$. Note that any $N_2$-embedding of $K_{3,5}$ has one hexagonal and six rectangular faces, and the vertices $I_{34}$ and $I_{34}^{\prime }$ are adjacent to $I_i,I_i^{\prime },I_{i^{\prime }}$ and $I_{i^{\prime }}^{\prime }$. So, to insert $I_{34}$ and $I_{34}^{\prime }$ into an $N_2$-embedding of $K_{3,5}$, we require two faces, say $F_1$ and $F_2$, which contains $I_i,I_i^{\prime },I_{i^{\prime }}$, and $I_{i^{\prime }}^{\prime }$. If either $F_1$ or $F_2$ is hexagonal, then the corresponding face may adopt one of the edges $(I_i,I_j)$ or $(I_i^{\prime },I_j)$. Let us take that the edge $(I_i,I_j)$ is embedded. Now, to insert an edge $(I_i^{\prime },I_j)$, a rectangular face containing $I_i^{\prime }$ and $I_j$ as diagonals is required. However, no such rectangular face exists because the edges $(I_i^{\prime },I_{i^{\prime }})$ and $(I_i^{\prime },I_{i^{\prime }}^{\prime })$ have been used twice in $F_1$ and $F_2$, which is a contradiction.

For all of the remaining cases, that is $|{\displaystyle \bigcup _{ij\ne 12,34}}{U}_{ij}|\le 2$ with $|U_{ij}|\le 1$ and $U_{{\left(pq\right)}^c}=\varnothing$ when $U_{pq}\ne \varnothing$ for $1\le p\ne q\le 4$, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ (refer to Figure 13a).

Figure 13. $|{\bigcup }_{n=1}^4{U}_n|=6$ with $|U_1|=2$.

Case 4.2.2 Assume that $|U_{34}|=1$. Let us take $ij\ne 12,34$.

Let $|U_{12}|\ge 3$, then the subgraph of $\mathbb{AG}\left(\mathcal{L}\right)$ induced by the sets $X=U_3\cup U_4\cup U_{34}$ and $Y=U_1\cup U_2\cup U_{12}$ contains $K_{3,7}$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Thus, $|U_{12}|\le 2$.

Let $|U_{12}|=2$. If $I\in {\displaystyle \bigcup _{ij\ne 12,34}}{U}_{ij}\cup U_{134}\cup U_{234}$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,6}\cup (K_4-e)$, so that, by Remark 1, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, ${\displaystyle \bigcup _{ij\ne 34,12}}{U}_{ij}\cup U_{134}\cup U_{234}=\varnothing$, and in this case, by Figure 13b, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Let $|U_{12}|=1$. If $|U_{ij}|\ge 2$, then the partite sets $X=U_{i^{\prime }}\cup U_{j^{\prime }}$ and $Y=\{I_i,I_i^{\prime },I_j,I_{ij},I_{ij}^{\prime },[I_{34},I_{12}]\}$ where $i^{\prime }\in \{1,2\}\setminus \left\{i\right\}$ and $j^{\prime }\in \{3,4\}\setminus \left\{j\right\}$ form a minor subgraph $K_{3,6}\cup (K_4-e)$ in $\mathbb{AG}\left(\mathcal{L}\right)$ so that, by Remark 1, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. If $U_{ij},U_{k\ell }\ne \varnothing$ for $ij,k\ell \ne 12,34$ where $\{i,j\}\cap \{k,\ell \}=j=\ell$, then the partite sets $X=\{I_i,[I_i^{\prime },I_{k\ell }],I_{\ell },I_{ij}\}$ and $Y=\{I_k,I_k^{\prime },I_m,[I_{34},I_{12}]\}$ where $m\notin \{i,j,k\}$ form $\left(H_4\cup (u_2,u_3)\right)-(u_1,u_4)$. A slight modification of the proof for $H_4$ in Lemma 3 yields $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Further, minor changes to the labels in Figure 13a give $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $|{\displaystyle \bigcup _{ij\ne 12,34}}{U}_{ij}|\le 1$.

