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

Comment #559 by Apurba Kumar Roy on

Is the tensor product of two complexes defined before this lemma?

Comment #562 by on

Yes, no, you are right! Hmm... what is a suitably general way of saying this? For example, something like: If $\otimes : \mathcal{A} \times \mathcal{B} \to \mathcal{C}$ is a functor where $\mathcal{A}, \mathcal{B}, \mathcal{C}$ are abelian categories, then given complexes $X^\bullet$, resp. $Y^\bullet$ in $\mathcal{A}$, resp. $\mathcal{B}$ the result of applying $\otimes$ to the pairs $(X^i, Y^j)$ is a double complex $X^\bullet \otimes Y^\bullet$?

Should be a remark in the chapter on homological algebra somewhere? What do you think.

Comment #566 by Apurba Kumar Roy on

This definition seems quite general. The chapter on homological algebra seems to be an appropriate place where it can be added.

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).