# External Product for Tor commutes with connecting maps

$$\newcommand{\Tor}{\operatorname{Tor}}$$I am trying to work through Weibel's Introduction to Homological Algebra. For modules $$A,A',B,B'$$ over a commutative ring $$R$$, he defines the external product map, $$\Tor_i^R(A,B) \otimes \Tor_j^R(A',B')\rightarrow \Tor_{i+j}^R(A\otimes_R A', B\otimes_R B')$$ and asks in exercise 2.7.5 part 3,

Show that the external product commutes with the connecting homomorphism $$\delta$$ in the long exact $$\Tor$$ sequences associated to $$0\rightarrow B_0\rightarrow B\rightarrow B_1\rightarrow 0$$.

I am not quite sure what I need to prove here. I don't see how the external product and the connecting maps for the mentioned $$\Tor$$ sequences fit into a square at all.

• You can probably focus on one argument, as most of your picture is "symmetric", say $B$, as Weibel says. Take a resolution of that SEC, as in the Horseshoe Lemma. This should allow you to define the connecting morphism for the Tor on the target of the external product. (Alternative: check Cartan and Eilenberg's book, where what "commutes with" is probably spelled out in detail!)
– Pedro
Commented Jun 2, 2021 at 16:10
• $\newcommand{\Tor}{\operatorname{Tor}}$I don't think I follow your suggestion. How do I get to the $\Tor$ on the target using a resolution of the given SEC? Would I not end up with something like $\Tor_i^R(A\otimes_R A'\otimes_R B',B_1)\rightarrow \Tor_{i-1}^R(A\otimes_R A'\otimes_R B',B_0)$ instead? I did check out Cartan and Eilenberg's book on a friend's suggestion, and assuming I understood it correctly, it seems that they assume that tensoring the given SEC with $B'$ preserves its exactness, and then prove that the external product commutes with the connecting maps. Commented Jun 3, 2021 at 4:40

I think you probably need to prove the following. Let $$0 \rightarrow B_0 \xrightarrow{f} B \xrightarrow{g} B_1 \rightarrow 0$$ be a SES, and suppose that $$B'$$ is some module such that $$0 \rightarrow B_0 \otimes B' \rightarrow B \otimes B' \rightarrow B_1 \otimes B' \rightarrow 0$$ is also a SES. I don't know how one could set this up if we don't assume the tensored sequence isn't exact. We need to show that the following square commutes
where the vertical maps are given by the exterior product. Here is a sketch. Let $$P, P', P''$$ be projective resolutions of $$A, A', A \otimes A'$$, respectively. Recall that there is a map $$P \otimes P' \rightarrow P''$$ that is unique up to chain homotopy equivalence, where $$P \otimes P'$$ denotes the total complex of the tensor double complex. Our goal is then to show that
commutes. The bottom square commutes by naturality of the connecting homomorphism. To check the commutativity of the top square, we argue directly on the level of chains and use the explicit description of the connecting homomorphism. Let $$\alpha = \sum_m p_m \otimes b_m$$ and $$\beta = \sum_n p_n' \otimes b_n'$$ be cycles representing elements of $$H_i(P \otimes B_1)$$ and $$H_j(P' \otimes B')$$, respectively. On one hand, we have $$(\delta \otimes \mathrm{id})([\alpha]\otimes[\beta]) = [\delta \alpha] \otimes [\beta] \mapsto [(\mathrm{id} \otimes f)^{-1}(\sum_m dp_m \otimes g^{-1}b_m) \otimes \beta].$$ On the other hand, \begin{align*} \delta[\alpha \otimes \beta] &= \delta[\sum_{m,n} p_m \otimes p_n' \otimes b_m \otimes b_n'] \\ &= [(\mathrm{id} \otimes \mathrm{id} \otimes f \otimes \mathrm{id})^{-1} (\sum_{m,n} dp_m \otimes p_n' \otimes g^{-1}b_m \otimes b_n' - p_m \otimes dp_n' \otimes g^{-1}b_m \otimes b_n')] \\ &= [(\mathrm{id} \otimes f \otimes \mathrm{id} \otimes \mathrm{id})^{-1} ((\sum_m dp_m \otimes g^{-1}b_m) \otimes \beta - (\sum_m p_m \otimes g^{-1}b_m) \otimes d \beta)] \\ &= [(\mathrm{id} \otimes f \otimes \mathrm{id} \otimes \mathrm{id})^{-1} (\sum_m dp_m \otimes g^{-1}b_m) \otimes \beta] \\ &= [(\mathrm{id} \otimes f)^{-1}(\sum_m dp_m \otimes g^{-1}b_m) \otimes \beta] \end{align*} where we have used the fact that $$d \beta = 0$$. Also note that I've been freely commuting the tensors. This proves that the upper square commutes.