The generator complexes are mutually inverse

The middle corner and the module $Q$. By [L3] and the tensor-unit and associativity isomorphisms of [L4] the graded abelian group $Q:={}_{i}P{\otimes }_{A_m}{P}_i$ is free with basis $u_1=e_i\otimes e_i$ in degree $0$ and $u_2=(i\mathrm{\mid }i-1\mathrm{\mid }i)\otimes e_i$ in degree $1$, so that $Q=\mathbb{Z}{u}_1\oplus \mathbb{Z}{u}_2$; by [L5] and [L4] the terms of the totalization $N:=R_i{\otimes }_{A_m}{R}_i^{-1}$ are $N^{-1}\cong U_i$, $N^0\cong A_m\oplus (P_i{\otimes }_{\mathbb{Z}}Q{\otimes }_{\mathbb{Z}}{}_{i}P\{-1\})$ and $N^1\cong U_i\{-1\}$.

The two maps of the source's square. Define the $A_m$-bimodule maps $\tau \text{\hspace{0.17em}⁣}:U_i\to P_i{\otimes }_{\mathbb{Z}}Q{\otimes }_{\mathbb{Z}}{}_{i}P\{-1\}$ and $\delta \text{\hspace{0.17em}⁣}:P_i{\otimes }_{\mathbb{Z}}Q{\otimes }_{\mathbb{Z}}{}_{i}P\{-1\}\to U_i\{-1\}$ by $\tau (x\otimes y):=x\otimes u_1\otimes (i\mathrm{\mid }i-1\mathrm{\mid }i)y+x\otimes u_2\otimes y$, $\delta (x\otimes u_1\otimes y):=x\otimes y$ and $\delta (x\otimes u_2\otimes y):=x(i\mathrm{\mid }i-1\mathrm{\mid }i)\otimes y$; both are bilinear because multiplication is, and both are degree zero: in the shifted middle object the $u_1$ and $u_2$ components have degrees $\mathrm{deg}x+\mathrm{deg}y-1$ and $\mathrm{deg}x+\mathrm{deg}y$, respectively, matching the degrees of their images $x\otimes y$ and $x(i\mathrm{\mid }i-1\mathrm{\mid }i)\otimes y$ in $U_i\{-1\}$. Each summand in $\tau (x\otimes y)$ has degree $\mathrm{deg}x+\mathrm{deg}y$; the natural balanced identification gives differentials $({\beta }_i,-\tau )$ and $({\gamma }_i,\delta )$. Negating the middle $P_i\otimes Q\otimes {}_{i}P\{-1\}$ coordinate gives the source’s signed chart, in which the differentials read ${\mathrm{\partial }}^{-1}={\beta }_i+\tau$, ${\mathrm{\partial }}^{-1}(u)=({\beta }_i(u),\tau (u))$, and ${\mathrm{\partial }}^0=({\gamma }_i,-\delta )$, ${\mathrm{\partial }}^0(a,z)={\gamma }_i(a)-\delta (z)$, the source's anticommutative square of Section 2 with the sign on $\delta$, and ${\mathrm{\partial }}^0{\mathrm{\partial }}^{-1}=0$ is the automatic square-zero condition of the totalization [L5].

The splitting of $N^0$. Let $\xi (a)$ be the image of ${\gamma }_i(a)={\sum }_j{x}_j\otimes y_j$ under $x\otimes y\mapsto x\otimes u_1\otimes y$, so that $\delta \xi (a)={\sum }_j{x}_j\otimes y_j={\gamma }_i(a)$ because $\delta$ removes the middle $u_1$; write $W:=P_i{\otimes }_{\mathbb{Z}}\mathbb{Z}{u}_1{\otimes }_{\mathbb{Z}}{}_{i}P\{-1\}$ for the $u_1$-component and define $\mathrm{\Psi }(a,w,u):=(a+{\beta }_i(u),\xi (a)+w+\tau (u))$. The map $\mathrm{\Psi }\text{\hspace{0.17em}⁣}:A_m\oplus W\oplus U_i\to N^0$ is an isomorphism of graded bimodules: its inverse sends $(a^{\mathrm{\prime }},z)$ to $u:={\tau }_2^{-1}(z_2)$, $a:=a^{\mathrm{\prime }}-{\beta }_i(u)$, $w:=z_1-{\xi }_1(a)-{\tau }_1(u)$, where $z=z_1+z_2$ decomposes along the $u_1$- and $u_2$-components, ${\tau }_2$ denotes the injective $u_2$-component $x\otimes y\mapsto x\otimes u_2\otimes y$ of $\tau$, and the subscript $1$ denotes the $u_1$-component; these four maps are well defined and degree zero. Consequently $N^0$ is the direct sum of ${\mathrm{\partial }}^{-1}(U_i)=\{({\beta }_i(u),\tau (u))\}$, the graph $T_0^0:=\{(a,\xi (a)):a\in A_m\}$ and the $u_1$-component $T_1^0:=\{(0,w):w\in W\}$, while $N^{-1}=U_i$ and $N^1=U_i\{-1\}$.

