local pictures of $(B^3, B^3\cap K_+)$ (resp. $(B^3, B^3\cap K_-)$) are given by Figure \[fig:crossingchange\].

![\[fig:crossingchange\] A crossing change. $K_+,K_-\subset S^3$ agree, except in the three-ball pictured here.](crossingchange.eps)

Corollary $1.5$ of [@FourBall] states that if $K_+$ and $K_-$ differ by a crossing change, then we have the following inequality: $$\label{eq:crossing}
\tau(K_+)-1\le \tau(K_-) \le \tau(K_+).$$ Now it is straightforward to see that $K_{p,pl+1}$ can be changed into $K_{p,p(l-1)+1}$ by a sequence of $\frac{p(p-1)}{2}$ crossing changes, each of which changes a positive crossing to a negative (to see this induct on $p$ and, for the induction step, change the first $p-1$ crossings of a full twist on $p$ strands). Thus $$\label{eq:ninequality} \tau(K_{p,pl+1}) -\frac{p(p-1)}{2} \le \tau(K_{p,p(l-1)+1})\le \tau(K_{p,pl+1}).$$ Theorems \[thm:filtY\] and \[thm:filtYneg\] tell us that Inequality is satisfied for $K_{p,pn+1}$ provided $|n|>N$. Combining this with Inequality yields for all $n$.

The proof of Inequality will easily extend to each of the invariants $\tau_\alpha(Y,K)$, once a generalization of the crossing change inequality is established for knots in $Y$. We defer the proof of this latter inequality to [@Slice], where it will follow from the fact that $|\tau_\alpha(Y,K)|$ bounds the genus of any smoothly embedded surface $$i: (F,\partial F) \hookrightarrow Y\times [0,1],$$ with $i|_{\partial F} = 