Proof. Let $G$ be a graph. Since every normal tree-decomposition of $G$ has at most $|V(G)|$ bags, there are finitely many normal tree-decompositions of $G$ . So there exists a normal tree-decomposition $\mathcal{D}$ of $G$ that lexicographically minimises $N({\mathcal{D}})$ among all normal tree-decompositions of $G$ . Suppose for contradiction that there is a tree-decomposition ${\mathcal{D}}'$ of $G$ such that $N({\mathcal{D}}')$ is lexicographically less than $N({\mathcal{D}})$ . Let ${\mathcal{D}}''$ be obtained by normalising ${\mathcal{D}}'$ . Then $N({\mathcal{D}}'')$ is lexicographically less than $N({\mathcal{D}})$ , contradicting our choice of $\mathcal{D}$ . This shows that $\mathcal{D}$ is an atomic tree-decomposition of $G$ .

Now suppose that $G$ has an atomic tree-decomposition $\mathcal{D}$ that is not normal, and let ${\mathcal{D}}'$ be the normalisation of $\mathcal{D}$ . Since normalisation only removes bags from $\mathcal{D}$ , $N({\mathcal{D}}')$ is lexicographically less than $N({\mathcal{D}})$ .

Proof. By Lemma5, every atomic tree-decomposition is normal. Suppose for contradiction that ${\mathcal{D}}'$ is a proper refinement of an atomic tree-decomposition $\mathcal{D}$ of a graph $G$ . Let ${\mathcal{D}}''$ be a tree-decomposition of $G$ obtained by normalising ${\mathcal{D}}'$ , and note that ${\mathcal{D}}''$ is also a proper refinement of $\mathcal{D}$ . We will show that $N({\mathcal{D}}'')$ is lexicographically less than $N({\mathcal{D}})$ , thus contradicting that $\mathcal{D}$ is atomic. To this end, let $B_x$ be the largest bag of $\mathcal{D}$ that is not a subset of any bag of ${\mathcal{D}}''$ . Now, for any bag $B_u$ of $\mathcal{D}$ larger than $B_x$ , there is some bag $B''_{v}$ of ${\mathcal{D}}''$ and some bag $B_w$ of $\mathcal{D}$ such that $B_u\subseteq B''_v\subseteq B_w$ . The second part of Lemma5 implies $u=w$ , and thus $B_u=B''_v=B_w$ .

Suppose for contradiction that there is a bag $B''_u$ of ${\mathcal{D}}''$ of size at least $|B_x|$ that is not a bag of $\mathcal{D}$ . Since ${\mathcal{D}}''$ is a refinement of $\mathcal{D}$ , there is a bag $B_v$ of $\mathcal{D}$ containing $B''_u$ as a proper subset. By the above argument, $B_v$ is also a bag of ${\mathcal{D}}''$ , contradicting that ${\mathcal{D}}''$ is normal. Hence, every bag of ${\mathcal{D}}''$ of size at least $|B_x|$ is also a bag of $\mathcal{D}$ . Since every bag of ${\mathcal{D}}''$ of size at least $|B_x|$ is also a bag of $\mathcal{D}$ but not vice versa and both are normal tree-decompositions, $N({\mathcal{D}}'')$ is lexicographically less than $N({\mathcal{D}})$ , a contradiction.

Proof. Since $\mathcal{D}$ is normal, $B_x$ is not a subset of $S$ unless $y=x$ and $S=B_y$ . Suppose for contradiction that $B_x$ intersects multiple components of $G-S$ , and let $H_1$ be one such component.

Let $T_1$ and $T_2$ be disjoint copies of $T$ , where $u_i$ is the copy of $u$ in $T_i$ , for each $u\in V(T)$ and $i\in \{1,2\}$ . Let $T'$ be obtained from $T_1\cup T_2$ by adding the edge $y_1y_2$ . Let $D_1\;:\!=\;V(H_1)\cup S$ and $D_2\;:\!=\;V(G)\setminus V(H_1)$ . For each node $u_i\in V(T_i)$ , let $B_{u_i}\;:\!=\; B_u \cap V(D_i)$ . So ${\mathcal{D}}'\;:\!=\;(B_u\;:\;u\in V(T'))$ is a tree-decomposition of $G$ and is a proper refinement of $\mathcal{D}$ since $B_x$ is not a subset of any bag of ${\mathcal{D}}'$ . This contradiction completes the proof.

Proof. Consider a separation $(A,B)$ of $G$ and a node $y\in V(T)$ such that $A\cap B\subseteq B_y$ and $B_y\setminus A\neq \varnothing$ . By Lemma7, there is a unique component $C$ of $G-(A\cap B)$ that intersects $B_y$ . Since $V(C)\cap B\supseteq B_y\setminus A\neq \varnothing$ , we have $V(C)\cap A=\varnothing$ , so $B_y\setminus B=\varnothing$ . Thus every bag of $(B_x\;:\;x\in V(T))$ is unbreakable.

