Three projective degrees suffice for twisted first-homology comparison

Construct comparison maps in just three degrees. Lift ${ϵ}_C:C_0\to S$ across the surjection ${ϵ}_D$ using projectivity of $C_0$, obtaining $f_0:C_0\to D_0$. The image of $f_0{d}_1$ lies in $\ker {ϵ}_D=\mathrm{im}{\delta }_1$, because ${ϵ}_D{f}_0{d}_1={ϵ}_C{d}_1=0$. Regard ${\delta }_1:D_1\to \ker {ϵ}_D$ as a surjection, and use projectivity of $C_1$ to lift this map to $f_1:C_1\to D_1$ with ${\delta }_1{f}_1=f_0{d}_1$. Now ${\delta }_1{f}_1{d}_2=f_0{d}_1{d}_2=0$, so $f_1{d}_2$ maps into $\ker {\delta }_1=\mathrm{im}{\delta }_2$. Lift it across $D_2\to \ker {\delta }_1$ by projectivity of $C_2$, obtaining $f_2$ with ${\delta }_2{f}_2=f_1{d}_2$. This gives all required chain and augmentation identities. Reversing $C,D$ produces a map $g$ in exactly the same finite way. No exactness in degree2 was used.

Uniqueness up to the homotopy needed in degree one. More generally let $f,f^{\mathrm{\prime }}:C\to D$ be any two comparison maps lifting the same $1_S$, and put $a_i=f_i-f_i^{\mathrm{\prime }}$. Since ${ϵ}_D{a}_0=0$, projectivity of $C_0$ and exactness at $D_0$ give $h_0:C_0\to D_1$ with ${\delta }_1{h}_0=a_0$. The map $a_1-h_0{d}_1:C_1\to D_1$ has zero composite with ${\delta }_1$, because ${\delta }_1{a}_1=a_0{d}_1$. By exactness at $D_1$ and projectivity of $C_1$, it therefore lifts to $h_1:C_1\to D_2$ with ${\delta }_2{h}_1=a_1-h_0{d}_1$. Thus $a_0={\delta }_1{h}_0$ and $a_1={\delta }_2{h}_1+h_0{d}_1$. No further homotopy degree is needed for first homology. This construction applies in particular to the two comparisons $gf$ and $1_C$, and also to $fg$ and $1_D$.

Tensor the actual identities, without a flatness assumption. For each left-linear map $u$ used above, define $1\otimes u$ on elementary tensors by $n\otimes c\mapsto n\otimes u(c)$. It respects the balanced relation since $u(rc)=ru(c)$, and it respects the additive relations in [F2]. Consequently these induced maps respect sums, differences, compositions and every chain identity above. If $z\in N{\otimes }_R{C}_1$ satisfies $(1\otimes d_1)z=0$, the second identity of step2.1 becomes $((1\otimes f_1)-(1\otimes f_1^{\mathrm{\prime }}))z=(1\otimes {\delta }_2)(1\otimes h_1)z$, an actual boundary. Chain maps preserve cycles and boundaries, so this proves equality of their induced first-homology maps. Apply it to $gf,1_C$ and $fg,1_D$ to show that the induced maps of $f$ and $g$ are mutually inverse. This uses preserved homotopy identities, not preserved exactness of the tensored augmented complexes. In particular no flatness of $N$ has entered.

Compatibility, finite choice use and limits. A composite of augmentation-preserving comparisons is again such a comparison, so the uniqueness in step3.1 identifies its first-homology map with the unique comparison to the third complex. For a homomorphism $q:N\to N^{\mathrm{\prime }}$ of right modules, the tensor maps $q\otimes 1$ commute with all $1\otimes f_i$ on each elementary tensor and hence on finite sums; this proves naturality. The same uniqueness shows that a separately supplied geometric or algebraic comparison induces exactly this isomorphism. Each lift in steps1.1–2.1 is supplied by the projectivity of its particular domain, and the proof takes only finitely many successive witnesses, including the reverse map and the two finite homotopies. No choice of lifts in all degrees, all objects or all coefficient modules is required: independence specifies a unique resulting map for each $N$. Zero modules, zero differentials compatible with the exactness assumptions, and the zero coefficient module obey the same equations. The proof neither constructs a topological spine nor assumes that arbitrary coefficient tensoring preserves exactness. This proves all the stated assertions with the prescribed left/right sides over an arbitrary ring.