Zero localized intersection rank gives isotropy on a finite boundary union

Write $M=\mathrm{\partial }W$. Under relative cap duality, the map $j:H^2(W,M;K^{\vee })\to H^2(W;K^{\vee })$ corresponds to the intersection map on $H_2(W;K)$. This map is zero under the given boundary-generation hypothesis. Indeed, if $z=i_{\ast }b$ for $b\in H_2(M;K)$, closed duality represents $\bar{b}$ by a degree-one cohomology class on $M$. Its pair connector is the relative dual class of $\bar{z}$, by the cap boundary identity, and its image under $j$ is zero by exactness. Every $z$ has this form by hypothesis. Thus $j=0$, with intrinsic dual systems identified using the supplied Hermitian form.

Extend $x$ by zero from $N$ to $H_1(M;R)$. Since it dies in $W$, the pair sequence supplies $a\in H_2(W,M;R)$ with $\mathrm{\partial }a=x$. Absolute cap duality sends its conjugate to $u\in H^2(W;R^{\vee })$, whose boundary restriction is the dual of $\bar{x}$, by the signed cap boundary formula in degree two. The image of $x$ in $H_1(M;K)$ is zero: on selected components this is the assumed vanishing and on other components $x$ was zero. Hence $u_K{\mathrm{\mid }}_M=0$. Pair exactness puts $u_K$ in the image of $j$, which is zero by step 1.1. Choose one $R^{\vee }$-valued two-cocycle representing $u$. Since its image represents zero over $K$, one whole $K^{\vee }$-valued one-cochain $b$ satisfies $\delta b=u$. Its quotient $\nu =b<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>R$ is a $(K\mathrm{/}R{)}^{\vee }$-valued cocycle on $W$, and on each selected component its coefficient Bockstein is the dual of $\bar{x}$. This constructs a whole cochain, rather than choosing lifts of infinitely many quotient coefficients.

On a selected connected component $N_j$, closed duality gives $H^2(N_j;K^{\vee })=0$ from $H_1(N_j;K)=0$. Also $H^1(N_j;K^{\vee })=0$. To see this without a universal-coefficient theorem, use the one-vertex model over $K$ from [F3], so $C_0=K$. Its $d_1$ image is either zero or $K$. In the first case $H_1=0$ says $C_1=\mathrm{im}{d}_2$, so every one-cocycle vanishes. In the second case one vector $v$ with $d_{1}v=1$ splits $C_1=\ker d_1\oplus vK$; a one-cocycle vanishes on $\ker d_1=\mathrm{im}{d}_2$, so equals a zero-cochain composed with $d_1$. The dual identification in [F2] preserves these assertions. Exactness [F3] therefore makes the degree-one coefficient Bockstein an isomorphism. Consequently the restriction of $[\nu ]$ in step 2.1 is exactly the inverse-Bockstein class defining the linking form of $x{\mathrm{\mid }}_{N_j}$. This applies to the finite component list separately.

Let $y$ also belong to the displayed kernel. Its image cycle in $W$ is the boundary of a single $R$-valued two-chain $c$. The sum of the component linking evaluations is therefore $\nu (y)=\nu (\mathrm{\partial }c)=\delta \nu (c)=0$. Different components do not have cross terms because the chain and cochain groups on the finite disjoint union split componentwise; their cap fundamental classes carry their induced orientations. This is precisely $B_N(x,y)$. The proof used only finite component operations, exactness and one individual cochain witness, and made no assertion about the other boundary components.