The Gordon-Luecke knot-complement theorem

For each $J\in \{K,K^{\mathrm{\prime }}\}$, the boundary $T_J$ is a circle bundle over the connected knot $J$ with connected fibres, hence is connected: a separation would split the boundary into saturated open sets whose projections, by local bundle charts, separate $J$. By [F10], $E_J$ is a compact smooth orientable $3$-manifold with boundary $T_J$. Since $T_J$ is connected, it lies in one component of $E_J$. Any other component is boundaryless, hence lies in $\mathrm{int}{E}_J$ and is open in $S^3$; it is also closed in $S^3$, since it is closed in $E_J$ and $E_J$ is closed in $S^3$ by [F10]. This contradicts connectedness of $S^3$. Thus both exteriors are connected and, being manifolds, path-connected.

Suppose both knots are nontrivial and let $\gamma$ be the slope represented by $h({\mu }_K)$. Let $W,W^{\mathrm{\prime }}$ be the source and target filling solid tori with product coordinates. If $h$ preserves orientation, set $g(z,w)=(z,w)$; if $h$ reverses orientation, set $g(z,w)=(z,\bar{w})$. In either case $g$ has the same orientation sign as $h$ and sends the source meridian to the target meridian. If $\varphi :\mathrm{\partial }W\to \mathrm{\partial }{E}_K$ is the source gluing map, define the target gluing map by ${\varphi }^{\mathrm{\prime }}=h\circ \varphi \circ g^{-1}$. It is orientation-reversing because $\varphi$ is and $g,h$ have the same sign; its meridian image is $h({\mu }_K)$, so it defines the filling along $\gamma$. The maps $h$ and $g$ glue to a homeomorphism of the filled manifolds. The source filling is $S^3$, hence the target filling along $\gamma$ is homeomorphic to $S^3$. By [F1], $\gamma$ is the target meridian slope; [F7] then gives the boundary isotopy in the Statement.

Suppose one knot is unknotted. By [F4] its standard radius-tube exterior is a solid torus with the displayed proper meridian disk. Use the ambient isotopy in the definition of unknotted to transport its chosen tube to a tube around the standard core $U$. Identify the transported and standard tube domains with the same normal bundle $\nu (U)$. For each tubular embedding ${\mathrm{\Phi }}_j$, its differential along the zero section induces an invertible smooth bundle map $L_j:\nu (U)\to \nu (U)$ on the normal quotient: invertibility follows because ${\mathrm{\Phi }}_j$ is an embedding and is the identity on the zero section. Precompose ${\mathrm{\Phi }}_j$ with the fibrewise map $L_j^{-1}$. After restricting to a common neighbourhood of the zero section, both induced normal maps are the identity; compactness of $U$ lets us choose one such neighbourhood for the whole core. Thus the hypotheses of [F5] hold. Under [A1], [L1] supplies its countable-choice premise, so an ambient isotopy identifies the tubes after shrinking. The standard disk transfers to the smaller chosen exterior. By [F6], the region between the smaller and original tube is a product boundary collar; trim or extend the disk across the product annulus so its boundary lies on the chosen exterior torus. The resulting proper disk still has essential boundary. Pull that disk across $h$ or $h^{-1}$ into the other exterior. Its parametrization is a continuous disk map with boundary in the target torus; the boundary class remains nontrivial because the boundary restriction is a homeomorphism. If the other knot were nontrivial, apply [F9] to its compact connected orientable smooth exterior (connectedness was proved in step 1.1), the whole smooth boundary torus as the boundary subsurface, and the trivial normal subgroup. The disk boundary lies in the relative interior of that torus, so [F9] gives a new smoothly properly embedded disk with essential boundary. This contradicts [F2]. No smoothing of the homeomorphic image disk and no replacement of $h$ by a diffeomorphism is required. Thus either both are unknotted or both are nontrivial.

If both are unknotted, [F8] gives ambient equivalences from each knot to the standard unknot; composing them gives an ambient equivalence between the knots. This conclusion is independent of whether the given $h$ preserves the knot-meridian slope.

Apply [F3] to the actual boundary restriction of $h$ and the two tubular solid tori. The extensions agree with $h$ on the common boundary and therefore glue to a homeomorphism of pairs $(S^3,K)\to (S^3,K^{\mathrm{\prime }})$. By [F3] the ambient orientation sign is the same as that of $h$. Together with steps 2.1, 1.2, and 3.1 this proves every case.