Proof. Let $\mathcal{D}$ be a refined tree-decomposition of $G$ with width $\mathrm {tw}(G)$ . Consider a bag $B$ of $\mathcal{D}$ . Suppose for the sake of contradiction that $|B|\geqslant \Delta +2$ and there is a vertex $v\in B$ with degree $\Delta$ in $G[B]$ . Thus, $N_G[v]\subseteq B$ and some vertex in $B$ is not adjacent to $v$ . Hence $(N_G[v], V(G)\setminus \{v\})$ is a separation of $G$ that breaks $B$ , which contradicts Lemma8.

Proof. Suppose for contradiction that for some $y\in V(T)$ , there is a separation $(A_1,A_2)$ of $G$ such that both $|(B_y \cup A_1)\cap A_2|$ and $|(B_y\cup A_2)\cap A_1|$ are strictly less than $|B_y|$ . Select such a separation of minimum order. In particular, there is no set $S'$ of size smaller than $|A_1\cap A_2|$ that separates $A_1\cap B_y$ from $A_1\cap A_2$ , or a smaller order separation can be obtained by taking $S'$ and the vertices in components of $G-S'$ intersecting $A_1\cap B_y$ on one side and taking $S'$ and all remaining vertices on the other. Thus, by Menger’s Theorem, there is a set $\mathcal{P}_1$ of $|A_1\cap A_2|$ disjoint paths from $A_1\cap A_2$ to $B_y\cap A_1$ in $G$ . By construction each of these paths is internally disjoint from $A_1\cap A_2$ and hence is contained in $G[A_1]$ . By symmetry, there is a set $\mathcal{P}_2$ of $|A_1\cap A_2|$ disjoint paths from $A_1\cap A_2$ to $B_y\cap A_2$ in $G[A_2]$ .

For each $a\in A_1\cap A_2$ , let $w_a$ be the closest vertex to $y$ in $T$ such that $a\in B_{w_a}$ (meaning that $w_a=y$ if $a\in B_y$ ). Now, for each $w\in V(T)$ , define

\begin{equation*}B'_w\;:\!=\;B_w\cup \{a\in A_1\cap A_2 \;:\;w\in V(yTw_a)\}.\end{equation*}

Here $yTw_a$ is the $yw_a$ -path in $T$ . Note that for every vertex $u\in V(G)$ , the set $\{w\in V(T)\;:$ $u\in B'_w\}$ is a superset of $\{w\in V(T)\;:\;u\in B_w\}$ and induces a subtree of $T$ , so $(B'_w\;:\;w\in V(T))$ is a tree-decomposition of $G$ .

Let $T''$ be the tree with vertex set $V(T)\times \{1,2\}$ such that vertices $(u,i)$ and $(w,j)$ are adjacent if $i=j$ and $uw\in E(T)$ , or $i\neq j$ and $u=w=y$ . For each $(u,i)\in V(T'')$ , define $B''_{(u,i)}\;:\!=\;B'_u\cap V(A_i)$ . For each $i\in \{1,2\}$ , let $T_i\;:\!=\;T[V(T)\times \{i\}]$ and note that $(B''_x\;:\;x\in V(T_i))$ is a tree-decomposition of $G[A_i]$ . Since $A_1\cap A_2\subseteq B''_{(y,1)}\cap B''_{(y,2)}$ , it follows that $(B''_x\;:\;x\in V(T''))$ is a tree-decomposition of $G$ .

Proof. Let $X_v\;:\!=\;B'_v\setminus B_v$ , and let $j\;:\!=\;3-i$ . By construction, $X_v\subseteq (A_1\cap A_2)\setminus B_y$ . For each $a\in X_v$ , consider the path $P_{a,j}$ in $\mathcal{P}_{j}$ that has $a$ as an endpoint. Since both $B_{w_a}$ and $B_y$ contain a vertex of $P_{a,j}$ and $v\in V(yTw_a)$ , it follows that $B_v$ contains a vertex of $P_{a,j}$ . In particular, since $a\notin B_v$ , we see that $B_v$ contains a vertex in $V(P_{a,j})\setminus (A_1\cap A_2)$ . Thus $B_v\setminus A_i$ contains at least $|X_v|$ vertices, and so $|B''_{(v,i)}|= |B_v\cap A_i|+|X_v|\leqslant |B_v|$ . If equality holds, then $|B_v\setminus A_i|=|X_v|\leqslant |(A_1\cap A_2)\setminus B_y|$ . This means

\begin{equation*}|B''_{(v,j)}|\leqslant |B_v\setminus A_i|+|A_1\cap A_2|\leqslant |(A_1\cap A_2)\setminus B_y| +|A_1\cap A_2|.\end{equation*}

Recall that $|(B_y \cup A_1)\cap A_2|\lt |B_y|$ , meaning $|(A_1\cap A_2)\setminus B_y|\lt |B_y\setminus A_2|$ . Thus

\begin{equation*}|B''_{(v,j)}|\lt |B_y\setminus A_2|+|A_1\cap A_2|=|(B_y\cup A_2)\cap A_1|\lt |B_y|,\end{equation*}

as required.