Let $U_{12}=\varnothing$. Then $U_{{\left(pq\right)}^c}=\varnothing$ for all $U_{pq}\ne \varnothing$; otherwise, $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_8-4e$, which is isomorphic to $\left(H_4\cup (u_1,u_3)\right)-(v_1,v_2)$, so Lemma 3 gives us $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. If $|U_{ij}|\ge 4$, then the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime }\in \{1,2\}\setminus \left\{i\right\}$ and $j^{\prime }\in \{3,4\}\setminus \left\{j\right\}$ contain $K_{7,3}$ in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction. Suppose $|U_{ij}|\in \{2,3\}$. If $|U_{k\ell }|\ge 2$ for some $k\ell \ne ij$ where $\{k,\ell \}\cap \{i,j\}=k=i,$ then the subgraph $G_{15}=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{34},I_{k\ell },I_{k\ell }^{\prime },(I_i,I_j),(I_i^{\prime },I_j)\}$ contains $K_{5,3}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime },j^{\prime }\notin \{i,j\}$. Note that any $N_2$-embedding of $K_{5,3}$ has one hexagonal and six rectangular faces. Further, in $\mathbb{AG}\left(\mathcal{L}\right)$, $I_{34}$ is adjacent to $I_i,I_i^{\prime },I_{i^{\prime }},I_{i^{\prime }}^{\prime }$, and, also, $I_{k\ell },I_{k\ell }^{\prime }$ are adjacent to $I_{i^{\prime }},I_{i^{\prime }}^{\prime },I_j$. So, to embed the vertices $I_{34},I_{k\ell }$, and $I_{k\ell }^{\prime },$ one hexagonal and two rectangular faces containing both $I_{i^{\prime }}$ and $I_{i^{\prime }}^{\prime }$ are required. In such a case, one cannot find two rectangular faces with the diagonal vertices $I_i,I_j$ and $I_i^{\prime },I_j$. So, either the edge $(I_i,I_j)$ or $(I_i^{\prime },I_j)$ cannot be drawn without crossing, which is a contradiction. Thus, we obtain the result as in the statement-(iv)[b2].

Case 4.2.3 Suppose $U_{34}=\varnothing$.

If $|U_{ij}\cup U_{{\left(ij\right)}^c}|\ge 4$ for some $ij\notin \{12,34\}$, then the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}\cup U_{{\left(ij\right)}^c}$ where $i^{\prime },j^{\prime }\notin \{i,j\}$ form a complete bipartite graph whose crosscap is more than two.

Let $|U_{ij}|\in \{2,3\}$ for some $ij\notin \{12,34\}$. Then, clearly, $U_{{\left(ij\right)}^c}$ must be empty. Let $k\ell \notin \{12,34,ij,{\left(ij\right)}^c\}$. If $|U_{ij}\cup U_{k\ell }\cup U_{{(k\ell )}^c}|\ge 5,$ then the sets $X=U_i\cup U_j\cup U_{ij}\cup \left\{[I_{k\ell },I_{{k\ell }^c}]\right\}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime }\in \{1,2\}\setminus \left\{i\right\}$ and $j^{\prime }\in \{3,4\}\setminus \left\{j\right\}$ form $K_{6,3}\cup (K_4-e)$ and, by Remark 1, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $2\le |U_{ij}\cup U_{k\ell }\cup U_{{(k\ell )}^c}|\le 4.$ Now, there are at most three possibilities:

(i).

$|U_{ij}|=3$ and $|U_{k\ell }|=1$; this case is pictured in Figure 14.

Figure 14. $|{\bigcup }_{n=1}^4{U}_n|=6$ with $|U_1|=|U_2|=2$.

(ii).

$|U_{ij}|=2$ and $|U_{k\ell }|=|U_{{(k\ell )}^c}|=1$; this case is pictured in Figure 15a.

Figure 15. $|{\bigcup }_{n=1}^4{U}_n|=6$ with $|U_1|=|U_2|=2$.

(iii).

$|U_{ij}|=|U_{k\ell }|=2$; this case is pictured in Figure 15b.

Thus, in all these cases, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$

Let $|U_{ij}|\le 1$ for all $ij\notin \{12,34\}$. Then, at least one $U_{ij}=\varnothing$ for $ij\notin \{12,34\}$. Otherwise, the graph induced by $\{I_1,I_1^{\prime },I_2,I_2^{\prime },I_3,I_4,[I_{13},I_{24}],[I_{14},I_{23}]\}$ forms $K_8-3e$ in $\mathbb{AG}\left(\mathcal{L}\right)$. Clearly, $\tilde{\gamma }(K_8-3e)\ge 3$ because the number of faces in the $N_2$-embedding of $K_8-3e$ is 17, which contradicts the well-known fact that $\frac{2|E|}{|F|}$ must be greater than the girth value (refer to Observation 1). Therefore, $|{\displaystyle \bigcup _{ij\ne 12,34}}{U}_{ij}|\le 3$. Thus, by [20], we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $|{\displaystyle \bigcup _{ij\ne 12,34}}{U}_{ij}|=3$.

