Lemma 15.70.5. 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

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

Our sign rules tell us that

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

We can describe the last term as follows

$(\beta ^{p, s} \circ \text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )})(f) = \beta ^{p, s}(\text{d}_ L \circ f - (-1)^{s + 1} f \circ \text{d}_ K)$

if $f \in \mathop{\mathrm{Hom}}\nolimits ^{-s - 1}(K^\bullet , L^\bullet )$. We conclude that in some unspecified sense $\text{d}(\beta ^{p, s})$ is a sum of three terms with signs as follows

15.70.5.1
$$\label{more-algebra-equation-beta} \text{d}(\beta ^{p, s}) = \text{d}_ M(\beta ^{p, s}) -(-1)^ n\text{d}_ L(\beta ^{p, s}) + (-1)^{p + 1}\text{d}_ K(\beta ^{p, s})$$

Next, we consider an element $\alpha$ of degree $n$ of the left hand side. We can write it like so

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

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

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

Our sign rules tell us that

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

where if we further write $\alpha ^{p, q, r} = \sum g_ i^{p, q} \otimes k_ i^ r$ then we have

$\text{d}_{\mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , M^\bullet )}(\alpha ^{p, q, r}) = \sum (\text{d}_ M \circ g_ i^{p, q}) \otimes k_ i^ r - (-1)^{p + q} \sum (g_ i^{p, q} \circ \text{d}_ L) \otimes k_ i^ r$

We conclude that in some unspecified sense $\text{d}(\alpha ^{p, q, r})$ is a sum of three terms with signs as follows

15.70.5.2
$$\label{more-algebra-equation-alpha} \text{d}(\alpha ^{p, q, r}) = \text{d}_ M(\alpha ^{p, q, r}) -(-1)^{p + q}\text{d}_ L(\alpha ^{p, q, r}) + (-1)^{p + q}\text{d}_ K(\alpha ^{p, q, r})$$

To define our map we will use the canonical maps

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

which sends $\varphi \otimes k$ to the map $\psi \mapsto \varphi (\psi (k))$. This is functorial in all three variables. With $s = q + r$ there is an inclusion

$\mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(K^ r, L^{-q}), M^ p) \subset \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits ^{-s}(K^\bullet , L^\bullet ), M^ p)$

coming from the projection $\mathop{\mathrm{Hom}}\nolimits ^{-s}(K^\bullet , L^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _ R(K^ r, L^{-q})$. Since $\alpha ^{p, q, r}$ is nonzero only for a finite number of $r$ we see that for a given $s$ there is only a finite number of $q, r$ with $q + r = s$. Thus we can send $\alpha$ to the element $\beta$ with

$\beta ^{p, s} = \sum \nolimits _{q + r = s} \epsilon _{p, q, r} c_{p, q, r}(\alpha ^{p, q, r})$

where where the sum uses the inclusions given above and where $\epsilon _{p, q, r} \in \{ \pm 1\}$. Comparing signs in the equations (15.70.5.1) and (15.70.5.2) we see that

1. $\epsilon _{p, q, r} = \epsilon _{p + 1, q, r}$

2. $-(-1)^ n\epsilon _{p, q, r} = -(-1)^{p + q}\epsilon _{p, q - 1, r}$ or equivalently $\epsilon _{p, q, r} = (-1)^ r\epsilon _{p, q - 1, r}$

3. $(-1)^{p + 1}\epsilon _{p, q, r} = (-1)^{p + q}\epsilon _{p, q, r + 1}$ or equivalently $(-1)^{q + 1}\epsilon _{p, q, r} = \epsilon _{p, q, r + 1}$.

A good solution is to take

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

The choice of this sign is explained in the remark following the proof. $\square$

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