The Stacks project

Lemma 12.25.5. Let $\mathcal{A}$ be an abelian category. Let $M^\bullet $ be a complex of $\mathcal{A}$. Let

\[ a : M^\bullet [0] \longrightarrow \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right) \]

be a homotopy equivalence in the category of complexes of complexes of $\mathcal{A}$. Then the map $\alpha : M^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ induced by $M^\bullet \to A^{0, \bullet }$ is a homotopy equivalence.

Proof. The statement makes sense as a complex of complexes is the same thing as a double complex. The assumption means there is a map

\[ b : \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right) \longrightarrow M^\bullet [0] \]

such that $a \circ b$ and $b \circ a$ are homotopic to the identity in the category of complexes of complexes. This means that $b \circ a$ is the identity of $M^\bullet [0]$ (because there is only one term in degree $0$). Also, observe that $b$ is given by a map $b^0 : A^{0, \bullet } \to M^\bullet $ and zero in all other degrees. Thus $b$ induces a map $\beta : \text{Tot}(A^{\bullet , \bullet }) \to M^\bullet $ and $\beta \circ \alpha $ is the identity on $M^\bullet $. Finally, we have to show that the map $\alpha \circ \beta $ is homotopic to the identity. For this we choose maps of complexes $h^ n : A^{n, \bullet } \to A^{n - 1, \bullet }$ such that $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$ which exist by assumption. Here $d_1 : A^{n, \bullet } \to A^{n + 1, \bullet }$ are the differentials of the complex of complexes. We will also denote $d_2$ the differentials of the complexes $A^{n, \bullet }$ for all $n$. Let $h^{n, m} : A^{n, m} \to A^{n - 1, m}$ be the components of $h^ n$. Then we can consider

\[ h' : \text{Tot}(A^{\bullet , \bullet })^ k = \bigoplus \nolimits _{n + m = k} A^{n, m} \to \bigoplus \nolimits _{n + m = k - 1} A^{n, m} = \text{Tot}(A^{\bullet , \bullet })^{k - 1} \]

given by $h^{n, m}$ on the summand $A^{n, m}$. Then we compute that the map

\[ d_{\text{Tot}(A^{\bullet , \bullet })} \circ h' + h' \circ d_{\text{Tot}(A^{\bullet , \bullet })} \]

restricted to the summand $A^{n, m}$ is equal to

\[ d_1^{n - 1, m} \circ h^{n, m} + (-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m} + h^{n + 1, m} \circ d_1^{n, m} + h^{n, m + 1} \circ (-1)^ nd_2^{n, m} \]

Since $h^ n$ is a map of complexes, the terms $(-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m}$ and $h^{n, m + 1} \circ (-1)^ nd_2^{n, m}$ cancel. The other two terms give $(\alpha \circ \beta )|_{A^{n, m}} - \text{id}_{A^{n, m}}$ because $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$. This finishes the proof. $\square$


Comments (0)

There are also:

  • 1 comment(s) on Section 12.25: Spectral sequences: double complexes

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