Circumcision by a framed core and zero-framed meridian preserves the boundary parametrization

Product exterior and actual slopes. Use the abstract product coordinates of $N$ to draw it as an unknotted solid torus in $S^3$. Let $A$ be the core of the complementary solid torus. Then $K$ is linked once with $A$ and $U$, while $A,U$ are disjoint meridians of $K$. The exterior of $A\cup K\cup U$ is $X=P\times S^1,$ where $P$ is a disk with two disjoint interior disks removed. One way to see this product without a diagram is to remove $K$ first: its exterior in $S^3$ is $D^2\times S^1$, where the circle factor is the meridian of $K$; the two meridians $A,U$ are fibers $\{a\}\times S^1$ and $\{u\}\times S^1$. Removing their product tubular neighborhoods removes two disks from the first factor. The boundary tori are $T_K,T_A,T_U$, corresponding to the three boundary circles of $P$. Write $h$ for the positive product fiber and $b_K,b_A,b_U$ for the three oriented base-boundary circles. In the usual meridian/longitude coordinates, ${\mu }_K=h,{\lambda }_K=b_K,{\mu }_U=b_U,{\lambda }_U=h\mathrm{.}$ The retained external torus is $T_A$; its solid-torus meridian in the original $N$ is $h$. Thus the two filling slopes are $b_K+fh$ and $h$, where $f$ is the integer framing on $K$ relative to the selected product longitude. All these are identifications of actual product annuli and circles; no homology-only recognition of a filled manifold is being used.

Framing reduction relative to the retained torus. There is a smooth map $\xi :P\to S^1$ of degree $f$ on the $K$ boundary, degree $-f$ on the $U$ boundary (with the common oriented-boundary convention), and constant one on a collar of the $A$ boundary. Explicitly draw $P$ as a planar disk punctured at $k,u$, where $k$ and $u$ are the removed disks around $K$ and $U$. With the induced boundary orientations, start with ${\xi }_0(z)={\left(\frac{z-u}{\mathrm{\mid }z-u\mathrm{\mid }}\frac{\mathrm{\mid }z-k\mathrm{\mid }}{z-k}\right)}^f$. Its degrees on the outer boundary $A$, the $K$ boundary and the $U$ boundary are respectively $0,f,-f$: on $A$ the two factors have winding $1$ and $-1$, on $K$ only the second winds and its winding is $+1$, and on $U$ only the first winds and its winding is $-1$. Thus ${\xi }_0$ has a smooth real argument on an $A$ collar; multiply by the inverse argument with a collar cutoff to make it one there. This changes no boundary degrees. Choose the boundary collar arguments similarly to make the restrictions standard rotations in their angular coordinates. The map $\mathrm{\Phi }:X\to X,\mathrm{\Phi }(x,t)=(x,\xi (x)t)$ is an explicit diffeomorphism, the identity on $T_A$, fixing $h$ and sending $b_K$ to $b_K+fh$. On $T_U$ it also preserves the filling slope $h$. It extends over the $U$ filling: a solid torus whose meridian is $h$ admits the explicit extension of the boundary map $b_U\mapsto b_U-fh$, $h\mapsto h$ obtained by rotating its meridian disk by $-f$ turns as its longitude runs once. In coordinates $(r{e}^{i\theta },e^{i\phi })$ this is $(r{e}^{i(\theta -f\phi )},e^{i\phi })$. Constants or zero-degree boundary phase changes extend by the same collar interpolation. Hence it suffices to prove the identity for the pair of slopes $b_K,h$. Equivalently one can choose the product framing on $N$ to agree with the supplied framing of $K$ from the outset; conjugating a boundary-relative diffeomorphism by that product identification is still boundary-relative.

The proper disk and the cut-open compression body. Fill $T_U$ with a solid torus $V=D^2\times S^1$ whose meridian $\mathrm{\partial }{D}^2\times \{1\}$ is identified with $h$. Choose an embedded base arc $\alpha$ from the $A$ boundary to the $U$ boundary, disjoint from the $K$ boundary. The vertical annulus $\alpha \times S^1\subset X$ has a fiber boundary on each of $T_A,T_U$. Cap its $T_U$ boundary by the displayed meridian disk of $V$. After rounding its one seam, this is an embedded proper disk $D$ with boundary $h\subset T_A$. The cut-open manifold $(X\cup V)\setminus \text{\hspace{0.17em}⁣}\setminus D$ is a punctured solid torus whose uncompressed torus boundary is $T_K$ and whose meridian there is $h$. Here is the cut-and-glue verification. Cut $P$ along $\alpha$; it becomes an annulus. Consequently the product cut along the vertical annulus is an annulus times $S^1$, namely $T^2\times I$. Cutting $V$ along its meridian disk gives $D^2\times I$. Its annular attaching face is $\mathrm{\partial }{D}^2\times I$, attached in the $h$ direction of the boundary created from the $A\mathrm{/}U$ sides of the cut. Thus the cut-open manifold is exactly $T^2\times I$ with one 3-dimensional 2-handle $D^2\times I$ attached along the primitive curve $h$ on one of its ends. That end compresses to a sphere, while the other end remains $T_K$. This compression body is $(D^2\times S^1)\setminus \mathrm{int}{B}^3$ with meridian $h$: in the explicit product solid torus, take a meridian disk $D^2\times \{1\}$, thicken it, and retain a boundary collar. Their union is the just-described $T^2\times I$ plus the meridian 2-handle. The complementary region is the interior of the product obtained by cutting the solid torus along that meridian disk, hence is a 3-ball. This identifies the compression body by its actual collar and disk, rather than inferring its type from its fundamental group. The compressed sphere consists of the original annular part of $T_A$ and the two copies of $D$, with their product attaching coordinates retained.

The remaining filling and boundary reconstruction. Now fill $T_K$ along $b_K$. Its filling meridian intersects the compression body's meridian $h$ once. The filled, punctured solid torus is therefore a 3-ball. Explicitly, the unpunctured union is the boundary of $D^2\times D^2$: the pieces $D^2\times S^1$ and $S^1\times D^2$ are its two solid tori, and their meridians are the two product circle factors $h,b_K$. Thus it is $S^3$ with its explicit standard product smoothing. Retaining the already specified puncture removes an ordinary product ball from that standard sphere and leaves a ball, with the annular boundary part inherited from $T_A$ and the two disk faces inherited from $D$. Finally restore the cut along $D$. This attaches one ordinary 3-dimensional 1-handle to that ball, along the two displayed disk faces. It is a solid torus. Its meridian disk is $D$, so its external meridian is exactly $h$. More precisely the ball-with-two-disk-faces just constructed has its remaining annular boundary in the original $T_A$ product coordinates; regluing these two product disk faces reconstructs the same $D^2\times S^1$ product neighborhood with the same annular side. Choose the product ball and the meridian disk in the original $N$ by cutting its specified meridian disk. The preceding product identifications match those cut coordinates and their annular side. They therefore give a diffeomorphism after regluing whose restriction to all of $T_A$ is the original identity. No arbitrary boundary diffeomorphism is discarded or extended by a classification theorem. Compatible corner smoothings are compared in their seam collars by the existing rounding lemma. Combining this boundary-relative identification with the explicit framing-reduction map $\mathrm{\Phi }$ proves the Statement for every integral $f$.