Based diagrams admit spherical long endpoint presentations

Straighten only a crossing-free marked germ. Translate the mark to $z=0$ and rotate its oriented tangent to the positive real axis, by paths of positive planar similarities. Since the mark is a regular non-double point, there is a small disk meeting the diagram in just one arc. The inverse-function theorem expresses that arc as $y=f(x)$, where $f(0)=f^{\mathrm{\prime }}(0)=0$. It can be straightened by the supported shear $(x,y)\mapsto (x,y-t\chi (x,y)f(x))$, $0\le t\le 1$: choose a cutoff equal to one near the smaller graph segment and supported in a rectangle of width and height comparable to a sufficiently small $r$. There $f(x)=O(r^2)$, whereas ${\mathrm{\partial }}_y\chi =O(r^{-1})$. Thus $1-t({\mathrm{\partial }}_y\chi )f>0$ after shrinking $r$. Each vertical coordinate map is strictly increasing and the identity outside a compact interval, so each shear is a global orientation-preserving diffeomorphism; its inverse is smooth by its positive vertical derivative. Choose nested neighborhoods so that after this shear the diagram in $\mathrm{\mid }z\mathrm{\mid }<2r$ is the real diameter and no other strand meets that disk. Shrink $r<1$ if necessary. The part not in the real interval $\mathrm{\mid }x\mathrm{\mid }<r$ consequently satisfies $\mathrm{\mid }z\mathrm{\mid }\ge r$. All these isotopies merely transport the mark and crossing decorations; their local supports meet no crossing.

An actual rotation recharting. Choose $0<ϵ<\min (r\mathrm{/}4,1\mathrm{/}4)$ and set $a=iϵ$. On the oriented stereographic sphere use the fractional linear coordinate $w=-\frac{1+\bar{a}z}{z-a}\mathrm{.}$ The matrix $\left(\begin{array}{cc}-\bar{a}& -1\\ 1& -a\end{array}\right)\mathrm{/}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}$ is unitary with determinant one: its columns have norm one and scalar product zero, and its unnormalized determinant is $1+\mathrm{\mid }a{\mathrm{\mid }}^2$. Hence its fractional linear action is the action of an orientation-preserving rotation of the round two-sphere. Its omitted old point is $a$, which lies off the real germ in the marked disk, hence in a face adjacent to the mark and outside all crossing disks. The old omitted point $\mathrm{\infty }$ is sent to $-\bar{a}$, which is an ordinary point of the new chart. The map is smooth on the entire spherical diagram and preserves the oriented local crossing charts and all transported decorations.

A quantitative exterior closing arc. For every old diagram point outside the interval $\mathrm{\mid }x\mathrm{\mid }<r$, step 1.1 and the triangle inequality give $\mathrm{\mid }w\mathrm{\mid }\le \frac{1+ϵ\mathrm{\mid }z\mathrm{\mid }}{\mathrm{\mid }z\mathrm{\mid }-ϵ}\le B:=\frac{1+ϵr}{r-ϵ}\mathrm{.}$ The second inequality holds because the rational function has derivative $-(1+{ϵ}^2)\mathrm{/}(s-ϵ{)}^2<0$ for $s\ge r$. The inequality $B<1\mathrm{/}ϵ$ is equivalent to $2ϵ<r(1-{ϵ}^2)$, which follows from $ϵ<r\mathrm{/}4$ and $ϵ<1\mathrm{/}4$. Choose a real $R$ strictly between $B$ and $1\mathrm{/}ϵ$. On the real germ $\mathrm{\mid }w(x){\mathrm{\mid }}^2=\frac{1+{ϵ}^2{x}^2}{x^2+{ϵ}^2},$ whose derivative with respect to $x^2$ is $({ϵ}^4-1)\mathrm{/}(x^2+{ϵ}^2{)}^2<0$. Since the value at $x=0$ is $1\mathrm{/}{ϵ}^2>R^2$ and at $\mathrm{\mid }x\mathrm{\mid }=r$ is at most $B^2<R^2$, the circle $\mathrm{\mid }w\mathrm{\mid }=R$ meets that germ at exactly $x=\pm c$, with $c^2=\frac{1-R^2{ϵ}^2}{R^2-{ϵ}^2},0<c<r\mathrm{.}$ Both intersections are transverse, since $c\mathrm{\ne }0$ and the derivative just computed is nonzero there. The only diagram points outside the closed disk $\mathrm{\mid }w\mathrm{\mid }\le R$ form the single open arc parametrized by $-c<x<c$; its closure contains the mark and has no crossing. The image is a round circular arc, because a fractional linear map sends a circle on the sphere to a circle and its pole $iϵ$ is not on the real circle. Every other strand and every crossing is strictly inside the disk. Thus this is the claimed exterior closing arc, with uniform crossing-free neighborhoods of it and of its two cut points.

