Category theory; homological algebra Definition

Octahedral axiom

Builds on Additive category Additive functor

Definition

Let $\mathcal{T}$ be an additive category with an additive automorphism $\Sigma$ . Let a class of distinguished triangles $X \to Y \to Z \to \Sigma X$ be given that satisfies the axioms TR1, TR2 and TR3 of a triangulated category. The class satisfies the octahedral axiom (TR4) if the following holds. Let $f \colon X \to Y$ and $g \colon Y \to Z$ be morphisms, and let $X \xrightarrow{f} Y \xrightarrow{i} Z' \xrightarrow{i'} \Sigma X, \qquad Y \xrightarrow{g} Z \xrightarrow{j} X' \xrightarrow{j'} \Sigma Y, \qquad X \xrightarrow{g f} Z \xrightarrow{k} Y' \xrightarrow{k'} \Sigma X$ be distinguished triangles. Then there are morphisms $a \colon Z' \to Y'$ and $b \colon Y' \to X'$ such that

$Z' \xrightarrow{a} Y' \xrightarrow{b} X' \xrightarrow{(\Sigma i) j'} \Sigma Z'$ is a distinguished triangle;
$a i = k g$ and $k' a = i'$ ;
$b k = j$ and $j' b = (\Sigma f) k'$ .

Remarks

Here $Z'$ , $X'$ and $Y'$ are cones of $f$ , $g$ and $g f$ . Example: take $f = \id_X$ , so that $Z' = 0$ with the triangle $X \xrightarrow{\id} X \to 0 \to \Sigma X$ , and take the second and third triangles equal. Then $a = 0$ and $b = \id_{X'}$ work: $0 \to X' \xrightarrow{\id} X' \to 0$ is distinguished, being a rotation of $X' \xrightarrow{\id} X' \to 0 \to \Sigma X'$ ; $a i = 0 = k g$ because $k g$ is a composite of consecutive maps of a distinguished triangle; and $k' a = 0 = i'$ , $b k = k = j$ , $j' b = j' = k' = (\Sigma f) k'$ . In the homotopy category of chain complexes the axiom holds, with $Y'$ and the maps $a$ and $b$ built from mapping cones. The axiom is a form of the isomorphism $(Z/X)/(Y/X) \cong Z/Y$ for $X \subseteq Y \subseteq Z$ : read $Z'$ as $Y/X$ , $Y'$ as $Z/X$ and $X'$ as $Z/Y$ .

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Tag	`MY2XG7XTS8`
URI	`m2://MY2XG7XTS8`
Kind	definition
Writing	W0 stub
MSC	18G80
Defines	Octahedral axiom
Uses	Additive category, Additive functor

Octahedral axiom

Definition

Remarks

Reading path

Further reading

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