Zero filling of a fibered knot caps its fiber

By [F2] and [A1], the needed countable-choice hypothesis [F1] holds. Let $F$ be the fiber from [F5]. Cutting the exterior along $F$ gives $F\times [0,1]$, with its interval ends reidentified by the monodromy $\phi$; because $\phi$ fixes $\mathrm{\partial }F$ pointwise, the boundary torus is identified with $\mathrm{\partial }F\times S^1$.

The curve $\mathrm{\partial }F$ on the peripheral torus is an essential parallel of the knot and bounds the fiber in the exterior, so its homology class maps to zero in $H_1(E_{K_8};\mathbb{Z})$. It is primitive because it is a simple essential curve. By uniqueness in [F4], it represents $\lambda$ up to orientation, hence it is exactly the unoriented filling slope $0$ by [F3].

For the model gluing, use the second coordinate of $D^2\times S^1$ as the mapping-torus direction and identify the meridian disk boundary with $\mathrm{\partial }F\times \{t\}$. The filled space is the mapping torus of $\bar{F}=F{\cup }_{\mathrm{\partial }F}{D}^2$: extend $\phi$ over the disk by the identity, which is continuous and fixes the seam since $\phi {\mathrm{\mid }}_{\mathrm{\partial }F}=\mathrm{i}\mathrm{d}$. The actual filling map has the same meridian slope by [F3]; [F7] extends the transition map over the filling solid torus, so the actual filling is oriented-homeomorphic to this model mapping torus.

The surface $\bar{F}$ is obtained from a once-punctured torus by capping its sole boundary component with a disk, so it is a closed torus. The mapping torus of the extended homeomorphism $\bar{\phi }$ is consequently a torus bundle over $S^1$. Inclusion $F\hookrightarrow \bar{F}$ induces an isomorphism on first homology, and the extension is the identity on the cap, so its matrix remains the matrix in [F5].