Characterizing the forbidden pairs for graphs to be super-edge-connected¹¹1The research is supported by National Natural Science Foundation of China (12261086).

Let $G$ be a connected $P_{3}$-free graph. Then $G$ is a complete graph, and thus $G$ is super-edge-connected.

Let $S$ be a connected graph such that every connected $S$-free graph is super-edge-connected. Since $G_{i}$ in Figure 2 is not super-edge-connected for $i\in\{1,\cdots,6\}$, then we know $G_{i}$ is not $S$-free and contains $S$ as an induced subgraph. We observe that the common induced subgraph of the graphs in ${G}_{3}$ and ${G}_{6}$ is a path and the longest induced path of the graphs in ${G}_{1}$ is $P_{3}$, then $S$ must be an induced subgraph of $P_{3}$. The proof is complete. ∎

We first prove the necessity. Assume $\mathcal{H}=\{R,S\}$ is a set of connected graphs such that both $R$ and $S$ are not an induced subgraph of $P_{3}$, and every connected $\mathcal{H}$-free graph is super-edge-connected. Let $G$ be a connected $\mathcal{H}$-free graph. Then $G\ncong P_{2}\Box P_{3}$ and $G\ncong P_{n},C_{n}\ (n\geq 4)$. Since $G_{i}$ in Figure 2 is not super-edge-connected, we obtain that $G_{i}$ is not $\mathcal{H}$-free and contains at least one of $R$ and $S$ as an induced subgraph for $i\in\{1,\cdots,6\}$.

By contradiction, assume that neither $R$ nor $S$ is an induced subgraph of $H_{0}$. Without loss of generality, assume that $R$ is an induced subgraph of $G_{1}$. Since $R$ is not an induced subgraph of $H_{0}$. We observe that $R$ must contain a $K_{t}$ as an induced subgraph with $t\geq 4$ . For $i\in\{2,3,4,5,6\}$, $G_{i}$ is $R$-free. Then they contain $S$ as an induced subgraph. We observe that the common induced subgraph of the graphs in ${G}_{3}$ and ${G}_{6}$ is a path and the longest induced path of $G_{2}$ is $P_{3}$, then $S$ should be an induced subgraph of $P_{3}$, a contradiction. Claim 1 is thus proved.

By Claim 1, assume, without loss of generality, that $R$ is an induced subgraph of $H_{0}$. We distinguish two cases as follows.

For $i\in\{3,4,5,6\}$, $G_{i}$ is $R$-free. Then they contain $S$ as an induced subgraph. We observe that the common induced subgraph of the graphs in ${G}_{3}$ and ${G}_{6}$ is a path and the longest induced path of $G_{4}$ is $P_{4}$, then $S$ should be an induced subgraph of $P_{4}$. So $\mathcal{H}=\{R,S\}\preceq\left\{H_{0},P_{4}\right\}$ .

Since $R$ is not an induced subgraph of $P_{3}$, we get that $R$ must contain a triangle. For $i\in\{3,4,5\}$, $G_{i}$ is $R$-free. Then they contain $S$ as an induced subgraph. We observe that the maximal common induced subgraph of the graphs in ${G}_{4}$ and ${G}_{5}$ is a $T_{1,1,2}$, then $S$ should be an induced subgraph of $T_{1,1,2}$. So $\mathcal{H}=\{R,S\}\preceq\left\{Z_{1},T_{1,1,2}\right\}$.

Now we are going to prove the sufficiency. By contradiction, assume $G$ is a connected $\mathcal{H}$-free graph, but $G$ is not super-edge-connected, where ($i$) $G\ncong C_{4}$ and $\mathcal{H}\preceq\left\{H_{0},P_{4}\right\}$, or ($ii$) $G\ncong P_{2}\Box P_{3}$, $G\ncong P_{n},C_{n}\ (n\geq 4)$ and $\mathcal{H}\preceq\left\{Z_{1},T_{1,1,2}\right\}$. Since $G$ is not super-edge-connected, there is a minimum edge-cut $F$ such that $G-F$ has no isolated vertices. Clearly, $G-F$ has only two components, say $X$ and $Y$. Let $X_{1}=V\left(X\right)\cap V(F)$ and $Y_{1}=V\left(Y\right)\cap V(F)$. Denote $X_{2}=V(X)\setminus X_{1}$ and $Y_{2}=V(Y)\setminus Y_{1}$. Note that $|V(X)|,|V(Y)|\geq 2$ and $|X_{1}|,|Y_{1}|\leq|F|=\kappa^{\prime}(G)\leq\delta(G)$.

