The boundary kernel of a rational homology ball is a linking metabolizer

The rational homology-pair sequence and [F1], with $H_i(W;\mathbb{Q})=0$ for $i>0$, give $H_1(Y;\mathbb{Q})=H_2(Y;\mathbb{Q})=0$; connectedness gives $H_0(Y;\mathbb{Q})=\mathbb{Q}$, and orientation gives $H_3(Y;\mathbb{Q})=\mathbb{Q}$. Thus $Y$ is a rational homology sphere. By [F4] and [A1], both $W$ and its smooth boundary $Y$ have finite CW type, so their integral homology groups are finitely generated. The groups $H_1(Y;\mathbb{Z})$, $H_1(W;\mathbb{Z})$ and $H_2(W;\mathbb{Z})$ have vanishing rationalizations and are therefore finite.

Put $A=H_1(Y;\mathbb{Z})$, $B=H_1(W;\mathbb{Z})$, and $C=H_2(W;\mathbb{Z})$. Since $A$ is finite by step 1.1, the universal coefficient sequence [F3] gives $H^1(Y;\mathbb{Z})=0$; [F7] then gives $H_2(Y;\mathbb{Z})=0$. Also $H_3(W;\mathbb{Z})=0$: [F1] identifies $H_4(W,Y)$ with $H^0(W)=\mathbb{Z}$, and its fundamental-class boundary maps to the generator of $H_3(Y)=\mathbb{Z}$; since $H_3(W,Y)\cong H^1(W)=0$, exactness forces $H_3(W)=0$.

By [F3], $H^2(W;\mathbb{Z})\cong \mathrm{Ext}(B,\mathbb{Z})$ and $H^3(W;\mathbb{Z})\cong \mathrm{Ext}(C,\mathbb{Z})$, because $B,C$ are finite and $H_3(W)=0$. Their orders are $\mathrm{\mid }B\mathrm{\mid }$ and $\mathrm{\mid }C\mathrm{\mid }$. Under [F1], these are the orders of $H_2(W,Y)$ and $H_1(W,Y)$, respectively. The pair sequence [F2] shows that $P=\ker (A\to B)$ is the cokernel of the injective map $H_2(W)\to H_2(W,Y)$, while $B\to H_1(W,Y)$ is onto because the next map $H_0(Y)\to H_0(W)$ is an isomorphism. Hence $\mathrm{\mid }P\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }C\mathrm{\mid }$ and $\mathrm{\mid }A\mathrm{/}P\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }C\mathrm{\mid }$, so $\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }P{\mathrm{\mid }}^2$.

To prove isotropy, [F2] and the naturality in [F7] identify $P$ under Poincaré duality with the image of $H^2(W;\mathbb{Z})\to H^2(Y;\mathbb{Z})$. For $x,y\in P$, choose lifts $\tilde{x},\tilde{y}\in H^2(W;\mathbb{Z})$. Since $W$ is a rational homology ball, the coefficient Bockstein $H^1(W;\mathbb{Q}\mathrm{/}\mathbb{Z})\to H^2(W;\mathbb{Z})$ is onto by [F8]: the adjacent groups $H^1(W;\mathbb{Q})$ and $H^2(W;\mathbb{Q})$ vanish. Choose $\gamma \in H^1(W;\mathbb{Q}\mathrm{/}\mathbb{Z})$ with Bockstein $\tilde{y}$. The Bockstein–duality formula in [F5] evaluates ${\lambda }_Y(y,x)$ as the restriction of $\gamma ⌣\tilde{x}$ on $[Y]$; symmetry of ${\lambda }_Y$ then gives the same value for ${\lambda }_Y(x,y)$. This cup class extends over $W$, and the inclusion sends $[Y]$ to zero in $H_3(W)$ because $Y=\mathrm{\partial }W$. Thus ${\lambda }_Y(x,y)=0$.

The isotropy from step 3.2 gives $P\subseteq P^{\perp }$. Nonsingularity [F5] implies $\mathrm{\mid }{P}^{\perp }\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }\mathrm{/}\mathrm{\mid }P\mathrm{\mid }=\mathrm{\mid }P\mathrm{\mid }$ by step 3.1. Therefore $P=P^{\perp }$, so $P$ is a metabolizer by [F6].