Proof. It suffices to show that consecutive vertices of $P'$ are at distance at most $d+2r+2\ell$ . If $P$ does not intersect $Z$ , $P$ is an $r$ -subpath of $P-Z$ , and hence $\textrm {dist}_G(x,y)\leqslant d\leqslant d+2r+2\ell$ as desired. So we may assume that $P$ intersects $Z$ , and that $v_1,\ldots ,v_n$ are defined with $n\geqslant 1$ .

Consider consecutive interior vertices $\iota (v_i),\iota (v_{i+1})$ of $P'$ ; we must show that $\textrm {dist}_G(\iota (v_i),\iota (v_{i+1}))\leqslant d+2r+2\ell$ ; it suffices to show $\textrm {dist}_G(v_i,v_{i+1})\leqslant d+2r$ . If $v_i,v_{i+1}$ are consecutive in $P$ , then $\textrm {dist}_G(v_i,v_{i+1})\leqslant r$ , as desired. Otherwise, let $u_i$ be the vertex directly after $v_i$ in $P$ , and let $w_{i+1}$ be the vertex directly before $v_{i+1}$ in $P$ ; it suffices to show that $\textrm {dist}_G(u_i,w_{i+1})\leqslant d$ . Consider the $r$ -subpath $P_i$ of $P$ going from $u_i$ to $w_{i+1}$ . $P_i\cap Z=\emptyset$ by definition of the subsequence $v_1,\ldots ,v_n$ , thus $P_i$ is an $r$ -subpath of $P-Z$ and has weak diameter in $G$ at most $d$ . Therefore, $\textrm {dist}_G(u_i,w_{i+1})\leqslant d$ , as desired.

It remains to show that $\textrm {dist}_G(x,\iota (v_1))$ and $\textrm {dist}_G(y,\iota (v_n))$ are at most $d+2r+2\ell$ . We show that $\textrm {dist}_G(x,\iota (v_1))\leqslant d+2r+2\ell$ ; the argument for $\textrm {dist}_G(y,\iota (v_n))$ is symmetric. If $x\in Z$ , then $v_1=x$ and thus $\textrm {dist}_G(x,\iota (v_1))=\textrm {dist}_G(x,\iota (x))\leqslant \ell \leqslant d+2r+2\ell$ . Otherwise, let $w_1$ be the vertex directly before $v_1$ in $P$ ; it suffices to show that $\textrm {dist}_G(x,w_1)\leqslant d+r+\ell$ . Let $P'$ be the $r$ -subpath of $P$ going from $x$ to $w_1$ ; we must have $P'\cap Z=\emptyset$ by definition of $v_1$ , hence $P'$ is an $r$ -subpath of $P-Z$ . Thus, $\textrm {wdiam}_G(P')\leqslant d$ ; in particular $\textrm {dist}_G(x,w_1)\leqslant d\leqslant d+2r+2\ell$ , as desired.

Proof. Let $S$ denote a centre of $Z$ , and define $\iota \,:\,Z\rightarrow S$ such that $\iota (v)\in N_G^{\ell }(v)\cap S$ for each $v\in Z$ . Since $P\subseteq Z$ , it is immediate that every $r$ -subpath of $P-Z$ has weak diameter in $G$ at most $0$ . We also know that for any $v\in Z$ , $\textrm {dist}_G(v,\iota (v))\leqslant \ell$ . Thus, we can apply Proposition 7 to obtain a $(2r+2\ell )$ -walk from $x$ to $y$ whose interior is in $S$ . Since $|S|\leqslant k$ , $P'$ can involve at most $k+2$ distinct vertices, and thus $\textrm {dist}_G(x,y)\leqslant (k+1)(2r+2\ell )$ , as desired.

