ensures that the rotation angles work correctly to allow an extension of $t$ to obtain an $h_0$ with $h_0^n = t_{{\operatorname{{\mathcal{C}}}}'}$ in ${\operatorname{Mod}}(S_g({\operatorname{{\mathcal{C}}}}\setminus {\operatorname{{\mathcal{C}}}}'))$, where ${\operatorname{{\mathcal{C}}}}'= \cup_{i=1}^r {\operatorname{{\mathcal{C}}}}_i'$.

Now write ${\operatorname{{\mathbb O}}}_1 = \{Q_{1,1}, Q_{1,2}, \ldots, Q_{1,m_1}\}$ and consider disks $D_{1,i}$ around $Q_{1,i}$ such that $t(D_{1,i}) = t(D_{1,i+1})$. Similarly, write ${\operatorname{{\mathbb O}}}_1'=\{Q_{1,1}',Q_{1,2}',\ldots, Q_{1,m_1}'\}$ and consider disks $D_{1,i}'$ as earlier. Then we attach $m_1$ annuli connecting $\partial D_{1,i}$ to $\partial D_{1,i}'$. Each such annulus contains a nonseparating curve $c_{1,i}$, which is unique unto isotopy. Repeating this process for $1\leq i\leq m_1$, we obtain the surface $S_g({\operatorname{{\mathcal{C}}}}\setminus ({\operatorname{{\mathcal{C}}}}'\cup{\operatorname{{\mathcal{C}}}}_1))$. Since $t(D_{1,i}) = D_{1,i+1}$, we may extend the homeomorphism $h_0$ to a homeomorphism $\widetilde{h_0} \in {\operatorname{Mod}}(S_g({\operatorname{{\mathcal{C}}}}'\cup {\operatorname{{\mathcal{C}}}}_1))$, which cyclically permutes the $c_{1,i}$. If $|{\operatorname{{\mathbb O}}}_1| = |{\operatorname{{\mathbb O}}}_1'| = n$, then define $h_1 := \widetilde{h_0}t_{c_{1,1}}$. Otherwise, since ${\operatorname{{\mathbb O}}}_1 \sim {\operatorname{{\mathbb O}}}_1'$, the difference in the turning angles around $Q_{1,i}$ and $Q_{1,i}'$ is $2\pi/n$. Let $\widetilde{h_1}$ be the $(1/n)^{th}$-twist around $c_{1,1}$. Now $h_1 := \widetilde{h_0}\widetilde{h_1}$ is an $(r,1)$-permuting root of $t_{{\operatorname{{\mathcal{C}}}}' \cup {\operatorname{{\mathcal{C}}}}_1}$ of degree $n$ in ${\operatorname{Mod}}(S_g({\operatorname{{\mathcal{C}}}}\setminus ({\operatorname{{\mathcal{C}}}}' \cup {\operatorname{{\mathcal{C}}}}_1)))$. We now repeat this process inductively to obtain an $(r,k)$-permuting root $h:= h_k\in {\operatorname{Mod}}(S_g)$ of $t_{{\operatorname{{\mathcal{C}}}}}$ of degree $n$.

It remains to show 