Degree-zero Tor is the tensor product in either construction

Write $d:P_1\to P_0$ and $ϵ:P_0\to M$. The complex is concentrated in nonnegative degrees, so its degree-zero homology is the abelian group $T=(N{\otimes }_R{P}_0)\mathrm{/}\mathrm{im}(1\otimes d)$, by Tor from a projective resolution of the left module. The balanced formula $n\otimes p\mapsto n\otimes ϵ(p)$ induces $a:T\to N{\otimes }_{R}M$ by Universal property of the tensor product for balanced maps into abelian groups, since $ϵd=0$.

For $n\in N$ and $m\in M$, define $b(n,m)=[n\otimes p]$ using any lift $ϵ(p)=m$. Any two lifts differ by $d(u)$, so their classes agree. Sums and scalar multiples of lifts show that $b$ is additive in both variables and $b(nr,m)=b(n,rm)$. Thus the tensor universal property gives $\bar{b}:N{\otimes }_{R}M\to T$. This definition uses unique classes, not a simultaneous selection of lifts.

The composites $a\bar{b}$ and $\bar{b}a$ fix respectively $n\otimes m$ and $[n\otimes p]$. These elements generate the two abelian groups, so the maps are inverse.

Write $e:Q_1\to Q_0$ and $\eta :Q_0\to N$. For the right-resolved construction in Tor from a projective resolution of the right module, use $U=(Q_0{\otimes }_{R}M)\mathrm{/}\mathrm{im}(e\otimes 1)$. The augmentation gives $U\to N{\otimes }_{R}M$; its inverse sends $n\otimes m$ to $[q\otimes m]$ for any $\eta (q)=n$. Differences of lifts belong to $\mathrm{im}e$, so the class is unique, and balancing follows from using $qr$ as a lift of $nr$. Both composites fix the generating tensors, proving the second canonical identification.