Proof. Let $S$ denote a centre of $Z$ , and let $Z'\,:\!=\,N_G^{r+\ell }(S)$ . For any $u,v\in V(G)\setminus Z'$ at distance at most $r$ in $G$ and any shortest path $Q$ from $u$ to $v$ in $G$ , observe that $Q$ is disjoint from $Z$ , as otherwise we would have $u,v\in N_G^r(Z)\subseteq Z'$ . Thus, $Q$ is also a path of length at most $r$ in $G-Z$ , and $\textrm {dist}_{G-Z}(u,v)\leqslant r$ . Therefore, for any monochromatic $r$ -path $P$ in $G$ under $c\cup c'$ , any $r$ -subpath $P'$ of $P-Z'$ is a monochromatic $r$ -path in $G-Z$ under $c$ . Thus, $P'$ has weak diameter in $G-Z$ (and $G$ ) at most $d$ . Define a map $\iota \,:\,Z'\rightarrow S$ by selecting $\iota (z)\in N_G^{r+\ell }(z)\cap S$ for each $z\in Z'$ ; by Proposition 7, we obtain a $(d+2r+2(r+\ell ))=(d+4r+2\ell )$ -walk which has the same endpoints as $P$ but whose interior is contained in $S$ . Since $|S|\leqslant k$ , this $(d+4r+2\ell )$ -walk involves at most $k+2$ distinct vertices, and thus has weak diameter at most $(k+1)(d+4r+2\ell )$ in $G$ , as desired.

Proof. If $a=0$ the claim is trivially true, as $G=G_0$ by (a) and $c=c_0$ is an $(m,r,D)$ -colouring of $G_0=G$ by (c), since $r'\geqslant r$ and $d\leqslant D$ . So assume $a\geqslant 1$ .

Observe that $c$ is a well-defined colouring of $G$ by (a), (b)(i) and (d)(ii). So it suffices to show that for every real number $r\gt 0$ and every monochromatic $r$ -component $M$ of $G$ under $c$ , $\textrm {wdiam}_G(V(M))\leqslant D$ .

By (a), for each $i\in \{1,\ldots ,a\}$ , $(G_i,\widetilde {G}_i)$ is a separation of $G$ with separator $S_i$ , which we use implicitly throughout this proof. Also, note that $\bigcap _{i=1}^a V(\widetilde {G}_i)=V(G_0)$ by (b)(i).

For all $i\in \{1,\ldots ,a\}$ and $j\in \{1,2,3\}$ , let $N_i^j\,:\!=\,N_{G_i}^{jr}(S_i)$ and $S_i^j\,:\!=\,N_{G_i}^{jr}(S_i)\setminus N_{G_i}^{(j-1)r}$ . Note that for each $i\in \{1,\ldots ,a\}$ , since $c_i$ has an $(S_i,r)$ -barrier, $S_i^2$ , $S_i^3$ are both monochromatic of different colours, and for each $v\in S_i^1$ there is an $s\in S_i\cap N_{G_i}^r(v)\subseteq S_i\cap N_G^r(v)$ with the same colour as $v$ . Let $V'\,:\!=\,V(G_0)\cup \bigcup _{i=1}^a N_i^2$ .

Proof. For any $i\in \{1,\ldots ,a\}$ , any $r$ -path $P$ from $u\in V(\widetilde {G}_i)\cup N_i^2$ to $v\in V(G_i)\setminus N_i^2$ must contain a vertex in $S_i^2$ and $S_i^3$ ; see Fig. 1. As $S_i^2$ and $S_i^3$ are monochromatic of different colours, $P$ cannot be monochromatic. It follows that for any monochromatic $r$ -component $M$ of $G$ , $V(M)$ must be contained in either $V(\widetilde {G}_i)\cup N_i^2$ or $V(G_i)\setminus N_i^2$ . Accounting for the fact that this holds for each $i\in \{1,\ldots ,a\}$ , we can conclude that $V(M)$ is either contained in $V(G_i)\setminus N_i^2\subseteq V(G_i)$ for some $i\in \{1,\ldots ,a\}$ , or in $\bigcap _{i=1}^a (V(\widetilde {G}_i)\cup N_i^2) = V'$ .