Case 5 Let $|{\bigcup }_{n=1}^4{U}_n|=5.$ Then, $|U_1|=2$. If $U_{ij}=\varnothing$ for all $1\le i<j\le 4$, then $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\le 1$. Observe that we do not want to consider the sets $U_{ij}$ for $i\ne 1$ whenever $U_{{\left(ij\right)}^c}=\varnothing$ because every vertex in $U_{ij}$ is adjacent to $I_i,I_j$ and $(I_i,I_j)\in E\left(\mathbb{AG}\left(\mathcal{L}\right)\right)$. If $|U_{ij}|\ge 5$ for some $i\ne 1$, then the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime },j^{\prime }\notin \{i,j\}$ form $K_{3,7}$ in $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction.

Case 5.1 Assume $|U_{ij}|=4$ for some $i\ne 1$. Then, $U_{{\left(mn\right)}^c}=\varnothing$ whenever $U_{mn}\ne \varnothing$; otherwise, the sets $X=U_i\cup U_j\cup U_{ij}\cup \left\{[I_{mn},I_{{\left(mn\right)}^c}]\right\}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}$ where $i^{\prime },j^{\prime }\notin \{i,j\}$ form $K_{7,3}$ as a minor of $\mathbb{AG}\left(\mathcal{L}\right)$. Similarly, $U_{{\left(ij\right)}^c}=\varnothing$; otherwise $K_{6,4}$ is a minor of $\mathbb{AG}\left(\mathcal{L}\right)$. If $|U_{k\ell }|\ge 2$ for some $k\ne 1$ and $k\ell \ne ij$, then the subgraph $G_{16}=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{k\ell }^{\prime },(I_i,I_j)\}$ contains $K_{6,3}$ with the partition sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_{i^{\prime }}$ where $i^{\prime }\notin \{1,i,j\}$. Since $\{i,j\}\cap \{k,\ell \}\ne \varnothing$, let $\{i,j\}\cap \{k,\ell \}=i=k$. Clearly, $j\in \{2,3,4\}\setminus \{k,\ell \}$. Note that each face in any $N_2$-embedding of $K_{6,3}$ is rectangular, and the vertices $I_{k\ell },I_{k\ell }^{\prime }$ are adjacent to $I_1,I_1^{\prime }$ and $I_j$. Therefore, to insert $I_{k\ell }$ and $I_{k\ell }^{\prime }$, two rectangular faces that contain $I_1,I_1^{\prime }$ and $I_j$ are required. Next, to insert the edge $(I_i,I_j)$, a rectangular face with the diagonals $I_i$ and $I_j$ is required. However, the edges $(I_1,I_j)$ and $(I_1^{\prime },I_j)$ have been used twice to form the first two rectangular faces. So, one cannot construct another rectangular face that contains $I_i$ and $I_j$ with a single left-out vertex of $Y,$ which is a contradiction.

Therefore, for the remaining case, that is, $|U_{k\ell }|\le 1$ for all $k\ne 1$ and $k\ell \ne ij$ with $U_{{\left(mn\right)}^c}=\varnothing$ whenever $U_{mn}\ne \varnothing$, by using Figure 16a, one can have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Figure 16. $|{\bigcup }_{n=1}^4{U}_n|=5$ with $|U_1|=2$.

Case 5.2 Assume $|U_{ij}|=3$ for some $i\ne 1$. Let $p\notin \{1,i,j\}$. Clearly, $U_{{\left(ij\right)}^c}=\varnothing$; otherwise, the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_p\cup U_{{\left(ij\right)}^c}$ form $K_{5,4}$.

If $|U_{k\ell }|=3$ for some $k\ne 1$ and $k\ell \ne ij$, then the subgraph $G_{15}^{\prime }=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{k\ell }^{\prime },I_{k\ell }^{″},(I_i,I_j),(I_1,I_p),(I_1^{\prime },I_p)\}$ has a similar structure of $G_{15}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_p$, and so $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Suppose $|U_{k\ell }|,|U_{mn}|=2$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. Let $\{i,j\}\cap \{k,\ell \}=i=k.$ Then, $G_{17}=\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{k\ell }^{\prime },I_{mn},I_{mn}^{\prime },(I_i,I_j)\}$ has $K_{5,3}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_{\ell }$. Any $N_2$-embedding of $K_{5,3}$ has one hexagonal and six rectangular faces. Notice that $I_{k\ell },I_{k\ell }^{\prime }$ are adjacent to $I_1,I_1^{\prime },I_j$, and $I_{mn},I_{mn}^{\prime }$ are adjacent to $I_1,I_1^{\prime },I_i$. So, to embed $I_{k\ell },I_{k\ell }^{\prime },I_{mn}$, and $I_{mn}^{\prime },$ one hexagonal and two rectangular faces containing both $I_1$ and $I_1^{\prime }$ are required. However, the edge $(I_i,I_j)$ cannot be drawn without crossing, which is a contradiction. Therefore, $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }|\le 3$ and $|U_{k\ell }|\ne 3$.

