A concordance annulus gives a zero-surgery cobordism

Put $E=\overline{W\setminus N}$, $C=A\times S^1$, and $E_i=E\cap S_i^3$, $N_i=N\cap S_i^3$, $C_i=C\cap S_i^3$. The supplied product identifies $N$ with $S^1\times I\times D^2$, $N_i$ with $S^1\times \{i\}\times D^2$, $C$ with $S^1\times I\times S^1$, and $C_i$ with its endpoint torus. Attach $F=D^2\times S^1\times I$ along $C$, identifying its disk-boundary circle with the annulus longitude and its other circle with the normal meridian. Each endpoint boundary is then $E_i$ filled by $F_i=D^2\times S^1\times \{i\}$ along the preferred longitude, precisely $M(K_i)$ by [F1]. The common annular-side collar gives a manifold after gluing, with the same smooth or topological category as the supplied data; in the smooth case round the product corners. This constructs $V=E{\cup }_{C}F$ with the prescribed oriented ends.

Here is the relative chain comparison in every degree. Thicken the pieces slightly along their supplied common collars to make excisive covers, using the same thickening at the ends; the thickenings retract onto the indicated pieces and intersections as pairs. For either decomposition $X=P{\cup }_{C}Q$ with corresponding endpoint $X_i=P_i{\cup }_{C_i}{Q}_i$, the small-chain sequence underlying [F3], quotiented by the endpoint sequence, is degreewise exact: $0\to C_{\ast }(C,C_i;R)\to C_{\ast }(P,P_i;R)\oplus C_{\ast }(Q,Q_i;R)\to C_{\ast }^{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}(X,X_i;R)\to 0$, with maps $c\mapsto (c,-c)$ and $(p,q)\mapsto p+q$. Exactness follows simplex by simplex, and subdivision and its prism preserve the endpoint chains, so the last complex computes $H_{\ast }(X,X_i;R)$. Its connecting map sends a relative cycle $p+q$ to the class of $\mathrm{\partial }p=-\mathrm{\partial }q$ in the intersection quotient. Thus it supplies the relative exact sequence in all degrees, including degrees zero and one, rather than only a comparison of absolute ranks.

The endpoint inclusions $N_i\to N$ and $C_i\to C$ are deformation retractions up to inverse inclusion, by linearly contracting the $I$ coordinate to $i$. Their relative homology groups therefore vanish in every degree. Apply step 2.1 to $(W,S_i^3)=(E,E_i)\cup (N,N_i)$: the exact sequence gives $H_k(E,E_i;R)\cong H_k(W,S_i^3;R)$ for every $k$. The latter groups vanish because $S_i^3\to W$ induces homology isomorphisms in every degree, as follows directly from the long exact sequence of this pair. Next $F_i\to F$ is also a product deformation retraction. Applying the same exact sequence to $(V,M(K_i))=(E,E_i)\cup (F,F_i)$ gives $H_k(V,M(K_i);R)\cong H_k(E,E_i;R)=0$ for every $k$. The pair sequence now proves that each $M(K_i)\to V$ is an $R$-homology isomorphism, including in degree zero and the top degrees.

Suppose finally that ${\pi }_1(W)=1$. The exterior $E$ is connected: $W$ is connected, $N$ and its side $C$ are connected, and any other component of the manifold $E$ would be a component of $W$ disjoint from the gluing side. Choose basepoint paths in $C$ and write $\mathrm{\ell },m$ for its annulus and meridian circles. The map ${\pi }_1(C)\to {\pi }_1(N)$ is surjective with kernel normally generated by $m$. The pushout [F4] consequently identifies ${\pi }_1(W)$ with ${\pi }_1(E)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨m⟩\text{\hspace{0.17em}⁣}⟩$; this follows also directly by eliminating the generator $\mathrm{\ell }$ of the second factor in its amalgamated presentation. Thus $m$ normally generates ${\pi }_1(E)$. For the second gluing, ${\pi }_1(C)\to {\pi }_1(F)$ is surjective with kernel generated by $\mathrm{\ell }$, so the same elimination gives ${\pi }_1(V)={\pi }_1(E)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\mathrm{\ell }⟩\text{\hspace{0.17em}⁣}⟩$. The image of $m$ still normally generates this quotient and lies in each endpoint knot-exterior group, hence in its zero-surgery group image. This proves the stated endpoint normal-generation conclusion. The proof uses the normal neighborhood already assumed in the Statement and makes no assertion that local flatness alone constructs it.