# Why the octahedral axiom?

My question is about the octahedral axiom (OA) in the definition of a triangulated category. For what I can understand so far (cf. Huybrechts, Fourier-Mukai in algebraic gometry, Definition 1.32), this axiom wants to roughly generalise the "double quotient" situation in the category of abelian groups, i.e. if $A\subset B\subset C$ are abelian groups then $C/B\cong(C/A)/(B/A)$.

I would like to know why people think that this axiom is superfluous.

I reported the "double quotient" situation because one may say that if it wants to generalise a situation which is natural in the non-generalised case, then one would expect this situation to be natural too. But it seems to me that this argument is too weak and probably there are better arguments...

Moreover, is it true that everyone is convinced about that?

As a motivation to this question I would like to say that: $1)$ last summer a paper by Maciocia appeared in which he proved the OA is a consequence of the previous ones. But unfortunately there was an error which is still not fixed. $2)$ In the Huybrechts book, I have cited before, he doesn't state the OA because he "will never use it explicitely and only once implicitely". $3)$ I am very curious about that.

Thank you all!

P.S.: I have tried to find something in the literature or in StackExchange as well but I was apparently unable. Sorry if it is a duplicate.

• What convinced be of the axiom is this observation: let's have composable morphisms $f$ and $g$. Then you can form distinguished triangles over $f$, $g$ and $g\circ f$, and obviously they should be somehow related. The axiom then says that there is a distinguished triangle such that everything commutes. – Ennar Dec 28 '15 at 19:35
• @Ennar, thank you for your comment. Sorry but I cannot see your point: it seems to me that you are just paraphrasing the definition. And actually this is my point: why this assumption (which seems strong) should be implied by the other ones? (Of course I don't want an answer, but just a "feeling"). – User3773 Dec 28 '15 at 22:23
• Sorry, I misunderstood your question (I should have clicked on the link before commenting). I'm afraid I cannot offer much help in this regard, but am rather interested in this as well. – Ennar Dec 28 '15 at 22:55