The Stacks project

Lemma 12.18.4. Let $\mathcal{A}$ be an abelian category. Let $(A, E, \alpha , f, g)$ be an exact couple. Let $(E_ r, d_ r)_{r \geq 1}$ be the spectral sequence associated to the exact couple. In this case we have

\[ 0 = B_1 \subset \ldots \subset B_{r + 1} = g(\mathop{\mathrm{Ker}}(\alpha ^ r)) \subset \ldots \subset Z_{r + 1} = f^{-1}(\mathop{\mathrm{Im}}(\alpha ^ r)) \subset \ldots \subset Z_1 = E \]

and the map $d_{r + 1} : E_{r + 1} \to E_{r + 1}$ is described by the following rule: For any (test) object $T$ of $\mathcal{A}$ and any elements $x : T \to Z_{r + 1}$ and $y : T \to A$ such that $f \circ x = \alpha ^ r \circ y$ we have

\[ d_{r + 1} \circ \overline{x} = \overline{g \circ y} \]

where $\overline{x} : T \to E_{r + 1}$ is the induced morphism.

Proof. Omitted. $\square$


Comments (0)

There are also:

  • 3 comment(s) on Section 12.18: Spectral sequences: exact couples

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Your email address will not be published.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 011T. Beware of the difference between the letter 'O' and the digit '0'.