Proof. Observation 3 handles the $c = 0$ case. Now assume that $c \geqslant 1$ . Lemma 6 below says that (a) implies (b). Since every $\ell$ -partition is an $\ell$ -covering, (b) implies (c), and (d) implies (e). Lemma 5 below says that (c) implies (d). Finally, Lemma 8 below says that (e) implies (a).

Proof. Fix $t \in{\mathbb{N}}$ . Let $B_1,\dots, B_c \in \beta [t]$ . Let $X$ be a component of $G-(B_1\cup \dots \cup B_c)$ . For each $i\in \{1,\dots,c\}$ , let $\mathcal{B}_i$ be a set of at most $t$ elements of $\beta$ whose union is $B_i$ . Let $\mathcal{F} \,:\!=\, \mathcal{B}_1 \times \dots \times \mathcal{B}_c$ , and for each $y = (A_1, \dots, A_c) \in \mathcal{F}$ , define $Q_y$ as follows. Let $X_y$ the component of $G - (A_1 \cup \dots \cup A_c)$ containing $X$ . Since $\beta$ is $(c,d)$ -disjointed, there exists $Q_y \subseteq V(X_y)$ of size at most $d$ such that for every component $Y$ of $X_y-Q_y$ there is some $i \in \{1, \dots, c\}$ such that $V(Y) \cap N_G(A_i \setminus (A_1 \cup \dots \cup A_{i-1})) = \varnothing$ . Now let $Q \,:\!=\, \bigcup _{y \in \mathcal{F}} Q_y$ , and note that $|Q|\leqslant d|\mathcal{F}| \leqslant dt^c$ .

Suppose for contradiction that for some component $Y$ of $X - Q$ and each $i \in \{1,\dots, c\}$ , there is a vertex $b_i \in N_G(Y) \cap B^{\prime}_i$ , where $B^{\prime}_i \,:\!=\, B_i \setminus (B_1 \cup \dots \cup B_{i-1})$ . Let $y = (A_1, \dots, A_c) \in \mathcal{F}$ be such that $(b_1, \dots, b_c) \in A_1 \times \dots \times A_c$ , and consider that component $Y^{\prime}$ of $X_y - Q_y$ containing $Y$ . By the definition of $Q_y$ , there is some $i \in \{1,\dots, c\}$ such that $Y^{\prime}$ contains no neighbour of a vertex in $A_i \setminus (A_1 \cup \dots \cup A_{i-1})$ . In particular, all neighbours of vertices of $Y$ are either vertices of $Y^{\prime}$ or neighbours of vertices of $Y^{\prime}$ , so $b_i$ is not a neighbour of any vertex of $Y$ , a contradiction.

Proof. By assumption, $G$ has an $H$ -partition $\beta = (V_h \,:\, h \in V(H))$ where $H$ is a graph of treewidth at most $c$ and $|V_h| \leqslant \ell$ for all $h$ . We first show that the singleton partition of $H$ is $(c,c)$ -disjointed. Let $v_1, \dotsc, v_c \in V(H)$ and let $X$ be a component of $H-\{v_1,\dots, v_c\}$ . Let $(W_x \,:\, x \in V(T))$ be a tree-decomposition of $H$ where $|W_x| \leqslant c + 1$ for all $x \in V(T)$ . We may assume that $W_x \neq W_y$ whenever $x \neq y$ . For each $i \in \{1,\dots,c\}$ , let $T_i$ be the subtree of $T$ induced by $\{x \in V(T) \,:\, v_i \in W_x\}$ .

First suppose that $V(T_i) \cap V(T_j) = \varnothing$ for some $i, j \in \{1,\dots,c\}$ . Let $z \in V(T_i)$ be the closest node (in $T$ ) to $T_j$ . Let $Q \,:\!=\, W_z \cap X$ . Note that $Q \subseteq W_z \setminus \{v_i\}$ so $|Q| \leqslant c$ . Any path from $v_i$ to $v_j$ in $H$ passes through $W_z$ , so each component of $X - Q$ is disjoint from $N_H(v_i)$ or $N_H(v_j)$ .