By $\kappa^{\prime}(G)\leq\delta(G)$, $|X_{1}|\leq|V(X)|$ and $|X_{1}|\leq|F|=\kappa^{\prime}(G)$, we have the following inequalities.

If $X_{2}=\emptyset$, then all the inequalities above will be equalities. Thus we obtain that $X$ is a complete graph with order $\delta(G)$, each vertex in $X$ is incident with exactly one edge in $F$, and $|X|=|X_{1}|=|F|=\kappa^{\prime}(G)=\delta(G)$.

By Claim 2, we consider two cases in the following.

Assume, without loss of generality, that $X_{2}=\emptyset$. Then, by Claim 2, $X$ is a complete graph with order $\delta(G)$, each vertex in $X$ is incident with exactly one edge in $F$, and $|X|=|X_{1}|=|F|=\kappa^{\prime}(G)=\delta(G)$. In addition, by $|V(X)|\geq 2$, we have $\delta(G)\geq 2$.

Suppose $Y_{2}=\emptyset$. Then $G$ is the union of two complete graphs on $\delta(G)$ vertices, together with the perfect matching $F$, that is, $G$ is isomorphic to the Cartesian product graph $K_{\delta(G)}\Box K_{2}$. If $\delta(G)\geq 3$, then there exists an induced $P_{4}$, a contradiction. Otherwise, if $\delta(G)=2$, then $G\cong C_{4}$, also a contradiction.

Suppose $Y_{2}\neq\emptyset$. Then by $Y$ is connected, there is a path $x_{1}x_{2}y_{1}y_{2}$, where $x_{1},$ $x_{2}\in V\left(X\right)$, $y_{1}\in Y_{1}$ and $y_{2}\in Y_{2}$. If $x_{1}y_{1}\notin E(G)$, then $x_{1}x_{2}y_{1}y_{2}$ is an induced path of $G$, a contradiction. Thus we assume $x_{1}y_{1}\in E(G)$. Since $d_{G}\left(y_{2}\right)\geq\delta(G)\geq 2$, $y_{2}$ has a neighbor $y_{3}$ different from $y_{1}$. If $y_{1}y_{3}\notin E(G)$, then $G\left[\{x_{1},y_{1},y_{2},y_{3}\}\right]\cong P_{4}$, a contradiction. Otherwise, if $y_{1}y_{3}\in E(G)$, then $G\left[\{x_{1},x_{2},y_{1},y_{2},y_{3}\}\right]\cong H_{0}$, also a contradiction.

Clearly, $|X_{1}|=|F|\geq|Y_{1}|$. If $|X_{1}|>|Y_{1}|$, then there exists a vertex in $Y_{1}$, say $y_{1}$, has at least two neighbors in $X_{1}$, say $x_{1}$ and $x_{2}$. By $Y$ is connected, $y_{1}$ has at least one neighbor, say $y_{2}$, in $V(Y)$. Since $X$ is a complete graph and each vertex in $X$ is incident with exactly one edge in $F$, we know that $G\left[\{x_{1},x_{2},y_{1},y_{2}\}\right]\cong Z_{1}$, a contradiction. Thus, we assume $|X_{1}|=|Y_{1}|$ in the following.

Since $|X_{1}|=|Y_{1}|=|F|$, we know that each vertex in $X_{1}$ has only one neighbor in $Y_{1}$ and each vertex in $Y_{1}$ has only one neighbor in $X_{1}$. If $|X_{1}|=|Y_{1}|\geq 3$, then the induced subgraph by any three vertex in $X_{1}$, say $x_{1},x_{2},x_{3}$, together with the neighbor of $x_{1}$ in $Y_{1}$ is isomorphic to $Z_{1}$, a contradiction. It is remaining to consider $|X_{1}|=|Y_{1}|=2$. Assume $X_{1}=\{x_{1},x_{2}\}$, $Y_{1}=\{y_{1},y_{2}\}$ and $F=\{x_{1}y_{1},x_{2}y_{2}\}$.

