Preferred longitudes of finite disk-band knots lie in the second derived subgroup

On the peripheral torus $T$, the supplied meridian and longitude give $H_1(T;\mathbb{Z})=\mathbb{Z}[\mu ]\oplus \mathbb{Z}[\lambda ]$, as its finite square cell model shows. The tube retracts onto its core and has $H_1(N(K);\mathbb{Z})=\mathbb{Z}$. The longitude bounds the trimmed parallel Seifert surface in $X_K$, so its class there is zero, while it maps to the core generator in $N(K)$; the meridian maps to zero in that tube. Apply [F1] to $S^3=X_K\cup N(K)$ with slightly thickened collars. The segment $0=H_2(S^3)\to H_1(T)\to H_1(X_K)\oplus H_1(N(K))\to H_1(S^3)=0$ makes its middle map an isomorphism. Its values on the two displayed generators therefore give $H_1(X_K)=\mathbb{Z}[\mu ]$ exactly, including the marked generator.

For normal generation apply [F3] to $S^3=X_K\cup N(K)$. In the tube group the meridian becomes $1$ and the longitude becomes its core $c$, so the pushout is $({\pi }_1(X_K)\ast ⟨c⟩)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\mu ,\lambda c^{-1}⟩\text{\hspace{0.17em}⁣}⟩$, naturally ${\pi }_1(X_K)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\mu ⟩\text{\hspace{0.17em}⁣}⟩$. This group is trivial: cover $S^3$ by the complements of its north and south poles; each is ${\mathbb{R}}^3$ and contracts, and their overlap is path-connected, so [F3] gives ${\pi }_1(S^3)=1$. This proves normal generation of the exterior group. In the zero filling the meridian becomes the filling core and the longitude becomes $1$; the identical pushout calculation gives ${\pi }_1(M(K))={\pi }_1(X_K)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\lambda ⟩\text{\hspace{0.17em}⁣}⟩$. It too is normally generated by the image of $\mu$. No incompressibility or injection assertion about a torus group is used.

Signed intersection with the properly trimmed Seifert surface defines a homomorphism $H_1(X_K;\mathbb{Z})\to \mathbb{Z}$ taking $\mu$ to one. In this finite model its chain-level justification is explicit. Undo the supplied finite local roundings, work with its finite polygonal faces, and move each fixed finite loop or chain into general position with these faces by finite subdivision and small piecewise-linear perturbations. Each perturbation excludes only finitely many improper affine incidences, so finite rational choices suffice. These operations preserve the homology class by the straight simplex prisms in sufficiently small convex coordinate boxes. For a two-chain in the interior of $X_K$, its transverse intersection with the oriented surface is a finite oriented one-chain. Its boundary is the intersection with the two-chain boundary: no boundary term comes from the surface boundary, which lies on $T$ outside that interior. The signed sum of this one-chain's endpoints is zero, so intersection counts agree on homologous one-cycles. Boundary collars move any fixed finite chain into the interior before this calculation. The meridian meets the surface once with positive sign by the supplied peripheral orientation. Thus step 1.1 makes this intersection homomorphism an isomorphism. All loops in a small positive parallel copy $F^{+}$ of the trimmed surface miss the original surface, so their images in $H_1(X_K)$ are zero. The parallel copy and its boundary longitude are supplied by the finite oriented ribbon charts and tube.

By [F2], step 2.1 implies that the entire image of ${\pi }_1(F^{+})$ in ${\pi }_1(X_K)$ lies in its commutator subgroup. The oriented boundary loop of $F^{+}$ represents zero in $H_1(F^{+};\mathbb{Z})$: a finite triangulation of the supplied disk-band surface gives an oriented two-chain whose boundary is exactly that boundary loop, with all internal edges cancelling. Apply [F2] to $F^{+}$ to conclude that its boundary element belongs to $[{\pi }_1(F^{+}),{\pi }_1(F^{+})]$. Its image in ${\pi }_1(X_K)$ therefore belongs to $[{\pi }_1(X_K{)}^{(1)},{\pi }_1(X_K{)}^{(1)}]={\pi }_1(X_K{)}^{(2)}$. The compatible tube identifies this boundary with the preferred longitude up to conjugation. Derived subgroups are normal, as conjugation sends each finite commutator product to another, so the marked $\lambda$ has the same conclusion.

The proof used the actual supplied finite surface model only to define and justify intersection counts and its finite fundamental two-chain. The marked tube, trimming and parallel copy were input or finite product constructions. All perturbations concerned one finite chain at a time; all homology comparisons and pushout maps were natural or uniquely defined. Consequently no simultaneous choice of representatives, triangulation of an arbitrary topological manifold, or locally flat four-dimensional normal-bundle assertion has been imported. This proves all stated peripheral and group conclusions in their full supplied-data scope.