Now assume that $V(T_i) \cap V(T_j) \neq \varnothing$ for all $i, j\in \{1,\dots, c\}$ . Let $T_X$ be the subgraph of $T$ induced by $\{x \in V(T) \,:\, V(X) \cap W_x \neq \varnothing \}$ . Since $X$ is connected, $T_X$ is a subtree of $T$ . Suppose that $V(T_i) \cap V(T_X) = \varnothing$ for some $i$ . Since $N_H(v_i) \subseteq \bigcup (W_x \,:\, x \in V(T_i))$ , it follows that $N_H(v_i) \cap V(X) = \varnothing$ and so we may take $Q \,:\!=\, \varnothing$ in this case. Now assume that $V(T_i) \cap V(T_X) \neq \varnothing$ for all $i \in \{1,\dots,c\}$ . By the Helly property, $\tilde{T} \,:\!=\, T_1 \cap \dotsb \cap T_c \cap T_X$ is a non-empty subtree of $T$ . For $x \in V(\tilde{T})$ , we have $|W_x| \leqslant c + 1$ and so $W_x = \{v_1, \dotsc, v_c, u\}$ for some $u \in V(X)$ . First suppose that $|V(\tilde{T})| \geqslant 2$ . Then there are adjacent $x, y \in V(\tilde{T})$ with $W_x = \{v_1, \dotsc, v_c, u\}$ and $W_y = \{v_1, \dotsc, v_c, v\}$ for $u, v \in V(X)$ . Since $W_x \neq W_y$ , we have $u \neq v$ and thus there is no $(u,v)$ -path in $H - \{v_1, \dotsc, v_c\}$ , contradicting the connectedness of $X$ . Hence $\tilde{T}$ consists of a single vertex $z$ ; thus $W_z = \{v_1, \dotsc, v_c, u\}$ for some $u \in V(X)$ . Let $Q \,:\!=\, \{u\}$ and consider a component $Y$ of $X - Q$ . Let $T_{Y}$ be the subtree of $T$ induced by $\{y\in V(T) \,:\, V(Y) \cap W_y \neq \varnothing \}$ . Since $T_{Y}$ is connected and does not contain $z$ , it is disjoint from some $T_i$ . As above, $N_H(v_i) \cap V(Y) = \varnothing$ , as required.

We have shown that the singleton partition of $H$ is $(c,c)$ -disjointed. Now focus on $G$ . By assumption, $\beta$ is an $\ell$ -partition of $G$ . Let $V_{v_1}, \dotsc, V_{v_c}$ be parts in $\beta$ , and let $X$ be a component of $G - (V_{v_1} \cup \dotsb \cup V_{v_c})$ . Then $X \subseteq \bigcup \{V_h \,:\, h \in X^{\prime}\}$ where $X^{\prime}$ is a component of $H - \{v_1, \dotsc, v_c\}$ . Since $H$ is $(c,c)$ -disjointed, there exists $Q^{\prime} \subseteq V(X^{\prime})$ of size at most $c$ such that each component $X^{\prime} - Q^{\prime}$ is disjoint from some $N_H(v_i)$ . Let $Q \,:\!=\, \bigcup \{V_h \,:\, h \in Q^{\prime}\}$ , which has size at most $c\ell$ . Each component of $X - Q$ is disjoint from some $N_G(V_{v_i})$ .

Proof. Fix $t \in{\mathbb{N}}$ . Let $\tilde{B}_1, \dots, \tilde{B}_c \in \tilde{\beta }[t]$ and let $\tilde{X}$ be a component of $\tilde{G}-\big(\tilde{B}_1\cup \dots \cup \tilde{B}_c\big)$ . For each $i \in \{1,\dots, c\}$ , there is a subset $S_i \subseteq \beta$ of size at most $t$ such that $\tilde{B}_i = \bigcup _{B \in S_i} (B\cap V(\tilde{G}))$ . Let $(B_1,\dots,B_c)\,:\!=\, (\bigcup S_1,\dots,\bigcup S_c)$ , and let $\beta^{\prime\prime}$ be the $t\ell$ -covering of $G$ given by $\beta \cup \{B_1,\dots,B_c\}$ . Let $X$ be the component of ${G-(B_1\cup \dots \cup B_c})$ which contains $\tilde{X}$ , and for each $i \in \{1,\dots, c\}$ let $B^{\prime}_i\,:\!=\, B_i\setminus (B_1\cup \dots \cup B_{i-1})$ . Since $\beta$ is $(c,f)$ -disjointed, there is a subset $Q$ of $V(X)$ of size at most $f(t)$ such that each component of $X-Q$ disjoint from $N_G\big(B^{\prime}_i\big)$ for some $i \in \{1,\dots,c\}$ . Let $\tilde{Q}\,:\!=\, Q\cap V(\tilde{X})$ , and note that $|\tilde{Q}|\leqslant |Q|\leqslant f(t)$ . Each component of $\tilde{X}-\tilde{Q}$ is contained in a component of $X - Q$ , and hence is disjoint from $N_{\tilde{G}}\big(\tilde{B}_i\setminus \big(\tilde{B}_1\cup \dots \cup \tilde{B}_{i-1}\big)\big)\subseteq N_G\big(B^{\prime}_i\big)$ for some $i \in \{1,\dots, c\}$ . Hence $\tilde{\beta }$ is $(c,f)$ -disjointed.

