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The Stacks project

15.103 Tricks with double complexes

This section continues the discussion in Homology, Section 12.26.

Lemma 15.103.1. Let A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots be a complex of complexes of abelian groups. Assume H^{-p}(A_ p^\bullet ) = 0 for all p \geq 0. Set A^{p, q} = A_ p^ q and view A^{\bullet , \bullet } as a double complex. Then H^0(\text{Tot}_\pi (A^{\bullet , \bullet })) = 0.

Proof. Denote f_ p : A_ p^\bullet \to A_{p + 1}^\bullet the given maps of complexes. Recall that the differential on \text{Tot}_\pi (A^{\bullet , \bullet }) is given by

\prod \nolimits _{p + q = n} (f^ q_ p + (-1)^ p\text{d}^ q_{A_ p^\bullet })

on elements in degree n. Let \xi \in H^0(\text{Tot}_\pi (A^{\bullet , \bullet })) be a cohomology class. We will show \xi is zero. Represent \xi as the class of an cocycle x = (x_ p) \in \prod A^{p, -p}. Since \text{d}(x) = 0 we find that \text{d}_{A_0^\bullet }(x_0) = 0. Since H^0(A_0^\bullet ) = 0 there exists a y_{-1} \in A^{0, -1} with \text{d}_{A_0^\bullet }(y_{-1}) = x_0. Then we see that \text{d}_{A_1^\bullet }(x_1 + f_0(y_{-1})) = 0. Since H^{-1}(A_1^\bullet ) = 0 we can find a y_{-2} \in A^{1, -2} such that -\text{d}_{A_1^\bullet }(y_{-2}) = x_1 + f_0(y_{-1}). By induction we can find y_{-p - 1} \in A^{p, -p - 1} such that

(-1)^ p\text{d}_{A_ p^\bullet }(y_{-p - 1}) = x_ p + f_{p - 1}(y_{-p})

This implies that \text{d}(y) = x where y = (y_{-p - 1}). \square

Lemma 15.103.2. Let

(A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots ) \longrightarrow (B_0^\bullet \to B_1^\bullet \to B_2^\bullet \to \ldots )

be a map between two complexes of complexes of abelian groups. Set A^{p, q} = A_ p^ q, B^{p, q} = B_ p^ q to obtain double complexes. Let \text{Tot}_\pi (A^{\bullet , \bullet }) and \text{Tot}_\pi (B^{\bullet , \bullet }) be the product total complexes associated to the double complexes. If each A_ p^\bullet \to B_ p^\bullet is a quasi-isomorphism, then \text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet }) is a quasi-isomorphism.

Proof. Recall that \text{Tot}_\pi (A^{\bullet , \bullet }) in degree n is given by \prod _{p + q = n} A^{p, q} = \prod _{p + 1 = n} A^ q_ p. Let C_ p^\bullet be the cone on the map A_ p^\bullet \to B_ p^\bullet , see Derived Categories, Section 13.9. By the functoriality of the cone construction we obtain a complex of complexes

C_0^\bullet \to C_1^\bullet \to C_2^\bullet \to \ldots

Then we see \text{Tot}_\pi (C^{\bullet , \bullet }) in degree n is given by

\prod _{p + q = n} C^{p, q} = \prod _{p + q = n} C^ q_ p = \prod _{p + q = n} (B^ q_ p \oplus A^{q + 1}_ p) = \prod _{p + q = n} B^ q_ p \oplus \prod _{p + q = n} A^{q + 1}_ p

We conclude that \text{Tot}_\pi (C^{\bullet , \bullet }) is the cone of the map \text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet }) (We omit the verification that the differentials agree.) Thus it suffices to show \text{Tot}_\pi (A^{\bullet , \bullet }) is acyclic if each A_ p^\bullet is acyclic. This follows from Lemma 15.103.1. \square


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