Proof. This is trivially true for any $v\in V(G_0)$ , and for any $v\in S_i^1$ for some $i\in \{1,\ldots ,a\}$ , the claim follows since $c_i$ has an $(S_i,r)$ -barrier and $r\leqslant \ell '$ . Since $V(M)\subseteq V'$ , we now only need to consider $v\in S_i^2$ for some $i\in \{1,\ldots ,a\}$ . Observe that $V(M)$ is not contained in $S_i^2$ , as it is not contained in $V(G_i)$ , thus there must exist some $u\in V(M)\setminus S_i^2$ ; since $V(M)\subseteq V'$ , this means that $u\in V(\widetilde {G}_i)\cup N_i^1$ . Because $u,v$ are both in $M$ , there exists a monochromatic $r$ -path $P$ from $v\in S_i^2$ to $u\in V(\widetilde {G}_i)\cup N_i^1$ in $G$ of the same colour. Observe that $P$ must contain a vertex in $S_i^1$ ; see Fig. 1. Let $x$ be the first (in the order of the sequence of $P$ ) such vertex in $S_i^1\cap P$ ; observe that every vertex before $x$ must be in $S_i^2$ . Thus, the $r$ -subpath $P'$ of $P$ from $v$ to $x$ is an $r$ -path contained in $S_i^1\cup S_i^2\subseteq N_i^2$ . Since $c_i$ has an $(S_i,r)$ -barrier, there must be an $s\in S_i\cap N_G^r(x)\subseteq V(G_0)$ with the same colour as $x$ . Hence, $v$ and $s$ have the same colour, and there is an $r$ -path contained in $N_i^2$ from $v$ to $s$ obtained by appending $s$ to the end of $P'$ . Since $S_i$ is a $(k,\ell )$ -centred set in $G$ , $N_i^2$ is a $(k,\ell +2r)$ -centred set in $G$ . Thus, we can apply Corollary 8 to find that $\textrm {dist}_G(v,s)\leqslant (k+1)(2r+2(2r+\ell ))=\ell '$ , as desired.

Proof. Let $f_0\,:\!=\,f$ , and for every integer $k\geqslant 1$ and real number $r\gt 0$ , let:

\begin{align*} g_k'(r)&\,:\!=\,8(k+1)r,\\ g_k^*(r)&\,:\!=\,2g_k'(r)+2r,\\ f_k^*(r)&\,:\!=\,f_{k-1}(g_k^*(r)),\\ f_k^{\#}(r)&\,:\!=\,(k+1)(f_k^*(r)+4g_k^*(r)+12r),\text{ and}\\ f_k(r)&\,:\!=\,f_k^{\#}(r)+2g_k^*(r). \end{align*}

Observe that if $f$ is a dilation, then for every integer $k\geqslant 0$ , $g_k'$ , $g_k^*$ , $f_k^*$ , $f_k^{\#}$ , and $f_i$ are all dilations. Also, note that for every integer $k\geqslant 0$ and real number $r\gt 0$ with $r\geqslant \ell$ , $g_k^*(r)\geqslant r\geqslant \ell$ .

Now, let $r$ be defined as in the statement of Lemma 15; we need to show that $G$ admits an $(n'+1,r,f_k(r))$ -colouring with colours $\{1,\ldots ,n'+1\}$ that extends $c_Z$ . We do this via induction, primarily on $k$ and secondarily on $|V(G)|$ .