Proof. We proceed by induction on $|V(G)|$ . Let $S \,:\!=\, S_1 \cup \dotsb \cup S_{c - 1}$ .

Case 0. $V(G) = R \cup S$ : Let $H$ be the complete graph on vertices $x_1, \dotsc, x_{c - 1}, y$ . Let $V_{x_i} \,:\!=\, S_i \setminus (S_1 \cup \dotsb \cup S_{i - 1})$ for each $i$ and let $V_y \,:\!=\, R$ . Then $(V_x \,:\, x \in V(H))$ is a $c$ -tree-partition of $G$ with width at most $W$ and $|V_y| = |R| \leqslant 2(|R| - 2k) \leqslant 2 \ell (|R| - 2k)$ . From now on assume that $G - (R \cup S)$ is non-empty.

Case 1. $4 k \leqslant |R|\leqslant 12 k$ : Since $\beta$ is an $\ell$ -covering, and $|R| \leqslant 12k$ , we can pick $S_c \in \beta [12k]$ such that $R \subseteq S_c$ and $|S_c| \leqslant \ell |R| \leqslant 2 \ell (|R| - 2k)$ .

Let $G_1, \dotsc, G_a$ be the connected components of $G - (S \cup S_c)$ . For each $i \in \{1,\dots,c\}$ , let $S^{\prime}_i \,:\!=\, S_i \setminus (S_1 \cup \dotsb \cup S_{i - 1})$ . To complete this case, we first prove the following.

Figure 2.

The graphs $F_1$ and $F_2$ in the case $c=2$ .

Proof. If $|V(G_j)| \lt 4 k$ , then take $H_j$ to be the complete graph on vertices $x_1, \dotsc, x_{c}, z$ with the partition $V^j_{x_i} \,:\!=\, S^{\prime}_i$ for $i \in \{1,\dots,c\}$ and $V^j_{z} \,:\!=\, V(G_j)$ . Then this gives us the desired $c$ -tree-partition of $G[V(G_j) \cup S \cup S_c]$ . So assume $|V(G_j)| \geqslant 4k$ . Note that $\beta [12k]$ is a $(c,f(12 k))$ -disjointed $12k\ell$ -covering containing $S_1, \dots, S_c$ , so there is a subset $Q^{\prime}_j \subseteq V(G_j)$ of size at most $f(12 k)$ and there is a partition $\{A_1, \dotsc, A_c\}$ of $V\big(G_j - Q^{\prime}_j\big)$ such that each $A_i$ is a union of vertex sets of components of $G_j - Q^{\prime}_j$ that do not intersect $N_{G}\big(S^{\prime}_i\big)$ . Let $Q_j$ be a set such that $Q^{\prime}_j \subseteq Q_j \subseteq V(G_j)$ and $4k \leqslant |Q_j| \leqslant f(12 k)$ . As illustrated in Figure 2, consider the subgraph

\begin{align*} F_i \,:\!=\, &\ G\left[A_i \cup Q_j \cup S_1 \cup \dotsb \cup S_{i - 1} \cup \left(S_{i + 1} \setminus S^{\prime}_i\right) \cup \dotsb \cup \left(S_c \setminus S^{\prime}_i\right)\right] \\ = &\ G\left[A_{i} \cup Q_{j} \cup S^{\prime}_{1} \cup \dotsb \cup S^{\prime}_{i - 1} \cup S^{\prime}_{i + 1} \cup \dotsb \cup S^{\prime}_{c}\right]. \end{align*}

