Planar wheel-like bricks ¹¹1 The research is partially supported by NSFC (No. 12271235) and NSF of Fujian Province (No. 2021J06029).
E-mail addresses: [email protected] (F. Lu), [email protected] (J. Xue)

Obviously, $d_{G/X}(x)=d_{G/\overline{X}}(\overline{x})$. Let $H_{1}=G/X-\{e,f\}$ and $H_{2}=G/\overline{X}-\{e,g\}$. As $e\in\partial(X)$, $d_{H_{1}}(x)=d_{H_{2}}(\overline{x})$. Then $G-\{e,f,g\}$ is isomorphic to $H_{1}(x)\odot H_{2}(\overline{x})$. By the definition of removable doubletons, $H_{1}$ and $H_{2}$ are matching covered bipartite graphs. Therefore, $G-\{e,f,g\}$ is matching covered by Proposition 2.3. Assume that $A_{i}$ and $B_{i}$ are the two color classes of $H_{i}$ such that both ends of $e$ lie in $B_{i}$ for $i=1,2$. Then $G-\{e,f,g\}$ is a bipartite graph with two color classes: $A_{1}\cup B_{2}\setminus\{\overline{x}\}$ and $A_{2}\cup B_{1}\setminus\{{x}\}$. Note that $E_{G}[B_{1},B_{2}]=\{e\}$. So $G-\{f,g\}$ is bipartite. And then it is matching covered by Theorem 4.1.1 in [10]. As two ends of $f$ lie in $A_{1}$, and two ends of $g$ lie in $A_{2}$, $G-f$ and $G-g$ are not matching covered by a simple computation. Therefore, the result follows. ∎

It can be checked that $\{u_{2},u_{3}\}$ is a barrier of $G-uu_{1}$. By Lemma 2.2, $u_{2}u_{3}$ is forbidden in $G-uu_{1}$. So the result holds. ∎

As shown in Figure 1, $\{e_{i},f_{i}\}$ is a removable doubleton for $i=1,2,3$, and the bold edge is a removable edge. So the result follows. ∎

If $|V(G)|=4$, then $G$ is isomorphic to $K_{4}$, which is wheel-like. Suppose that $|V(G)|\geq 6$. By Proposition 3.2$,\overline{C_{6}}$ and $R_{8}$ are not wheel-like. If $G$ is distinct from $\overline{C_{6}}$ and $R_{8}$, then $G$ has two nonadjacent removable edges by Theorem 2.4, so it is not wheel-like. ∎

By Theorem 3.4, $G$ contains no subgraphs that is a subdivision of either $K_{3,3}$ or $K_{5}$. As $G/{\overline{X}}$ is matching covered, it is 2-connected (see Remark 2 on Page 145 of [10]). Then $G[{X}]$ is connected. So $G/X$ contains no subgraphs that is a subdivision of either $K_{3,3}$ or $K_{5}$. Therefore, $G/X$ is planar. So does $G/\overline{X}$. ∎

By Proposition 2.7, $G$ is either nonsolid or $W_{5}$. As $W_{5}$ is wheel-like, we may assume that $G$ is a nonsolid brick. Then $G$ has a nontrivial separating cut $C$. Since $G$ is 3-connected and both $C$-contractions of $G$ are matching covered, both $C$-contractions of $G$ are isomorphic to $K_{4}$’s (up to multiple edges). It can be checked that $G$ is isomorphic to $\overline{C_{6}}$ or one of the graphs in Figure 2. By Proposition 3.2, $\overline{C_{6}}$ is not wheel-like. It can be seen that all the graphs in Figure 2 are not wheel-like (the bold edges are removable). Therefore, the result holds. ∎

As $H-e\cong G\cong H-f$ and $G$ is matching covered, both $e$ and $f$ are removable in $H$. Obviously, an edge that is removable in $G$ is removable in $H$. If $\{e,e^{\prime}\}$ is a removable doubleton of $G$, then every perfect matching of $H$ containing $e$ or $f$ contains $e^{\prime}$. So $e$ and $f$ are forbidden in $H-e^{\prime}$, that is, $e^{\prime}$ is not removable in $H$. Therefore, if $G$ does not contain any vertex $v$ such that each removable class of $G$ has an edge in $\partial(v)$, neither does $H$. ∎

