$\mathrm{Tor}_i(-, -)$ is symmetric

The $\mathrm{Tor}$ functors, which are ubiquitous in algebraic topology and more generally homological algebra, encode the failure of the tensor product to be an exact functor. As a derived functor, one typically defines $\mathrm{Tor}_i^R(M, N)$ by taking a projective resolution of the first argument $M$ and computing the homology of the tensored chain complex. Surprisingly, this seemingly asymmetric process produces a symmetric bifunctor; in this short note we explain why this is so and construct a natural isomorphism $\mathrm{Tor}_i^R(M, N) \cong \mathrm{Tor}_i^R(N, M)$.

Throughout this post, let $R$ be a commutative ring.

Projective resolutions and $\mathrm{Tor}$ functors

In order to define the $\mathrm{Tor}$ functors, we first need to fix a projective resolution $\mathscr{P}_{\bullet}(M)$ for every $R$-module $M$. For the unacquainted, this means we first regard $M$ as the chain complex $M_{\bullet}$ given by

$$M_{i} = \begin{cases} M &\text{if } i = 0 \\ 0 &\text{if } i \neq 0 \end{cases}$$

and choose a complex of projective modules $\mathscr{P}_{\bullet}(M)$ quasi-isomorphic to $M_{\bullet}$ (which we denote using $\sim$). By quasi-isomorphic, we mean that both complexes have isomorphic homology, which in particular implies that $\mathscr{P}_\bullet(M)$ is exact at all positive degrees. To obtain a sequence which is exact at every term, we may augment $\mathscr{P}_\bullet(M)$ by replacing the $-1$st degree with $M$ and defining an augmentation map $\mathrm{d}_0 \colon \mathscr{P}_0(M) \to M$ (typically denoted by $\epsilon$) using the quasi-isomorphism as follows:

$$\mathscr{P}_0(M) \twoheadrightarrow \mathrm{H}_0(\mathscr{P}_\bullet(M)) \overset{\cong}{\to} \mathrm{H}_0(M_\bullet) \cong M.$$

By construction, $\ker \mathrm{d}_0 = \operatorname{im} \mathrm{d}_1$, so the augmented complex $\mathscr{P}_\bullet(M) \twoheadrightarrow M \to 0$, which we now denote by $\widehat{\mathscr{P}}_\bullet(M)$, is exact at each degree.

Given an $R$-module $N$, we define $\mathrm{Tor}_{i}(M, N) = \mathrm{H}_{i}(\mathscr{P}_{\bullet}(M) \otimes_R N)$ to be the homology of the tensored complex (which is no longer necessarily exact). It is a fundamental result of homological algebra that this process does not depend on the specific choice of projective resolutions: any two choices result in naturally isomorphic functors. Furthermore, since $- \otimes_R -$ is symmetric, one obtains a naturally isomorphic functor if one chooses to use the left tensor product $N \otimes_R -$ instead.

The failure for $\mathscr{P}_{\bullet}(M) \otimes_R N$ to be exact is due to the fact that $- \otimes_R N$ is typically only right exact. As it is both instructive and useful for later arguments, we highlight the specific obstruction to a right exact functor preserving long exact sequences.

Proof. By inserting the images of each differential, we obtain a short exact sequence $0 \to \mathrm{d}A_{i+1} \to A_i \to \mathrm{d}A_i \to 0$ fitting into the diagram

Since $F$ is right exact, $F(A_{i+1}) \twoheadrightarrow F(\mathrm{d}A_{i+1})$ is still surjective, and $F(\mathrm{d}A_{i+1}) \to F(A_i) \to F(\mathrm{d}A_i) \to 0$ is exact. Consider the diagram

By exactness and surjectivity, we have $\ker \big(F(A_i) \to F(\mathrm{d}A_i)\big) = \operatorname{im} \big(F(\mathrm{d}A_{i+1}) \to F(A_i)\big) = \operatorname{im} F(\mathrm{d}_{i+1})$. Finally, since $F(A_i) \to F(\mathrm{d}A_{i})$ is also surjective, it follows that

$$\ker\big(F(\mathrm{d} A_i) \to F(A_{i-1})\big) \cong \ker\left(\frac{F(A_i)}{\operatorname{im} F(\mathrm{d}_{i+1})} \to F(A_{i-1})\right) = \frac{\ker F(\mathrm{d}_i)}{\operatorname{im} F(\mathrm{d}_{i+1})} = \mathrm{H}_i(F(A_\bullet)).$$

In other words, for all $i > 0$, the preceding proposition says that $\mathrm{Tor}_i^R(M, N)$ measures the failure of $- \otimes_R N$ to preserve the injectivity of the map $\mathrm{d}P_i \hookrightarrow P_{i-1}$ for any given projective resolution $P_\bullet$ of $M$.

