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
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
Each $\alpha ^{p, q}$ is an element
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
On the other hand, if $\beta $ is an element of degree $n$ of the left hand side, then
and we can write $\beta ^{p, q} = \sum \gamma _ i^ p \otimes \delta _ i^ q$ with $\gamma _ i^ p \in K^ p$ and
By our sign rules we have
We send the element $\beta $ to $\alpha $ with
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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Comment #7130 by Hao Peng on
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