The Stacks project

Lemma 12.26.4. Let $M^\bullet $ be a complex of abelian groups. Let

\[ \ldots \to A_2^\bullet \to A_1^\bullet \to A_0^\bullet \to M^\bullet \to 0 \]

be an exact complex of complexes of abelian groups. Set $A^{p, q} = A_{-p}^ q$ to obtain a double complex. Let $\text{Tot}_\pi (A^{\bullet , \bullet })$ be the product total complex associated to the double complex (see proof). Then the map $\text{Tot}_\pi (A^{\bullet , \bullet }) \to M^\bullet $ induced by $A_0^\bullet \to M^\bullet $ is a quasi-isomorphism.

Proof. Abbreviating $T^\bullet = \text{Tot}_\pi (A^{\bullet , \bullet })$ we define

\[ T^ n = \prod \nolimits _{p + q = n} A^{p, q} = \prod \nolimits _{p + q = n} A_{-p}^ q \quad \text{with}\quad \text{d}_{T^\bullet }^ n = \prod \nolimits _{n = p + q} (f_{-p}^ q + (-1)^ pd_{A_{-p}^\bullet }^ q) \]

where $f_ p^\bullet : A_ p^\bullet \to A_{p - 1}^\bullet $ are the maps of complexes in the lemma. We will show that $T^\bullet $ is acyclic when $M^\bullet $ is the zero complex. This will suffice by the following trick. Set $B_ n^\bullet = A_{n + 1}^\bullet $ and $B_0^\bullet = M^\bullet $. Then we have an exact sequence

\[ \ldots \to B_2^\bullet \to B_1^\bullet \to B_0^\bullet \to 0 \to 0 \]

as in the lemma. Let $S^\bullet = \text{Tot}_\pi (B^{\bullet , \bullet })$. Then there is an obvious short exact sequence of complexes

\[ 0 \to M^\bullet \to S^\bullet \to T^\bullet [1] \to 0 \]

and we conclude by the long exact cohomology sequence. Some details omitted.

Assume $M^\bullet = 0$. We will show $H^0(T^\bullet ) = 0$. The same argument works for other degrees. Let $x =(x_ n) \in \mathop{\mathrm{Ker}}(d_{T^\bullet })$ map to $\xi \in H^0(T^\bullet )$ with $x_ n \in A^{-n, n} = A_ n^ n$. Since $M^0 = 0$ we find that $x_0 = f_1^0(y_0)$ for some $y_0 \in A_1^0$. Then $x_1 - d^0_{A_1^\bullet }(y_0) = f_2^1(y_1)$ because it is mapped to zero by $f_1^1$ as $x$ is a cocycle. for some $y_1 \in A_2^1$. Continuing, using induction, we find $y = (y_ i) \in T^{-1}$ with $d_{T^\bullet }(y) = x$ as desired. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E1R. Beware of the difference between the letter 'O' and the digit '0'.