Assume that the underlying simple graph of $G$ is $H$. If $G=H$, the result holds obviously. So we assume that $G\neq H$. If $H$ is near-bipartite, then $H$ is $K_{4}$ by Lemma 3.3. By Proposition 3.7, all the multiple edges of $H$ are incident with a common vertex. If $H$ is not near-bipartite, then every removable class of $H$ is a removable edge. By Theorem 1.1, $u_{h}$ is incident with at least three removable edges. By Proposition 3.7 again, every multiple edge is incident with $u_{h}$. ∎

1) By Theorem 1.1, $G_{1}$ and $G_{2}$ contain at least three removable classes, respectively. Suppose that $R_{1}$ and $R_{2}$ are removable classes of $G_{1}$ and $G_{2}$, respectively, such that $R_{1}\cap\partial(u)=\emptyset$ and $R_{2}\cap\partial(v)=\emptyset$. Then $R_{1}\cup R_{2}$ is a matching in $G$. For $i\in\{1,2\}$, if $R_{i}$ is a removable edge, then $R_{i}$ is also a removable edge in $G$ by Lemma 2.9, and if $R_{i}$ is a removable doubleton, then one edge of $R_{i}$ is also removable in $G$ by Theorem 2.10. It means that $R_{1}\cup R_{2}$ contains two nonadjacent removable edges of $G$, contradicting the fact that $G$ is wheel-like. So at least one of $G_{1}$ and $G_{2}$ is wheel-like, such that $u$ or $v$ is its hub.

2) Suppose, to the contrary, that $G_{2}$ is not a wheel-like brick. Let $F_{1}$, $F_{2}$ and $F_{3}$ be removable classes of $G_{2}$ such that there does not exist any vertex in $G_{2}$ incident with one edge in $F_{1}$, $F_{2}$ and $F_{3}$, respectively. For $i=1,2,3$, if $F_{i}\cap\partial(v)=\emptyset$, or $F_{i}\cap\partial(v)=\{e_{i}\}$ and $\theta(e_{i})$ is removable in $G_{1}$, then $F_{i}$ contains a removable edge in $G$ by Lemma 2.9 and Theorem 2.10; if $F_{i}\cap\partial(v)=\{e_{i}\}$, and $\{\theta(e_{i}),e_{i}^{\prime}\}$ is a removable doubleton in $G_{1}$, then by Theorem 2.10, the edge $e_{i}^{\prime}$ is a removable edge in $G$. (If $e_{i}\in\partial(v)$, $e_{i}$ lies in a removable doubleton of $G_{2}$, and $\theta(e_{i})$ lies in a removable doubleton in $G_{1}$, then $G$ has a removable doubleton by Lemma 2.11, and so the underlying simple graph of $G$ is $K_{4}$ by Lemma 3.3, contradicting the assumption that $G$ is a splicing of two bricks.) So, in above both alternatives, for each $F_{i}$ ($i=1,2,3$), we have a removable class $F_{i}^{\prime}$ of $G$ such that $F_{i}^{\prime}\subset F_{i}$ or $F_{i}^{\prime}\subset(E(G_{1})\setminus\partial(u))$. Therefore, there does not exist any vertex in $G$ incident with one edge in $F_{1}^{\prime}$, $F_{2}^{\prime}$ and $F_{3}^{\prime}$, respectively, contradicting the fact that $G$ is wheel-like. ∎

Assume that $u_{i}u_{i+1}\in E(G)$ for $i=1,2,\ldots,s$ (the subscript is modulo $s$), and $v_{i}v_{i+1}\in E(H)$ for $i=1,2,\ldots,t$ (the subscript is modulo $t$). Let $W=G(u)\odot H(v)$ and $C=\partial_{W}(V(G)\setminus\{u\})$.

We will prove the sufficiency firstly. By Statement 2), every multiple edge is incident with the hub. Then $|E[v,v_{h}]|=d_{G}(u)-2$. We will prove the edge in $E(W)\setminus\partial(v_{h})$ is nonremovable. As $u_{1}u_{r}\notin E(W)$, that is $r\neq 2$ and $r\neq s$, we consider the following alternatives.

Case 1. $r=3~{}\mbox{or}~{}s-1$.