Suppose $y_{1}y_{2}\in E(G)$. Since $G$ is not a cycle, $y_{1}$ or $y_{2}$, say $y_{2}$ has a neighbor $y_{3}$ in $Y_{2}$. If $y_{1}y_{3}\in E(G)$, then $G[\{x_{1},y_{1},y_{2},y_{3}\}]\cong Z_{1}$, a contradiction. So $y_{1}y_{3}\notin E(G)$. Since $\delta(G)\geq 2$, $y_{3}$ has another neighbor $y_{4}$ different from $y_{2}$. If $y_{2}y_{4}\in E(G)$, then $G[\{x_{2},y_{2},y_{3},y_{4}\}]\cong Z_{1}$, a contradiction. So $y_{2}y_{4}\notin E(G)$. If $y_{1}y_{4}\notin E(G)$, then $G[\{x_{2},y_{1},y_{2},y_{3},y_{4}\}]\cong T_{1,1,2}$, a contradiction. So $y_{1}y_{4}\in E(G)$. If $V(G)=\{x_{1},x_{2},y_{1},y_{2},y_{3},y_{4}\}$, then $G\cong P_{2}\Box P_{3}$, a contradiction. Thus, by $G$ is connected, there is a vertex $y_{5}$ adjacent to $y_{1}$, $y_{2}$, $y_{3}$, or $y_{4}$. We only consider the case $y_{1}y_{5}\in E(G)$ here, others can be analysed similarly. If $y_{4}y_{5}\in E(G)$, then $G[\{x_{1},y_{1},y_{4},y_{5}\}]\cong Z_{1}$, a contradiction. Otherwise, if $y_{4}y_{5}\notin E(G)$, then $G[\{x_{1},x_{2},y_{1},y_{4},y_{5}\}]\cong T_{1,1,2}$, also a contradiction.

Suppose $y_{1}y_{2}\notin E(G)$. Since $Y$ is connected, there is a path $z_{1}z_{2}\cdots z_{t}$ connecting $y_{1}$ and $y_{2}$ in $Y$, where $z_{1}=y_{1}$, $z_{t}=y_{2}$ and $t\geq 3$. Furthermore, we choose this path $z_{1}z_{2}\cdots z_{t}$ to be a shortest path connecting $y_{1}$ and $y_{2}$ in $Y$. Since $G$ is not a cycle, there is a vertex $w\in Y\setminus\{z_{1},z_{2},\cdots,z_{t}\}$ such that $w$ is adjacent to some vertex in $\{z_{1},z_{2},\cdots,z_{t}\}$. Assume $wz_{i}\in E(G)$ and $wz_{j}\notin E(G)$ for $j<i$. Let $z_{0}=x_{1}$ and $z_{-1}=x_{2}$ . If $wz_{i+1}\in E(G)$, then $G[\{w,z_{i-1},z_{i},z_{i+1}\}]\cong Z_{1}$, a contradiction. Otherwise, if $wz_{i+1}\notin E(G)$, then $G[\{w,z_{i-2},z_{i-1},z_{i},z_{i+1}\}]\cong T_{1,1,2}$, also a contradiction.

Since the distance between a vertex in $X_{2}$ and a vertex in $Y_{2}$ is at least three, $G$ must contains $P_{4}$ as an induced path. So the case $G\ncong C_{4}$ and $\mathcal{H}\preceq\left\{H_{0},P_{4}\right\}$ can not happen. Thus we only consider the case that $G\ncong P_{2}\Box P_{3}$, $G\ncong P_{n},C_{n}\ (n\geq 4)$ and $\mathcal{H}\preceq\left\{Z_{1},T_{1,1,2}\right\}$ in the following. We distinguish two subcases as follows.

