The Stacks project

Lemma 12.26.2. 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 such that for all $p \in \mathbf{Z}$ the complexes

\[ \ldots \to \mathop{\mathrm{Ker}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{M^\bullet }^ p) \to 0 \]

are exact as well. Set $A^{p, q} = A_{-p}^ q$ to obtain a double complex. Then $\text{Tot}(A^{\bullet , \bullet }) \to M^\bullet $ induced by $A_0^\bullet \to M^\bullet $ is a quasi-isomorphism.

Proof. Using the short exact sequences $0 \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to A_ n^ p \to \mathop{\mathrm{Im}}(d^ p_{A_ n^\bullet }) \to 0$ and the assumptions we see that

\[ \ldots \to \mathop{\mathrm{Im}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{M^\bullet }^ p) \to 0 \]

is exact for all $p \in \mathbf{Z}$. Repeating with the exact sequences $0 \to \mathop{\mathrm{Im}}(d^{p - 1}_{A_ n^\bullet }) \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to H^ p(A_ n^\bullet ) \to 0$ we find that

\[ \ldots \to H^ p(A_2^\bullet ) \to H^ p(A_1^\bullet ) \to H^ p(A_0^\bullet ) \to H^ p(M^\bullet ) \to 0 \]

is exact for all $p \in \mathbf{Z}$.

Write $T^\bullet = \text{Tot}(A^{\bullet , \bullet })$. We will show that $H^0(T^\bullet ) \to H^0(M^\bullet )$ is an isomorphism. The same argument works for other degrees. Let $x \in \mathop{\mathrm{Ker}}(\text{d}_{T^\bullet }^0)$ represent an element $\xi \in H^0(T^\bullet )$. Write $x = \sum _{i = n, \ldots , 0} x_ i$ with $x_ i \in A_ i^ i$. Assume $n > 0$. Then $x_ n$ is in the kernel of $d_{A_ n^\bullet }^ n$ and maps to zero in $H^ n(A_{n - 1}^\bullet )$ because it maps to an element which is the boundary of $x_{n - 1}$ up to sign. By the first paragraph of the proof, we find that $x_ n \bmod \mathop{\mathrm{Im}}(d^{n - 1}_{A_ n^\bullet })$ is in the image of $H^ n(A_{n + 1}^\bullet ) \to H^ n(A_ n^\bullet )$. Thus we can modify $x$ by a boundary and reach the situation where $x_ n$ is a boundary. Modifying $x$ once more we see that we may assume $x_ n = 0$. By induction we see that every cohomology class $\xi $ is represented by a cocycle $x = x_0$. Finally, the condition on exactness of kernels tells us two such cocycles $x_0$ and $x_0'$ are cohomologous if and only if their image in $H^0(M^\bullet )$ are the same. $\square$

Comments (0)

There are also:

  • 1 comment(s) on Section 12.26: Double complexes of abelian groups

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 09IZ. Beware of the difference between the letter 'O' and the digit '0'.