Winding-zero infection preserves the classical Blanchfield pairing

Use the actual product gluing from [F1]. Put $S=N(\eta )=V_K$, $T_0=\mathrm{\partial }S$, $A=M(P)\setminus \mathrm{int}N(\eta )$ and $X_K=S^3\setminus \mathrm{int}N(K)$. Its spine is simultaneously $Z=M(P){\cup }_{S}M(K)$ and $Z=(A{\cup }_{T_0}{X}_K){\cup }_{T_0}S=M(P(K)){\cup }_{T_0}S$. The product-coordinate deformation to this spine fixes the remaining bottom faces $A$ and $X_K$; the rounded torus collars identify their union with the outgoing boundary. Hence it gives an equivalence of pairs $(E,M(P(K)))\simeq (Z,M(P(K)))$. Excision in these collars identifies its relative homology with $H_{\ast }(S,T_0)$. These are actual subspaces and product homotopies, rather than an inference from ordinary homology.

The ordinary generator of $H_1(E;\mathbb{Z})=\mathbb{Z}$ defines the specified infinite-cyclic system; the same integral peripheral computation in [F1] shows that it sends ${\mu }_P$ and ${\mu }_{P(K)}$ to $t$ and ${\mu }_K$ to $1$. Its restriction to $S$ is trivial, since the core represents the zero-winding axis. It is also trivial on $M(K)$ and $X_K$: their ordinary first homology is generated by ${\mu }_K$, so their map to $\mathbb{Z}$ is zero. Trivial coefficient groups on these pieces are their ordinary $k$ homology tensored with $\mathrm{\Lambda }$. This assertion does not require an infinite vector-space splitting: the singular chain complex with trivial coefficients is its $k$ complex tensored with the free $k$ module $\mathrm{\Lambda }={⨁}_{n\in \mathbb{Z}}k{t}^n$; cycles, boundaries and exactness commute with this explicit direct sum. In particular $H_1(S;\mathrm{\Lambda })=\mathrm{\Lambda }$, $H_1(M(K);\mathrm{\Lambda })=\mathrm{\Lambda }[{\mu }_K]$, and the core inclusion into $M(K)$ is this generator.

Mayer–Vietoris for the first description of $Z$ gives the degree-one difference map $H_1(S;\mathrm{\Lambda })\to H_1(M(P);\mathrm{\Lambda })\oplus H_1(M(K);\mathrm{\Lambda })$ as $a\mapsto ([\eta ]a,-[{\mu }_K]a)$. It is injective because its second coordinate is minus the identity of $\mathrm{\Lambda }$. The following degree-zero difference map is also injective: it is $\mathrm{\Lambda }\to \mathrm{\Lambda }\mathrm{/}(t-1)\oplus \mathrm{\Lambda }$, $a\mapsto (a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>(t-1),-a)$. Exactness therefore identifies $H_1(E;\mathrm{\Lambda })$ with the quotient of $H_1(M(P);\mathrm{\Lambda })\oplus \mathrm{\Lambda }$ by the displayed graph. The first summand maps isomorphically onto that quotient; explicitly $(x,a)$ maps to $x+[\eta ]a$. In particular the core of $S$, equivalently the companion meridian, maps to the image of $[\eta ]$ under this isomorphism.

The group $\mathbb{Z}$ is PTFA, its $k$ group ring is the stated $\mathrm{\Lambda }$, and its fraction division ring is $F$. All three boundary components of $E$ have ordinary first Betti number one. The ordinary boundary-cokernel rank is zero by the incoming second-homology generation in [F1]. Applying the full rank comparison in [F2] therefore gives $H_2(E;F)=\mathrm{im}{H}_2(\mathrm{\partial }E;F)$. On $M(P)$ and $M(P(K))$ the restricted infinite-cyclic maps are nontrivial and their first Betti numbers are one, so the other assertion in [F2] gives $H_1(-;F)=0$ there. No vanishing on $M(K)$ is claimed or needed. Consequently [F3] applies to the selected two-component oriented boundary $N=M(P)\bigsqcup (-M(P(K)))$.

In the pair from step 1.1, the trivial coefficient relative solid-torus calculation gives $H_1(S,T_0;\mathrm{\Lambda })=0$ and $H_2(S,T_0;\mathrm{\Lambda })=\mathrm{\Lambda }$, generated by a meridian disk. Indeed $H_1(T_0;\mathrm{\Lambda })\to H_1(S;\mathrm{\Lambda })$ is onto with kernel the meridian summand, and the degree-zero inclusion is an isomorphism. Under the gluing, the meridian boundary of that disk is ${\mu }_{\eta }={\lambda }_K$. This longitude represents zero in $H_1(X_K;\mathrm{\Lambda })$: the coefficients on $X_K$ are trivial, and the preferred longitude has zero ordinary class in the knot exterior by its zero-linking property, as explicitly computed in [F1]. Its class is therefore zero in $H_1(M(P(K));\mathrm{\Lambda })$. The relative connector $H_2(E,M(P(K));\mathrm{\Lambda })\to H_1(M(P(K));\mathrm{\Lambda })$ is zero on the generator and hence zero. Pair exactness now makes $i_{P(K)}$ an isomorphism, since the following relative group is zero. Together with step 3.1 this proves both asserted isomorphisms and identifies $T[\eta ]$ with $[{\mu }_K]$.

For $x,y\in H_1(M(P);\mathrm{\Lambda })$, the classes $(x,-Tx)$ and $(y,-Ty)$ lie in the kernel of $H_1(N;\mathrm{\Lambda })\to H_1(E;\mathrm{\Lambda })$ by the definition of $T$. The orthogonal sum of the boundary linking forms on this kernel is zero by step 3.2. Reversing the orientation of a closed component negates its cap fundamental class, hence negates inverse-cap duality and the resulting Blanchfield form; its Bockstein and evaluation are unchanged. Bilinearity gives $B_{-M(P(K))}(-Tx,-Ty)=-B_{P(K)}(Tx,Ty)$. Thus kernel isotropy says precisely $B_P(x,y)-B_{P(K)}(Tx,Ty)=0$, proving the claimed isometry. This uses the actual Bockstein linking forms and does not assume any general nonsingularity theorem.

Apply step 5.1 to $x=y=[\eta ]$ and use the class identification in step 4.1. It gives $B_{P(K)}([{\mu }_K],[{\mu }_K])=B_P([\eta ],[\eta ])$, so nonzero self-value is preserved. Altering a base path multiplies the based class by some $t^a$; conjugate linearity in the first variable and right linearity in the second multiply its self-value by $t^{-a}$ and $t^a$, whose product is one in the commutative quotient module $F\mathrm{/}\mathrm{\Lambda }$. All constructions used supplied smooth peripheral products, finite product gluings and the choice-free local homological suppliers. They introduce neither a locally flat annulus-tube assertion nor an unproved arbitrary twisted boundary-generation claim.