Adjacent-index handles with zero intersection do not cancel

Take $W_1=S^1\times D^2$ with its boundary torus $S^1\times S^1$; the belt sphere of the $1$-handle is the meridian $\{p\}\times S^1$. The attaching circle $\gamma$ bounds a closed disk in $\mathrm{\partial }{W}_1$ that avoids $\{p\}\times S^1$, so $\gamma$ is disjoint from the belt sphere and, bounding a disk in the boundary torus, is null-homotopic in $W_1$.

Since the two circles are disjoint, the attaching-belt matrix entry is $0$ over both $\mathbb{Z}$ and ${\mathbb{Z}}_2$; in particular the pair is not geometrically cancelling, because a geometrically cancelling pair has a unit entry by [F2].

Thicken the old stage and handle slightly across their seam to obtain open path-connected sets $U,V$, with $U\simeq W_1$, $V\simeq D^2\times D^1$, and $U\cap V\simeq S^1\times D^1$. By [F3], ${\pi }_1(U\cup V)$ is the pushout of ${\pi }_1(W_1)\leftarrow \mathbb{Z}\to 1$; the map into ${\pi }_1(W_1)$ is represented by $\gamma$, which is null-homotopic by step 1.1. The pushout is therefore ${\pi }_1(W_1)$. The product contraction $S^1\times D^2\to S^1\times \{0\}$ gives ${\pi }_1(W_1)\cong \mathbb{Z}$, so ${\pi }_1(W_1\cup h^2)\cong \mathbb{Z}$. This is a direct handle-gluing argument and does not assume an unspecified one-critical-point Morse presentation.

By [F4] a cancellation of the pair would give a diffeomorphism $W_1\cup h^2\cong W_0=D^3$ relative to the lower stage, hence an isomorphism of fundamental groups $\mathbb{Z}\cong 1$, which is impossible. Therefore the pair does not cancel, even though both the algebraic and the geometric intersection counts are zero.

Consequently the single-point criterion of the cancellation theorem cannot be replaced by a count-only condition: an adjacent-index pair with vanishing (algebraic and geometric) intersection need not cancel, and no diffeomorphism relative to the lower stage removes the pair.