$N$ splits as a direct sum of three subcomplexes. Put $T_{-1}:=[U_i\stackrel{{\mathrm{\partial }}^{-1}}{\to }{\mathrm{\partial }}^{-1}(U_i)]$, $T_0:=\{(a,\xi (a)):a\in A_m\}$ concentrated in degree $0$, and $T_1:=[W\stackrel{-\delta }{\to }{U}_i\{-1\}]$, with differentials the restrictions of ${\mathrm{\partial }}^{-1}$ and ${\mathrm{\partial }}^0$; these are subcomplexes of $N$ because ${\mathrm{\partial }}^0{\mathrm{\partial }}^{-1}=0$ on $U_i$ by step 2.1, ${\mathrm{\partial }}^0(a,\xi (a))={\gamma }_i(a)-\delta \xi (a)=0$ by step 3.1 and ${\mathrm{\partial }}^0(0,w)=-\delta (w)$, and by the direct sum decomposition of step 3.1 the objects of $N$ are the degreewise direct sums $N^j=T_{-1}^j\oplus T_0^j\oplus T_1^j$. Hence $N=T_{-1}\oplus T_0\oplus T_1$ as complexes of graded bimodules, and $a\mapsto (a,\xi (a))$ identifies $T_0$ with the diagonal bimodule $A_m$ concentrated in degree $0$.

The two outer summands are contractible. The restriction ${\mathrm{\partial }}^{-1}:U_i\to {\mathrm{\partial }}^{-1}(U_i)$ is surjective by construction and injective because $\tau$ is injective (its $u_2$-component ${\tau }_2$ alone is already injective, as observed in step 3.1); hence it is an isomorphism, and $T_{-1}$ is a two-term complex with invertible differential, contractible by [L6]. The restriction $-\delta :W\to U_i\{-1\}$ is an isomorphism, with inverse $x\otimes y\mapsto -x\otimes u_1\otimes y$, so $T_1$ is contractible by [L6] as well.

The first homotopy equivalence. By steps 4.1 and 5.1 the complex $N$ is the direct sum of $T_0\cong A_m$ with two contractible complexes; a finite direct sum of contractible complexes is contractible, the contracting homotopy of a biproduct being the biproduct of the given homotopies, so the projection $N\to T_0\cong A_m$ and the inclusion $T_0\to N$ are inverse homotopy equivalences. This proves $R_i{\otimes }_{A_m}{R}_i^{-1}\simeq A_m$, and since the action of a complex with two-sided finite graded projective terms on $C_m$ is well defined on homotopy classes [L2], these maps induce natural isomorphisms $R_i{R}_i^{-1}\cong {\mathrm{Id}}_{C_m}$.

The opposite order. Write $c_i=(i\mathrm{\mid }i-1\mathrm{\mid }i)$. The middle corner in $N^{\mathrm{\prime }}=R_i^{-1}{\otimes }_{A_m}{R}_i$ is again $Q={}_{i}P{\otimes }_{A_m}{P}_i$, not the oppositely typed tensor $P_i{\otimes }_{A_m}{}_{i}P$. Its terms are $U_i$ in degree $-1$, $A_m\oplus (P_i\otimes Q\otimes {}_{i}P\{-1\})$ in degree $0$, and $U_i\{-1\}$ in degree $1$. Define ${\tau }^{\mathrm{\prime }}(x\otimes y)=x{c}_i\otimes u_1\otimes y+x\otimes u_2\otimes y,$ ${\delta }^{\mathrm{\prime }}(x\otimes u_1\otimes y)=x\otimes y,{\delta }^{\mathrm{\prime }}(x\otimes u_2\otimes y)=x\otimes c_{i}y\mathrm{.}$ These formulas are obtained by inserting ${\gamma }_i$ on the left and multiplying on the right in the tensor totalization; in particular ${\mathrm{\partial }}^{\mathrm{\prime }-1}=({\beta }_i,{\tau }^{\mathrm{\prime }})$ and ${\mathrm{\partial }}^{\mathrm{\prime }0}=({\gamma }_i,-{\delta }^{\mathrm{\prime }})$. They are bimodule-linear and homogeneous, and their composite is zero by [L5]. The $u_2$-component of ${\tau }^{\mathrm{\prime }}$ is the identity under its shift identification, while ${\delta }^{\mathrm{\prime }}$ is the identity from the $u_1$-component to $U_i\{-1\}$. With ${\xi }^{\mathrm{\prime }}$ given by inserting $u_1$ in ${\gamma }_i(a)$, one has ${\delta }^{\mathrm{\prime }}{\xi }^{\mathrm{\prime }}={\gamma }_i$. Hence the same explicit coordinate map ${\mathrm{\Psi }}^{\mathrm{\prime }}(a,w,u)=(a+{\beta }_i(u),{\xi }^{\mathrm{\prime }}(a)+w+{\tau }^{\mathrm{\prime }}(u))$ and its componentwise inverse from step 3.1 split $N^{\mathrm{\prime }}$ into its diagonal $A_m$ and two identity-pivot pairs. Their inverse differentials are the contracting homotopies, proving $R_i^{-1}{\otimes }_{A_m}{R}_i\simeq A_m$. Applying the action as in step 6.1 gives the opposite functor identity.

Conclusion. The complexes $R_i$ and $R_i^{-1}$ are mutually inverse up to the homotopy equivalences of steps 6.1 and 7.1, hence are inverse isomorphisms in $C_m$ and induce two-sided inverse functor isomorphisms on $C_m$; in particular both are equivalences of $C_m$. No choice principle is used, all the identifications being explicit finite formulas.