Note that $|B''_{(y,1)}|=|(A_1\cap A_2)\cup (B_y\setminus A_2)|\lt |B_y|$ , and similarly $|B''_{(y,2)}|\lt |B_y|$ . Let $k$ be the maximum integer such that for some $v\in V(T)$ with $|B_v|=k$ we have $|B''_{v,1}|\lt k$ and $|B''_{(v,2)}|\lt k$ (so $k\geqslant |B_y|$ ).

By the above claim, the number of bags of ${\mathcal{D}}''$ of size $k'$ equals the number of bags of $\mathcal{D}$ of size $k'$ for all $k'\gt k$ , and ${\mathcal{D}}''$ has strictly fewer bags of size $k$ than $\mathcal{D}$ . This contradicts the fact that $\mathcal{D}$ is atomic.

Proof. Homotopy between paths in $\mathcal{P}$ is an equivalence relation. If some equivalence class has $c$ paths then we are done. Assume each equivalence class at most $c-1$ paths. The number of equivalence classes is at least $2g+1$ . Taking one path from each equivalence class gives $2g+1$ pairwise non-homotopic $ab$ -paths, contradicting Lemma16.

Proof. Let $n\;:\!=\;4g+2$ . Let $a,b$ be the two vertices of degree $n$ in $K_{2,n}$ . There are $n$ internally disjoint $ab$ -paths $P_1,\ldots ,P_n$ in $K_{2,n}$ . By Lemma17, three of $P_1,\ldots ,P_n$ are pairwise homotopic. Without loss of generality, $P_1,P_2,P_3$ are pairwise homotopic, and there is a disc $D$ in $\Sigma$ bounded by $P_1\cup P_2$ with the internal vertex of $P_3$ in the interior of $D$ . Choose $D$ to be minimal with this property. Then no other vertex is in the interior of $D$ . Since $n\geqslant 4$ , $P_1\cup P_2$ is separating.

Proof. Fix an embedding of $G$ in a surface $\Sigma$ of Euler genus $g$ . Let $H$ be obtained from $G[S]$ by contracting each branch set of the $K_{2,4g+2}$ minor. We obtain an embedding of $H$ in $\Sigma$ . By Lemma18, there is a cycle $C_0$ in $H$ bounding a disc and separating two vertices in $H$ . The branch sets corresponding to $C_0$ thus contain a cycle $C$ in $G[S]$ bounding a disc $D$ and separating two vertices in $S$ . Let $A$ be the set of all vertices of $G$ embedded in $D$ (including $C$ ). Let $B$ be the set of all vertices of $G$ embedded in $\Sigma -D$ plus $V(C)$ . So $(A,B)$ is a separation of $G$ that breaks $S$ .

Proof. Lemmas8 and 19 imply that $G[B_x]$ is $K_{2,4g+2}$ -minor-free for each $x\in V(T)$ . Since $K_{2,4g+2}$ is an apex-forest, by Lemma20, every $K_{2,4g+2}$ -minor-free graph has treewidth at most $4g+2$ . The result follows.

Proof. Let $t\;:\!=\;\lceil {\sqrt {g+1}\ }\rceil$ . So $g\leqslant t^2-1$ , and $G$ embeds in a surface $\Sigma$ of Euler genus at most $t^2-1$ . Let $k\;:\!=\;3$ and $n\;:\!=\;4t$ . By assumption, the $n\times n$ grid is a minor of $G[S]$ . Let $G'$ be the $n\times n$ grid, embedded in $\Sigma$ , obtained from $G[S]$ by appropriate contractions and deletions. By Lemma22, some $3 \times 3$ subgrid $G''$ of $G'$ is embedded in a closed disc $D_0$ in $\Sigma$ such that the boundary cycle $C_0$ of the subgrid is the boundary of $D_0$ . Note that $G'-V(G'')$ is connected, and that at least $3$ or the four corner vertices of $G''$ have neighbours in $G'-V(G'')$ . Thus $G'-V(G'')$ is embedded in $\Sigma -D_0$ . The branch sets corresponding to $C_0$ contain a cycle $C$ in $G[S]$ bounding a disc $D$ , where there is at least one vertex of $S$ in the interior of $D$ (corresponding to the internal vertex of the $3\times 3$ grid), and there is at least one vertex of $S$ in $\Sigma -D$ . Let $A$ be the set of all vertices of $G$ embedded in $D$ (including the boundary). Let $B$ be the set of all vertices of $G$ embedded in $\Sigma -D$ plus the vertices on the boundary of $D$ . So $(A,B)$ is a separation of $G$ that breaks $S$ .

