The Stacks project

Definition 13.3.1. Let $\mathcal{D}$ be an additive category. Let $[1] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[1]$ be an additive functor which is an auto-equivalence of $\mathcal{D}$.

  1. A triangle is a sextuple $(X, Y, Z, f, g, h)$ where $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ and $f : X \to Y$, $g : Y \to Z$ and $h : Z \to X[1]$ are morphisms of $\mathcal{D}$.

  2. A morphism of triangles $(X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$ is given by morphisms $a : X \to X'$, $b : Y \to Y'$ and $c : Z \to Z'$ of $\mathcal{D}$ such that $b \circ f = f' \circ a$, $c \circ g = g' \circ b$ and $a[1] \circ h = h' \circ c$.

Comments (4)

Comment #1392 by sdf on

The second line is badly phrased. Also, the correct word to use is "sextuple" rather than "sixtuple". May I suggest instead

"In this situation we define a {\it triangle } to be a sextuple..."

Comment #1409 by on

Dear sdf, many thanks for this and your other comments. See here.

Comment #4327 by Christian Q. on

I just want to remark, that this definition is not stable under equivalence, i.e. given an equivalence between a triangulated category and an additive category does not necessarily induce a triangulated structure on the additive category. This is because the strict equality requirement implies, that the translations are invertible as maps. I would prefer a unique isomorphism instead of an equality here, but it may not make any difference in practice, since passing to a skeleton resolves the issue.

Comment #4483 by on

Well, the usual practice when doing what you say is to have a comment after the definition saying something like "using our chosen isomorphisms we are going to identify with from now on". So there is practically no difference. If more people agree with making this change then we will do so.

There are also:

  • 7 comment(s) on Section 13.3: The definition of a triangulated category

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View Definition 13.3.1 as pdf