The homological universal-coefficient Tor obstruction map

AC is used in Boundaries and cycles in a free complex over a PID are free to make $B_j$ and $Z_j$ free, hence flat, even for infinitely generated $C_j$. Thus $0\to B_{n-1}\to Z_{n-1}\to H_{n-1}C\to 0$ is a supplied free resolution of length one. Tensoring it with $G$ identifies ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$ with $T:=\ker (B_{n-1}{\otimes }_{R}G\to Z_{n-1}{\otimes }_{R}G)$. AC also implies the DC premise used for comparison of supplied resolutions in The balanced Tor bifunctor, so this kernel represents the stated balanced Tor group naturally.

Corestrict $d_n:C_n\twoheadrightarrow B_{n-1}$ before tensoring. Let ${\bar{d}}_n\otimes 1:C_n\otimes G\to B_{n-1}\otimes G$ denote the resulting surjection. If $c\in C_n\otimes G$ is a cycle, then the composite of ${\bar{d}}_n\otimes 1$ with $B_{n-1}\otimes G\to Z_{n-1}\otimes G\to C_{n-1}\otimes G$ is zero. The last arrow is injective: tensor $0\to Z_{n-1}\to C_{n-1}\to B_{n-2}\to 0$ and use flatness of $B_{n-2}$. Hence $({\bar{d}}_n\otimes 1)c\in T$. If $c=(d_{n+1}\otimes 1)b$, then ${\bar{d}}_n{d}_{n+1}=0$, so this element is zero. We obtain a homomorphism on homology.

A chain map $f:C\to C^{\mathrm{\prime }}$ restricts to maps of the corresponding $B$ and $Z$ modules; together with a coefficient map $G\to G^{\mathrm{\prime }}$ (over the same ring), these maps commute with the corestricted differentials and the inclusions defining $T$. Thus the cycle formula commutes with them. Naturality of the free-resolution kernel identification with Tor follows from the comparison and coherence built into The balanced Tor bifunctor. The map is surjective as well: $C_n\otimes G\twoheadrightarrow B_{n-1}\otimes G$ is surjective by right exactness, and any lift of an element of $T$ is a cycle since $Z_{n-1}\otimes G\hookrightarrow C_{n-1}\otimes G$. Surjectivity is recorded to identify the obstruction map used by the universal-coefficient exact sequence, though no splitting is claimed here.