By Lemma 7, the restriction of $\beta$ to $V(F_i)$ is a $(c,f)$ -disjointed $\ell$ -covering of $F_i$ . Apply induction to $F_i$ with the sets $S_1,\dotsc,S_{i - 1},S_{i + 1} \setminus S^{\prime}_i,\dotsc,S_c\setminus S^{\prime}_i$ in place of the sets $S_1,\dotsc,S_{c-1}$ and the set $Q_{j}$ in the place of $R$ . For each $i \in \{1,\dotsc,c\}$ , this gives a graph $L_i$ of treewidth at most $c$ containing a $c$ -clique $\{x_1, \dotsc, x_{i-1}, x_{i+1},\dotsc, x_c, y\}$ such that $F_i$ has an $L_i$ -partition $\big( V^{j,i}_x \,:\, x \in V(L_i) \big)$ with $V^{j,i}_{x_m} = S^{\prime}_m$ for all $m \in \{1,\dots, i-1,i+1,\dots, c\}$ and $Q_j \setminus \big((S\cup S_c) \setminus S^{\prime}_i)\big) \subseteq V_y^{j,i}$ where

\begin{equation*} |V^{j,i}_y| \leqslant 2 \ell (|Q_j| - 2k) \leqslant 2 \ell f(12 k). \end{equation*}

Let $L_i^+$ be $L_i$ together with a vertex $x_i$ adjacent to the clique $\{x_1, \dotsc, x_{i - 1}, x_{i + 1}, \dotsc, x_c, y\}$ . So $\textrm{tw}(L^+_i)\leqslant c$ . Set $V^{j,i}_{x_i} \,:\!=\, S^{\prime}_i$ . Then $L^+_i$ contains the $(c + 1)$ -clique $K^+ \,:\!=\, \{x_1, \dotsc, x_c, y\}$ and $\big(V^{j,i}_h \,:\, h \in V(L^+_i)\big)$ is an $L^+_i$ -partition of $G[A_i \cup Q_j \cup S \cup S_c]$ . Now we may assume that $V(L^+_1), \dotsc, V(L^+_c)$ pairwise intersect in exactly the clique $K^+$ . Let $H_j \,:\!=\, L_1^+\cup \dots \cup L_c^+$ . Since each $L_i^+$ has treewidth at most $c$ , so does $H_j$ . For $x \notin K^+$ , set $V^j_x \,:\!=\, V^{j,i}_x$ for the unique $i$ for which $x \in V(L_i^+)$ , for $i \in \{1, \dotsc, c \}$ set $V^j_{x_i} \,:\!=\, S^{\prime}_i$ , and set $V^j_y \,:\!=\, \bigcup _{i \in \{1, \dotsc, c\}} V^{j,i}_y$ . Since $|V^{j}_y| \leqslant 2c \ell f(12 k)$ , the partition $\big( V^j_x \,:\, x \in V(H_j) \big)$ has width at most $W$ . Setting $K \,:\!=\, \{ x_1 \dotsc, x_c \}$ , the claim follows.

We may assume that $V(H_1), \dotsc, V(H_a)$ pairwise intersect in exactly the clique $K$ . Let $H \,:\!=\,{H_1\cup \dotsc \cup H_a}$ . For $x \in V(H)$ , setting $V_x \,:\!=\, V^j_x$ if $x \in V(H_j)$ is well defined, and yields an $H$ -partition $(V_h \,:\, h \in V(H))$ of $G$ . Since each $H_i$ has treewidth at most $c$ , so does $H$ . Let $y \,:\!=\, x_c$ . Then $R \setminus S \subseteq S^{\prime}_c = V_{y}$ , and, as noted at the start of this case, $|V_{y}| = |S^{\prime}_c| \leqslant |S_c| \leqslant 2 \ell (|R| - 2k)$ . Hence, the width of this partition is at most $W$ , as required.

Case 2. $12 k \lt |R| \leqslant f(12 k)$ : Since $\textrm{tw}(G) \lt k$ , by the separator lemma of Robertson and Seymour [Reference Robertson and Seymour65, (2.6)], there is a partition $(A, B, C)$ of $V(G)$ with no edges between $A$ and $B$ , where $|C| \leqslant k$ and $|A \cap R|, |B \cap R| \leqslant \tfrac{2}{3} |R \setminus C|$ . Let $G_1 \,:\!=\, G[A \cup C]$ and $G_2 \,:\!=\, G[B \cup C]$ . Let $R_1 \,:\!=\, (R \cap A) \cup C$ and $R_2 \,:\!=\, (R \cap B) \cup C$ . Since $|R| \geqslant 12 k$ ,

\begin{align*} |R_1| & = |A \cap R| + |C| \leqslant \tfrac{2}{3} |R| + k \lt |R| \quad \textrm{and}\\ |R_1| & \geqslant |R| - |B \cap R| \geqslant |R| - \tfrac{2}{3} |R \setminus C| \geqslant \tfrac{1}{3}|R|\geqslant 4k. \end{align*}