Proof. Let $r,k\geqslant 0$ be integers as in Theorem26, let $p^*=\max \{\binom {p}{2},2r(k+1)+(k+1)(p+2)\}$ , and let $c_p=f(p^*)$ , where $f$ is from Theorem25. Assume that $\mathrm {tw}(G[S])\geqslant c_p$ . Our goal is to show that $S$ is reducible. By Theorem25, $G[S]$ contains a $p^*$ -wall $W$ as a subgraph. As illustrated in Figure 2, in $W$ there are:

• • $k+1$ pairwise disjoint subwalls $H_1,\ldots , H_{k+1}$ , each of height $p+2$ and disjoint from the perimeter of $W$ , and

• • for each $i\in \{1,\ldots ,k+1\}$ there is a set $\mathcal{C}_i$ of $r$ pairwise disjoint cycles in $W$ , each disjoint from $H_1\cup \ldots \cup H_{k+1}$ , such that each $C\in \mathcal{C}_i$ separates $H_i$ and $\bigcup \{H_j\;:\;j\neq i]\}$ .

Since $W$ is a subdivision of a 3-connected planar graph, by a theorem of Whitney [Reference Whitney85], $W$ has a unique embedding in the plane (up to the choice of the outerface). In this embedding, any curve in the plane from $H_i$ to $H_j$ (for $i\neq j$ ) must intersect each cycle $C\in \mathcal{C}_i$ . Thus $\hat {d}_W(H_i,H_j)\geqslant r$ , and so by Theorem26 there is some $X\subseteq V(G)$ of size at most $\binom {p}{2}$ and some $i\in \{1,\ldots , k+1\}$ such that $H_i$ is dividing in $(G-X)\cup W$ . Let $S'$ be the union of $X$ and the vertex set $\mathrm {per}(H_i)$ of the perimeter of $H_i$ . Let $A$ be the union of $S'$ and the vertex sets of all components of $G-S'$ that intersect the perimeter of $W$ . Let $B\;:\!=\; V(G)\setminus (A\setminus S')$ . So $(A,B)$ is a separation of $G$ with $A\cap B=S'$ and $V(H_i)\subseteq B$ . Let $X_W$ be the vertex set of the perimeter of $W$ , and note that $|(S\cup A)\cap B|\leqslant |S|-|X_W|+|X|$ , which is less than $|S|$ since $|X_W|\gt p^*\geqslant \binom {p}{2}\geqslant |X|$ . Likewise $|(S\cup B)\cap A|\leqslant |S|-|V(H_i)|+|S'|$ , which is less than $|S|$ since there are at least $p^2$ vertices of $H_i$ that are not in the perimeter of $H_i$ , there are at most $\binom {p}{2}$ vertices of $S'$ that are not in the perimeter of $H_i$ , and all vertices in the perimeter of $H_i$ are in $S\cap S'$ . Thus $S$ is reducible, as required.

Proof. Consider any tree-decomposition ${\mathcal{D}}=(B_x\;:\;x\in V(T))$ of $G$ in which no bag contains $S$ . Our goal is to show that $\mathcal{D}$ has width at least $|S|+c$ . For every edge $xy\in E(T)$ , orient $xy$ towards $x$ if $G_{x:y}$ has more vertices of $S$ than $G_{y:x}$ , orient $xy$ towards $y$ if $G_{y:x}$ has more vertices of $S$ than $G_{x:y}$ , and otherwise orient $xy$ arbitrarily. Let $t_0$ be a sink of the resulting oriented tree, and let $t_1,\ldots ,t_d$ be the neighbours of $t_0$ . For each $i\in \{1,\ldots , d\}$ , let $S_i\;:\!=\;S\cap V(G_{t_i:t_0})$ , and let $S_0\;:\!=\;S\setminus \bigcup \{S_1,\ldots ,S_d\}$ . Without loss of generality, $|S_1|=\max \{|S_1|,\ldots ,|S_d|\}$ .

