The torsion linking pairing of a rational homology sphere is nonsingular

Since $H_1(Y;\mathbb{Z})$ is finite, $H^1(Y;\mathbb{Q})=0$ by [F2], and $H^2(Y;\mathbb{Q})=0$ by [F1]. Exactness of [F3] therefore makes $\beta :H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})\to H^2(Y;\mathbb{Z})$ an isomorphism. Write $\mathrm{PD}:H_1(Y;\mathbb{Z})\to H^2(Y;\mathbb{Z})$ for inverse cap duality.

We first fix the intersection sign and verify independence directly. For transverse oriented $2$-chains $C,T$ in an oriented $3$-manifold, orient their intersection so that $\mathrm{\partial }(C\cap T)=(\mathrm{\partial }C)\cap T-C\cap (\mathrm{\partial }T)\mathrm{.}$ This sign follows by the outward-normal-first boundary convention in an oriented coordinate ball: taking a boundary in the second sheet passes its outward normal past the one normal direction of the first sheet. Consequently, the total signed endpoint count gives $C\cdot \mathrm{\partial }T=(\mathrm{\partial }C)\cdot T\mathrm{.}$ These identities extend linearly to chains; the terms on paired interior faces cancel. For a closed $2$-chain $E$ and torsion cycle $y$, choose $m>0$ and $\mathrm{\partial }T=my$. Then $m(E\cdot y)=E\cdot \mathrm{\partial }T=0$, so $E\cdot y=0$. Thus if $\mathrm{\partial }C=nx$ and $\mathrm{\partial }{C}^{\mathrm{\prime }}=n^{\mathrm{\prime }}x$, the closed chain $n^{\mathrm{\prime }}C-n{C}^{\mathrm{\prime }}$ has zero intersection with $y$, giving $(C\cdot y)\mathrm{/}n=(C^{\mathrm{\prime }}\cdot y)\mathrm{/}{n}^{\mathrm{\prime }}$. Replacing $x$ by $x+\mathrm{\partial }S$ permits replacing $C$ by $C+nS$ and changes the quotient by the integer $S\cdot y$. Replacing $y$ by $y+\mathrm{\partial }T$ changes it by $(C\cdot \mathrm{\partial }T)\mathrm{/}n=x\cdot T$, also an integer. A change of transverse position is a chain homotopy: its prism has the moved-chain difference as boundary, so these same boundary identities apply. Hence the expression defines a pairing on the stated homology classes, independent of all these choices modulo $\mathbb{Z}$.

To identify that pairing with the Bockstein, use [F4] and [F5] and represent $x$ by a primal cellular $1$-cycle $a$, and $y$ by a dual cellular $1$-cycle $b$. Since $n[a]=0$, there is an integral primal cellular $2$-chain $C$ with $\mathrm{\partial }C=na$. The dual-cell pairing gives the rational cellular $1$-cochain $u(e)=\frac{C\cdot e}{n}$ on every oriented dual edge $e$. Let $z_a$ be the integral cellular $2$-cochain on the dual complex defined by $z_a(\sigma )=a\cdot \sigma$. The sign convention in step 1.2 gives, for every dual $2$-cell $\sigma$, $(\delta u)(\sigma )=\frac{C\cdot \mathrm{\partial }\sigma }{n}=\frac{(\mathrm{\partial }C)\cdot \sigma }{n}=a\cdot \sigma =z_a(\sigma )\mathrm{.}$ There is no extra minus sign: in an oriented chart with $a$ tangent to the first coordinate and its dual $2$-cell tangent to the second and third coordinates, $a\cdot \sigma =+1$. A primal $2$-cell followed by its positive transverse dual edge likewise has intersection $+1$. These are precisely the oriented dual-cell representatives of inverse cap duality, so $[z_a]=\mathrm{PD}(x)$.

Reducing $u$ modulo $\mathbb{Z}$ gives a cocycle $\bar{u}$, because $\delta u=z_a$ is integral. Under the natural comparison [F5], the connecting map in [F3] is constructed by lifting this cocycle to a rational cochain and taking its coboundary; therefore $\beta [\bar{u}]=[\delta u]=\mathrm{PD}(x),⟨[\bar{u}],y⟩=u(b)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathbb{Z}=\frac{C\cdot b}{n}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathbb{Z}\mathrm{.}$ By step 1.2 this is also the bounding-chain expression for any representatives and any bounding multiple in the definition. Cellular representatives cause no restriction by [F5]: cellular and singular chain models represent the same classes, and changes of representatives and transverse positions have already been accounted for. Thus the precise identity is ${\lambda }_Y(x,y)=⟨{\beta }^{-1}\mathrm{PD}(x),y⟩\mathrm{.}$ The evaluation is on the homology class $y$, not on a cohomology class called its Poincaré dual; equivalently it is the corresponding degree-$3$ cup-product evaluation on $[Y]$.

Under [F2], evaluation identifies $H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})$ with the entire character group $\mathrm{Hom}(H_1(Y;\mathbb{Z}),\mathbb{Q}\mathrm{/}\mathbb{Z})$. The displayed identity in step 3.1 identifies the adjoint of the bounding-chain pairing with the composite of this isomorphism, ${\beta }^{-1}$, and $\mathrm{PD}$. All three are isomorphisms, so the pairing is nonsingular. In particular its bilinearity also follows from these homomorphisms and evaluation.

For torsion classes $x,y$, choose transverse representatives and chains $C,D$ with $\mathrm{\partial }C=nx$ and $\mathrm{\partial }D=my$. The boundary identity of step 1.2 applied to $C\cap D$ gives $n(x\cdot D)=m(C\cdot y)$. Interchanging a $1$-cycle and a $2$-chain in an oriented $3$-manifold changes the intersection sign by $(-1{)}^{(3-1)(3-2)}=+1$, so $x\cdot D=D\cdot x$. Divide by $nm$ and reduce modulo $\mathbb{Z}$ to obtain ${\lambda }_Y(x,y)={\lambda }_Y(y,x)$. This proves symmetry with the signed bounding-chain convention of the definition.