The base case occurs when $k=0$ ; in this case, for each connected component $C$ of $G$ , $V(C)\subseteq \bigcup _{P\in B_t}P$ for some $t\in V(T)$ . Furthermore, for this $t$ , $C$ is isometric in $G\langle \widehat {\bigcup _{P\in B_t}P}\rangle$ . Since $G\langle \widehat {\bigcup _{P\in B_t}P}\rangle \in \mathcal{H}$ and $r\geqslant \ell$ , $G\langle \widehat {\bigcup _{P\in B_t}P}\rangle$ admits an $(n'+1,r,f(r))$ -colouring, which we may assume is with colours $\{1,\ldots ,n'+1\}$ ; the restriction of this colouring to $V(C)$ then gives an $(n'+1,r,f(r))$ -colouring of $C$ with colours $\{1,\ldots ,n'+1\}$ . As this holds for every connected component of $G$ , taking the union of these colourings gives an $(n'+1,r,f(r))$ -colouring of $G$ with colours $\{1,\ldots ,n'+1\}$ . Since $f(r)=f_0(r)$ and $Z=\emptyset$ as $|S|=0$ , this is the desired colouring. So we may now assume that $k\geqslant 1$ , and that the lemma holds for all smaller values of $k$ and when $k$ is the same but $|V(G)|$ is smaller.

We begin by making a few assumptions. Notice that $(\bigcup _{P\in B_t}P : t\in V(T))$ is a tree-decomposition of $G$ ; let $\widehat {G}$ be the corresponding completion of $G$ . We may assume that $G=\widehat {G}$ . Otherwise, observe that $\mathcal{P}$ is still a partition of $\widehat {G}$ , and that $(B_t\,:\,t\in V(T))$ is still a tree-decomposition for $\widehat {G}/\mathcal{P}$ of adhesion at most $k$ , since the extra edges added to $G/\mathcal{P}$ to make $\widehat {G}/\mathcal{P}$ only go between parts that share a bag. Furthermore, note that edges of $G$ have weight in $\widehat {G}$ no larger than their weight in $G$ ; since no vertices or edges are deleted going from $G$ to $\widehat {G}$ , this means for any $P\in \mathcal{P}$ , $\textrm {rad}(\widehat {G}[P])\leqslant \textrm {rad}(G[P])\leqslant \ell$ . Thus, $\mathcal{P}$ is also an $\ell$ -shallow partition for $\widehat {G}$ . Additionally, observe that $(\bigcup _{P\in B_t}P : t\in V(T))$ is still a tree-decomposition of $\widehat {G}$ , and that the completion of $\widehat {G}$ with respect to $(\bigcup _{P\in B_t}P : t\in V(T))$ is still $\widehat {G}$ . Also, for each $t\in V(T)$ , observe that $\widehat {G}\langle \widehat {\bigcup _{P\in B_t}P}\rangle =\widehat {G}[\bigcup _{P\in B_t}P]=G\langle \widehat {\bigcup _{P\in B_t}P}\rangle \in \mathcal{H}$ ; thus for each $\mathcal{P}_t\subseteq B_t$ , $\widehat {G}\langle \widehat {\bigcup _{P\in B_t}P}\rangle [\bigcup _{P\in \mathcal{P}_t}P]=\widehat {G}[\bigcup _{P\in \mathcal{P}_t}P]=G\langle \widehat {\bigcup _{P\in B_t}P}\rangle [\bigcup _{P\in \mathcal{P}_t}P]\in \mathcal{H}$ . So $(\mathcal{P},(B_t\,:\,t\in V(T)))$ is still a $(k,\ell ,\mathcal{H})$ -strong-construction for $\widehat {G}$ . Finally, as observed when we first defined the completion, we have $\textrm {dist}_G=\textrm {dist}_{\widehat {G}}$ , hence $N_{\widehat {G}}^{3r}(S)=N_{G}^{3r}(S)$ , and any $(n'+1,r,f_k(r))$ -colouring of $\widehat {G}$ is an $(n'+1,r,f_k(r))$ -colouring of $G$ . Thus, we can proceed by setting $G\,:\!=\,\widehat {G}$ .

