Lemma 15.71.4. Let $R$ be a ring. Given complexes $K^\bullet , L^\bullet , M^\bullet $ of $R$-modules there is a canonical morphism

of complexes of $R$-modules functorial in all three complexes.

Lemma 15.71.4. Let $R$ be a ring. Given complexes $K^\bullet , L^\bullet , M^\bullet $ of $R$-modules there is a canonical morphism

\[ \text{Tot}(K^\bullet \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet )) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , \text{Tot}(K^\bullet \otimes _ R L^\bullet )) \]

of complexes of $R$-modules functorial in all three complexes.

**Proof.**
Via the discussion in Remark 15.71.2 the existence of such a canonical map follows from Categories, Remark 4.43.12. We also give a direct construction.

Let $\alpha $ be an element of degree $n$ of the right hand side. Thus

\[ \alpha = (\alpha ^{p, q}) \in \prod \nolimits _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, \text{Tot}^ p(K^\bullet \otimes _ R L^\bullet )) \]

Each $\alpha ^{p, q}$ is an element

\[ \alpha ^{p, q} = (\alpha ^{r, s, q}) \in \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, \bigoplus \nolimits _{r + s + q = n} K^ r \otimes _ R L^ s) \]

where we think of $\alpha ^{r, s, q}$ as a family of maps such that for every $x \in M^{-q}$ only a finite number of $\alpha ^{r, s, q}(x)$ are nonzero. By our sign rules we get

\begin{align*} \text{d}(\alpha ^{r, s, q}) & = \text{d}_{\text{Tot}(K^\bullet \otimes _ R L^\bullet )} \circ \alpha ^{r, s, q} - (-1)^ n \alpha ^{r, s, q} \circ \text{d}_ M \\ & = \text{d}_ K \circ \alpha ^{r, s, q} + (-1)^ r \text{d}_ L \circ \alpha ^{r, s, q} - (-1)^ n \alpha ^{r, s, q} \circ \text{d}_ M \end{align*}

On the other hand, if $\beta $ is an element of degree $n$ of the left hand side, then

\[ \beta = (\beta ^{p, q}) \in \bigoplus \nolimits _{p + q = n} K^ p \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^ q(M^\bullet , L^\bullet ) \]

and we can write $\beta ^{p, q} = \sum \gamma _ i^ p \otimes \delta _ i^ q$ with $\gamma _ i^ p \in K^ p$ and

\[ \delta _ i^ q = (\delta _ i^{r, s}) \in \prod \nolimits _{r + s = q} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-s}, L^ r) \]

By our sign rules we have

\begin{align*} \text{d}(\beta ^{p, q}) & = \text{d}_ K(\beta ^{p, q}) + (-1)^ p \text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet )}(\beta ^{p, q}) \\ & = \sum \text{d}_ K(\gamma _ i^ p) \otimes \delta _ i^ q + (-1)^ p \sum \gamma _ i^ p \otimes (\text{d}_ L \circ \delta _ i^ q - (-1)^ q \delta _ i^ q \circ \text{d}_ M) \end{align*}

We send the element $\beta $ to $\alpha $ with

\[ \alpha ^{r, s, q} = c^{r, s, q}(\sum \gamma _ i^ r \otimes \delta _ i^{s, q}) \]

where $c^{r, s, q} : K^ r \otimes _ R \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, L^ s) \to \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ r \otimes _ R L^ s)$ is the canonical map. For a given $\beta $ and $r$ there are only finitely many nonzero $\gamma _ i^ r$ hence only finitely many nonzero $\alpha ^{r, s, q}$ are nonzero (for a given $r$). Thus this family of maps satisfies the conditions above and the map is well defined. Comparing signs we see that this is compatible with differentials. $\square$

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## Comments (2)

Comment #7130 by Hao Peng on

Comment #7288 by Johan on