An S3 filling detects the meridian slope

The chosen tubular neighbourhood is a disk bundle over the connected knot $K$, so its boundary $T_K$ is a circle bundle over $K$ with connected fibres and is connected. Indeed, a separation of $T_K$ would, since every fibre is connected, split it into two saturated open sets; their projections are open by the local bundle charts, disjoint, and cover the connected base $K$. Since $T_K$ is a connected smooth manifold, it is path-connected. The exterior $E_K$ is a compact smooth $3$-manifold with boundary $T_K$. Because $T_K$ is connected, it lies in one component of $E_K$. Any other component is boundaryless, hence lies in $\mathrm{int}{E}_K$ and is open in $S^3$; it is also closed in $S^3$, since it is closed in $E_K$ and $E_K$ is closed in $S^3$ by [F4]. This contradicts connectedness of $S^3$. Thus $E_K$ is connected and, being a smooth manifold, path-connected.

Full AC [A1] supplies the countable-choice hypothesis of [F1] by [L1], and step 1.1 verifies its connected-exterior hypothesis. If $E_K(p\mathrm{/}q)\cong S^3$, its first homology vanishes. By [F1], this excludes $p=0$ and every $\mathrm{\mid }p\mathrm{\mid }\ge 2$, leaving $\mathrm{\mid }p\mathrm{\mid }=1$.

If additionally $q\mathrm{\ne }0$, the slope is nonmeridional, so [F2] gives a nontrivial fundamental group, contradicting ${\pi }_1(S^3)=1$. Hence $q=0$; coprimality then gives $p=\pm 1$, which is the meridional slope.

Conversely, if $q=0$, coprimality forces $p=\pm 1$. By [F3], filling along this meridian slope is the original decomposition $S^3=E_K\cup N(K)$ up to the permitted solid-torus gluing, hence is homeomorphic to $S^3$.