First consider the case where $2|S_1|\geqslant |S\setminus S_0|$ . Since $t_0$ is a sink, $|S_1|\leqslant |S\setminus B_{t_1}|$ . Thus, there are disjoint pairs $\{a_1,b_1\},\ldots ,\{a_{|S_1|},b_{|S_1|}\}$ , where each $a_i$ is in $S_1$ and each $b_i$ is in $S\setminus B_{t_1}$ . By assumption, there is a set of internally disjoint paths $\{P_{i,j}\;:\;i\in \{1,\ldots ,|S_1|\},j\in \{1,\ldots , 2c+4\}\}$ , each internally disjoint from $S$ , such that for each $i\in \{1,\ldots ,|S_1|\}$ and $j\in \{1,\ldots , 2c+4\}$ the endpoints of $P_{i,j}$ are $a_i$ and $b_i$ . By the definition of a tree-decomposition, $B_{t_0}\cap B_{t_1}$ separates $S_1$ from $S\setminus B_{t_1}$ in $G$ , and so $B_{t_0}\cap B_{t_1}$ contains a vertex of each of these paths. Since $\{a_1,\ldots ,a_{|S_1|}\}\cap B_{t_0}=\varnothing$ and $\{b_1,\ldots ,b_{|S_1|}\}\cap B_{t_1}=\varnothing$ and each path is internally disjoint from $S$ , we have $|(B_{t_0}\cap B_{t_1})\setminus S|\geqslant (2c+4)|S_1|$ . Thus, $|B_{t_0}|\geqslant |S_0|+(2c+4)|S_1|\geqslant |S_0|+(c+2)|S\setminus S_0|$ . By assumption $S$ is not a subset of $B_{t_0}$ , so $|S\setminus S_0|\geqslant 1$ and $|B_{t_0}|\geqslant |S|+c+1$ .

Now consider the case where $2|S_1|\leqslant |S\setminus S_0|-1$ . Let $\mathcal{X}=\{\{a_1,b_1\},\ldots ,\{a_z,b_z\}\}$ be a maximum sized collection of disjoint pairs of vertices such that each set $\{a_i,b_i\}$ intersects exactly two sets in $\{S_1,\ldots ,S_d\}$ . Suppose for contradiction that $2z\leqslant |\bigcup \{S_1,\ldots ,S_d\}|-2$ . Then for some $i\in \{1,\ldots ,d\}$ there are at least two vertices $v$ and $w$ in $S_i$ that are not in a pair in $\mathcal{X}$ , and all vertices in $S\setminus (S_0\cup S_i)$ are in a pair in $\mathcal{X}$ . Since $|S_i|\leqslant |S_1|$ and $2|S_1|\lt |S\setminus S_0|$ , we have $|S_i|\lt |S\setminus (S_0\cup S_i)|$ , there is a pair $\{a_j,b_j\}$ in $\mathcal{X}$ that is disjoint from $S_i$ . Replacing $\{a_j,b_j\}$ by $\{v,a_j\}$ and $\{b_j,w\}$ increases the size of $\mathcal{X}$ , yielding the desired contradiction. Thus, $2z\geqslant | S_1\cup \ldots \cup S_d|-1=|S\setminus S_0|-1$ . By construction, there is a set of internally disjoint paths $\{P_{i,j}\;:\;i\in \{1,\ldots ,z\},j\in \{1,\ldots , 2c+4\}\}$ , each internally disjoint from $S$ , such that for each $i\in \{1,\ldots ,|S_1|\}$ and $j\in \{1,\ldots , 2c+4\}$ the endpoints of $P_{i,j}$ are $a_i$ and $b_i$ . Note that for distinct $i,j\in \{1,\ldots ,d\}$ , the set $B_{t_0}$ separates $S_i$ from $S_j$ in $G$ . Thus, $B_{t_0}$ contains an internal vertex of each of these paths, and so $|B_{t_0}|\geqslant |S_0|+(2c+4)z\geqslant |S_0|+(c+2)(|S\setminus S_0|-1)$ . By assumption $S\not \subseteq B_{t_0}$ , so $|S_1|\geqslant 1$ and $|S\setminus S_0|\geqslant 2|S_1|+1\geqslant 3$ . Thus, $|B_{t_0}|\geqslant (|S|-1)+2(c+1)\gt |S|+c+1$ .

In both cases, $\mathcal{D}$ has width at least $|B_{t_0}|-1\geqslant |S|+c$ . Thus, every tree-decomposition of $G$ with width less than $|S|+c$ has a bag containing $S$ .

Proof. Draw $G_0$ in the plane (allowing crossings), and let $H$ be the planarisation of this drawing (that is, introduce a new vertex in $H$ at each crossing). We introduce $H$ simply to make it easier to talk about the drawing, it has not no essential role in the construction. Let $T^*$ be a rooted spanning tree of the dual of $H$ . Let $f_0,f_1,\ldots , f_s$ be the faces of $H$ (equivalently, vertices of $T^*$ ) ordered so that $f_s$ is the root of $T^*$ , and for each edge $f_if_j$ in $T^*$ , if $i\lt j$ then $f_j$ is the parent of $f_i$ ; we denote this edge $f_if_j$ by $e_i$ .