Suppose $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }|=3$. Since $|U_{k\ell }|\ne 3$ for all $k\ne 1$ and $k\ell \ne ij,$ we have $|U_{k\ell }|=2$ and $|U_{mn}|=1$ for some $m\ne 1$ and $mn\ne ij,k\ell .$ Next, we claim that $U_{{(k\ell )}^c}=U_{{\left(mn\right)}^c}=\varnothing .$ If $U_{{(k\ell )}^c}\ne \varnothing ,$ then by letting $\{i,j\}\cap \{k,\ell \}=i=k$, $K_{7,3}$ can be formed by the sets $X=U_i\cup U_j\cup U_{ij}\cup U_{k\ell }$ and $Y=U_1\cup \left\{[I_{\ell },I_{{(k\ell )}^c}]\right\}$. If $U_{{\left(mn\right)}^c}\ne \varnothing ,$ then $\mathbb{AG}\left(\mathcal{L}\right)$ has a similar structure to $G_{15}$, so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3.$

Suppose $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }|\le 2$. As mentioned, $U_{{(k\ell )}^c}=\varnothing$ when $|U_{k\ell }|=2$ for $k\ne 1$ and $k\ell \ne ij.$ Suppose $|U_{k\ell }|=1$ and $|U_{{(k\ell )}^c}|\ge 2$. Then, $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime },(I_i,I_j),(I_1,I_{\ell })$, $and(I_1^{\prime },I_{\ell })\}$ has $K_{5,3}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_{\ell }$. Note that any $N_2$-embedding of $K_{5,3}$ has one hexagonal and six rectangular faces, $I_{k\ell }$ is adjacent to $I_1,I_1^{\prime },I_j,I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime }$, and $I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime }$ are adjacent to $I_k,I_{\ell },I_{k\ell }$. So, the three vertices $I_{k\ell },I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime }$ together with the edges $(I_i,I_j),(I_1,I_{\ell }),(I_1^{\prime },I_{\ell })$ cannot be embedded, and, also, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $|U_{k\ell }\cup U_{{(k\ell )}^c}|\le 2.$ Further, if $|U_{k\ell }\cup U_{{(k\ell )}^c}|=|U_{\ell m}\cup U_{{(\ell m)}^c}|=2$ for $k\ell \ne ij$ and $\ell m\ne ij,k\ell$, then $\mathbb{AG}\left(\mathcal{L}\right)$ contains $K_{3,7}$, which is a contradiction.

Thus, an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ can be retrieved from Figure 16a for $|{\displaystyle \bigcup _{pq\ne ij}}{U}_{pq}|\le 3$ with $U_{{\left(pq\right)}^c}=\varnothing$ if $|U_{pq}|=2$.

Case 5.3 Assume $|U_{ij}|=2$ for some $i\ne 1$. Clearly, $|U_{{\left(ij\right)}^c}|\le 1$; otherwise, the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_p\cup U_{{\left(ij\right)}^c}$ where $p\notin \{1,i,j\}$ form $K_{5,4}$.

If $|U_{k\ell }|,|U_{mn}|=2$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$, then $U_{{\left(ij\right)}^c},U_{{(k\ell )}^c},U_{{\left(mn\right)}^c}=\varnothing$. Further, an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ in the case of $|U_{ij}|=|U_{k\ell }|=|U_{mn}|=2$ is given in Figure 16b so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