Assume that $u_{r}=u_{3}$ (the case when $u_{r}=u_{s-1}$ is the same by symmetry). Let $B_{u_{1}u_{2}}=\{v_{h},u_{3},v_{t-2}\}$. Then $W-B_{u_{1}u_{2}}-{u_{1}u_{2}}$ has exactly three components: $W[\{u_{2}\}],W[\{v_{t-1}\}]$ and $W[V(G)\setminus(\{u_{2},v_{t-1}\}\cup B_{u_{1}u_{2}})]$, all of which are odd. Therefore, $B_{u_{1}u_{2}}$ is a barrier of $W-{u_{1}u_{2}}$ containing both ends of the edge $v_{t-2}v_{h}$. By Lemma 2.2, $u_{1}u_{2}$ is nonremovable in $W$. By symmetry, $u_{2}u_{3}$ is also nonremovable in $W$.

Assume that $s=5$. It can be checked $\{v_{h},u_{3},v_{t-2}\}$ is a barrier of $W-u_{4}u_{5}$ containing $v_{t-2}v_{h}$. So $u_{4}u_{5}$ is nonremovable in $W$. If $t=3$, $\{v_{h},u_{1},u_{s-1}\}$ is a barrier of $W-v_{1}v_{2}$ containing $u_{s-1}v_{h}$. So $v_{1}v_{2}$ are nonremovable in $W$. Every edge of $E(W)\setminus\partial(v_{h})\setminus\{u_{1}u_{2},u_{2}u_{3},u_{4}u_{5},v_{1}v_{2}\}$ satisfies the triangle condition, and then it is not removable by Proposition 3.1. If $t\geq 5$, then every edge in $E(W)\setminus\partial(v_{h})\setminus\{u_{1}u_{2},u_{2}u_{3},u_{4}u_{5}\}$ satisfies the triangle condition; so it is not removable by Proposition 3.1 again.

Now we assume that $s\geq 7$. If $t=3$, it can be checked $\{v_{h},u_{1},u_{s-1}\}$ is a barrier of $W-v_{1}v_{2}$ containing $u_{s-1}v_{h}$. So $v_{1}v_{2}$ are nonremovable in $W$. Every edge of $E(W)\setminus\partial(v_{h})\setminus\{u_{1}u_{2},u_{2}u_{3},v_{1}v_{2}\}$ satisfies the triangle condition, and then it is not removable by Proposition 3.1. If $t\geq 5$, then every edge of $E(W)\setminus\partial(v_{h})\setminus\{u_{1}u_{2},u_{2}u_{3}\}$ satisfies the triangle condition; so it is not removable by Proposition 3.1 once more.

Case 2. $r=4~{}\mbox{or}~{}s-2$.

Without loss of generality, assume that $u_{r}=u_{4}$ (the case when $u_{r}=u_{s-2}$ is the same by symmetry). Let $B_{u_{2}u_{3}}=\{v_{h},u_{4},v_{t-2}\}$. Then $W-B_{u_{2}u_{3}}-{u_{2}u_{3}}$ has exactly three components: $W[\{u_{3}\}],W[\{v_{t-1}\}]$ and $W[V(G)\setminus(\{u_{3},v_{t-1}\}\cup B_{u_{2}u_{3}})]$, which are odd. Therefore, $B_{u_{2}u_{3}}$ is a barrier of $W-{u_{2}u_{3}}$ containing both ends of the edge $v_{t-2}v_{h}$. By Lemma 2.2, $u_{2}u_{3}$ is nonremovable in $W$. If $t\geq 5$, then every edge of $E(W)\setminus\partial(v_{h})\setminus\{u_{2}u_{3}\}$ satisfies the triangle condition; so it is not removable by Proposition 3.1. Then we consider the case when $t=3$. It can be checked $\{v_{h},u_{1},u_{s-1}\}$ is a barrier of $W-v_{1}v_{2}$ containing $u_{s-1}v_{h}$. So $v_{1}v_{2}$ is nonremovable in $W$. Note that every edge of $E(W)\setminus\partial(v_{h})\setminus\{u_{2}u_{3},v_{1}v_{2}\}$ satisfies the triangle condition, and then it is not removable by Proposition 3.1 again.

Case 3. $5\leq r\leq s-3$.

If $t=3$, it can be checked $\{v_{h},u_{1},u_{s-1}\}$ is a barrier of $W-v_{1}v_{2}$ containing $u_{s-1}v_{h}$. So $v_{1}v_{2}$ is nonremovable in $W$. Note that every edge of $E(W)\setminus\partial(v_{h})\setminus\{v_{1}v_{2}\}$ satisfies the triangle condition, and then it is not removable by Proposition 3.1. If $t\geq 5$, then every edge of $E(W)\setminus\partial(v_{h})$ satisfies the triangle condition; so it is not removable by Proposition 3.1 again. Therefore, $W$ is wheel-like.