Assume $\left|N_{G}\left(x_{2}\right)\cap X_{2}\right|\geq 2$. Then there exist two vertices $x_{2}^{\prime}$, $x_{2}^{\prime\prime}\in X_{2}$ such that $x_{2}x_{2}^{\prime}$, $x_{2}x_{2}^{\prime\prime}\in E(G)$. If $x_{2}^{\prime}x_{2}^{\prime\prime}$, $x_{2}^{\prime}x_{1}$, $x_{2}^{\prime\prime}x_{1}\notin E(G)$, then $G\left[\{x_{1},x_{2},x_{2}^{\prime},x_{2}^{\prime\prime},y_{1}\}\right]\cong T_{1,1,2}$, a contradiction. If at least one of these three edges exists, then $G$ contains an induced subgraph isomorphic to $Z_{1}$, also a contradiction. For example, if $x_{2}^{\prime}x_{1}\in E(G)$, then $G\left[\{x_{1},x_{2},x_{2}^{\prime},y_{1}\}\right]\cong Z_{1}$. The case $\left|N_{G}\left(y_{2}\right)\cap Y_{2}\right|\geq 2$ can be proved similarly.

Let $x_{3}\in N_{G}\left(x_{2}\right)\cap X_{2}$ and $y_{3}\in N_{G}\left(y_{2}\right)\cap Y_{2}$. If $x_{1}x_{3}$ or $y_{1}y_{3}\in E(G)$, then $G$ contains an induced $Z_{1}$. For example, if $x_{1}x_{3}\in E(G)$, then $G\left[\{x_{1},x_{2},x_{3},y_{1}\}\right]\cong Z_{1}$. Hence, $x_{1}x_{3}\notin E(G)$ and $y_{1}y_{3}\notin E(G)$.

Suppose $d_{G}\left(x_{1}\right)\geq 3$. Let $w\in N_{G}\left(x_{1}\right)\setminus\{x_{2},y_{1}\}$. If $w\in X_{2}$, then $G\left[\{x_{1},x_{2},y_{1},w\}\right]\cong Z_{1}$ when $wx_{2}\in E(G)$, and $G\left[\{x_{1},x_{2},y_{1},y_{2},w\}\right]\cong T_{1,1,2}$ when $wx_{2}\notin E(G)$, contradicts to the assumption. By a similar argument, we can obtain contradictions for $w\in X_{1}$ and $w\in Y_{1}$. Thus $d_{G}\left(x_{1}\right)=2$. Similarly, $d_{G}\left(y_{1}\right)=2$. Since $G\ncong P_{n},C_{n}\ (n\geq 4)$, then there exists a vertex in $G$ such that its degree is at least three. Assume, without loss of generality, $V(X)$ has a vertex with degree at least three. Choose such a vertex $z_{1}$ such that its distance to $x_{1}$ is minimum in the component $X$. Let $z_{1}z_{2}\cdots z_{r}(z_{r}=x_{1})$ be the shortest path between $z_{1}$ and $x_{1}$. Then $d_{G}(z_{i})=2$ for $2\leq i\leq r$. Let $z_{1}^{\prime}$, $z_{1}^{\prime\prime}\in N_{G}\left(z_{1}\right)\setminus\{z_{2}\}$. Then we can find an induced $Z_{1}$ or $T_{1,1,2}$ according to $z_{1}^{\prime}z_{1}^{\prime\prime}$ is an edge in $G$ or not.

Without loss of generality, assume that$\left|N_{G}\left(x_{2}\right)\cap X_{2}\right|=0$ and $\left|N_{G}\left(y_{2}\right)\cap Y_{2}\right|\leq 1$. Since $N_{G}\left(x_{2}\right)\subseteq X_{1}$ and $|X_{1}|\leq|F|=\kappa^{\prime}(G)\leq\delta(G)$, we have $d_{G}\left(x_{2}\right)=|X_{1}|=|F|=\kappa^{\prime}(G)=\delta(G)$. Note that $|Y_{1}|\leq|F|=|X_{1}|$.

