Lemma 15.71.5. Let $R$ be a ring. Given complexes $K^\bullet , L^\bullet$ of $R$-modules there is a canonical morphism

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

of complexes of $R$-modules functorial in both 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

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

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

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

where we think of $\alpha ^{r, s, q}$ as a family of maps such that for every $x \in L^{-q}$ 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}_ L \\ & = \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}_ L \end{align*}

Now an element $\beta \in K^ n$ we send to $\alpha$ with $\alpha ^{n, -q, q} = \beta \otimes \text{id}_{L^{-q}}$ and $\alpha ^{r, s, q} = 0$ if $r \not= n$. This is indeed an element as above, as for fixed $q$ there is only one nonzero $\alpha ^{r, s, q}$. The description of the differential shows this is compatible with differentials. $\square$

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A62. Beware of the difference between the letter 'O' and the digit '0'.