Finite-free chain evaluation under the Blanchfield coefficient hypotheses

A cochain $f:C_1\to N$ is a cocycle precisely when $f(B_1)=0$. Restricting to $Z_1$ therefore induces a map $H_1=Z_1\mathrm{/}{B}_1\to N$. A coboundary $g{d}_1$ vanishes on $Z_1$, so this induces the stated well-defined map $\kappa$. It is onto if every map $Z_1\to N$ vanishing on $B_1$ extends to $C_1$. It is injective if every map $B_0\to N$ extends to $C_0$: indeed a cocycle restricting to zero on $Z_1$ factors uniquely through $d_1:C_1\to B_0$, and such an extension expresses it as a coboundary.

For the principal-right-ideal alternative, $N$ is right divisible: given $n=q+R\in N$ and $0\mathrm{\ne }a\in R$, the element $q{a}^{-1}+R$ multiplied on the right by $a$ equals $n$. We prove by induction on $m$ that any right-linear map $f:S\to N$ from any right submodule $S\subseteq R^m$ extends to $R^m$. For $m=0$ this is immediate. Let the projection of $S$ to the last coordinate be the right ideal $J$. If $J=0$, apply induction on the first $m-1$ coordinates and give the last basis vector value zero. Otherwise write $J=aR$ with $a\mathrm{\ne }0$ and take $z=(z^{\mathrm{\prime }},a)\in S$. Then $S=S^{\mathrm{\prime }}\oplus zR$, where $S^{\mathrm{\prime }}=S\cap (R^{m-1}\times 0)$: subtract $zr$ from an element with last coordinate $ar$, and uniqueness follows from the absence of zero divisors.

If $H_0=0$, then $d_1:C_1\to C_0$ is onto. Lift the finitely many given basis vectors of $C_0$ and extend linearly to a section $s:C_0\to C_1$. This gives $C_1=Z_1\oplus s(C_0)$. Extend a map from $Z_1$ by zero on the second summand to prove surjectivity in step 1.1. Here $B_0=C_0$, so the extension required for injectivity is automatic. The section uses only finitely many selections.

Extend $f{\mathrm{\mid }}_{S^{\mathrm{\prime }}}$ to $g:R^{m-1}\to N$ by induction. Right divisibility supplies $v\in N$ with $va=f(z)-g(z^{\mathrm{\prime }})$. Define the extension on $R^m$ by $(u,r)\mapsto g(u)+vr$. It agrees with $f$ on $S^{\mathrm{\prime }}$ and on $z$, hence on all of $S$. This completes the finite induction, using one ideal generator, one lift, and one division at each of finitely many stages. It does not use a maximal extension argument or an infinite family of choices.

Apply the extension result of step 2.2 first to $Z_1\subseteq C_1$ and then to $B_0\subseteq C_0$, using the supplied finite bases. These are exactly the surjectivity and injectivity requirements of step 1.1. The first alternative was established in step 2.1, so $\kappa$ is an isomorphism in either case. Restriction and passage to cycles commute with every compatible left coefficient action; thus the inverse, being unique, also respects that action.