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Lemma 15.57.1. Let $R$ be a ring. Let $P^\bullet $ be a complex of $R$-modules. Let $\alpha , \beta : L^\bullet \to M^\bullet $ be homotopy equivalent maps of complexes. Then $\alpha $ and $\beta $ induce homotopy equivalent 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 $\text{Tot}(L^\bullet \otimes _ R P^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R P^\bullet )$. $\square$


Comments (4)

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 is a functor where are abelian categories, then given complexes , resp. in , resp. the result of applying to the pairs is a double complex ?

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.

There are also:

  • 2 comment(s) on Section 15.57: Derived tensor product

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