Hence, $4 k \leqslant |R_1| \leqslant f(12k)$ and similarly $4 k \leqslant |R_2| \leqslant f(12k)$ . Also $|V(G) - V(G_1)| = |B| \geqslant |R_2| - |C| \geqslant 4k - k \gt 0$ , so $|V(G_1)| \lt |V(G)|$ and likewise for $G_2$ . Fix $j \in \{1, 2\}$ . Let $\beta _j$ be the restriction of $\beta$ to $V(G_j)$ . By Lemma 7, $\beta _j$ is a $(c, f)$ -disjointed $\ell$ -covering of $G_j$ . Let $S^j_i \,:\!=\, S_i \cap V(G_j)$ for each $i \in \{1,\dots,c-1\}$ ; note that each set $S^j_i$ is a union of at most $12k$ elements of $\beta _j$ .

Apply induction to $G_j$ with $S_i^j$ in place of $S_i$ and $R_j$ in place of $R$ . Thus, there is a graph $H_j$ of treewidth at most $c$ , a $c$ -clique $\{x_1, \dotsc, x_{c - 1}, y\}$ of $H_j$ , and an $H_j$ -partition $(V^j_x \,:\, x \in V(H_j))$ of $G_j$ of width at most $W$ such that $V^j_{x_i} = S^j_i \setminus \big(S^j_1 \cup \dotsb \cup S^j_{i - 1}\big)$ for each $i$ and $R_j \setminus \big(S_{1}^j \cup \dotsb \cup S_{c - 1}^j\big) \subseteq V^j_{y}$ with $|V_{y}| \leqslant 2 \ell (|R_j| - 2k)$ . We may assume that the intersection of $V(H_1)$ and $V(H_2)$ is equal to the clique $K \,:\!=\, \{x_1, \dotsc, x_{c - 1}, y\}$ . Let $H$ be the union of $H_1$ and $H_2$ and consider the $H$ -partition $(V_x \,:\, x \in V(H))$ of $G$ given by $V_x \,:\!=\, V^j_x$ for $x \in V(H_j) \setminus V(H_{3-j})$ and $V_x \,:\!=\, V^1_x \cup V^2_x$ for $x \in K$ . As $H_1$ and $H_2$ both have treewidth at most $c$ , so does $H$ .

Now $\{x_1, \dotsc, x_{c - 1}, y\}$ is a $c$ -clique, $V_{x_i} = S_{i} \setminus (S_1 \cup \dotsb \cup S_{i - 1})$ and $R \setminus S \subseteq V_y$ with

\begin{align*} |V_y| \;\leqslant \; |V_{y}^1| + |V_{y}^2| \;\leqslant \; 2 \ell (|R_1| + |R_2| - 4k) \;\leqslant \; 2 \ell (|R| + 2 |C| - 4k) \;\leqslant \; 2 \ell (|R| - 2k). \end{align*}

The other parts do not change and so we have the desired $H$ -partition of $G$ .

Proof of Lemma 8. Let $\beta$ be a $(c,f)$ -disjointed $\ell$ -covering of $G$ . If $|V(G)| \lt 4k$ , then the trivial partition $\{V(G)\}$ satisfies the claim. Otherwise $|V(G)| \geqslant 4k$ . Let $R \subseteq V(G)$ with $|R| = 4k$ . Let $S_1,\dots,S_{c-1}$ be arbitrary elements of $\beta$ . Since $\beta$ is an $\ell$ -covering, $|S_i| \leqslant \ell \leqslant 12 \ell k$ for each $i$ . Now Lemma 8 follows from Lemma 9.

Proof. Since $\textrm{tw}(\widehat{\ell \, G}) = \textrm{tw}(G)+1$ for any graph $G$ and $\ell \in{\mathbb{N}}$ , part (i) follows by induction.

