Lemma 15.58.16. Let $R$ be a ring. Let $K^\bullet , L^\bullet $ be complexes of $R$-modules. There is a canonical isomorphism

\[ K^\bullet \otimes _ R^\mathbf {L} L^\bullet \longrightarrow L^\bullet \otimes _ R^\mathbf {L} K^\bullet \]

functorial in both complexes which uses a sign of $(-1)^{pq}$ for the map $K^ p \otimes _ R L^ q \to L^ q \otimes _ R K^ p$ (see proof for explanation).

**Proof.**
Replace the complexes by K-flat complexes $K^\bullet , L^\bullet $. Then we consider the map

\[ \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \longrightarrow \text{Tot}(L^\bullet \otimes _ R K^\bullet ) \]

given by using $(-1)^{pq}$ times the canonical map $K^ p \otimes _ R L^ q \to L^ q \otimes _ R K^ p$. This is an isomorphism. To see that it is a map of complexes we compute for $x \in K^ p$ and $y \in L^ q$ that

\[ \text{d}(x \otimes y) = \text{d}_ K(x) \otimes y + (-1)^ px \otimes \text{d}_ L(y) \]

Our rule says the right hand side is mapped to

\[ (-1)^{(p + 1)q}y \otimes \text{d}_ K(x) + (-1)^{p + p(q + 1)} \text{d}_ L(y) \otimes x \]

On the other hand, we see that

\[ \text{d}((-1)^{pq}y \otimes x) = (-1)^{pq} \text{d}_ L(y) \otimes x + (-1)^{pq + q} y \otimes \text{d}_ K(x) \]

These two expressions agree by inspection and the lemma is proved.
$\square$

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