We may also assume that $S^{\mathcal{P}}$ is nonempty. Otherwise, if $G$ is empty, we are clearly done, and if $G$ is nonempty, pick any part $P'$ of $\mathcal{P}$ and pick some vertex $q'\in V(T)$ with $P'\in B_{q'}$ . Let $S'^{\mathcal{P}}\,:\!=\,\{P\}$ , let $S'\,:\!=\,\bigcup _{P\in S'^{\mathcal{P}}} P=P'$ , let ${Z^{\mathcal{P}}}'$ be the set of parts that intersect $N_G^{3r}(S')$ , and let $c_{Z'}$ be an arbitrary $(n'+1)$ -colouring of $Z'\,:\!=\,\bigcup _{P\in {Z^{\mathcal{P}}}'} V(P)$ with colours $\{1,\ldots ,n'+1\}$ . Observe that $Z$ must be empty, thus any colouring that extends $c_{Z'}$ also extends $c_{Z}$ . Therefore, since $|S'^{\mathcal{P}}|=1\leqslant k$ , we may proceed by setting $S^{\mathcal{P}}\,:\!=\,S'^{\mathcal{P}}$ , $S\,:\!=\,S'$ , $Z^{\mathcal{P}}\,:\!=\,Z'^{\mathcal{P}}$ , $Z\,:\!=\,Z'$ and $c_{Z}\,:\!=\,c_{Z'}$ . Observe also that since parts are nonempty, $S^{\mathcal{P}}$ being nonempty implies that $S$ and hence $Z$ are nonempty.

Henceforth, we can now assume that $S^{\mathcal{P}}$ , and consequently $Z$ , are nonempty, and that $G=\widehat {G}$ . We use the latter assumption implicitly throughout the remainder of the proof. Note that this gives, as mentioned when we argued that we could take $G=\widehat {G}$ , that for each $t\in V(T)$ and each $\mathcal{P}_t\subseteq B_t$ , $G[\bigcup _{P\in \mathcal{P}_t}P]\in \mathcal{H}$ .

For each $e=tt'\in E(T)$ , let $S_e^{\mathcal{P}}\,:\!=\,B_t\cap B_{t'}$ and let $S_e\,:\!=\,\bigcup _{P\in S_e^{\mathcal{P}}}P$ . Note that $|S_e^{\mathcal{P}}|\leqslant k$ ; consequently, $S_e$ is a $(k,r)$ -centred set in $G$ , as $\ell \leqslant r$ .

For each $P\in S^{\mathcal{P}}$ , let $T_P$ be the subgraph of $T$ induced by the vertices $t\in V(T)$ such that $B_t\cap \textrm {Parts}_{\mathcal{P}}(N_G^{3r}(P))\neq \emptyset$ . Since $P$ induces a nonempty connected subgraph of $G$ , $N_G^{3r}(P)$ also induces a nonempty connected subgraph of $G$ , and thus $\textrm {Parts}_{\mathcal{P}}(N_G^{3r}(P))$ induces a nonempty connected subgraph of $G/\mathcal{P}$ . Hence, $T_P$ is nonempty and connected, and for each $tt'\in E(T_P)$ , $B_t\cap B_{t'}\cap \textrm {Parts}_{\mathcal{P}}(N_G^{3r}(P))\neq \emptyset$ . Additionally, $T_P$ contains $q$ as a vertex, as $P\in S^{\mathcal{P}}\subseteq B_q$ .

Let $T'\,:\!=\,\bigcup _{P\in S^{\mathcal{P}}}T_P$ . Notice that $T'$ is connected, as each $T_P$ , $P\in S^{\mathcal{P}}$ , is connected and contains $q$ , and nonempty, as $|S^{\mathcal{P}}|\gt 0$ . Further, notice that $t\in V(T)$ is in $V(T')$ if and only if $B_t\cap Z^{\mathcal{P}}\neq \emptyset$ , and that $B_t\cap B_{t'}\cap Z^{\mathcal{P}}\neq \emptyset$ for any $tt'\in E(T')$ . Let $G'\,:\!=\,\bigcup _{t\in V(T')}G[\bigcup _{P\in B_t} P]$ . Since $T'$ is a subtree of $T$ , by a prior observation we made when we first defined the completion, $G'$ is isometric in $G$ . Also, notice that for any part $P\in \mathcal{P}$ , $G[P]$ is either a subgraph of $G'$ , or $P$ is disjoint from $V(G')$ ; let $\mathcal{P}'\,:\!=\,\textrm {Parts}_{\mathcal{P}}(V(G'))$ , we thus have that $\mathcal{P}'$ is an $\ell$ -shallow partition for $G'$ . Finally, notice that $Z\subseteq V(G')$ .

