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. $\square$

Comment #1584 by Darij Grinberg on

Does "In this case $\mathcal{T}'$" mean "In this case, any $\mathcal{T}'$ satisfying (1)" or "In this case, there exists some $\mathcal{T}'$ satisfying (1) such that $\mathcal{T}'$?

Comment #8863 by on

In the statement, second sentence, one can weaken "$\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$" to just "$\mathcal{D}'$ is a non-empty preadditive full subcategory of $\mathcal{D}$" (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 $\mathcal{D}$:

Pick any object $X$ in $\mathcal{D}$. By (TR1), the triangle $X\xrightarrow{1_X}X\to 0\to X[1]$ is distinguished. Thus, (2) implies that $\mathcal{D}'$ has a zero object. On the other hand, by Lemma 13.4.11(3) and (TR2), the triangle $Y[-1]\xrightarrow{0}X\to X\oplus Y\to Y$ is distinguished. Now (2) implies that $\mathcal{D}'$ has the direct sum for $X$ and $Y$. This shows that $\mathcal{D}'$ is additive.

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