We now turn to the necessary. By Lemma 3.9, at least one of $G$ and $H$ is wheel-like such that $u$ or $v$ is its hub. Without loss of generality, assume that $G$ is wheel-like and $u=u_{h}$ (in fact, $G$ is an odd wheel). Since every edge in $\partial_{G}(u_{h})$ is a removable class in $G$, $H$ is wheel-like by Lemma 3.9. By Lemma 3.8, every multiple edge of $G$ or $H$ is incident with the hubs, that is, Statement 2) holds. We will obtain a contradiction that $W$ is not wheel-like if it does not satisfy Statements 1)-3).

Case A. $v=v_{i}$ for $i\in\{1,2,\ldots,t\}$.

Without loss of generality, assume that $v_{i}=v_{t}$ and $u_{1}v_{1}\in E(W)$. Assume that $u_{r}v_{t-1}\in E(W)$. As $W$ is a brick, it is 3-connected. Then $u_{r}\neq u_{1}$. Otherwise, $\{u_{1},v_{h}\}$ is 2-vertex cut in $W$.

If $s=t=3$, then the underlying simple graph of $W$ is isomorphic to $\overline{C_{6}}$ or the first two graphs in Figure 2. By Proposition 3.2, $\overline{C_{6}}$ is not wheel-like. Note that the first two graphs in Figure 2 are not wheel-like. So the underlying simple graph of $W$ is not wheel-like. Therefore, $W$ is not wheel-like by Proposition 3.7, a contradiction.

Next, we consider $s=3$ and $t\geq 5$. Since $(\partial(v_{h})\setminus E[v_{t},v_{h}])\cap C=\emptyset$ and every edge in $\partial(v_{h})$ is removable in $H$, every edge in $\partial(v_{h})\setminus E[v_{t},v_{h}]$ is removable in $W$ by Lemma 2.9. Without loss of generality, assume that $r=2$. Then it can be checked that $W-u_{1}u_{2}$ can be gotten from an odd wheel by inserting a vertex into two edges on the rim of the odd wheel, respectively, up to multiple edges (incident with the hub). So $W-u_{1}u_{2}$ is matching covered, and then $u_{1}u_{2}$ is a removable edge in $W$. Then $u_{1}u_{2}$ and an edge of $\partial(v_{h})\setminus E[v_{t},v_{h}]$ are two nonadjacent removable edges in $W$, a contradiction.

Now we consider $s\geq 5$ and $t\geq 3$. Suppose that $r=2~{}\mbox{or}~{}s$. Similar to last paragraph (when $s=3$ and $t\geq 5$), we can show that $u_{1}u_{r}$ is removable. Therefore, $u_{1}u_{r}$ and an edge of $\partial(v_{h})\setminus E[v_{t},v_{h}]$ are two nonadjacent removable edges in $W$, a contradiction. Therefore, $2<r<s$, that is, Statements 1) and 3) hold.

Case B. $v=v_{h}$.

If $s\geq 5$ and $t\geq 5$, then every edge in $\partial(u_{h})$ and $\partial(v_{h})$ is removable in $G$ and $H$, respectively, and hence the edges in $C$ are removable in $W$ by Lemma 2.9. As $s\geq 5$ and $t\geq 5$, there exist two nonadjacent edges in $C$, contradicting the fact that $W$ is wheel-like.

Assume that $s=3$ and $t\geq 5$ (the case when $t=3$ and $s\geq 5$ is the same by exchanging the roles of $G$ and $H$). If exactly one vertex in $\{u_{1},u_{2},u_{3}\}$, say $u_{1}$, satisfies $|E[u_{h},u_{1}]|\geq 2$, then, similar to the case when $t=3$ and $s\geq 5$ in Case A (by exchanging the roles of $G$ and $H$), we have $W$ satisfies Statements 1)-3). If there exist multiple edges between $u_{h}$ and at least two different vertices in $\{u_{1},u_{2},u_{3}\}$, then this case is similar to last paragraph (when $s\geq 5$ and $t\geq 5$ in Case B). So there exist two nonadjacent removable edges in $W$, a contradiction.

