Unit-slope torus knots are unknots

Suppose $p=\epsilon \in \{1,-1\}$. Write the original oriented curve as $K(s)=(r{e}^{2\pi i\epsilon s},\sin {\theta }_0{e}^{2\pi iqs})$, where $r=\cos {\theta }_0\in (0,1)$. Put ${\theta }_t=(1-t){\theta }_0$ and define $H_t(s)=(\cos {\theta }_t{e}^{2\pi i\epsilon s},\sin {\theta }_t{e}^{2\pi iqs})$. Each $H_t$ lies in $S^3$; its first coordinate is an injective circle parameterization because $\mathrm{\mid }\epsilon \mathrm{\mid }=1$ and $\cos {\theta }_t>0$, and the family is smooth including $t=1$. It begins with the given orientation and ends at the core circle with orientation sign $\epsilon$.

If $q=\epsilon \in \{1,-1\}$, the path $(z_1,z_2)\mapsto (\cos t{z}_1+\sin t{z}_2,-\sin t{z}_1+\cos t{z}_2)$, $0\le t\le \pi \mathrm{/}2$, lies in $SO(4)$ and isotopes the identity to $(z_1,z_2)\mapsto (z_2,-z_1)$. The image of the curve is $(\sin {\theta }_0{e}^{2\pi i\epsilon s},-r{e}^{2\pi ips})$. Let ${\eta }_0\in (0,\pi \mathrm{/}2)$ satisfy $\cos {\eta }_0=\sin {\theta }_0$ and $\sin {\eta }_0=r$, put ${\eta }_t=(1-t){\eta }_0$, and use $H_t(s)=(\cos {\eta }_t{e}^{2\pi i\epsilon s},-\sin {\eta }_t{e}^{2\pi ips})$. Its first coordinate has winding $\epsilon =\pm 1$, so the family is a smooth isotopy through embeddings to the core with orientation sign $\epsilon$. The initial coordinate rotation also preserves the knot orientation because it is an ambient isotopy.

By [F3], each smooth isotopy of the compact embedded circle constructed above extends to an ambient isotopy of $S^3$. Its terminal core is the standard unknot, with orientation sign $\epsilon$. If $\epsilon =-1$, write $z_j=x_j+i{y}_j$ and define $R_{\tau }(x_1,y_1,x_2,y_2)=(x_1,y_1\cos \tau +y_2\sin \tau ,x_2,-y_1\sin \tau +y_2\cos \tau )$. This is a rotation in the $(y_1,y_2)$-plane, so $R_{\tau }$ is an orientation-preserving isometry of $S^3$ from the identity to $R_{\pi }(z_1,z_2)=({\bar{z}}_1,{\bar{z}}_2)$. On the core, $R_{\pi }(e^{2\pi is},0)=(e^{-2\pi is},0)$, so this ambient isotopy reverses its orientation. In either sign case the original oriented curve is ambient isotopic to the standard oriented unknot.