Smooth-to-piecewise-linear knot exterior approximation

If a projection diagram $D$ is supplied, work in its fixed ${\mathbb{R}}^3$ chart and projection coordinates; the compact knot has a smaller tubular neighborhood contained in that chart. Without a supplied diagram, choose a point outside the compact tube image and use its stereographic chart. In either case all chords and their roundings are chosen in a sufficiently small tube about $K$, inside the supplied tube. Reparameterize $K$ by arclength on the circle. The tubular projection to $K$ is smooth and has derivative equal to the identity along $K$. Compactness of $K$ and continuity of this derivative give a uniform $\delta >0$ such that every chord joining $K(s)$ to $K(s+h)$, with $0<h<\delta$, lies in the tube and has strictly increasing tubular projection from $s$ to $s+h$. Indeed, the chord direction divided by $h$ converges uniformly to $K^{\mathrm{\prime }}(s)$ as $h\to 0$, and the differential of the projection converges uniformly to its value on the tangent. Choose a uniform finite partition of the parameter circle with mesh less than $\delta$ and join successive samples by these chords. Strict increase of the projection on each chord makes their union an embedded polygonal circle $P$ and makes its projection to $K$ a homeomorphism. If $D$ is supplied, its finitely many disjoint crossing disks and separated compact arc portions from [F6] give positive margins away from the transverse crossing neighborhoods; choose the mesh still finer so each polygonal strand stays within those margins and remains transverse with the same over/under height order in each crossing disk. Thus no new crossing occurs off those disks and exactly the declared crossing occurs in each disk.

For a polygonal knot $P$, choose pairwise disjoint small balls around its finitely many vertices, disjoint from the marked meridian point and any crossing charts. In each ball the knot is a single embedded polygonal arc with two straight end segments. Choose a linear coordinate along which both segments project monotonically; then the arc is the graph of a piecewise-linear transverse section over an interval. Smooth that section inside a smaller interval, fixed near its endpoints, to obtain a rounded arc. The changed arcs fit the unchanged edges smoothly, so the resulting $P^{\mathrm{♯}}$ is a smooth knot and its diagram code is unchanged when crossing charts are avoided.

In tube coordinates $P$ is the graph of a continuous, piecewise-smooth normal section $u:S^1\to D_{ϵ\mathrm{/}4}^2$. Smooth each of its finitely many corners inside pairwise disjoint short parameter intervals, leaving the section unchanged outside those intervals, to obtain a smooth section $q$ uniformly close to $u$. The graph $K^{\mathrm{\prime }}(s)=\mathrm{\Phi }(s,q(s))$ is a smooth embedding because its first coordinate is $s$. The family $K_t(s)=\mathrm{\Phi }(s,tq(s))$ consists of embeddings for the same reason, starts at $K$, ends at $K^{\mathrm{\prime }}$, and stays inside the supplied tube.

Apply [F5] to the smooth isotopy $K_t$ with an open support neighborhood whose closure lies inside the radius-$2ϵ$ tube. It gives an ambient smooth isotopy supported there carrying $K$ to $K^{\mathrm{\prime }}$. Its time-one diffeomorphism transports the chosen tube and its fibre meridians. This is the only use of ${\mathrm{A}\mathrm{C}}_{\omega }$ in the argument.

The two sections $u$ and $q$ lie in the interior quarter-disk. In each fibre disk, rescale to the unit disk and use the orientation-preserving disk automorphisms $M_a(z)=(z-a)\mathrm{/}(1-\bar{a}z)$: the map $M_{q(s)}^{-1}\circ M_{u(s)}$ sends $u(s)$ to $q(s)$. These maps depend continuously on $s$ and send each boundary circle to itself. On the outer radial collar between radii $ϵ$ and $2ϵ$, lift each boundary-circle map to an increasing angle function $h_s(\theta )$ close to $\theta$ and interpolate it linearly with $\theta$ as the radius increases; convex combinations of increasing lifts remain increasing. The resulting fibre maps agree at radius $ϵ$, are the identity at radius $2ϵ$, and are homeomorphisms on every fibre. Their union, extended by the identity outside the larger tube, is a homeomorphism of $S^3$ carrying $P$ to $K^{\mathrm{\prime }}$. It preserves the radius-$ϵ$ boundary torus and each fibre circle as a set, so it transports the marked meridian and identifies the compact exteriors.

In each such ball, the old and rounded arcs are graphs of two close sections in a product chart. A fibrewise disk homeomorphism taking one section to the other, extended across a collar to be the identity near the ball boundary, gives a homeomorphism supported in the ball. The finitely many disjoint homeomorphisms paste together; they carry the regular neighborhood, exterior, boundary torus and meridian class to those of $P^{\mathrm{♯}}$. This proves the converse rounding clause. Combining it with steps 3.1–3.2 proves all stated comparison clauses.