We will construct a $1$ -planar graph $G$ and a set $S\subseteq V(G)$ such that $G[S]$ is a subdivision of $G_0$ . We will then show that $\mathrm {tw}(G)\leqslant |S|$ and that $S$ satisfies the conditions of Lemma29. Let $V_0$ be the set of vertices of $G_0$ incident to $f_0$ . For each $i\in \{1,\ldots ,s\}$ , let $V_i$ be the set of vertices incident with $f_i$ and not incident with any face in $\{f_1,\ldots ,f_{i-1}\}$ . Let $E_0$ be the set of edges of $H$ incident to $f_0$ and not dual to edges in $T^*$ , and for each $i\in \{1,\ldots ,s\}$ , let $E_i$ be the set of edges of $E(H)\setminus (E_0\cup E_1\cup \cdots \cup E_{i-1})$ that are incident with $f_i$ and not dual to the edge $f_if_j$ of $T^*$ with $j\gt i$ . Thus the sets $E_i$ partition $E(H)$ , and for each $i\in \{1,\ldots , s-1\}$ the edge of $H$ dual to $e_i$ is in $E_j$ where $e_i=f_if_j$ with $i\lt j$ .

Iteratively construct $G$ from $G_0$ as follows. First, for each $e\in E_0$ , subdivide the edge-segment of $G_0$ corresponding to $e$ once. Let $X_0$ be the set of all subdivision vertices introduced in this step, together with $V_0$ . Now, for each $v\in X_0$ , and each $j\in \{0,1,\ldots , s\}$ , create $2c+4$ new vertices $a_{v,j,1},a_{v,j,2},\ldots , a_{v,j,2c+4}$ , and draw these vertices in $f_j$ . Draw an edge from $v$ to each of these new vertices that does not cross any edge in $H$ except those dual to the edges of the path from $f_0$ to $f_j$ in $T^*$ . For each pair of distinct vertices $v,w\in X_0$ and each $y\in \{1,\ldots , 2c+4\}$ , draw an edge from $v$ to $a_{w,0,y}$ in $f_0$ . At the final step, we will freely subdivide all edges incident to these newly created vertices, so the number of crossings on these edges is unimportant.

For $i=1,2,\ldots ,s$ apply the following step. First, for each $e=pq\in E_i$ , subdivide the edge-segment of $G_0$ corresponding to $e$ once between $p$ and the first crossing on $e$ starting at $p$ , once between $q$ and the first crossing on $e$ starting at $q$ , and once between each pair of consecutive crossings on $e$ . No future edges will cross $e$ . Let $X_i$ be the the set of all subdivision vertices added to edge-segments corresponding to edges in $E_i$ , together with all vertices in $V_i$ . For each $v\in X_i$ , each $j\in \{i,i+1,\ldots , s\}$ and each $y\in \{1,\ldots ,2c+4\}$ , create a new vertex $a_{v,j,y}$ and draw this vertex in $f_j$ . Draw an edge from $v$ to each of these new vertices that does not cross any edge in $H$ except those dual to edges of the path from $f_i$ to $f_j$ in $T^*$ . For each pair of distinct vertices $v\in X_i$ and $w\in X_0 \cup \ldots \cup X_i$ and each $y\in \{1,\ldots , 2c+4\}$ , draw an edge from $v$ to $a_{w,i,y}$ in $f_i$ . These steps may produce arbitrarily many crossings between edges which are not drawn in the image of $H$ , but we will freely subdivide all such edges in the final step of the construction.

Let $G'$ be the graph constructed so far. Let $S$ be the set of all vertices in $V(G_0)$ together with all subdivision vertices introduced so far (that is, all vertices not of the form $a_{v,j,y}$ ). By construction, the subgraph of $G'$ induced by $S$ is a subdivision of $G_0$ , and every edge in this subgraph is crossed at most once in the entire graph. Subdivide the remaining edges (those incident to vertices of the form $a_{v,j,y}$ ) until the resulting graph is $1$ -planar, and call the final graph $G$ .

We first construct a tree-decomposition of $G'$ . Let $T$ be the star with central vertex $r$ and leaf-set $V(G')\setminus S$ . Set $B'_r\;:\!=\;S$ and $B'_{w}\;:\!=\;N_{G'}[w]$ for each $w\in V(G')\setminus S$ . Since $V(G')\setminus S$ is an independent set in $G'$ , $(B'_x\;:\;x\in V(T))$ is a tree-decomposition of $G'$ with width $|S|$ (since $N_{G'}(w)\subseteq S$ and $|B'_{w}| =|N_{G'}[w]| \leqslant |S|+1$ for each $w\in V(G')\setminus S$ ). Since $G$ is a subdivision of $G'$ , $\mathrm {tw}(G) = \mathrm {tw}(G')\leqslant |S|$ .