Record spatial coordinate identifications. A planar shear in step 1.1 lifts on a spatial projection chart by $(x,y,h)\mapsto ({\mathrm{\Phi }}_t(x,y),h)$. Its compact planar support can also be cut off in height on a larger box containing the link: use its compactly supported time-dependent planar velocity multiplied by a height cutoff equal to one on that box, with zero height velocity. The flow is complete, agrees with the displayed lift near the link, and extends by the identity to $S^3$. The initial translations and rotations have their usual orientation-preserving similarity lifts and smooth conformal extensions at the point at infinity. For the rechart, realize the projection sphere as the equator of $S^3\subset {\mathbb{R}}^3\oplus \mathbb{R}$. If $Q\in SO(3)$ is the rotation from step 2.1, then $(v,s)\mapsto (Qv,s)$ is an orthogonal determinant-one map of ${\mathbb{R}}^4$ preserving $S^3$ and preserving its orientation. On the equatorial collar it transports the projection direction, the signed height, and the over/under datum. The Euclidean projection and the equatorial product projection require an explicit chart identification. Write $q=(0,0,1,0)$ for the spatial stereographic point, $z=x+iy$, $X(z)=\frac{(2x,2y,\mathrm{\mid }z{\mathrm{\mid }}^2-1)}{1+\mathrm{\mid }z{\mathrm{\mid }}^2},d(z)=\frac{2}{1+\mathrm{\mid }z{\mathrm{\mid }}^2}\mathrm{.}$ Standard inverse spatial stereographic projection is $E(z,h)=\frac{(2x,2y,\mathrm{\mid }z{\mathrm{\mid }}^2+h^2-1,2h)}{1+\mathrm{\mid }z{\mathrm{\mid }}^2+h^2}\mathrm{.}$ The product-collar chart with the same first normal derivative is $T(z,h)=(\sqrt{1-d(z{)}^2{h}^2}X(z),d(z)h)\mathrm{.}$ On $h=0$ these charts agree and have identical full first derivatives: both normal derivatives are $d(z)e_4$, while their tangential derivatives are those of $X$. On a compact planar neighborhood containing the diagram, the coordinate transition $\mathrm{\Psi }=E^{-1}T$ therefore satisfies $\mathrm{\Psi }-\mathrm{i}\mathrm{d}=O(h^2)$ and $D\mathrm{\Psi }-I=O(\mathrm{\mid }h\mathrm{\mid })$ uniformly. Choose a slightly larger compact planar box and cut off this difference there and in a height slab $\mathrm{\mid }h\mathrm{\mid }<\eta$. A height cutoff has derivative $O({\eta }^{-1})$, multiplying an error $O({\eta }^2)$; thus the resulting global compactly supported displacement $V$ has $\mathrm{\parallel }DV\mathrm{\parallel }<1\mathrm{/}2$ for small $\eta$. Every $\mathrm{i}\mathrm{d}+tV$ is injective by the Lipschitz bound, is a local orientation-preserving diffeomorphism, and is onto: it is proper, its image is closed and open, and the plane-height space is connected. Its inverse is smooth by the inverse-function theorem. This is an actual ambient coordinate isotopy, extending by the identity to $S^3$. First shrink any existing Euclidean height realization by a positive constant so it lies where both cutoffs equal one. Height scaling is an embedding isotopy: at every double point both branch heights are multiplied by the same positive factor. The new spatial representative is then exactly $T(z,h)$, with product-collar base coordinate $z$ and signed height $d(z)h$. Because $d(z)>0$, at a double point it preserves the strict over/under ordering. Hence the displayed Euclidean diagram is identified with the product-collar diagram by this explicit spatial map; they are not assumed literally equal off the equator. These are actual marked link isotopies and coordinate identifications with specified endpoints. Composing these coordinate maps and height isotopies gives the asserted spatial identification; no claim that the rotation is a planar isotopy is used.

Compatibility with protected long ends. The exterior arc and two slightly longer joining collars have injective projection and are disjoint in projection from the remaining compact part, except at their adjacent ends. On a narrow planar neighborhood of that arc its spatial height is a smooth function $h(x,y)$, extended from the arc in tubular coordinates. A supported vertical translation $(x,y,z)\mapsto (x,y,z-t\eta (x,y)h(x,y))$ sets its height to zero on a smaller neighborhood; this is a diffeomorphism for every $t$, since its derivative in $z$ is one. It cannot reverse an over/under ordering, and there are no double points in its smaller marked support. Choose the joining neighborhoods slightly longer before cutting, so the resulting cut collars are fixed smooth product collars. The remaining diagram is a compact tangle in the round disk, and the single planar exterior arc is its marked closure. These are precisely endpoint data that may be retained as protected long endpoints in a relative movie construction. Every operation has preserved the transported mark, and none passes it through a crossing. Finite composition proves the assertion, including its recorded spatial coordinate identification. No algebraic comparison of the two charts, no arbitrary mark-slide identity, and no purely planar comparison have been assumed.