Meridian-disk criterion for unknot recognition

Suppose first that $K$ is smoothly ambient isotopic to the unknot. Carry the standard smooth spanning disk back along that isotopy, obtaining a smooth embedded spanning disk $B$ for $K$. In the product coordinates of the supplied tube, a sufficiently small collar of $\mathrm{\partial }B=K$ has positive radial distance away from its boundary, with nonzero inward derivative there: the disk's inward tangent is linearly independent of the knot tangent, hence its normal projection is nonzero. Compactness of $K$ makes this estimate uniform. A compact portion of $B$ outside this collar is disjoint from $K$, so it misses all sufficiently small radius tubes. Thus for a sufficiently small $\delta >0$, $B$ meets the radius-$\delta$ tube in precisely that disk collar, and the radial level $\delta$ cuts a simple boundary circle transversely. The remaining disk $B_{\delta }$ is a smooth transverse proper disk in the radius-$\delta$ exterior. Its boundary has degree one along the core, because the original disk collar projects to $K$ with degree one; it is therefore non-null-homotopic on the radius-$\delta$ torus.

Conversely, let $D$ be a proper PL exterior disk with essential boundary $C$. By [F1], $[C]=\pm [\lambda ]$, so its degree under the product projection to the core is $1$ or $-1$. Apply [F2] to obtain an annulus $A$ on the tube side of the torus with boundaries $C$ and $K$, with smooth collar at $K$. Its interior misses $D$, because the interiors lie on opposite sides of the torus. The union $B=D{\cup }_{C}A$ is consequently an embedded disk with boundary exactly $K$.

The closed radius-$ϵ$ disk bundle is compact and is contained in the open tubular domain $\mathrm{\Omega }$. In the supplied product trivialization, compactness gives a number $R>ϵ$ for which the closed radius-$R$ disk bundle is still contained in $\mathrm{\Omega }$. Choose $\delta <ϵ$ as in step 1.1. Choose a smooth strictly increasing radial function $\rho :[0,R]\to [0,R]$ with $\rho (0)=0$, $\rho (\delta )=ϵ$, linear near zero and equal to the identity near $R$. Interpolation of positive derivatives with the prescribed interval integrals constructs such a function. The fibrewise map $z\mapsto \rho (\mathrm{\mid }z\mathrm{\mid })z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, extended at zero by its positive linear derivative, and the identity outside the larger tube give a smooth ambient diffeomorphism carrying the radius-$\delta$ exterior to the supplied exterior. It takes $B_{\delta }$ to a smooth transverse proper disk and preserves core degree on the boundary. Apply the PL-approximation clause of [F3] in the compatible structure [F5]. Its PL proper disk has boundary isotopic to the original degree-one circle and therefore essential. This proves the unknot-to-PL-disk direction for the actual supplied tube.

The disk $B$ satisfies the finite-rounding hypothesis in [F3]. On $D$ the required simplex extensions and injective fans come from [F5]. Along $C$, use the product collar of the torus: an exterior disk collar and the first annulus strip lie on its opposite sides. Their half-fans meet precisely in the tangent fan of $C$, so the combined fan is injective. Subdivide along the finitely many corner and isotopy seams of the annulus; its graph strips, radial-level annuli and product core collar have the piecewise smooth regular charts constructed in [F2]. Any join can be given matching collar parametrizations, and the graph-coordinate corner rounding is performed away from the protected core collar. Thus the union has a finite whole-star piecewise smooth surface triangulation with an already smooth collar of $K$. The relative smoothing clause of [F3] gives a smooth embedded disk with the very same boundary $K$.

By [F4], this smooth spanning disk makes the Seifert genus zero, and the given countable choice extends the disk-boundary isotopy to a smooth ambient isotopy to the standard oriented unknot. Together with step 2.1 this proves the asserted equivalence. For every essential disk, independently of its choice or of the smoothing construction, [F1] already gives $[\mathrm{\partial }D]=\pm [\lambda ]$ and odd mod-two intersection with the original marked meridian $\mu$. All clauses of the Statement follow under the original countable-choice hypothesis.