and $A'$ for the outermost intersection points.

\[lemma:pindown\] Let $\x=\{x_0,\y\}$ be an outermost intersection point with $\y\in C(i)$. Then

1.  The $\SpinC$ structure $\spinc\in \SpinC(Y)$ associated to $\x$ is independent of $z$ or $z'$, and agrees with the $\SpinC$ structure associated to the corresponding generator for $H$, $\{p,\y\}$.

2.  $A(\{x_0,\y\})=i$. That is, the Alexander grading of $\{x_0,\y\}$ with respect to $(w,z)$ agrees with that of $\{p,\y\}$.

3.  $A'(\{x_0,\y\})= pA(\{x_0,\y\})+\frac{(pn)(p-1)}{2} $

The first two parts follow from the fact that $(1)$ the map $\spinc_w$ does not depend on $z$ or $z'$ and $(2)$ there is an isotopy taking $\tilde{\beta}$ to $\mu$ in the complement of $z$, under which the generator $\{x_0,\y\}$ becomes identified with $\{p,\y\}$. The third is the content of Lemma $3.6$ of [@OSThurston] which establishes the result in the context of multi-pointed Heegaard diagrams for links. One can pass from their result to the present case by setting $l=1$, $p_1=p$, and $q_1=pn+1$ and then considering Chern classes of relative $\SpinC$ structures to determine the Alexander grading.

Next, we have Lemmas $3.3$ and $3.4$ of [@Cabling]

Let $H(p,n)$ be as above. Then for odd integers $i < 2n$, we have $$\begin{gathered}
A(x_{i-1} ,\y)-A(x_{i},\y)=A'(x_{i-1},\y)-A'(x_{i},\y)=1\\
A(x_i,\y)-A(x_{i+1},\y)=0\\
A'(x_i,\y)-A'(x_{i+1},\y)=p-1.\end{gathered}$$

\[lemma:varytuple\] Suppose $\y \in C(j), 