Last, we consider the case when $s=t=3$. Then the underlying simple graph of $W$ is isomorphic to $\overline{C_{6}}$, or one of graphs in Figures 2 and 3. Recall that $\overline{C_{6}}$ is not wheel-like. It can be checked that all the graphs in Figures 2 and 3 are not wheel-like. So the underlying simple graph of $W$ is not wheel-like. Therefore, $W$ is not wheel-like by Proposition 3.7, a contradiction. ∎

By Lemma 3.10, $W$ satisfies Statements 1)-3) of Lemma 3.10. Using the notations in the proof of Lemma 3.10, $W-\{u_{1},u_{r},v_{h}\}$ has three components $C_{1}=W[\{u_{2},\ldots,u_{r-1}\}]$, $C_{2}=W[\{u_{r+1},\ldots,u_{s}\}]$ and $C_{3}=W[\{v_{1},\ldots,v_{t-1}\}]$. Then, $C_{i}$ ($i\in\{1,2,3\}$), $u_{1}$,$u_{r}$,$v_{h}$ and suitable paths between them form a subdivision of $K_{3,3}$ (the vertex set $\{u_{1},u_{r},v_{h}\}$ is one of its color class). By Theorem 3.4, $W$ is nonplanar. ∎

Let $A$ and $B$ be the two color classes of $H$. Without loss of generality, assume that $\overline{x}\in A$. Suppose, to the contrary, that $\overline{x}b$ is nonremovable in $H$ ($b$ and $\overline{y}$ may be the same vertex). Then by Lemma 2.12 there exists an edge cut $\partial(Z)$ in $H$, such that $\overline{x}\in Z\cap A$, $b\in\overline{Z}\cap B$, $|Z\cap A|=|Z\cap B|$ and $\{\overline{x}b\}=E[Z\cap A,\overline{Z}\cap B]$.

As $\{\overline{x}b\}=E[Z\cap A,\overline{Z}\cap B]$, $N((\overline{Z}\cap B)\setminus\{b\})\subseteq\overline{Z}\cap A$. Let $S=\overline{Z}\cap A$ and $Q=Z\cup\{b\}$. As $G/{X}$ and $G/{Y}$ are bricks, $G/{X}$ and $G/{Y}$ are 3-connected. Therefore, both $G[\overline{X}]$ and $G[\overline{Y}]$ are connected and have odd numbers of vertices, respectively. If $\overline{y}\notin Q$, then every vertex of $(\overline{Z}\cap B)\setminus\{b,\overline{y}\}$, $G[Q]$ and $G[\overline{Y}]$ are the components of $G-S$. If $\overline{y}\in Q$, then $|(Q\cup\overline{Y})\setminus\{\overline{y}\}|$ is also odd in $G$, and every vertex of $(\overline{Z}\cap B)\setminus\{b\}$ and $Q$ are the components of $G-S$. We have $o(G-S)=|S|$ whether $\overline{y}$ is in $Q$ or not, as $|\overline{Z}\cap A|=|\overline{Z}\cap B|$. So $S$ is a barrier of $G$. Then $\partial(Q)$ is a tight cut by a simple computation. Noting $G$ is a brick, $G$ is free of nontrivial tight cuts. As $\{b,\overline{x}\}\subset Q$, we have $|\overline{Q}|=1$, that is, $|\overline{Z}\cap A|=1$. Since $|\overline{Z}\cap A|=|\overline{Z}\cap B|$, we have $|\overline{Z}\cap B|=1$, that is, $\{b\}=\overline{Z}\cap B$.

Assume that $\overline{y}\neq b$. Note that $\overline{x}b$ is nonremovable in $H$. Recalling $|\overline{Z}\cap A|=1$, we have $|N_{G}(b)|=2$, contradicting the assumption that $G$ is a brick. Now we consider the case when $\overline{y}=b$. It can be checked that $Z\cap B$ is also a barrier of $G$ by symmetry. So $|Z\cap B|=1$. Let $\{b^{\prime}\}=Z\cap B$. Similar to the case when $\overline{y}\neq b$, we have $|N_{H}(b^{\prime})|=2$, and so $|V(H)|=4$. In $G$, assume that $e$ is the corresponding edge of the edge $\overline{x}\,\overline{y}$. Then the end of $e$ in $\overline{X}$ and $b^{\prime}$ is a 2-vertex cut of $G$, contracting the fact that $G$ is a brick. ∎