We establish (ii) by induction on $c$ . In the case $c = 1$ , every $\ell$ -partition of $P_{\ell + 1}$ or $K_{1, \ell }$ contains an edge whose endpoints are in different parts, and we are done. Now assume the claim for $c - 1$ ( $c \geqslant 2$ ) and let $G \in \{ G_{c - 1, \ell }, C_{c - 1, \ell }\}$ . Consider an $\ell$ -partition of $\widehat{\ell \, G}$ . At most $\ell - 1$ copies of $G$ contain a vertex in the same part as the dominant vertex $v$ . Thus, some copy $G_0$ of $G$ contains no vertices in the same part as $v$ . By induction, $G_0$ contains a $c$ -clique $K$ whose vertices are in distinct parts. Since $v$ is dominant, $K \cup \{v\}$ satisfies the induction hypothesis.

Let $G \in \{G_{c, \ell }, C_{c, \ell } \}$ . Consider an $H$ -partition of $G$ of width at most $\ell$ . By (ii), $G$ contains a $(c+1)$ -clique whose vertices are in distinct parts. So $\omega (H) \geqslant c+1$ , implying $\textrm{tw}(H) \geqslant c$ . This establishes (iii).

Observe that $G_{2, \ell }$ is outerplanar (called a fan graph). The disjoint union of outerplanar graphs is outerplanar and the graph obtained from any outerplanar graph by adding a dominant vertex is planar; thus $G_{3, \ell }$ is planar. The disjoint union of planar graphs is planar, and the graph obtained from any planar graph by adding a dominant vertex is linklessly embeddable; thus $G_{4, \ell }$ is linklessly embeddable. This proves (iv).

We next show that $G_{c, \ell }$ is $K_{c, \max \{c, 3\}}$ -minor-free. $G_{1, \ell }$ is a path and so has no $K_{1, 3}$ -minor. $G_{2, \ell }$ is outerplanar and so has no $K_{2, 3}$ -minor. Let $c \geqslant 3$ and assume the result holds for smaller $c$ . Suppose that $G_{c, \ell }$ contains a $K_{c,c}$ -minor. Since $K_{c,c}$ is $2$ -connected, some copy of $G_{c-1, \ell }$ in $G_{c, \ell }$ contains a $K_{c-1, c}$ -minor. This contradiction establishes (v).

We show that $C_{c, \ell }$ contains no induced $P_{4}$ by induction on $c$ . First, $C_{1, \ell } = K_{1, \ell }$ does not contain $P_{4}$ . Next, suppose that $C_{c-1,\ell }$ does not contain an induced $P_{4}$ . $P_{4}$ does not have a dominant vertex and so any induced $P_{4}$ in $C_{c, \ell }$ must lie entirely within one copy of $C_{c-1,\ell }$ . In particular, $C_{c, \ell }$ does not contain an induced $P_{4}$ . This proves (vii).

Finally, Nešetřil and Ossona de Mendez [Reference Nešetřil and de Mendez57] proved (vii).

Proof. Suppose that $\{G_{c,\ell } \,:\, c,\ell \in{\mathbb{N}}\}$ has underlying treewidth $b$ . Thus, for some function $f$ , for all $c, \ell \in{\mathbb{N}}$ , we have $\textrm{tpw}_b(G_{c, \ell }) \leqslant f(\textrm{tw}(G_{c,\ell })) = f(c)$ . In particular, with $c \,:\!=\, b + 1$ and $\ell \,:\!=\, f(c)$ , we have $\textrm{tpw}_{c-1}(G_{c,\ell }) \leqslant \ell$ , which contradicts Lemma 12 (iii). The proof for $\{C_{c,\ell } \,:\, c,\ell \in{\mathbb{N}}\}$ is analogous.