Suppose $|U_{k\ell }|=2,|U_{mn}|\le 1$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. If $U_{{\left(ij\right)}^c},U_{{(k\ell )}^c}\ne \varnothing$, then the sets $X=U_1\cup U_p\cup U_{{\left(ij\right)}^c}$ and $Y=U_i\cup U_j\cup U_{ij}\cup \left\{[I_{(k\ell )},I_{{(k\ell )}^c}]\right\}$ where $p\notin \{1,i,j\}$ form $K_{5,4}-e$ in $\mathbb{AG}\left(\mathcal{L}\right)$ so that, by Remark 1, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Further, since $|U_{k\ell }|=2$, we have $|U_{{(k\ell )}^c}|\le 1$. Therefore, $|U_{{\left(ij\right)}^c}\cup U_{{(k\ell )}^c}|\le 1.$ Suppose $|U_{{\left(ij\right)}^c}\cup U_{{(k\ell )}^c}|=1,$ say $U_{{\left(ij\right)}^c}\ne \varnothing$. Then, $U_{{\left(mn\right)}^c}=\varnothing$; otherwise, $X=U_1\cup U_p\cup U_{{\left(ij\right)}^c}$ and $Y=U_i\cup U_j\cup U_{ij}\cup \left\{[I_{\left(mn\right)},I_{{\left(mn\right)}^c}]\right\}$ where $p\notin \{1,i,j\}$ form $K_{4,5}-e$ in $\mathbb{AG}\left(\mathcal{L}\right)$. So, $|U_{mn}\cup U_{{\left(mn\right)}^c}|\le 1.$ Suppose not, that is, $U_{{\left(ij\right)}^c},U_{{(k\ell )}^c}=\varnothing ,$ then $|U_{{\left(mn\right)}^c}|\le 1$; otherwise, $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{ij},I_{ij}^{\prime },I_{k\ell },I_{k\ell }^{\prime },(I_m,I_n),(I_1,I_{m^{\prime }}),(I_1^{\prime },I_{m^{\prime }})\}\cong K_{5,3}$ with the partite sets $X=U_1\cup U_{m^{\prime }}\cup U_{{\left(mn\right)}^c}$ and $Y=U_m\cup U_n\cup U_{mn}$ where $m^{\prime }\notin \{1,m,n\}$ is a similar structure to $G_{17}$ which has a crosscap of at least three. So, $|U_{mn}\cup U_{{\left(mn\right)}^c}|\le 2.$

Suppose $|U_{k\ell }|,|U_{mn}|\le 1$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. Then, by Theorem 4(ii), $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ provided $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }|=2$ with $|{\displaystyle \bigcup _{p\ne 1}}{U}_{{\left(pq\right)}^c}|=1$ or $|{\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }|=1$ with $|U_{{\left(ij\right)}^c}|=1,U_{{(k\ell )}^c}=\varnothing$ or $U_{{\left(ij\right)}^c}=\varnothing ,|U_{{(k\ell )}^c}|\le 2$ or ${\displaystyle \bigcup _{k\ne 1;k\ell \ne ij}}{U}_{k\ell }=\varnothing$ with $|U_{{\left(ij\right)}^c}|=1$.

Hence, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $4\le |{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}\cup U_{{\left(ij\right)}^c}|\le 6$ with $|{\displaystyle \bigcup _{i\ne 1}}{U}_{{\left(ij\right)}^c}|\le 1$ or $|{\displaystyle \bigcup _{i\ne 1}}{U}_{ij}|=3$ with $|U_{ij}\cup U_{{\left(ij\right)}^c}|\le 3$ and a unique $U_{{\left(ij\right)}^c}\ne \varnothing$ or ${\displaystyle \bigcup _{i\ne 1}}{U}_{ij}=2$ with $|U_{{\left(ij\right)}^c}|=1$.

Case 5.4 Assume $|U_{ij}|=1$ for all $i\ne 1$. Then, $|U_{{\left(ij\right)}^c}|\le 3$; otherwise, the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_1\cup U_{i^{\prime }}\cup U_{{\left(ij\right)}^c}$ where $i^{\prime }\notin \{1,i,j\}$ form $K_{3,7}$.

