Show \(\tor_R^*(\cdot, \cdot)\) is symmetric in the following way: Given \(M, N\), take free resolutions, view \(M_* \into M\) as a chain map and tensor with \(N_*\) to get a chain map\(\psi: M_* \tensor_R N_* \into M \tensor_R N_*\). Show that \(\psi\) is a quasi-isomorphism using the exact sequence \(0 \into (Z_n, 0) \into (N_n, 0) \into (B_{n-1}, 0) \into 0\), then switch the roles of \(M, N\).
Solution
1. Double complex setup:
1.1. Let \(P_* \xrightarrow{\varepsilon_P} M \to 0\) and \(Q_* \xrightarrow{\varepsilon_Q} N \to 0\) be projective resolutions of \(R\)-modules \(M\) and \(N\) respectively.
existence of projective resolutions in \(R\text{-}\mathbf{Mod}\).
1.2. Form the first-quadrant double complex \(C_{p,q} = P_p \otimes_R Q_q\) with horizontal differential \(d^h = d^P \otimes \operatorname{id}_Q\) and vertical differential \(d^v = (-1)^p \operatorname{id}_P \otimes d^Q\).
definition of tensor product double complex.
1.3. The total complex \(\operatorname{Tot}(P_* \otimes_R Q_*)\) has degree-\(n\) term \(T_n = \bigoplus_{p+q=n} P_p \otimes_R Q_q\) and boundary operator \(D = d^h + d^v\) with \(D^2 = 0\).
\((d^h + d^v)^2 = (d^h)^2 + (d^h d^v + d^v d^h) + (d^v)^2 = 0 + 0 + 0 = 0\).
2. Quasi-isomorphism \(\operatorname{Tot}(P_* \otimes_R Q_*) \xrightarrow{\sim} M \otimes_R Q_*\):
2.1. For each fixed \(q \ge 0\), \(Q_q\) is projective, hence flat as an \(R\)-module.
projective modules are flat.
2.2. Since \(Q_q\) is flat, tensoring the resolution \(P_* \to M \to 0\) with \(Q_q\) preserves exactness:\[ H_p(P_* \otimes_R Q_q) \cong H_p(P_*) \otimes_R Q_q \cong \begin{cases} M \otimes_R Q_q & p = 0 \\ 0 & p > 0. \end{cases} \]
flatness commutes with homology.
2.3. Viewing \(M \otimes_R Q_*\) as a double complex concentrated in row \(p = 0\), the augmentation map \(\varepsilon_P \otimes \operatorname{id}: P_* \otimes_R Q_* \to M \otimes_R Q_*\) is a quasi-isomorphism on each column.
2.
2.4. By the Acyclic Assembly Lemma (or the spectral sequence of a first-quadrant bicomplex), an augmentation that is exact on rows/columns induces a quasi-isomorphism on total complexes:\[ \operatorname{Tot}(P_* \otimes_R Q_*) \xrightarrow{\sim} \operatorname{Tot}(M \otimes_R Q_*) = M \otimes_R Q_*. \]
Weibel Theorem 2.7.2 (Acyclic Assembly Lemma).
2.5. Taking homology gives an isomorphism:\[ H_n(\operatorname{Tot}(P_* \otimes_R Q_*)) \cong H_n(M \otimes_R Q_*) \cong \operatorname{Tor}_n^R(M, N). \]
definition of \(\operatorname{Tor}_n^R(M, N)\) via projective resolution of \(N\).
3. Quasi-isomorphism \(\operatorname{Tot}(P_* \otimes_R Q_*) \xrightarrow{\sim} P_* \otimes_R N\):
3.1. Symmetrically, for each fixed \(p \ge 0\), \(P_p\) is projective, hence flat.
projective modules are flat.
3.2. Tensoring the resolution \(Q_* \to N \to 0\) with \(P_p\) preserves exactness:\[ H_q(P_p \otimes_R Q_*) \cong P_p \otimes_R H_q(Q_*) \cong \begin{cases} P_p \otimes_R N & q = 0 \\ 0 & q > 0. \end{cases} \]
flatness of \(P_p\).
3.3. By the Acyclic Assembly Lemma applied to the vertical filtration:\[ \operatorname{Tot}(P_* \otimes_R Q_*) \xrightarrow{\sim} P_* \otimes_R N. \]
Acyclic Assembly Lemma on vertical augmentation.
3.4. Taking homology gives:\[ H_n(\operatorname{Tot}(P_* \otimes_R Q_*)) \cong H_n(P_* \otimes_R N) \cong \operatorname{Tor}_n^R(N, M), \]via the natural isomorphism \(P_* \otimes_R N \cong N \otimes_R P_*\).
definition of \(\operatorname{Tor}_n^R(N, M)\) via projective resolution of \(M\).
4. Conclusion:Combining 2 and 3 yields canonical isomorphisms:\[ \operatorname{Tor}_n^R(M, N) \cong H_n(\operatorname{Tot}(P_* \otimes_R Q_*)) \cong \operatorname{Tor}_n^R(N, M). \]Thus \(\operatorname{Tor}_*^R\) is symmetric. Q.E.D.
2 and 3.