Proof. The following proof technique is well-known [Reference Liu and Wood53, Reference de Mendez, Oum and Wood59]. Assign vertices in $B$ to pairs of vertices in $A$ as follows. Initially, no vertices in $B$ are assigned. If there is an unassigned vertex $v \in B$ adjacent to distinct vertices $x,y \in A$ and no vertex in $B$ is already assigned to $\{x, y\}$ , then assign $v$ to $\{x,y\}$ . Repeat this operation until no more vertices in $B$ can be assigned. Let $B_1$ and $B_2$ be the sets of assigned and unassigned vertices in $B$ , respectively. Let $G^{\prime}$ be the graph obtained from $G[A\cup B_1]$ by contracting $vx$ into $x$ , for each vertex $v \in B_1$ assigned to $\{x, y\}$ . Thus $G^{\prime}$ is a minor of $G$ with $V(G^{\prime}) = A$ and $|E(G^{\prime})| \geqslant |B_1|$ . For each vertex $v \in B_2$ , the set $N_G(v) \cap A$ is a clique in $G^{\prime}$ (otherwise $v$ would have been assigned to a pair of non-adjacent neighbours) of size at least $s$ . For each $s$ -clique $C$ in $G^{\prime}$ , there are at most $t-1$ vertices $v \in B_2$ with $C \subseteq N_G(v)$ , otherwise $G$ has a minor in $\mathcal{K}_{s,t}$ (since $G[B]$ is connected). Since $G^{\prime}$ has no minor in $\mathcal{K}_{s,t}$ , we have $|E(G^{\prime})| \leqslant \delta |A|$ for some $\delta = \delta (s,t)$ ; see [Reference Kostochka45–Reference Kühn and Osthus50, Reference Reed and Wood61, Reference Thomason71, Reference Thomason72] for explicit bounds on $\delta$ . Thus $|B_1| \leqslant \delta |A|$ . Moreover, $G^{\prime}$ is $2 \delta$ -degenerate. Every $d$ -degenerate graph on $n$ vertices has at most $\binom{d}{s - 1} n$ cliques of order $s$ [Reference Wood78, Lemma 18]. Thus $|B_2| \leqslant (t - 1)\binom{2\delta }{s - 1}|A|$ . Hence $|B| = |B_1| + |B_2| \leqslant \big(\delta + (t - 1) \binom{2\delta }{s - 1}\big)|A|$ .

Proof. By Lemma 8 it suffices to show that the singleton partition of $G$ is $(s,f)$ -disjointed, where $f(n) \,:\!=\, \delta sn (k + 1)$ and $\delta \,:\!=\, \delta (s,t)$ from Lemma 15. Let $S_1,\dots,S_s$ be subsets of $V(G)$ of size at most $n$ , let $S \,:\!=\, S_1 \cup \dots \cup S_s$ , and for each $i \in \{1,\dots, s\}$ let $S^{\prime}_i \,:\!=\, S_i \setminus (S_1 \cup \dots \cup S_{i-1})$ . Let $X$ be a connected component of $G-S$ . Let $\mathcal{H}$ be the set of connected subgraphs $H$ of $X$ such that $H \cap N\big(S^{\prime}_i\big) \neq \varnothing$ for each $i \in \{1,\dots,s\}$ . Say $\mathcal{R}$ is a maximum-sized set of pairwise disjoint subgraphs in $\mathcal{H}$ . We may assume that $\bigcup \{ V(R) \,:\, R \in{\mathcal{R}} \} = V(X)$ . Let $X^{\prime}$ be the graph obtained from $G[S \cup V(X)]$ by contracting each subgraph $R \in{\mathcal{R}}$ into a vertex $v_R$ . So $V(X^{\prime}) = S\cup \{v_R \,:\, R\in{\mathcal{R}}\}$ . Since $X$ is connected, $\{v_R \,:\, R \in{\mathcal{R}}\}$ induces a connected subgraph of $X^{\prime}$ . By construction, in $X^{\prime}$ , each vertex $v_R$ has at least $s$ neighbours in $S$ . By Lemma 15, $|{\mathcal{R}}| \leqslant \delta |S|$ . By Lemma 16, there is a set $Q \subseteq V(X)$ of size at most $\delta |S| (k + 1) \leqslant f(n)$ such that $X-Q$ contains no graph in $\mathcal{H}$ . Thus each component $Y$ of $X-Q$ satisfies $V(Y) \cap N_G\big(S^{\prime}_i\big) = \varnothing$ for some $i \in \{1,\dots,s\}$ . Hence, the singleton partition of $G$ is $(s, f)$ -disjointed.

Proof. Since $K_t$ is a minor of every graph in $\mathcal{K}_{t-2,t}$ , Lemma 17 implies that every $K_t$ -minor-free graph of treewidth $k$ has $(t-2)$ -tree-partition-width $O(k^2)$ . Thus the underlying treewidth of the class of $K_t$ -minor-free graphs is at most $t-2$ . Suppose for contradiction that equality does not hold. That is, for some function $f$ , every $K_t$ -minor-free graph $G$ has $(t-3)$ -tree-partition-width at most $f(\textrm{tw}(G))$ . Let $\ell \,:\!=\, f(t-2)$ . The graph $G_{t - 2, \ell }$ in Lemma 12 has treewidth $t-2$ and is thus $K_t$ -minor-free. However, by Lemma 12, $\textrm{tpw}_{t-3}(G_{t - 2, \ell })\gt \ell = f(t - 2) = f(\textrm{tw}(G_{t - 2, \ell }))$ , which is the required contradiction.