Suppose $|U_{k\ell }|=|U_{mn}|=1$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. If $U_{{\left(ij\right)}^c},U_{{(k\ell )}^c},U_{{\left(mn\right)}^c}\ne \varnothing$, then the sets $X=U_1\cup U_2\cup U_3$ and $Y=\{I_4,[I_{ij},I_{{\left(ij\right)}^c}],[I_{k\ell },I_{{(k\ell )}^c}],[I_{mn},I_{{\left(mn\right)}^c}]\}$ form $H_4$ as a minor of $\mathbb{AG}\left(\mathcal{L}\right)$, which is a contradiction. Assume that $|U_{{\left(ij\right)}^c}|=3$. If $I\in U_{{(k\ell )}^c}\cup U_{{\left(mn\right)}^c}$, then $G_{18}=\mathbb{AG}\left(\mathcal{L}\right)-\{I,I_{k\ell },I_{mn},(I_i,I_j),(I_1,I_{i^{\prime }}),(I_1^{\prime },I_{i^{\prime }})\}$ contains $K_{6,3}$ with the partite sets $X=U_1\cup U_{i^{\prime }}\cup U_{{\left(ij\right)}^c}$ and $Y=U_i\cup U_j\cup U_{ij}$ and any $N_2$-embedding of $K_{3,6}$ has nine rectangular faces. Here, it is not hard to verify that all the left-out vertices and edges cannot be embedded into the nine rectangular faces so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $U_{{(k\ell )}^c}\cup U_{{\left(mn\right)}^c}=\varnothing$. Here, the graph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{mn}\}$ is a subgraph of the graph in Figure 2a, and the suitable labels in Figure 2b give two different faces in the $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{k\ell },I_{mn}\}$ that contains the vertices $N\left(I_{k\ell }\right)$ and $N\left(I_{mn}\right)$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. Assume $|U_{{\left(ij\right)}^c}|\le 2$. If $|U_{{\left(ij\right)}^c}\cup U_{{(k\ell )}^c}|\ge 4$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime },I_{mn},(I_i,I_j),(I_1,I_{i^{\prime }}),(I_1^{\prime },I_{i^{\prime }})\}$ has a similar structure to $G_{15}$ so that we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Additionally, by Theorem 4ii, $\mathbb{AG}\left(\mathcal{L}\right)$ is projective when $|{\displaystyle \bigcup _{i\ne 1}}{U}_{{\left(ij\right)}^c}|\le 1$. For all of the remaining cases, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ can be verified by drawing the $N_2$-embedding. Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ when $2\le |{\displaystyle \bigcup _{i\ne 1}}{U}_{{\left(ij\right)}^c}|\le 3$ with at least one of the sets’ $U_{{\left(ij\right)}^c}=\varnothing$.

Suppose $|U_{k\ell }|=1$ and $U_{mn}=\varnothing$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. If $|U_{{\left(ij\right)}^c}|=3$ and $U_{{(k\ell )}^c}\ne \varnothing$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{{(k\ell )}^c},I_{k\ell },(I_i,I_j),(I_1,I_{i^{\prime }}),(I_1^{\prime },I_{i^{\prime }})\}$ has a similar structure to $G_{18}$, and, if $|U_{{\left(ij\right)}^c}|=|U_{{(k\ell )}^c}|=2$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_{{(k\ell )}^c},I_{{(k\ell )}^c}^{\prime },I_{k\ell },(I_i,I_j),(I_1,I_{i^{\prime }}),(I_1^{\prime },I_{i^{\prime }})\}$ has a similar structure to $G_{15}$ so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3.$ Further, $\mathbb{AG}\left(\mathcal{L}\right)$ is projective if $|U_{{\left(ij\right)}^c}\cup U_{{(k\ell )}^c}|\le 1$. Thus, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $|U_{{\left(ij\right)}^c}\cup U_{{(k\ell )}^c}|\in \{2,3\}.$

Suppose $U_{k\ell },U_{mn}=\varnothing$ for $k,m\ne 1$ and $k\ell ,mn\ne ij$. Then, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$ whenever $2\le |U_{{\left(ij\right)}^c}|\le 3.$

Case 6 Let $|{\bigcup }_{n=1}^4{U}_n|=4.$ Then, by Theorem 4(i), $|U_{ij}\cup U_{{\left(ij\right)}^c}|\ge 3$ for some $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing .$ Further, if $|U_{ij}\cup U_{{\left(ij\right)}^c}|\ge 6$ with $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$, then the subgraph induced by the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y={\displaystyle \bigcup _{k\ne i,j}}{U}_k\cup U_{{\left(ij\right)}^c}$ contains one of the graph’s $K_{3,7}$, $K_{4,6}$, or $K_{5,5}$ as a subgraph so that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $3\le |U_{ij}\cup U_{{\left(ij\right)}^c}|\le 5$ for some $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing .$