Symmetry of $\mathrm{Tor}^R_i(-, -)$

To show that $\mathrm{Tor}_i$ is symmetric, we split into cases based on whether or not $i$ is positive or not. For $i = 0$, note that by right-exactness, we have

$$\mathrm{Tor}_0^R(M, N) = \operatorname{coker}(\mathscr{P}_1(M) \otimes_R N \to \mathscr{P}_0(M) \otimes_R N) = \operatorname{coker}(\mathscr{P}_1(M) \to \mathscr{P}_0(M)) \otimes_R N \cong M \otimes_R N,$$

since $\mathscr{P}_\bullet(M) \sim M$. Since the tensor product is symmetric, it follows that

$$\mathrm{Tor}_0^R(M, N) \cong M \otimes_R N \cong N \otimes_R M \cong \mathrm{Tor}_0^R(N, M).$$

For $i > 0$, it suffices to show that $\mathrm{H}_i(\mathscr{P}_\bullet(M) \otimes_R N)$ is naturally isomorphic to $\mathrm{H}_i(M \otimes_R \widehat{\mathscr{P}}_\bullet(N))$, since $- \otimes_R -$ is symmetric, and any complex is identical to its augmentation for positive degrees. For brevity, denote the projective resolution of $M$ by $P_\bullet$ and the augmented projective resolution of $N$ by $Q_\bullet$. By the previous proposition, for all $i > 0$ we have

$$\mathrm{H}_i(M \otimes_R Q_\bullet) \cong \ker \big(M \otimes_R \mathrm{d}Q_i \to M \otimes_R Q_{i-1}\big).$$

The trick now is to realize that $\mathrm{Tor}$ can also measure the failure for $M \otimes_R -$ to preserve injections in a second way: the $\mathrm{Tor}$ long exact sequence. To see, take the inclusion $\mathrm{d}Q_{i} \hookrightarrow Q_{i-1}$ and "complete" it to a short exact sequence. Then tensoring with $P_\bullet$ yields a short exact sequences of complexes

$$0 \to P_\bullet \otimes_R \mathrm{d}Q_{i} \to P_\bullet \otimes_R Q_{i-1} \to P_\bullet \otimes_R \mathrm{d}Q_{i-1} \to 0. \tag{$\star$}$$

Applying the zig-zag lemma yields the $\mathrm{Tor}$ long exact sequence, which ends with

$$\cdots \to \underbrace{\mathrm{Tor}_{1}^R(M, Q_{i-1})}_{= \, 0} \to \mathrm{Tor}_{1}^R(M, \mathrm{d}Q_{i-1}) \to M \otimes_R \mathrm{d}Q_{i} \to M \otimes_R Q_{i-1} \to M \otimes_R \mathrm{d}Q_{i-1} \to 0.$$

By flatness, the first term vanishes. It follows by exactness that ${\ker \big(M \otimes_R \mathrm{d}Q_i \to M \otimes_R Q_{i-1}\big) \cong \mathrm{Tor}_1^R(M, \mathrm{d}Q_{i-1})}$. To finish the argument, we prove a simple lemma involving the suspension of a chain complex. Recall that for a chain complex $(M_\bullet, \mathrm{d}_\bullet)$, the suspension $(\Sigma M_\bullet, \Sigma \mathrm{d}_\bullet)$ is defined by $\Sigma M_i = M_{i-1}$ and $\Sigma \mathrm{d}_i = \mathrm{d}_{i-1}$.

Proof. The connecting homomorphism of the zig-zag lemma provides a natural isomorphism $\delta \colon \mathrm{H}_{i}(N_\bullet) \to \mathrm{H}_{i-1}(L_\bullet)$, since $H_i(M_\bullet) = 0$.

Applying this lemma to $(\star)$ yields a chain of quasi-isomorphisms

$$P_\bullet \otimes_R N = P_\bullet \otimes_R \mathrm{d}Q_0 \sim \Sigma (P_\bullet \otimes_R \mathrm{d}Q_1) \sim \Sigma^2 (P_\bullet \otimes_R \mathrm{d}Q_{2}) \sim \cdots \sim \Sigma^{i-1}(P_\bullet \otimes_R \mathrm{d}Q_{i-1}) \sim \cdots$$

Finally, taking the $i$th homology of the first and $(i-1)$th term yields

$$\mathrm{H}_i(P_\bullet \otimes_R N) \cong \mathrm{H}_1(P_\bullet \otimes_R \mathrm{d}Q_{i - 1}) = \mathrm{Tor}_1^{R}(M, \mathrm{d}Q_{i-1}) \cong \ker \big(M \otimes_R \mathrm{d}Q_i \to M \otimes_R Q_{i-1}\big) \cong \mathrm{H}_i(M \otimes_R Q_\bullet),$$

so $\mathrm{Tor}_i^R(-, -)$ is symmetric for all $i > 0$.