Tubular exterior and boundary torus

By [F1], $K$ is continuous and $K(S^1)$ is compact because $S^1$ is compact. Since $S^3$ is Hausdorff, $K(S^1)$ is closed; [F1] also identifies it as a smooth embedded submanifold. Thus the closed smooth embedded hypothesis of [F4] holds, and [A1] permits its application. Take a tubular map $\mathrm{\Phi }:\mathrm{\Omega }\to U\subset S^3$ defined on an open neighbourhood of the zero section of $\nu (K)$.

Use the trivialization [F3] and its product Euclidean norm as the chosen smooth fibre norm in [F2], so $\mathrm{\Omega }$ is open in $S^1\times {\mathbb{R}}^2$. For each $s\in S^1$, openness gives an arc $I_s$ about $s$ and a radius $r_s>0$ with $I_s\times D_{r_s}^2\subset \mathrm{\Omega }$. Compactness of $S^1$ gives a finite subcover by such arcs; taking $ϵ$ smaller than half the least of their radii yields $S^1\times D_{2ϵ}^2\subset \mathrm{\Omega }$. The closed disk subbundle $S^1\times D_{ϵ}^2$ therefore lies in $\mathrm{\Omega }$.

Put $N(K)=\mathrm{\Phi }(S^1\times D_{ϵ}^2)$. The restricted map is a smooth embedding, so $N(K)$ is a compact embedded solid torus with boundary $T_K=\mathrm{\Phi }(S^1\times \mathrm{\partial }{D}_{ϵ}^2)\cong S^1\times S^1$. Since $S^3$ is compact and $E_K=S^3\setminus \mathrm{int}N(K)$ is closed, $E_K$ is compact. The annulus $S^1\times \{z:ϵ\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<2ϵ\}$ maps to a collar of $T_K$; in these coordinates $E_K$ is the side $\mathrm{\mid }z\mathrm{\mid }\ge ϵ$. Thus $E_K$ is a smooth three-manifold with boundary exactly $T_K$, oriented by restriction of the orientation of $S^3$ as in [F2].

On $S^3\setminus K$, define $H_t$ to be the identity outside $N(K)$ and, in the punctured disk fibres with $0<\mathrm{\mid }z\mathrm{\mid }\le ϵ$, set $H_t(\mathrm{\Phi }(s,z))=\mathrm{\Phi }(s,((1-t)\mathrm{\mid }z\mathrm{\mid }+tϵ)z\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$ for $0\le t\le 1$. The formula is continuous for $z\mathrm{\ne }0$ and is the identity at $\mathrm{\mid }z\mathrm{\mid }=ϵ$, so it glues continuously to the identity outside $N(K)$. It fixes every point of $E_K$, stays in $S^3\setminus K$, and at $t=1$ sends every point of the punctured tube to $T_K\subset E_K$. Hence $H_t$ is a strong deformation retraction of $S^3\setminus K$ onto $E_K$.