Lemma 15.67.3. Let $R$ be a ring. Given complexes $K^\bullet , L^\bullet , M^\bullet$ of $R$-modules there is a canonical morphism

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

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

Proof. Consider an element $\beta$ of degree $n$ of the right hand side. Then

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

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

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

We can apply the differentials $\text{d}_ M$ and $\text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )}$ to the element $\beta ^{p, q}$ and we can apply the differentials $\text{d}_ K$, $\text{d}_ L$, $\text{d}_ M$ to the element $\beta ^{p, r, s}$. We omit the precise definitions. The our sign rules tell us that

\begin{align*} \text{d}(\beta ^{p, r, s}) & = \text{d}_ M(\beta ^{p, r, s}) - (-1)^ n \text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )}(\beta ^{p, r, s}) \\ & = \text{d}_ M(\beta ^{p, r, s}) - (-1)^ n \left( \text{d}_ L(\beta ^{p, r, s}) - (-1)^{r + s} \text{d}_ K(\beta ^{p, r, s}) \right) \\ & = \text{d}_ M(\beta ^{p, r, s}) - (-1)^ n \text{d}_ L(\beta ^{p, r, s}) + (-1)^ p \text{d}_ K(\beta ^{p, r, s}) \end{align*}

On the other hand, an element $\alpha$ of degree $n$ of the left hand side looks like

$\alpha = (\alpha ^{t, s}) \in \bigoplus \nolimits _{t + s = n} \mathop{\mathrm{Hom}}\nolimits ^ t(L^\bullet , M^\bullet ) \otimes K^ s$

Each $\alpha ^{t, s}$ maps to an element

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

By our sign rules and with conventions as above we get

\begin{align*} \text{d}(\alpha ^{p, r, s}) & = \text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , M^\bullet )}(\alpha ^{p, r, s}) + (-1)^{p + r} \text{d}_ K(\alpha ^{p, r, s}) \\ & = \text{d}_ M(\alpha ^{p, r, s}) - (-1)^{p + r} \text{d}_ L(\alpha ^{p, r, s}) + (-1)^{p + r} \text{d}_ K(\alpha ^{p, r, s}) \end{align*}

To define our map we will use the canonical maps

$c_{p, r, s} : \mathop{\mathrm{Hom}}\nolimits _ R(L^{-r}, M^ p) \otimes _ R K^ s \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(K^ s, L^{-r}), M^ p)$

which sends $\varphi \otimes k$ to the map $\psi \mapsto \varphi (\psi (k))$. This is functorial in all three variables. However, since the signs above do not match we need to use instead some map

$\epsilon _{p, r, s} c_{p, r, s}$

for some sign $\epsilon _{p, r, s}$. Looking at the signs above we find that we need to find a solution for the equations

$\epsilon _{p, r, s} = \epsilon _{p + 1, r, s}, \quad \epsilon _{p, r, s} (-1)^ s = \epsilon _{p, r + 1, s}, \quad \epsilon _{p, r, s} (-1)^ r = \epsilon _{p, r, s + 1}$

A good solution is to take $\epsilon _{p, r, s} = (-1)^{rs}$. The choice of this sign is explained in the remark following the proof. $\square$

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