Lemma 12.25.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex. Let $A^{\bullet , \bullet }$ be a double complex. Let $\alpha ^ p : K^ p \to A^{p, 0}$ be morphisms. Assume that

1. For every $n \in \mathbf{Z}$ there are only finitely many nonzero $A^{p, q}$ with $p + q = n$.

2. We have $A^{p, q} = 0$ if $q < 0$.

3. The morphisms $\alpha ^ p$ give rise to a morphism of complexes $\alpha : K^\bullet \to A^{\bullet , 0}$.

4. The complex $A^{p, \bullet }$ is exact in all degrees $q \not= 0$ and the morphism $K^ p \to A^{p, 0}$ induces an isomorphism $K^ p \to \mathop{\mathrm{Ker}}(d_2^{p, 0})$.

Then $\alpha$ induces a quasi-isomorphism

$K^\bullet \longrightarrow \text{Tot}(A^{\bullet , \bullet })$

of complexes. Moreover, there is a variant of this lemma involving the second variable $q$ instead of $p$.

Proof. The map is simply the map given by the morphisms $K^ n \to A^{n, 0} \to \text{Tot}^ n(A^{\bullet , \bullet })$, which are easily seen to define a morphism of complexes. Consider the spectral sequence $({}'E_ r, {}'d_ r)_{r \geq 0}$ associated to the double complex $A^{\bullet , \bullet }$. By Lemma 12.25.3 this spectral sequence converges and the induced filtration on $H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is finite for each $n$. By Lemma 12.25.1 and assumption (4) we have ${}'E_1^{p, q} = 0$ unless $q = 0$ and $'E_1^{p, 0} = K^ p$ with differential ${}'d_1^{p, 0}$ identified with $d_ K^ p$. Hence ${}'E_2^{p, 0} = H^ p(K^\bullet )$ and zero otherwise. This clearly implies $d_2^{p, q} = d_3^{p, q} = \ldots = 0$ for degree reasons. Hence we conclude that $H^ n(\text{Tot}(A^{\bullet , \bullet })) = H^ n(K^\bullet )$. We omit the verification that this identification is given by the morphism of complexes $K^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ introduced above. $\square$

## 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Your email address will not be published.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.

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 0133. Beware of the difference between the letter 'O' and the digit '0'.