Recall that the usual definition of a triangulated category is an additive category equipped with a self equivalence called $[1]$ in which certain diagrams, of the form $X \to Y \to Z \to X[1]$ are called "exact", satisfying certain axioms. Two of these axioms are that

(1) Given $X \to Y$, it can be extended to an exact triangle $X \to Y \to Z \to X[1]$ and

(2) Given a commuting diagram $$\begin{matrix} X & \to & Y \\ \downarrow & & \downarrow \\ X' & \to & Y' \end{matrix}$$ and exact triangles $X \to Y \to Z \to X[1]$ and $X' \to Y' \to Z' \to X'[1]$, there is a map $Z \to Z'$ making the obvious diagram commute.

And, as every source on triangulated categories points out, one of the standard problems with the theory is that there is no uniqueness statement in these axioms. So, why not make one?

I envision a definition as follows: For any category $C$, let $Ar(C)$ be the category whose objects are diagrams $X \stackrel{f}{\longrightarrow} Y$ in $C$ and whose morphisms are commuting squares in $C$. Note that there are obvious functors $\mathrm{Source}$ and $\mathrm{Target}$ from $Ar(C) \to C$, and a natural transformation $\mathrm{Source} \to \mathrm{Target}$. Define a conical category to be an additive category $C$ with a self-equivalence $[1]$ and a functor $\mathrm{Cone} : Ar(C) \to C$, equipped with a natural transformations $\mathrm{Target} \to \mathrm{Cone} \to \mathrm{Source}[1]$, obeying certain axioms.

I noticed one place you have to be careful. In a triangulated category, if $X \to Y \to Z \to X[1]$ is exact, then so is $Y \to Z \to X[1] \to Y[1]$ (with a certain sign flip). The most obvious analoguous thing in a conical category would be for $\mathrm{Cone}(Y \to \mathrm{Cone}(X \to Y))$ to equal $X[1]$; the right thing is to ask for a natural isomorphism instead.

But everything else seems work out OK, at least in the case of the homotopy category of chain complexes. And it seems much more natural. What goes wrong if you try this?

I know that this is the sort of subject where people tend to mention $\infty$-categories; please bear in mind that I don't understand those very well. Everything I've said above just used ordinary $1$-categories, as far as I can tell.

dérivateurs. $\endgroup$ – Mariano Suárez-Álvarez Mar 9 '11 at 2:45