Now consider an arbitrary collection $\{\{a_1,b_1\},\ldots ,\{a_z,b_z\}\}$ of disjoint pairs of vertices in $S$ . By construction, for each $i\in \{1,\ldots , z\}$ , there are at least $2c+4$ common neighbours $a_{v,j,y}$ of $a_i$ and $b_i$ in $G'$ with $v\in \{a_i,b_i\}$ . Thus there is a collection $\{P_{i,j}\;:\;i\in \{1,\ldots , z\},j\in \{1,\ldots , 2c+4\}\}$ of internally disjoint paths in $G$ , each internally disjoint from $S$ , such that for each $i\in \{1,\ldots , z\}$ and $j\in \{1,\ldots , 2c+4\}$ the endvertices of $P_{i,j}$ are $a_i$ and $b_i$ . By Lemma29, every tree-decomposition of $G$ of width less than $|S|+c$ (and thus every tree-decomposition of $G$ of width less than $\mathrm {tw}(G)+c$ ) has a bag containing $S$ .

Proof. We may assume $c\lt n$ , or else the lemma is satisfied by the trivial tree-decomposition with a single node. Let $(V_0,V_1,\ldots ,V_m)$ be a layering of $G$ , and let ${\mathcal{D}}\;:\!=\;(B_x\;:\;x\in V(T))$ be a tree-decomposition of $G$ , such that $|B_x \cap V_i|\leqslant c$ for each $x\in V(T)$ and $i\in \{0,1,\ldots ,m\}$ . Let $p\;:\!=\;\lceil {\sqrt {n/c}}\rceil$ , which is at least $2$ since $c\lt n$ . For $i\in \{0,1,\ldots ,p-1\}$ let $\widehat {V}_i\;:\!=\;\bigcup \{V_j:j\equiv i\pmod {p}\}$ . So $\widehat {V}_0,\widehat {V}_1,\ldots ,\widehat {V}_{p-1}$ is a partition of $V(G)$ , and $|\widehat {V}_i|\leqslant \frac {n}{p}$ for some $i\in \{0,1,\ldots ,p-1\}$ . For each component $C$ of $G-V_i$ , let ${\mathcal{D}}_C$ be the tree-decomposition of $C$ induced by $\mathcal{D}$ (so that each of bag in the intersection of some $B_x$ with $V(C)$ . Each component of $G-\widehat {V}_i$ is contained within $p-1$ layers, so ${\mathcal{D}}_C$ has width at most $c(p-1)-1$ . These tree-decompositions can be combined to obtain a tree-decomposition ${\mathcal{D}}'$ of $G-{V}_i$ of width at most $c(p-1)-1$ . By adding $V_i$ to every bag of ${\mathcal{D}}'$ , we obtain a tree-decomposition ${\mathcal{D}}''$ of $G$ of width at most $\frac {n}{p}+c(p-1)-1\leqslant 2\sqrt {cn}$ .

Consider arbitrary bags $C_1,\ldots ,C_k$ of ${\mathcal{D}}''$ . By construction, there is a set $S\subseteq V(T)$ with $|S|\leqslant k$ such that $C_1\cup \ldots \cup C_k\subseteq X\;:\!=\; \widehat {V}_i\cup \bigcup \{B_x\;:\;x\in S\}$ . We now construct a tree-decomposition of $G[X]$ with width at most $(3k+1)c-1$ , which implies that $\mathrm {tw}( G[C_1\cup \ldots \cup C_k] )\leqslant (3k+1)c-1$ as desired.

Fix a vertex $r$ of $T$ . Let $T_j$ be a copy of $T$ for each $j\equiv i \pmod {p}$ , where these copies are pairwise disjoint. For each $x\in V(T)$ , let $x_j$ be the copy of $x$ in $T_j$ . Let $P\;:\!=\;(y_0,y_1,\ldots ,y_m)$ be a path, where $y_j$ is identified with the vertex $r_j$ whenever $j\equiv i\pmod {p}$ . We obtain a tree $U\;:\!=\;P\cup \bigcup \{T_j:j\equiv i\pmod {p}\}$ . For each node $x_j\in V(U)$ , let

\begin{equation*}A_{x_j}\;:\!=\; (B_x\cap V_j)\cup \bigcup \{ ( V_{j-1} \cup V_j \cup V_{j+1}) \cap B_z \;:\; z\in S\}.\end{equation*}

For each node $y_j\in V(U)$ with $j\not \equiv i\pmod {p}$ , let

\begin{equation*}A_{y_j}\;:\!=\; \bigcup \{ (V_{j-1} \cup V_j \cup V_{j+1} )\cap B_z \;:\; z\in S\}.\end{equation*}

We now prove that $(A_u\;:\;u\in V(U))$ is a tree-decomposition of $G[X]$ .

