Definition 13.3.1. Let $\mathcal{D}$ be an additive category. Let $[n] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[n]$ be a collection of additive functors indexed by $n \in \mathbf{Z}$ such that $[n] \circ [m] = [n + m]$ and $[0] = \text{id}$ (equality as functors). In this situation we define a triangle to be 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 $\mathcal{D}$. 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 $a[1] \circ h = h' \circ c$.

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 $X[n][m]$ with $X[n + m]$ from now on". So there is practically no difference. If more people agree with making this change then we will do so.

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

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