Twisted right-chain duals realize local coefficient duality

Use the finite-atlas paths of [F1] to transport the supplied base vector $1$ to a frame $e_x$ in every fiber. Two paths to $x$ give frames $e_x$ and $e_x^{\mathrm{\prime }}=e_{x}v$ where $v$ is unitary, because their difference is a based loop. Define $h_x(e_{x}a,e_{x}b)=\bar{a}b$. In the second frame the coordinates are $v^{-1}a,v^{-1}b$, and $\overline{v^{-1}a}{v}^{-1}b=\bar{a}(\bar{v}{)}^{-1}{v}^{-1}b=\bar{a}b\mathrm{.}$ Thus this definition is path-independent and every transport preserves $h$. The finite-atlas rule supplies all the paths by a formula, so no infinite family of choices is required. The resulting frames are unitary and their transition units are unitary.

Use the paths from [F1] to put a specified basis on every fiber, and let $u_{01}$ be the transition unit along the first edge of a singular simplex. In this gauge its right-chain boundary has first term $e_{{\mathrm{\partial }}_0\sigma }{u}_{01}$ and its remaining terms $(-1{)}^i{e}_{{\mathrm{\partial }}_i\sigma }$ for $i>0$. A right-linear cochain $f$ consequently has coboundary first term $f(e_{{\mathrm{\partial }}_0\sigma })u_{01}$, with the same remaining signed terms. Intrinsic local cochains use inverse first-edge transport. Thus these are exactly the cochain formulas for the coefficient system whose transport is right multiplication by $u_{01}^{-1}$; that transport commutes with left multiplication by $R$. Sending $f$ to its values on these supplied simplex basis elements is a bijection, in every degree, with the intrinsic cochains, and the displayed formulas show that it commutes with coboundaries. No values in an infinite product are lifted or chosen.

Intrinsically send $\bar{v}$ to the right-linear functional $w\mapsto h(v,w)$. In a unitary frame this is $J:\bar{R}\to R$, $J(\bar{a})=\bar{a}$, where the bar on the right is the ring involution. It is left-linear: $J(r\cdot \bar{a})=\overline{a\stackrel{ˉ}{r}}=r\bar{a}$. It commutes with transport, since $\overline{ua}=\bar{a}\bar{u}=\bar{a}{u}^{-1}$. Hence it identifies the conjugate chain-coefficient local system with the dual system of step 1.2. Conjugation changes neither underlying abelian-group cycles nor boundaries and is involutive, so the homology of that dual system identifies with $\overline{H_{\ast }(C_{\ast })}$. When $M$ is oriented and closed, apply [F2] to the actual dual local system from step 1.2, viewed as a local system of abelian groups with a commuting left-$R$ action. Its cap isomorphism gives precisely the displayed cohomology-to-conjugate-homology identification.

For $K$ coefficients, every right-linear $f:C_j\to K$ extends uniquely to the right-$K$ linear map $C_j{\otimes }_{R}K\to K$, $c\otimes k\mapsto f(c)k$. Balancing proves that formula well-defined; its inverse is restriction to $c\otimes 1$. These maps commute with the differentials and left $K$ actions. The chain coefficient system is now the scalar extension with fiber $K$, and its transport units have the same unitary involution property. Steps 1.2 and 2.1 applied to it give the asserted $K$ conclusion, including its left-$K$ linearity. This uses explicit tensor adjunction and no basis selection over $K$.

The quotient $K\mathrm{/}R$ is an $R$-bimodule because $R$ is a subring, and the involution descends because $\bar{R}=R$. Replace the value module in step 1.2 by this bimodule. Right multiplication by $u^{-1}$ is still well-defined and commutes with left $R$, so exactly the same simplex formula identifies its dual cochains. Inclusion and quotient of the three value modules commute with those multiplications, and therefore with the chain/cochain identifications. The dual-to-conjugate identification is intrinsic relative to $h$. Explicitly, under a unitary frame change $e^{\mathrm{\prime }}=ev$, vector coordinates change to $a^{\mathrm{\prime }}=v^{-1}a$ and right-linear functional coordinates to $q^{\mathrm{\prime }}=qv$. Then $\overline{a^{\mathrm{\prime }}}=\bar{a}(\bar{v}{)}^{-1}=\bar{a}v$, which is exactly the functional transformation. Equivalently its inverse sends $f$ to $\overline{e\stackrel{‾}{f(e)}}$; replacing $e$ by $ev$ leaves this unchanged since $v\bar{v}=1$. The same standard Hermitian structure extends to $K$, and these unitary frame identities hold there as well. All constructions used only the finite atlas data of [F1] and formula-defined maps on supplied generators. This proves every claimed compatibility without a choice principle.