Definition 13.9.9. Let $\mathcal{A}$ be an additive category. A termwise split exact sequence of complexes of $\mathcal{A}$ is a complex of complexes

$0 \to A^\bullet \xrightarrow {\alpha } B^\bullet \xrightarrow {\beta } C^\bullet \to 0$

together with given direct sum decompositions $B^ n = A^ n \oplus C^ n$ compatible with $\alpha ^ n$ and $\beta ^ n$. We often write $s^ n : C^ n \to B^ n$ and $\pi ^ n : B^ n \to A^ n$ for the maps induced by the direct sum decompositions. According to Homology, Lemma 12.13.10 we get an associated morphism of complexes

$\delta : C^\bullet \longrightarrow A^\bullet $

which in degree $n$ is the map $\pi ^{n + 1} \circ d_ B^ n \circ s^ n$. In other words $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ forms a triangle

$A^\bullet \to B^\bullet \to C^\bullet \to A^\bullet $

This will be the triangle associated to the termwise split sequence of complexes.

## Comments (3)

Comment #294 by arp on

Typo: In the definition of $\delta$, I think it should say in degree $n$ it is given by the map $\pi^{n + 1} \circ d_B^n \circ s^n$ (i.e. replace $C$ with $B$).

Comment #3250 by William Chen on

To make the terminology consistent with the next proposition (Tag 05SS), maybe this should be called a "termwise split exact sequence of 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Your email address will not be published.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 014I. Beware of the difference between the letter 'O' and the digit '0'.