Step 1: Let $P_\bullet \to M$ and $Q_\bullet \to N$ be projective resolutions. We show: $H_n(P_\bullet \otimes_R N) \cong H_n(M \otimes_R Q_\bullet) \cong H_n(P_\bullet \otimes_R Q_\bullet)$

Step 2: Consider the double complex $P_\bullet \otimes_R Q_\bullet$ with:

• Horizontal differential: $d^h = \partial^P \otimes \text{id}_Q$
• Vertical differential: $d^v = (-1)^p \text{id}_P \otimes \partial^Q$ at bidegree $(p, q)$

These anti-commute: $d^h d^v + d^v d^h = 0$.

Step 3: The total complex $\text{Tot}(P_\bullet \otimes_R Q_\bullet)$ has: $\text{Tot}_n = \bigoplus_{p+q=n} P_p \otimes_R Q_q$

with differential $d = d^h + d^v$.

Step 4: We can compute homology of $\text{Tot}$ in two ways via spectral sequences:

First spectral sequence: Fix $q$ and take homology in the $p$ direction first. ${}^I E^1_{p,q} = H_p^h(P_\bullet \otimes_R Q_q)$

Since $Q_q$ is projective (flat) and $P_\bullet \to M$ is a resolution: $H_p^h(P_\bullet \otimes_R Q_q) = \begin{cases} M \otimes_R Q_q & p = 0 \\ 0 & p > 0 \end{cases}$

Thus the spectral sequence degenerates at $E^1$: ${}^I E^2_{0,q} = H_q(M \otimes_R Q_\bullet)$

and converges to $H_n(\text{Tot})$.

Second spectral sequence: Fix $p$ and take homology in the $q$ direction first. ${}^{II} E^1_{p,q} = H_q^v(P_p \otimes_R Q_\bullet)$

By the same reasoning (using that $P_p$ is projective): ${}^{II} E^1_{p,q} = \begin{cases} P_p \otimes_R N & q = 0 \\ 0 & q > 0 \end{cases}$

Thus: ${}^{II} E^2_{p,0} = H_p(P_\bullet \otimes_R N)$

and converges to $H_n(\text{Tot})$.

Step 5: Since both spectral sequences converge to the same thing: $H_n(P_\bullet \otimes_R N) \cong H_n(\text{Tot}) \cong H_n(M \otimes_R Q_\bullet)$

This proves Tor can be computed using a resolution of either variable.

Step 6: Symmetry follows from the symmetry of tensor product $M \otimes_R N \cong N \otimes_R M$, which induces an isomorphism on Tor groups.