Algebraic cancellation does not yet give geometric cancellation

In the model $W=D^4\cup h^2$ the outgoing boundary is the boundary of the manifold obtained by attaching the standard $2$-handle; by [F2] this manifold is $S^2\times D^2$, so $N\cong S^2\times S^1$ and the belt sphere of $h^2$ is $B=\{p\}\times S^1$ for a point $p\in S^2$.

Let $A_0=S^2\times \{u_0\}\subseteq N$. Then $A_0$ meets $B$ transversely in the single point $(p,u_0)$, whose local sign is $+1$ for the product orientation of $S^2\times S^1$; the oriented and mod-2 intersection numbers of $A_0$ with $B$ are both $1$.

Perform a finger move of $A_0$ across $B$: choose a small embedded disk $D\subseteq A_0$ disjoint from $(p,u_0)$ and replace $D$ by a thin finger disk along a short arc starting normally at $D$, with interior off $A_0$, and passing across a short segment of $B$, the finger is the lateral boundary and end cap of a thin tubular cylinder, joined to $\mathrm{\partial }D$ and smoothed, producing an embedded $2$-sphere $A$ that agrees with $A_0$ outside a small neighbourhood of $D$ and crosses $B$ in two new transverse points. The two new intersections have opposite local signs, because $B$ enters and exits the finger cylinder through its two lateral walls; their induced outward normal directions are opposite, so the ordered tangent determinants have opposite signs; no other intersections are created or destroyed.

Hence $A\cap B$ consists of the original point, of sign $+1$, together with the finger pair of opposite signs; after orienting $A$ so that the original point keeps sign $+1$, the three local signs are $+1,+1,-1$ up to the order of the pair. By [F4] the oriented intersection number is $I(A,B)=1$ and the mod-2 number is $1$, while $A\cap B$ has three points.

The finger is an isotopy of the original sphere, so its product normal line framing is transported and gives an attaching embedding $A\times D^1\to N$. Reading the configuration as handle data, $A$ is the attaching sphere of a $3$-handle attached to $N$ and $B$ is the belt sphere of the $2$-handle $h^2$; by [F5] the attaching-belt matrix entry is $I(A,B)=1$, a unit, yet the spheres do not meet in exactly one point. The single-point hypothesis of the cancellation theorem is therefore strictly stronger than a unit or odd algebraic count, and the extra pair in this example can be removed by the inverse finger isotopy. The example proves the failure of the converse for the displayed configuration, not an obstruction to cancellation after further isotopy.