A compact manifold's twisted singular chains admit a one-vertex reduction without choice

For a singular simplex $\sigma :[v_0,\dots ,v_n]\to M$, put its coefficient in ${\mathcal{R}}_{\sigma (v_0)}$ and take the direct sum of those right modules over all singular simplices. For $n\ge 1$ define $d(c\sigma )=T_{01}(c)\sigma [1,\dots ,n]+{\sum }_{i=1}^n(-1{)}^{i}c\sigma [0,\dots ,\widehat{i},\dots ,n]$, and let the degree-zero differential be zero. Every face map is right-linear because transport is, so this defines right-$R$ linear differentials without assuming that $R$ is commutative. To check $d^2=0$, pair the two terms deleting vertices $i<j$. Their signs are $(-1{)}^{i+j-1}$ and $(-1{)}^{i+j}$. If $i>0$, both coefficients are $c$. If $i=0<1<j$, both coefficients are $T_{01}(c)$. If $(i,j)=(0,1)$, the coefficients are $T_{12}{T}_{01}(c)$ and $T_{02}(c)$, which agree because the two-edge path is endpoint-fixed homotopic to the direct edge in the simplex and transport is a groupoid functor. Thus every pair cancels. This includes degenerate simplices and degree one, where the next differential is zero. These are the underlying abelian local chains of [F1], now equipped with their verified right-$R$ structure.

Choose a finite cover by coordinate balls $U_1,\dots ,U_m$ with charts onto convex Euclidean balls, including a ball centered at $o$. This uses compactness and only a finite selection. Their intersection graph is connected: otherwise the unions belonging to different graph components would be disjoint nonempty open sets partitioning connected $M$. For each edge choose one point in the intersection, a finite selection; join it to each chart center by its straight coordinate segment. Choose a spanning tree by the finite index order and concatenate these segments to get paths from $o$ to every chart center. For $x\in M$, take the least-index chart containing it and follow its center path by the straight segment to $x$; at $o$ use the constant path. Denote this explicitly prescribed path by ${\gamma }_x$. This defines paths for all points from one finite list of charts and edge points, rather than selecting an infinite family of paths.

Transport the supplied basis $1\in {\mathcal{R}}_o$ along ${\gamma }_x$ to obtain a basis at each $x$. Thus [F1] makes every $C_j$ the free right $R$-module on its actual singular $j$-simplices. Write $e_x$ for the degree-zero basis and $e_{\sigma }$ for the degree-one basis of a path $\sigma :x\to y$. The degree-one boundary is $d_1{e}_{\sigma }=e_y{g}_{\sigma }-e_x$, where $g_{\sigma }\in H$ is the transition unit obtained by transporting first along ${\gamma }_x$, then along $\sigma$, and back along ${\gamma }_y$. For $x\mathrm{\ne }o$, the path ${\gamma }_x$ is itself one singular one-simplex; denote its basis element by $p_x$. These basis elements are distinct because their terminal points are distinct, and $d_1{p}_x=e_x-e_o$. Put $p_o=0$.

Change the degree-zero basis to $e_o$ and $b_x=e_x-e_o$ for $x\mathrm{\ne }o$. This is invertible even with infinitely many vertices: the inverse is $e_x=b_x+e_o$, and every vector has finite support. Retain the degree-one basis elements $p_x$. For every other one-simplex $\sigma :x\to y$, replace its basis element by $q_{\sigma }=e_{\sigma }-p_y{g}_{\sigma }+p_x$. This is another invertible basis change, with inverse $e_{\sigma }=q_{\sigma }+p_y{g}_{\sigma }-p_x$. Direct substitution gives $d_1{q}_{\sigma }=e_o(g_{\sigma }-1)$, whereas $d_1{p}_x=b_x$.

Let $Q_1$ be free on the $p_x$, $Q_0$ free on the $b_x$, and $Q_j=0$ in other degrees, with differential $p_x\mapsto b_x$. Let $C_1^{\mathrm{\prime }}$ be free on the remaining $q_{\sigma }$, $C_0^{\mathrm{\prime }}=R{e}_o$, and $C_j^{\mathrm{\prime }}=C_j$ for $j\ge 2$. The differential of a degree-two chain has zero $p_x$ coefficient for every $x$: applying $d_1$ would otherwise give a nonzero corresponding $b_x$ coefficient, contrary to $d_1{d}_2=0$ proved in step 1.1. Thus $d_2$ lands in $C_1^{\mathrm{\prime }}$, and higher differentials already stay in $C^{\mathrm{\prime }}$. This proves the claimed direct sum of complexes. Its contractible summand has the explicit homotopy $b_x\mapsto p_x$, zero on all other degrees. Dualizing this actual homotopy gives the cochain equivalences for any coefficient module; no exactness assertion about an infinite product or choice of primitives is needed.

The image of $d_1^{\mathrm{\prime }}$ is the right ideal generated by the elements $g_{\sigma }-1$. All transition units belong to $H$, so this ideal is contained in ${\sum }_{h\in H}(h-1)R$. Conversely every $h\in H$ is realized by transport along a based loop at $o$. That loop is one singular one-simplex and is not one of the removed $p_x$, whose endpoints differ. Its reduced boundary is $e_o(h-1)$. Thus every generator of the displayed ideal occurs, and equality follows.

If $H=⟨t⟩$ is cyclic, $t^k-1$ is a right multiple of $t-1$ for every integer $k$: use the finite geometric sum for $k>0$, zero for $k=0$, and $t^{-k}-1=-t^{-k}(t^k-1)$ for negative powers. Since powers of $t$ commute with $t-1$, this last expression is also a right multiple. The ideal is therefore $(t-1)R$, including $H=\{1\}$ with zero ideal. The other two sufficient conditions for principality are literal. All basis changes and homotopies were explicit finite-support formulas, and the only initial choices were finite. This proves the stated choice scope.