Next, let $E'$ denote the set of edges between $T'$ and $T-V(T')$ . For each $e\in E'$ , let $T_e$ be the connected component of $T-V(T')$ incident to $e$ ; note that $e$ is the only edge between $T_e$ and $T-V(T_e)$ . Thus, $T_e$ is disjoint from $T_{e'}$ , $e'\in E'\setminus \{e\}$ , and there is a unique vertex $q_e\in V(T_e)$ adjacent to $T-V(T_e)$ ; in particular $q_e$ is adjacent to $T'$ and $e$ is incident with $q_e$ . Additionally, notice that $V(T')\cup \bigcup _{e\in E'}V(T_e)=V(T)$ ; in particular, for each $e\in E'$ , $V(T)\setminus V(T_e)=V(T')\cup \bigcup _{e'\in E'\setminus \{e\}}V(T_{e'})$ . Finally, notice that because $V(T_e)\subseteq V(T)\setminus V(T')$ , for each $t\in V(T_e)$ , $B_t\cap Z^{\mathcal{P}}=\emptyset$ .

Now, for each $e\in E'$ , let $G_e\,:\!=\,\bigcup _{t\in V(T_e)}G[\bigcup _{P\in B_t}P]$ , and let $\widetilde {G}_e\,:\!=\,\bigcup _{t\in V(T)\setminus V(T_e)}G[\bigcup _{P\in B_t}P]$ . Since the only edge between $T_e$ and $T-V(T_e)$ was $e$ , whose endpoints are $q_e$ and some vertex of $T'$ , note that $V(G_e\cap \widetilde {G}_e)=S_e\subseteq V(G')$ and that $S_e^{\mathcal{P}}\subseteq B_{q_e}$ . Additionally, using the same reasoning as with $G'$ , notice that $G_e$ is an isometric subgraph of $G$ , and that $\mathcal{P}_e\,:\!=\,\textrm {Parts}_{\mathcal{P}}(V(G_e))$ is an $\ell$ -shallow partition of $G_e$ . Finally, notice that $Z\cap V(G_e)=\emptyset$ , as $Z^{\mathcal{P}}\cap B_t=\emptyset$ for each $t\in V(T_e)\subseteq V(T)\setminus V(T')$ .

We now seek to apply Lemma 10 on $G$ , using $G'$ as $G_0$ and $(G_e:e\in E')$ as $G_1,\ldots ,G_a$ . We use $r$ as both $r$ and $\ell$ , $n'+1$ for $m$ , $\{1,\ldots ,n'+1\}$ for $C$ , $k$ as itself, $f_k^{\#}(r)$ for $d$ , and $f_k(r)$ for $D$ . Notice that $\ell '$ is $g_k'(r)$ and $r'$ is $g_k^*(r)$ , hence $D=f_k(r)\geqslant d+2r'=f_k^{\#}(r)+2g_k^*(r)$ , as required.

To begin, notice that each vertex or edge of $G$ is contained in $G[\bigcup _{P\in B_t}P]$ for some $t\in V(T)$ . Since $V(T')\cup \bigcup _{e\in E'} V(T_e) = V(T)$ , for each $t\in V(T)$ , $G[\bigcup _{P\in B_t}P]$ is a subgraph of either $G'$ , or $G_e$ for some $e\in E'$ . Thus, $G=G'\cup \bigcup _{e\in E'}G_e$ , and we have satisfied Lemma 10 (a).

