Lemma 15.71.6. 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. 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.

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

15.71.6.1
$$\label{more-algebra-equation-beta} \text{d}(\beta ^{p, s}) = \text{d}_ M(\beta ^{p, s}) -(-1)^ n\text{d}_ L(\beta ^{p, s}) + (-1)^{p + 1}\text{d}_ K(\beta ^{p, s})$$

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

15.71.6.2
$$\label{more-algebra-equation-alpha} \text{d}(\alpha ^{p, q, r}) = \text{d}_ M(\alpha ^{p, q, r}) -(-1)^{p + q}\text{d}_ L(\alpha ^{p, q, r}) + (-1)^{p + q}\text{d}_ K(\alpha ^{p, q, r})$$

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 (15.71.6.1) and (15.71.6.2) 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. $\square$

Comment #7131 by Hao Peng on

Just to point out this seems to be a direct consequence of tag 0A5Y.

Comment #7135 by Hao Peng on

By taking $H^0$ of tag 0A5Y we can get $Hom(K, Hom^\bullet(L, M))\cong Hom(Tot(K\otimes L), M)$ functorially.

We can first get a map $Tot(K\otimes Hom^\bullet(K, L))\to L$ functorially for any $K, L$: It is just the adjunction map corresponding to $id: Hom^\bullet(K, L)\to Hom^\bullet(K, L)$.

Then to get a map from $Tot(Hom^\bullet(L, M)\otimes K)\to Hom^\bullet(Hom^\bullet(K, L), M)$, it suffices to show a map from $Tot(Tot(Hom^\bullet(L, M)\otimes K)\otimes Hom^\bullet(K, L))$ to $M$. By associality of $Tot(-\otimes -)$, the map is given by $Tot(Tot(Hom^\bullet(L, M)\otimes K)\otimes Hom^\bullet(K, L))\cong Tot(Hom^\bullet(L, M)\otimes Tot(K\otimes Hom^\bullet(K, L))\to Tot(Hom^\bullet(L, M)\otimes L)\to M$

Comment #7136 by on

Very good, thanks. In your construction of the evaluation map $Tot(K \otimes Hom(K, L)) \to L$ there is a sign (I think) because your are switching the order of the tensors. Then later you use this 2x so hopefully this will match the signs chosen in the proof above.

Comment #7137 by Hao Peng on

Why do I think there is no sign matter? Because $Tot(\cdot\otimes \cdot)$ is strctly associative(the two orders of tensor both corresponds to a complex with differentials $d^n=\bigoplus_{p+q+r=n}(d_1^p+(-1)^pd_2^q+(-1)^{p+q}d_3^r)$). I guess the sign matter terminates beyond the proof of tag 0A5Y.

Comment #7141 by Hao Peng on

Sorry, I finally realized the sign problem. There is truly a sign. But the same method work good for other propositions in this section.

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