A boundary-parallel disk fails the meridian test

The link $L{k}_E(v)$ of a boundary vertex in a triangulated $3$-manifold is a closed disk: its local link is a $2$-ball. In each tetrahedron containing $v$ the link meets that tetrahedron in the small triangle parallel to the opposite face; it misses tetrahedra not containing $v$. Across each shared face these triangles have matching normal arcs, so [F1] makes the link a normal surface. Its boundary is $L{k}_T(v)$, a simple closed curve in the small boundary-star disk of $v$, hence contractible in $T$. The local half-ball star shows the disk is boundary-parallel.

The link disk is connected, has nonempty boundary, and has Euler characteristic one. Its boundary is null-homotopic, so its projection to the product-circle factor has degree zero. After a transverse PL perturbation, the parity of intersections with a fibre meridian equals the parity of preimages of a regular point under that projection; a lift with zero net degree crosses the level an even number of times. Hence the intersection parity is zero.

The disk therefore passes the first three tests in [F2] but fails the odd-intersection test, so the filter rejects this candidate. This conclusion concerns this boundary-parallel disk only; it does not rule out any other normal essential disk in $E$ and therefore does not classify the knot.