Next, for each $e\in E'$ , recall that $S_e=V(G_e\cap \widetilde {G}_e)\subseteq V(G')$ and that $S_e$ is a $(k,r)$ -centred set in $G$ . Since $V(T)\setminus V(T_e)=V(T')\cup \bigcup _{e'\in E'\setminus \{e\}}V(T_{e'})$ , we have $\widetilde {G}_e=G'\cup \bigcup _{e'\in E'\setminus \{e\}}G_{e'}$ . Thus, the $\widetilde {G}_e$ [resp. the $S_e$ ], $e\in E'$ , are the $\widetilde {G}_i$ [resp. the $S_i$ ], $i\in \{1,\ldots ,a\}$ , in the statement of Lemma 10, so we have satisfied Lemma 10 (b).

Now, consider Lemma 10 (c). Notice that $\mathcal{P}'\setminus Z^{\mathcal{P}}$ is an $\ell$ -shallow partition of $G'-Z$ , and that $(B_t\setminus Z^{\mathcal{P}}:t\in V(T'))$ is tree-decomposition for $(G'-Z)/\mathcal{P}'$ ; since $B_t\cap B_{t'}\cap Z^{\mathcal{P}}\neq \emptyset$ for each $tt'\in E(T')$ , we find that $(B_t\setminus Z^{\mathcal{P}}:t\in V(T'))$ has adhesion at most $k-1$ . Also, notice that $(\bigcup _{P\in B_t\setminus Z^{\mathcal{P}}}P:t\in V(T'))$ is a tree-decomposition of $G'-Z$ ; fix $t\in V(T')$ and consider $(G'-Z)\langle \widehat {\bigcup _{P\in B_t \setminus Z^{\mathcal{P}}}P}\rangle$ . For any $t'\in V(T')$ adjacent to $t$ and any $u,v\in \bigcup _{P\in (B_t\cap B_{t'})\setminus Z^{\mathcal{P}}}P$ , observe that $uv\in E(G)$ and hence $uv\in E(G'-Z)$ . Additionally, notice that for each $uv\in E(G'-Z)$ , $uv$ has weight $\textrm {dist}_G(u,v)$ in both $G$ and $G'-Z$ ; this forces $\textrm {dist}_G(u,v)=\textrm {dist}_{G'-Z}(u,v)$ . It follows that $(G'-Z)\langle \widehat {\bigcup _{P\in B_t \setminus Z^{\mathcal{P}}}P}\rangle =(G'-Z)[\bigcup _{P\in B_t\setminus Z^{\mathcal{P}}}P]=G[\bigcup _{P\in B_t\setminus Z^{\mathcal{P}}}P]$ . Thus, for any $\mathcal{P}_t\subseteq B_t\setminus Z^{\mathcal{P}}\subseteq B_t$ , $(G'-Z)\langle \widehat {\bigcup _{P\in B_t \setminus Z^{\mathcal{P}}}P}\rangle [\bigcup _{P\in \mathcal{P}_t}P]=G[\bigcup _{P\in \mathcal{P}_t}P]\in \mathcal{H}$ . Therefore, $(\mathcal{P}'\setminus Z^{\mathcal{P}},(B_t\setminus Z^{\mathcal{P}}:t\in V(T'))$ is a $(k-1,\ell ,\mathcal{H})$ -strong-construction for $G'-Z$ .

