## 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 $y = (y_{-p - 1})$. $\square$

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

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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