Tor is symmetric over a commutative ring

Use the supplied resolution $P_{\bullet }\to M$. Over a commutative ring, turning a left module into a right module by $pr=rp$ preserves module maps, surjections and their lifting diagrams, so it preserves projectivity and exactness. Thus $P$ is a projective resolution on either side. The termwise swaps $P_j{\otimes }_{R}N\to N{\otimes }_R{P}_j$, $p\otimes n\mapsto n\otimes p$, are inverse to the opposite swaps. Both differential composites send $p\otimes n$ to $n\otimes d_{P}p$, so these are chain maps.

The first complex computes ${\mathrm{Tor}}_i^R(M,N)$ through Tor from a projective resolution of the right module, and the second computes ${\mathrm{Tor}}_i^R(N,M)$ through Tor from a projective resolution of the left module. Taking homology of the chain isomorphism gives the desired isomorphism for these supplied presentations. Only one factor has a nonzero chain degree, so no double-complex sign is inserted in this termwise swap.

Given maps $M\to M^{\mathrm{\prime }}$ and $N\to N^{\mathrm{\prime }}$, DC supplies the comparison maps used in Each resolution-defined Tor construction is covariant in both variables. A comparison $f:P\to P^{\mathrm{\prime }}$ and a map $v:N\to N^{\mathrm{\prime }}$ commute with the swaps since both composites send $p\otimes n$ to $v(n)\otimes f(p)$. Homology therefore gives naturality. Independence of lifts follows from the cited covariance proposition, and The Tor balance isomorphism is natural and coherent under a change of resolutions identifies these presentations with The balanced Tor bifunctor coherently. This proves the natural balanced Tor symmetry.