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.

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    $\begingroup$ You will end up needing more and more choices: apart from assigning cones to arrows, you need to make choices (at least!) for squares and cubes and so on. Bernhard Keller studied a similar situation, if I recall correctly, in his thesis. The idea converges to dérivateurs. $\endgroup$ Mar 9 '11 at 2:45
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    $\begingroup$ The key point is that $Ho(Ar(C))\to Ar(Ho(C))$ is not an equivalence. It is the first category that has a cone functor, not the second. That's Neil's answer. But you can turn it around and keep track of $Ho(Ar(C))$ and similar categories--that is a derivateur. $\endgroup$ Mar 9 '11 at 3:42

Let's take $C$ to be the category of chain complexes of abelian groups, and homotopy classes of maps. What do you want the cone functor to be? An object of $Ar(C)$ consists of a pair of chain complexes $X$ and $Y$, together with a homotopy class of chain maps between them. To construct the mapping cone in the usual way, you need an actual chain map, not just a homotopy class. I don't think that this $C$ supports the structure that you describe.

You could instead take $C$ to be the more rigid category of chain complexes and chain maps, and try to axiomatise the structure that allows you to form $Ho(C)$ and to prove that $Ho(C)$ is triangulated. One approach is to demand that $C$ is a model category in the sense of Quillen, with an extra stability axiom. This is discussed in Hovey's book on model categories. If you want something closer to the approach outlined in your question, you could look at the work of Baues. He has axioms for a category with a cylinder functor, and he develops a rich theory of unstable homotopy based on that, including Puppe sequences. I don't think he ever discusses the additional stability condition needed to give a triangulation, but it can't be too hard. A third approach would be to use stable infinity categories. That route involves a lot of statements and constructions that look very simple on the surface, but there tend to be a lot of elaborate technicalities hidden by the formalism that can jump out and bite you unexpectedly.

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    $\begingroup$ Thank you, I was being dumb. I checked that a homotopy class of maps between arrows gave me a homotopy class of class of cones, but didn't notice that this meant I was going from (Arrows in chain complexes) to (Homotopy chain complexes). Yeah, that's going to get messy to fix... $\endgroup$ Mar 9 '11 at 2:34

Verdier, Astérisque 239, Ch. II, Prop. 1.2.13 (p. 104) says that a triangulated category (with countable coproducts or products) equipped with a cone functor has to split.

Ben Wieland is right - you get a cone functor when working with dérivateurs (or filtered derived categories), but that functor is no longer intrinsic to the base category, and you have to carry diagram categories along.


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