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

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