Classical and localized PTFA Blanchfield nonsingularity criteria

Work first on one nonempty connected component. Put $C_{\ast }=C_{\ast }(M;R)$ in the right-module convention. By [F2], $R$ has no zero divisors, satisfies both Ore conditions, and has fraction division ring $K$. The supplied homomorphism has unitary transport units for the given involution. By [F4], replace $C_{\ast }$ by its explicitly homotopy equivalent free complex $C_{\ast }^{\mathrm{\prime }}$ with $C_0^{\mathrm{\prime }}=R$. This replacement preserves homology, dual cohomology and the natural evaluation maps; the discarded summand has an explicitly contracted identity differential.

By [F3] and $H_1(M;K)=0$, one has $H_1(C_{\ast }){\otimes }_{R}K=0$. The elementwise zero criterion of [F3] says that every element of $H_1(C_{\ast })$ has a nonzero right annihilator. Thus the whole module equals its torsion submodule. This uses arbitrary-complex localization, so no finite CW or triangulation assumption is introduced.

The degree-zero term of $C_{\ast }^{\mathrm{\prime }}{\otimes }_{R}K$ is $K$, and the image of its degree-one differential is either zero or $K$. In the first case $d_1=0$, and $H_1(M;K)=0$ implies that every degree-one chain is a boundary, so every degree-one dual cocycle vanishes. In the second case choose $z$ with $d_{1}z=1$; then $C_1^{\mathrm{\prime }}{\otimes }_{R}K=\ker d_1\oplus zK$. The same homology vanishing says $\ker d_1=\mathrm{im}{d}_2$, so every dual cocycle factors through $d_1$ and is the coboundary of its value on $z$. Hence $H^1({\mathrm{Hom}}_R(C_{\ast },K))=0$, using the tensor adjunction in [F5]. Also [F5] directly identifies $H^2({\mathrm{Hom}}_R(C_{\ast },K))$ with $\overline{H_1(M;K)}=0$. The exact segment [F6], applied to the compatible dual local systems of [F5], therefore gives a Bockstein isomorphism $\beta :H^1({\mathrm{Hom}}_R(C_{\ast },K\mathrm{/}R))\to H^2({\mathrm{Hom}}_R(C_{\ast },R))$.

Restriction of a cocycle to cycles always defines the left-$R$ linear evaluation map $\kappa :H^1({\mathrm{Hom}}_R(C_{\ast },K\mathrm{/}R))\to {\mathrm{Hom}}_R(H_1(C_{\ast }),K\mathrm{/}R)$. Under the additional alternatives it is an isomorphism by [F4]. Indeed the reduced boundary image $B_0\subseteq C_0^{\mathrm{\prime }}=R$ is a right ideal, so it is principal if $R$ is right principal. If $H_0(M;R)=0$, it is all of $R$. If the coefficient image is cyclic generated by $t$, it is $(t-1)R$, including zero for trivial image. These verify precisely the hypothesis of the principal-boundary evaluation lemma, whose proof allows arbitrary $C_1^{\mathrm{\prime }}$. Thus no finite generation or principal-ideal assumption is silently imposed in the other two branches.

By [F5], the inverse oriented duality map is an isomorphism $\overline{H_1(C_{\ast })}\to H^2({\mathrm{Hom}}_R(C_{\ast },R))$. Compose it with ${\beta }^{-1}$ from step 2.2 and the evaluation map $\kappa$ from step 2.3. Evaluating the resulting functional on $y$ defines $B(x,y)\in K\mathrm{/}R$. The composition is left-$R$ linear on the conjugate module and its values are right-$R$ linear functions of $y$, so $B(xr,y)=\bar{r}B(x,y)$ and $B(x,ys)=B(x,y)s$. This is exactly [F1]'s Bockstein–duality pairing, defined before any claim that $\kappa$ is invertible. Under the alternatives of step 2.3, all three maps in its adjoint are isomorphisms. Therefore $\overline{H_1(C_{\ast })}\to {\mathrm{Hom}}_R(H_1(C_{\ast }),K\mathrm{/}R)$ is an isomorphism, which is nonsingularity in the stated right-module convention.

A compact manifold has finitely many components, since its connected components are open and cover it. Apply the argument to each, using only finitely many basepoint and finite-atlas choices. Chains split over those components, and cochains are a finite product, so their cycle/boundary and Bockstein decompositions need no infinite primitive selection. The oriented fundamental class is the sum of the component classes, and the pairing is their orthogonal direct sum. Under the stated componentwise alternatives, each adjoint is an isomorphism, hence so is their finite direct sum. Empty $M$ gives the unique pairing on the zero module. In particular a cyclic integral coefficient image satisfies the third alternative even when $R=\mathbb{Z}[t^{\pm 1}]$ is not right principal. This proves the full stated scope without a choice axiom.