(i) Suppose $|U_{ij}\cup U_{{\left(ij\right)}^c}|=5$ for $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$. If either $|U_{ij}|=3$ or $|U_{{\left(ij\right)}^c}|=3$, then the sets $X=U_i\cup U_j\cup U_{ij}$ and $Y={\displaystyle \bigcup _{k\ne i,j}}{U}_k\cup U_{{\left(ij\right)}^c}$ form $K_{4,5}$, which is a contradiction. So, either $|U_{ij}|=4$ or $|U_{{\left(ij\right)}^c}|=4$. With no loss of generality, assume that $|U_{ij}|=4$. If $U_{k\ell },U_{{(k\ell )}^c}\ne \varnothing$ for $k\ell \ne ij,{\left(ij\right)}^c$, then clearly $|\{i,j\}\cap \{k,\ell \}|=1$ and $|\{m,n\}\cap \{k,\ell \}|=1$ where $m,n\in \{1,2,3,4\}\setminus \{i,j\}$. So, let us take $\{i,j\}\cap \{k,\ell \}=\left\{j\right\}$ and $\{m,n\}\cap \{k,\ell \}=\left\{m\right\}.$ This implies that $(I_{k\ell },I_i),(I_{{(k\ell )}^c},I_m)\in E\left(\mathbb{AG}\left(\mathcal{L}\right)\right)$. Then, the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{I_i,I_{k\ell },I_{{(k\ell )}^c}\}$ contains $K_{5,3}$ with the partite sets $X=U_j\cup U_{ij}$ and $Y=U_m\cup U_n\cup U_{{\left(ij\right)}^c}$. Now, the path $I_i-I_{k\ell }-I_{{(k\ell )}^c}$ has to be embedded into a single face of any $N_2$-embedding of $K_{5,3}$. Further, the vertices $I_i$ and $I_{{(k\ell )}^c}$ are adjacent to $I_j$ and $I_m.$ So, after embedding these four edges, the edge $(I_{k\ell },I_n)$ cannot be embeded, which means $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $U_{{(k\ell )}^c}=\varnothing$ when $U_{k\ell }\ne \varnothing$ for all $k\ell \ne ij,{\left(ij\right)}^c$, and, in such cases, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$.

(ii) Suppose $|U_{ij}\cup U_{{\left(ij\right)}^c}|=4$ for $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$. If $|U_{k\ell }\cup U_{{(k\ell )}^c}|\ge 3$ for $k\ell \ne ij$, then the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{U_{k\ell }\cup U_{{(k\ell )}^c}\}$ contains a crosscap two graph $K_{5,3}$ or $K_{4,4}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y={\displaystyle \bigcup _{m\ne i,j}}{U}_m\cup U_{{\left(ij\right)}^c}$. Since $|U_{k\ell }\cup U_{{(k\ell )}^c}|\ge 3$, we can take $|U_{k\ell }|\ge 2$. Notice that the path $I_{k\ell }-I_{{(k\ell )}^c}-I_{k\ell }^{\prime }$ together with the edges $(I_{k\ell },I_m),(I_{k\ell },I_i),(I_{k\ell }^{\prime },I_m)$, and $(I_{k\ell }^{\prime },I_i)$ should be embedded into a single face of an $N_2$-embedding of $K_{5,3}$. Thereafter, the face cannot adopt the edges $(I_{{(k\ell )}^c},I_j)$ and $(I_{{(k\ell )}^c},I_n)$ where $n\notin \{i,j,m\}$, which implies that $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. Therefore, $|U_{k\ell }\cup U_{{(k\ell )}^c}|=2$ for all $U_{k\ell },U_{{(k\ell )}^c}\ne \varnothing$ with $k\ell \ne ij$ and $1\le i,j\le 4$.

If $|U_{ij}|=3$, then, by Figure 17a, we obtain $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2$. If not, then $|U_{ij}|=2$. Suppose $|U_{k\ell }\cup U_{{(k\ell )}^c}|=|U_{mn}\cup U_{{\left(mn\right)}^c}|=2$ for $U_{k\ell },U_{{(k\ell )}^c},U_{mn},U_{{\left(mn\right)}^c}\ne \varnothing$ with $k\ell ,mn\ne ij$. Then, the subgraph $\mathbb{AG}\left(\mathcal{L}\right)-\{[I_{k\ell },I_{{(k\ell )}^c}],[I_{mn},I_{{\left(mn\right)}^c}]\}$ contains $K_{4,4}$ with the partite sets $X=U_i\cup U_j\cup U_{ij}$ and $Y=U_{i^{\prime }}\cup U_{j^{\prime }}\cup U_{{\left(ij\right)}^c}$, where $i^{\prime },j^{\prime }\notin \{i,j\}$. Note that every face of any $N_2$-embedding of $K_{4,4}$ is rectangular, and the vertices $[I_{k\ell },I_{{(k\ell )}^c}]$ and $[I_{mn},I_{{\left(mn\right)}^c}]$ are adjacent to the four vertices $I_i,I_j,I_{i^{\prime }}$, and $I_{j^{\prime }}$. So, to embed the vertices $[I_{k\ell },I_{{(k\ell )}^c}]$ and $[I_{mn},I_{{\left(mn\right)}^c}]$, two distinct rectangular faces with boundaries $I_i,I_j,I_{i^{\prime }}$, and $I_{j^{\prime }}$ are required, which is a contradiction. Therefore, at least one $U_{{(k\ell )}^c}=\varnothing$ when $U_{k\ell }\ne \varnothing$ for $k\ell \ne ij$ and $1\le i\ne j\le 4$. In this case, an $N_2$-embedding of $\mathbb{AG}\left(\mathcal{L}\right)$ is given in Figure 17b.