Thus, using $g_k^*(r)\geqslant \ell$ as $r$ and recalling that $f_{k-1}(g_k^*(r))=f_k^*(r)$ , we may apply the induction hypothesis, keeping $q$ unchanged and letting $S^{\mathcal{P}}$ be empty, to find an $(n'+1,g_k^*(r),f_k^*(r))$ -colouring $c'$ of $G'-Z$ with colours $\{1,\ldots ,n'+1\}$ . Now, recall that $\mathcal{P}'$ is an $\ell$ -shallow partition for $G'$ ; thus, for each $P\in \mathcal{P}'$ , $\textrm {wdiam}_{G'}(P)\leqslant 2\ell \leqslant 2r$ . Also, as $N_G^{3r}(S)\subseteq Z\subseteq V(G')$ notice that every $v\in N_G^{3r}(S)$ is still a distance at most $3r$ from $S$ in $G'$ . Let $S^*$ be a set containing, for each $P\in S^{\mathcal{P}}$ , exactly one vertex $v$ which is at distance at most $\ell \leqslant r$ from every other vertex in $G'[P]$ (which exists, since $\mathcal{P}$ is $\ell$ -shallow). Then for any $z\in Z$ , $\textrm {dist}_{G'}(z,N^{3r}_G(S))\leqslant 2r$ , $\textrm {dist}_{G'}(z,S)\leqslant 5r$ , and $\textrm {dist}_{G'}(z,S^*)\leqslant 6r$ . Since $|S^{\mathcal{P}}|\leqslant k$ , $|S^*|\leqslant k$ , and thus $Z$ is $(k,6r)$ -centred set in $G'$ . Since $(k+1)(f_k^*(r)+4g_k^*(r)+12r)=f_k^{\#}(r)$ , we thus have that the colouring $c''\,:\!=\,c'\cup c_Z$ of $G'$ is an $(n'+1,g_k^*(r),f_k^{\#}(r))$ -colouring with colours $\{1,\ldots ,n'+1\}$ via Corollary 9. So Lemma 10 (c) is satisfied with $c''$ as $c_0$ . Finally, note that $c''$ also extends $c_Z$ , by definition.

Now, for each $e\in E'$ , define $c_{S_e}\,:\!=\,c''\big |_{V(S_e)}$ . Let $Z_e^{\mathcal{P}}\,:\!=\,\textrm {Parts}_{\mathcal{P}}(N_{G_e}^{3r}(S_e))$ and $Z_e\,:\!=\,\bigcup _{P\in Z_e^{\mathcal{P}}} P$ , we can find a colouring $c_{Z_e}$ of $Z_e$ with colours $\{1,\ldots ,n'+1\}$ that extends $c_{S_e}$ and has an $(S_e,r)$ -barrier in $G_e$ . Recall that $\mathcal{P}_e$ is an $\ell$ -shallow partition of $G_e$ , and notice that $(B_t\,:\,t\in V(T_e))$ is a tree-decomposition for $G_e/\mathcal{P}_e$ of adhesion at most $k$ , and that $(\bigcup _{P\in B_t}P:t\in V(T_e))$ is a tree-decomposition for $G_e$ . For any $t\in V(T_e)$ , a similar argument to the one used for the weighted torsos of $G'-Z$ shows that $G_e\langle \widehat {\bigcup _{P\in B_t}P}\rangle =G_e[\bigcup _{P\in B_t}P]=G[\bigcup _{P\in B_t}P]$ ; thus, for any $\mathcal{P}_t\subseteq B_t$ , $G_e\langle \widehat {\bigcup _{P\in B_t}P}\rangle [\bigcup _{P\in \mathcal{P}_t}P]=G[\bigcup _{P\in \mathcal{P}_t}P]\in \mathcal{H}$ . It follows that $(\mathcal{P}_e, (B_t\,:\,t\in V(T_e)))$ is a $(k,\ell ,\mathcal{H})$ -strong-construction for $G_e$ . Finally, note that $|V(G_e)|\leqslant |V(G)-Z|\lt |V(G)|$ , as $Z\cap V(G_e)=\emptyset$ and $Z$ is nonempty. Hence, we can apply induction on $G_e$ , using $r$ as itself, $q_e$ as $q$ , $S_e^{\mathcal{P}}\subseteq B_{q_e}$ as $S^{\mathcal{P}}$ , and $c_{Z_e}$ as $c_Z$ . This allows us to extend $c_{Z_e}$ to an $(n'+1,r,f_k(r))$ -colouring $c_e$ of $G_e$ with colours $\{1,\ldots ,n'+1\}$ . Note that since $c_e$ extends $c_{Z_e}$ , $c_e$ has an $(S_e,r)$ -barrier in $G_e$ , and that $c_e=c''$ on $S_i$ , by definition of $c_{S_e}$ . Hence, Lemma 10 (d) is satisfied with the various $c_e$ , $e\in E'$ , acting as $c_1,\ldots ,c_a$ .