Proof. Since $K_{s,t}$ is a subgraph of every graph in $\mathcal{K}_{s,t}$ , Lemma 17 implies that every $K_{s,t}$ -minor-free graph $G$ of treewidth $k$ has $s$ -tree-partition-width $O(k^2)$ . Thus the underlying treewidth of the class of $K_{s,t}$ -minor-free graphs is at most $s$ . Suppose that it is at most $s - 1$ . Thus for some function $f$ , every $K_{s,t}$ -minor-free graph $G$ satisfies $\textrm{tpw}_{s-1}(G) \leqslant f(\textrm{tw}(G))$ . Let $\ell \,:\!=\, f(s)$ . The graph $G_{s,\ell }$ given by Lemma 12 has treewidth $s$ and is $K_{s,\max \{s,3\}}$ -minor-free, implying it is $K_{s,t}$ -minor-free (since $t \geqslant \max \{s,3\}$ ). However, by Lemma 12, $\textrm{tpw}_{s - 1}(G_{s, \ell }) \gt \ell = f(s) = f(\textrm{tw}(G_{s, \ell }))$ , which is the required contradiction.

Proof. Every $K_{1,1}$ -minor-free graph $G$ has no edges, and so $\textrm{tpw}_0(G) \leqslant 1$ . Every $K_{1,2}$ -minor-free graph $G$ has at most one edge in each component, and so $\textrm{tpw}_0(G) \leqslant 2$ . Each block of any $K_{2,2}$ -minor-free graph $G$ is a triangle, an edge or an isolated vertex; so $\textrm{tpw}_1(G) \leqslant 2$ .

Proof. By Lemma 8 it suffices to show that the singleton partition of $G$ is $(p,f)$ -disjointed, where $f(n) \in O(kn)$ . To this end, let $S_1,\dots,S_p$ be subsets of $V(G)$ of size at most $n$ , let $S \,:\!=\, S_1 \cup \dots \cup S_p$ , and for each $i \in \{1, \dots, p\}$ let $S^{\prime}_i \,:\!=\, S_i \setminus (S_1 \cup \dots \cup S_{i-1})$ . We may assume that $V(X) = \{1,\dots,p\}$ . Let $\mathcal{H}$ be the set of connected subgraphs $H$ of $G-S$ such that $V(H) \cap N_G\big(S^{\prime}_i\big) \neq \varnothing$ for each $i \in \{1,\dots,p\}$ .

Consider any set $\mathcal{J}$ of pairwise vertex-disjoint graphs in $\mathcal{H}$ of maximum size. Assign subgraphs in $\mathcal{J}$ to pairs of vertices in $S$ as follows. Initially, no subgraphs in $\mathcal{J}$ are assigned. If there is an unassigned subgraph $H \in \mathcal{J}$ adjacent to vertices $x \in S^{\prime}_i$ and $y \in S^{\prime}_j$ , for some distinct $i,j \in \{1,\dots,p\}$ , and no subgraph in $\mathcal{J}$ is already assigned to $\{x,y\}$ , then assign $H$ to $\{x,y\}$ . Repeat this operation until no more subgraphs in $\mathcal{J}$ can be assigned.

Let $\mathcal{J}_1$ and $\mathcal{J}_2$ be the sets of assigned and unassigned subgraphs in $\mathcal{J}$ respectively. Let $G^{\prime}$ be the graph obtained from $G$ as follows: for each $H \in \mathcal{J}_1$ assigned to $\{x,y\}$ , contract an $(x,y)$ -path through $H$ down to a single edge $xy$ . Delete any remaining vertices not in $S$ . Thus $G^{\prime}$ is a topological minor of $G$ with $V(G^{\prime}) = S$ and ${|E(G^{\prime})| \geqslant |\mathcal{J}_1}|$ . Consider a subgraph $H \in \mathcal{J}_2$ . Since $H \in{\mathcal{H}}$ there are neighbours $v_1,\dots,v_p$ of $H$ with $v_i \in S^{\prime}_i$ for each $i \in \{1,\dots,p\}$ . Since $H \in \mathcal{J}_2$ , the set $\{v_1,\dots,v_p\}$ is a clique in $G^{\prime}$ (otherwise $H$ could have been assigned to some non-adjacent $v_i$ and $v_j$ ). Charge $H$ to $(v_1,\dots,v_p)$ .