Figure 17. $|{\bigcup }_{n=1}^4{U}_n|=4$ with $|U_{12}\cup U_{34}|=4$.

(iii) Suppose $2\le |U_{ij}\cup U_{{\left(ij\right)}^c}|\le 3$ for all $U_{ij},U_{{\left(ij\right)}^c}\ne \varnothing$ with $1\le i\ne j\le 4$. Then, by Theorem 4i, there exists $U_{k\ell }$ such that $U_{k\ell },U_{{(k\ell )}^c}\ne \varnothing$ with $|U_{k\ell }\cup U_{{(k\ell )}^c}|=3$ and ${\displaystyle \bigcup _{mn\ne k\ell ,{(k\ell )}^c;U_{mn}\ne \varnothing }}{U}_{{\left(mn\right)}^c}\ne \varnothing .$

Suppose $|U_{ij}\cup U_{{\left(ij\right)}^c}|=3$ for all $1\le i\ne j\le 4$. That is, $|U_{12}\cup U_{34}|=|U_{13}\cup U_{24}|=|U_{14}\cup U_{23}|=3$. Without a loss of generality, we let $|U_{12}|=|U_{13}|=|U_{14}|=2$. Now, consider the bipartite graph $G_{19}=\mathbb{AG}\left(\mathcal{L}\right)-\{(I_2,I_3),(I_2,I_4),(I_3,I_4),(I_2,I_{34}),(I_3,I_{24}),(I_4,I_{23})\}$ with the partite sets $X=U_1\cup U_{12}\cup U_{13}\cup U_{14}$ and $Y=U_2\cup U_3\cup U_4\cup U_{34}\cup U_{24}\cup U_{23}$. Note that $\tilde{\gamma }\left(G_{19}\right)=2$ and the faces of any $N_2$-embedding of $G_{19}$ have one of the following possibilities:

• Nine rectangular and two hexagonal faces;
• Ten rectangular faces and one octagonal face.

Since, in $G_{19}$, the only common neighbor for $I_2$ and $I_{34}$ in X is $I_1$, no rectangular face has both $I_2$ and $I_{34}$. Therefore, the edge $(I_2,I_{34})$ should be embedded in a face of a length of more than four; so the edges are $(I_3,I_{24})$ and $(I_4,I_{23})$. Thus, we have to embed the three mutually disjoint edges of $⟨Y⟩$ in either two hexagonal faces or one octagonal face. However, in any case, the faces may adopt at most two mutually disjoint edges of $⟨Y⟩$, and, so, $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)\ge 3$. For the remaining cases, we have $\tilde{\gamma }\left(\mathbb{AG}\left(\mathcal{L}\right)\right)=2.$ □

Remark 2. 

As an illustration, we consider the case (v)[a] in Theorem 5. Let $|U_1|=|U_2|=|U_3|=|U_4|=1$ and $|U_{23}|=4$. If $|U_{24}|=|U_{34}|=1$, then the corresponding five-partite graph, as in Figure 18a, has a crosscap two. Additionally, if $|U_{24}|=2$, then the crosscap of the corresponding five-partite graph, given in Figure 18b, is not equal to two. Moreover, the five-partite graph G in Figure 18b is minimal with respect to $\tilde{\gamma }\left(G\right)\ne 2.$

Figure 18. Five-partite graphs.

6. Conclusions

The forbidden subgraphs for a crosscap two surface (a Klein bottle) are not known yet. In this regard, an open problem will be to determine a family of graphs that has a crosscap number two. This paper provides a class of r-partite graphs, where $2\le r\le 5$, that can be both embedded and not embedded in a crosscap two surface. This was completed by using the classification of all lattices with at most four atoms whose annihilating-ideal graph has a crosscap two.