The Stacks project

Lemma 13.4.16. Let $\mathcal{D}$ be a pre-triangulated category. Assume that $\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$. The following are equivalent

  1. there exists a set of triangles $\mathcal{T}'$ such that $(\mathcal{D}', \mathcal{T}')$ is a pre-triangulated subcategory of $\mathcal{D}$,

  2. $\mathcal{D}'$ is preserved under $[1]$ and $[1] : \mathcal{D}' \to \mathcal{D}'$ is an auto-equivalence and given any morphism $f : X \to Y$ in $\mathcal{D}'$ there exists a distinguished triangle $(X, Y, Z, f, g, h)$ in $\mathcal{D}$ such that $Z$ is isomorphic to an object of $\mathcal{D}'$.

In this case $\mathcal{T}'$ as in (1) is the set of distinguished triangles $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ such that $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$. Finally, if $\mathcal{D}$ is a triangulated category, then (1) and (2) are also equivalent to

  1. $\mathcal{D}'$ is a triangulated subcategory.

Proof. Omitted. $\square$

Comments (3)

Comment #1584 by Darij Grinberg on

Does "In this case " mean "In this case, any satisfying (1)" or "In this case, there exists some satisfying (1) such that ?

Comment #8863 by on

In the statement, second sentence, one can weaken " is an additive full subcategory of " to just " is a non-empty preadditive full subcategory of " (where a preadditive subcategory of a preadditive category is a subcategory with a preadditive structure such that the inclusion functor is additive). Condition (2) implies additivity for :

Pick any object in . By (TR1), the triangle is distinguished. Thus, (2) implies that has a zero object. On the other hand, by Lemma 13.4.11(3) and (TR2), the triangle is distinguished. Now (2) implies that has the direct sum for and . This shows that is additive.

There are also:

  • 13 comment(s) on Section 13.4: Elementary results on triangulated categories

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Your email address will not be published.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 05QX. Beware of the difference between the letter 'O' and the digit '0'.