Lemma 10.75.5. Let $R$ be a ring. For any $i \geq 0$ the functors $\text{Mod}_ R \times \text{Mod}_ R \to \text{Mod}_ R$, $(M, N) \mapsto \text{Tor}_ i^ R(M, N)$ and $(M, N) \mapsto \text{Tor}_ i^ R(N, M)$ are canonically isomorphic.

Proof. Let $F_\bullet$ be a free resolution of the module $M$ and let $G_\bullet$ be a free resolution of the module $N$. Consider the double complex $(A_{i, j}, d, \delta )$ defined as follows:

1. set $A_{i, j} = F_ i \otimes _ R G_ j$,

2. set $d_{i, j} : F_ i \otimes _ R G_ j \to F_{i-1} \otimes G_ j$ equal to $d_{F, i} \otimes \text{id}$, and

3. set $\delta _{i, j} : F_ i \otimes _ R G_ j \to F_ i \otimes G_{j-1}$ equal to $\text{id} \otimes d_{G, j}$.

This double complex is usually simply denoted $F_\bullet \otimes _ R G_\bullet$.

Since each $G_ j$ is free, and hence flat we see that each row of the double complex is exact except in homological degree $0$. Since each $F_ i$ is free and hence flat we see that each column of the double complex is exact except in homological degree $0$. Hence the double complex satisfies the conditions of Lemma 10.75.3.

To see what the lemma says we compute $R(A)_\bullet$ and $U(A)_\bullet$. Namely,

\begin{eqnarray*} R(A)_ i & = & \mathop{\mathrm{Coker}}(A_{1, i} \to A_{0, i}) \\ & = & \mathop{\mathrm{Coker}}(F_1 \otimes _ R G_ i \to F_0 \otimes _ R G_ i) \\ & = & \mathop{\mathrm{Coker}}(F_1 \to F_0) \otimes _ R G_ i \\ & = & M \otimes _ R G_ i \end{eqnarray*}

In fact these isomorphisms are compatible with the differentials $\delta$ and we see that $R(A)_\bullet = M \otimes _ R G_\bullet$ as homological complexes. In exactly the same way we see that $U(A)_\bullet = F_\bullet \otimes _ R N$. We get

\begin{eqnarray*} \text{Tor}_ i^ R(M, N) & = & H_ i(F_\bullet \otimes _ R N) \\ & = & H_ i(U(A)_\bullet ) \\ & = & H_ i(R(A)_\bullet ) \\ & = & H_ i(M \otimes _ R G_\bullet ) \\ & = & H_ i(G_\bullet \otimes _ R M) \\ & = & \text{Tor}_ i^ R(N, M) \end{eqnarray*}

Here the third equality is Lemma 10.75.3, and the fifth equality uses the isomorphism $V \otimes W = W \otimes V$ of the tensor product.

Functoriality. Suppose that we have $R$-modules $M_\nu$, $N_\nu$, $\nu = 1, 2$. Let $\varphi : M_1 \to M_2$ and $\psi : N_1 \to N_2$ be morphisms of $R$-modules. Suppose that we have free resolutions $F_{\nu , \bullet }$ for $M_\nu$ and free resolutions $G_{\nu , \bullet }$ for $N_\nu$. By Lemma 10.71.4 we may choose maps of complexes $\alpha : F_{1, \bullet } \to F_{2, \bullet }$ and $\beta : G_{1, \bullet } \to G_{2, \bullet }$ compatible with $\varphi$ and $\psi$. We claim that the pair $(\alpha , \beta )$ induces a morphism of double complexes

$\alpha \otimes \beta : F_{1, \bullet } \otimes _ R G_{1, \bullet } \longrightarrow F_{2, \bullet } \otimes _ R G_{2, \bullet }$

This is really a very straightforward check using the rule that $F_{1, i} \otimes _ R G_{1, j} \to F_{2, i} \otimes _ R G_{2, j}$ is given by $\alpha _ i \otimes \beta _ j$ where $\alpha _ i$, resp. $\beta _ j$ is the degree $i$, resp. $j$ component of $\alpha$, resp. $\beta$. The reader also readily verifies that the induced maps $R(F_{1, \bullet } \otimes _ R G_{1, \bullet })_\bullet \to R(F_{2, \bullet } \otimes _ R G_{2, \bullet })_\bullet$ agrees with the map $M_1 \otimes _ R G_{1, \bullet } \to M_2 \otimes _ R G_{2, \bullet }$ induced by $\varphi \otimes \beta$. Similarly for the map induced on the $U(-)_\bullet$ complexes. Thus the statement on functoriality follows from the statement on functoriality in Lemma 10.75.3. $\square$

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