Equal surgery manifolds need not have isotopic diagrams

By [F1], surgery on $D_0$ leaves the standard oriented $S^3$ unchanged.

Give $S^3$ its unit-sphere model in ${\mathbb{C}}^2$ and let $U(t)=(e^{2\pi it},0)$. With $N(U)=\{\mathrm{\mid }{z}_2\mathrm{\mid }\le 1\mathrm{/}\sqrt{2}\}$, the exterior is $D^2\times S^1$ with disk coordinate $z_1$ and core coordinate $w=z_2\mathrm{/}\mathrm{\mid }{z}_2\mathrm{\mid }$. The preferred longitude ${\lambda }_U$ is the disk-meridian class and ${\mu }_U$ is the core-longitude class by [F2]–[F3]. The diffeomorphism $g(z,w)=(w^{-1}z,w)$ preserves orientation and sends $[{\mu }_U]+[{\lambda }_U]$ to $[{\mu }_U]$. Let $f_0$ be the original $S^3$ gluing, which sends the meridian of the removed unknot neighbourhood to $[{\mu }_U]$, and choose the $+1$-filling map $f_1=g^{-1}\circ f_0$. It sends the attached meridian to $[{\mu }_U]+[{\lambda }_U]$, the required slope by [F1]. The map equal to $g$ on the exterior and the identity on the attached solid torus is well-defined because $g\circ f_1=f_0$; it is a smooth orientation-preserving diffeomorphism to the original $S^3$. Thus $D_1$ also presents $S^3$.

The diagrams are not ambient isotopic: $D_0$ has zero components and $D_1$ has one, while [F5] implies ambient isotopy preserves component count. Yet they have the same oriented surgery result by steps 1.1–1.2. Moreover [F4] identifies their difference as a permitted stabilization. Hence the false claim fails.