Consider a vertex $v\in X$ . Say $v\in V_j$ . If $v\in \bigcup \{B_z\;:\;z\in S\}$ and $j\bmod {p}$ is in $\{i-1,i,i+1\}$ then $v\in A_{x_j}$ for every $x\in V(T)$ , and $v\in A_{y_{j-1}} \cup A_{y_{j+1}}$ , and $v$ is in no other bag $A_u$ with $u\in V(U)$ . If $v\in \bigcup \{B_z\;:\;z\in S\}$ and $j\bmod {p}$ is not in $\{i-1,i,i+1\}$ then $v\in A_{y_{j-1}} \cup A_{y_j} \cup A_{y_{j+1}}$ , and $v$ is in no other bag $A_u$ with $u\in V(U)$ . Otherwise, $v\not \in \bigcup \{B_z\;:\;z\in S\}$ . Since $v\in X$ , we have $j\equiv i \pmod {p}$ , implying $v\in A_{x_j}$ whenever $v\in B_x$ with $x\in V(T)$ , and $v$ is in no other bag $A_u$ with $u\in V(U)$ . In each case, the set of bags $A_u$ that contain $v$ correspond to a connected subtree of $U$ . Hence $(A_u\;:\;u\in V(U))$ satisfies the vertex-property of tree-decompositions.

Consider an edge $vw$ of $G[X]$ . So $v,w\in B_x$ for some node $x\in V(T)$ . Say $v\in V_j$ and $w\in V_\ell$ . If $j,\ell \equiv i \pmod {p}$ , then since $p\geqslant 2$ , we have $\ell =j$ and $v,w\in A_{x_j}$ . If $j\equiv i \pmod {p}$ and $\ell \not \equiv i \pmod {p}$ , then $|j-\ell |=1$ and $w\in \cup \{ B_z\;:\;z\in S\}$ , implying that $v,w\in A_{x_j}$ . If $j\not \equiv i \pmod {p}$ and $\ell \not \equiv i \pmod {p}$ , then $|j-\ell |\leqslant 1$ and $v,w\in \cup \{B_z\;:\;z\in S\}$ , implying that $v,w\in A_{y_j}$ . Hence $(A_u\;:\;u\in V(U))$ satisfies the edge-property of tree-decompositions.

Therefore, $(A_u\;:\;u\in V(U))$ is a tree-decomposition of $G[X]$ . Observe that $|A_{x_j}|\leqslant (3k+1)c$ and $|A_{y_j}|\leqslant 3ck$ . Thus, the width of $(A_u\;:\;u\in V(U))$ is at most $(3k+1)c-1$ . Hence, $G[X]$ and $G[C_1\cup \ldots \cup C_k]$ have treewidth at most $(3k+1)c-1$ .

Proof. Let $(V_0,V_1,\ldots ,V_m)$ be a layering of $G$ , and let $(B_x\;:\;x\in V(T))$ be a tree-decomposition of $G$ , such that $|B_x \cap V_i|\leqslant c$ for each $x\in V(T)$ and $i\in \{0,1,\ldots ,m\}$ . Let $p\;:\!=\;\lceil {\sqrt {n/c}}\rceil$ . Since $n\gt c$ , we have $p\geqslant 2$ . For $i\in \{0,1,\ldots ,p-1\}$ , let $\widehat {V}_i\;:\!=\;\bigcup \{V_j:j\equiv i\pmod {p}\}$ . So $\widehat {V}_0,\widehat {V}_1,\ldots ,\widehat {V}_{p-1}$ is a partition of $V(G)$ , and $|\widehat {V}_i|\leqslant \frac {n}{p}$ for some $i\in \{0,1,\ldots ,p-1\}$ . Let $S\;:\!=\;\widehat {V}_i$ . Each component $C$ of $G-S$ is contained within $p-1$ layers, and thus ${\mathcal{D}}_C\;:\!=\;(B_x\cap V(C): x \in V(T))$ is a tree-decomposition of $C$ with width at most $c(p-1)-1$ . Hence $G-S$ has a tree-decomposition ${\mathcal{D}}'$ of width at most $c(p-1)-1\leqslant \sqrt {cn}$ , obtained by combining these tree-decompositions. Let $X$ be the union of $k$ bags from ${\mathcal{D}}'$ . Then $X\subseteq \widehat {X}\;:\!=\;B_{x_1}\cup \ldots \cup B_{x_k}$ for some $x_1,\ldots ,x_k\in V(T)$ . Observe that $( \widehat {X} \cap (V_0\cup V_1), \widehat {X} \cap (V_1\cup V_2),\ldots , \widehat {X}\cap (V_{m-1}\cup V_m) )$ is a path-decomposition of $G[\widehat {X}]$ with width at most $2ck-1$ . The result follows.