Standard infection adjunction kills precisely the companion longitude

Put $A=M(P)\setminus \mathrm{int}N(\eta )$, $X_K=M(K)\setminus \mathrm{int}{V}_K$ and $S=N(\eta )=V_K$. Collapse the interval coordinates in the two defining products to their identified bottom ends. The homotopies agree on $S$, so descend to a deformation to $Z=M(P){\cup }_{S}M(K)$. Rewriting $M(P)=A\cup S$ and $M(K)=X_K\cup S$ gives $Z=(A{\cup }_{\mathrm{\partial }S}{X}_K){\cup }_{\mathrm{\partial }S}S=M(P(K)){\cup }_{\mathrm{\partial }S}S\mathrm{.}$ The deformation fixes $A$ and $X_K$ at their bottom ends; the explicit rounded torus collars identify this fixed union with the outgoing boundary. It therefore supplies the relative homotopy equivalence needed for the actual outgoing inclusion, rather than just an unmarked homotopy equivalence of spaces.

Apply [F2] to product-collared open neighborhoods in this last spine. On its overlap torus the two generators, written using companion peripheral coordinates, are ${\mu }_K,{\lambda }_K$. The solid torus $S$ has infinite-cyclic group generated by its core $c$, and the gluing sends ${\mu }_K$ to $c$ and ${\lambda }_K$ to $1$. Thus the pushout group has presentation relative to $G={\pi }_1(M(P(K)))$ as $(G\ast ⟨c⟩)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨{\mu }_K{c}^{-1},{\lambda }_K⟩\text{\hspace{0.17em}⁣}⟩\mathrm{.}$ Eliminate $c$: the map from $G\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨{\lambda }_K⟩\text{\hspace{0.17em}⁣}⟩$ to this group is induced by $G$, while the reverse map is induced by the quotient on $G$ and $c\mapsto [{\mu }_K]$. They respect the displayed relations and are inverse on both generating factors. This proves the first asserted isomorphism and surjectivity for the actual inclusion. The core $c$ was precisely ${\lambda }_{\eta }$, the oriented axis core; hence it also proves the marked meridian-axis identity. Neither torus-to-manifold map was assumed injective.

For the manifold adjunction, the supplied boundary collars give open neighborhoods with the same homotopy types as $W$, $E$ and their common surgery component. Apply [F2] again. Put $B={\pi }_1(W)$ and let $f:G\to B$ be its marked boundary inclusion map. By step 2.1, the other pushout factor is the quotient $G\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨{\lambda }_K⟩\text{\hspace{0.17em}⁣}⟩$. Every compatible pair of homomorphisms from these two factors is uniquely determined by a homomorphism from $B$ that kills $f({\lambda }_K)$: its values on the second factor are forced by its values on $G$, and they descend exactly when that relator is killed. The group $B\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨f({\lambda }_K)⟩\text{\hspace{0.17em}⁣}⟩$ has precisely this universal property, with the actual inclusion from $B$. Explicitly send its coset of $b$ to the inclusion image of $b$; in the reverse direction use the quotient on $B$ and the homomorphism from $G\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨{\lambda }_K⟩\text{\hspace{0.17em}⁣}⟩$ induced by $f$. The composites are identities on the two pushout factors. This proves the second isomorphism and its stated map without an injectivity hypothesis on $f$.

Replacing a chosen path to a peripheral loop by another changes its based group element by conjugation. The normal closure of a conjugate is the same normal subgroup, so both quotient descriptions and their inclusion-induced maps are independent of that change in the asserted sense. All geometric operations above are the actual supplied product homotopies and collar neighborhoods; the group calculations use finite words and the unique maps of the pushout property. The van Kampen supplier constructs these maps from finite loop and homotopy subdivisions rather than a global family of representatives. Thus no choice principle, smooth structure on $W$, or locally flat annulus product theorem has entered the proof.