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
Comments (2)
Comment #7130 by Hao Peng on
Comment #7288 by Johan on