Suppose $|X_{1}|>|Y_{1}|$. Then $\left|N_{G}\left(y_{2}\right)\cap Y_{2}\right|\geq 1$. Let $y_{3}\in N_{G}\left(y_{2}\right)\cap Y_{2}$. If $y_{1}y_{3}\in E(G)$, then $G\left[\{x_{1},y_{1},y_{2},y_{3}\}\right]\cong Z_{1}$, a contradiction. So assume $y_{1}y_{3}\notin E(G)$. If $\left|N_{G}\left(y_{1}\right)\cap X_{1}\right|\geq 2$, then we can find an induced $Z_{1}$ or $T_{1,1,2}$ according to any two vertices in $N_{G}\left(y_{1}\right)\cap X_{1}$ are adjacent or not. So assume $y_{1}$ has exactly one neighbor $x_{1}$ in $X_{1}$. By $|X_{1}|>|Y_{1}|\geq 1$, we get $|X_{1}|\geq 2$. Since $y_{1}$ has exactly one neighbor $x_{1}$ in $X_{1}$, we have $|Y_{1}|\geq 2$, thus $|X_{1}|\geq 3$. If there is at least one edge in $G[X_{1}]$, then we can find an induced $Z_{1}$, a contradiction. Otherwise, we can find an induced $T_{1,1,2}$ rooted at $x_{2}$, also a contradiction.

Suppose $|X_{1}|=|Y_{1}|$. Then by $|X_{1}|=|F|=\kappa^{\prime}(G)=\delta(G)$, we know that each vertex in $X_{1}$ has only one neighbor in $Y_{1}$ and each vertex in $Y_{1}$ has only one neighbor in $X_{1}$. That is, $F$ is perfect matching in $G[X_{1}\cup Y_{1}]$ connecting $X_{1}$ and $Y_{1}$.

Suppose $|X_{1}|=|Y_{1}|\geq 3$. Then according to there is an edge in $G[X_{1}]$ or not, we can find an induced $Z_{1}$ or $T_{1,1,2}$, a contradiction. Suppose $|X_{1}|=|Y_{1}|=2$. By a similar argument as Subcase 1.2, we can obtain a contradiction. Suppose $|X_{1}|=|Y_{1}|=1$. Since $G\ncong P_{n},C_{n}\ (n\geq 4)$, then there exists a vertex in $G$ such that its degree is at least three. By a similar argument as Subcase 2.1.2, we can also obtain a contradiction.

Let $X_{1}^{1}=\left\{x\in X_{1}:N_{G}(x)\cap X_{2}\neq\emptyset\right\}$ and $Y_{1}^{1}=\left\{y\in Y_{1}:N_{G}(y)\cap Y_{2}\neq\emptyset\right\}$. Denote $X_{1}^{2}=X_{1}-X_{1}^{1}$ and $Y_{1}^{2}=Y_{1}-Y_{1}^{1}$. Since $G$ contains no path $x_{2}x_{1}y_{1}y_{2}$ satisfying $x_{i}\in X_{i}$ and $y_{i}\in Y_{i}$ for $i=1,2$, we obtain that $X_{1}^{2}\neq\emptyset$, $Y_{1}^{2}\neq\emptyset$ and there is no edge connecting $X_{1}^{1}$ and $Y_{1}^{1}$. Since $X$ is connected, there is an edge connecting $X_{1}^{1}$ and $X_{1}^{2}$, say $x_{1}^{1}x_{1}^{2}\in E(G)$, where $x_{1}^{1}\in X_{1}^{1}$ and $x_{1}^{2}\in X_{1}^{2}$. Let $x_{2}$ be a neighbor of $x_{1}^{1}$ in $X_{2}$ and $y_{1}$ be a neighbor of $x_{1}^{1}$ in $Y_{1}$. Since $|X_{1}^{1}|<\delta(G)$, $x_{2}$ has a neighbor, say $x_{3}$, in $X_{2}$. If $x_{1}^{1}x_{3}\in E(G)$, then $G\left[\{x_{1}^{1},x_{2},x_{3},y_{1}\}\right]\cong Z_{1}$, a contradiction. So assume $x_{1}^{1}x_{3}\notin E(G)$. If $x_{1}^{2}y_{1}\in E(G)$, then $G\left[\{x_{1}^{1},x_{1}^{2},x_{2},y_{1}\}\right]\cong Z_{1}$, a contradiction. So assume $x_{1}^{2}y_{1}\notin E(G)$. Thus $G\left[\{x_{1}^{1},x_{1}^{2},x_{2},x_{3},y_{1}\}\right]\cong T_{1,1,2}$, also a contradiction.

Since all cases lead to contradictions, the proof is thus complete. ∎