## 15.58 Tensor products of complexes

Let $R$ be a ring. The category $\text{Comp}(R)$ of complexes of $R$-modules has a symmetric monoidal structure. Namely, suppose that we have two complexes of $R$-modules $L^\bullet$ and $M^\bullet$. Using Homology, Example 12.18.2 and Homology, Definition 12.18.3 we obtain a third complex of $R$-modules, namely

$\text{Tot}(L^\bullet \otimes _ R M^\bullet )$

Clearly this construction is functorial in both $L^\bullet$ and $M^\bullet$. The associativity constraint will be the canonical isomorphism of complexes

$\text{Tot}(\text{Tot}(K^\bullet \otimes _ R L^\bullet ) \otimes _ R M^\bullet ) \longrightarrow \text{Tot}(K^\bullet \otimes _ R \text{Tot}(L^\bullet \otimes _ R M^\bullet ))$

constructed in Homology, Remark 12.18.4 from the triple complex $K^\bullet \otimes _ R L^\bullet \otimes _ R M^\bullet$. The commutativity constraint is the canonical isomorphism

$\text{Tot}(L^\bullet \otimes _ R M^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R L^\bullet )$

which uses the sign $(-1)^{pq}$ on the summand $L^ p \otimes _ R M^ q$. To see that it is a map of complexes we compute for $x \in L^ p$ and $y \in M^ q$ that

$\text{d}(x \otimes y) = \text{d}_ L(x) \otimes y + (-1)^ px \otimes \text{d}_ M(y)$

Our rule says the right hand side is mapped to

$(-1)^{(p + 1)q}y \otimes \text{d}_ L(x) + (-1)^{p + p(q + 1)} \text{d}_ M(y) \otimes x$

On the other hand, we see that

$\text{d}((-1)^{pq}y \otimes x) = (-1)^{pq} \text{d}_ M(y) \otimes x + (-1)^{pq + q} y \otimes \text{d}_ L(x)$

These two expressions agree by inspection as desired.

Lemma 15.58.1. Let $R$ be a ring. The category $\text{Comp}(R)$ of complexes of $R$-modules endowed with the functor $(L^\bullet , M^\bullet ) \mapsto \text{Tot}(L^\bullet \otimes _ R M^\bullet )$ and associativity and commutativity constraints as above is a symmetric monoidal category.

Proof. Omitted. Hints: as unit $\mathbf{1}$ we take the complex having $R$ in degree $0$ and zero in other degrees with obvious isomorphisms $\text{Tot}(\mathbf{1} \otimes _ R M^\bullet ) = M^\bullet$ and $\text{Tot}(K^\bullet \otimes _ R \mathbf{1}) = K^\bullet$. to prove the lemma you have to check the commutativity of various diagrams, see Categories, Definitions 4.43.1 and 4.43.9. The verifications are straightforward in each case. $\square$

Lemma 15.58.2. Let $R$ be a ring. Let $P^\bullet$ be a complex of $R$-modules. Let $\alpha , \beta : L^\bullet \to M^\bullet$ be homotopic maps of complexes. Then $\alpha$ and $\beta$ induce homotopic maps

$\text{Tot}(\alpha \otimes \text{id}_ P), \text{Tot}(\beta \otimes \text{id}_ P) : \text{Tot}(L^\bullet \otimes _ R P^\bullet ) \longrightarrow \text{Tot}(M^\bullet \otimes _ R P^\bullet ).$

In particular the construction $L^\bullet \mapsto \text{Tot}(L^\bullet \otimes _ R P^\bullet )$ defines an endo-functor of the homotopy category of complexes.

Proof. Say $\alpha = \beta + dh + hd$ for some homotopy $h$ defined by $h^ n : L^ n \to M^{n - 1}$. Set

$H^ n = \bigoplus \nolimits _{a + b = n} h^ a \otimes \text{id}_{P^ b} : \bigoplus \nolimits _{a + b = n} L^ a \otimes _ R P^ b \longrightarrow \bigoplus \nolimits _{a + b = n} M^{a - 1} \otimes _ R P^ b$

Then a straightforward computation shows that

$\text{Tot}(\alpha \otimes \text{id}_ P) = \text{Tot}(\beta \otimes \text{id}_ P) + dH + Hd$

as maps $\text{Tot}(L^\bullet \otimes _ R P^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R P^\bullet )$. $\square$

Lemma 15.58.3. Let $R$ be a ring. The homotopy category $K(R)$ of complexes of $R$-modules endowed with the functor $(L^\bullet , M^\bullet ) \mapsto \text{Tot}(L^\bullet \otimes _ R M^\bullet )$ and associativity and commutativity constraints as above is a symmetric monoidal category.

Proof. This follows from Lemmas 15.58.1 and 15.58.2. Details omitted. $\square$

Lemma 15.58.4. Let $R$ be a ring. Let $P^\bullet$ be a complex of $R$-modules. The functors

$K(R) \longrightarrow K(R), \quad L^\bullet \longmapsto \text{Tot}(P^\bullet \otimes _ R L^\bullet )$

and

$K(R) \longrightarrow K(R), \quad L^\bullet \longmapsto \text{Tot}(L^\bullet \otimes _ R P^\bullet )$

are exact functors of triangulated categories.

Proof. This follows from Derived Categories, Remark 13.10.9. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).