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

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

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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)