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Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural functor $$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ My question is:

Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?

That is, is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$

If this isn't a "universal triangulated category," does there exist such a construction?

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    $\begingroup$ I think there's some confusions here that are worth sorting through. (1) 'associated simplicial category' would usually mean C back again, but you seem to use it later to mean the category of simplicial objects in C, with its simplicial structure? (2) the htpy category of a stable $\infty$-category has a canonical triangulated structure, not just any old $\infty$-category, (3) the homotopy category of sA, for A abelian, is not $D(A)$, or triangulated, but rather $D_{\ge 0}(A)$... I think there are some other issues, but that's a start $\endgroup$ Aug 9, 2020 at 12:33
  • $\begingroup$ @DylanWilson Thank you very much! I'll change my question accordingly. $\endgroup$ Aug 9, 2020 at 12:36
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    $\begingroup$ more stuff: if you'd like to take an ordinary category $C$ and then view $sC$ as simplicially enriched, I think you need $C$ to be complete or cocomplete (otherwise how do you define the simplicial structure on hom-objects?). Also, another issue you might run into is that the construction 'sC' isn't very universal, as far as I know... for specific examples of C it ends up having a universal property as a sifted cocompletion of a certain subcategory of C, but I don't know a good invariant description in general $\endgroup$ Aug 9, 2020 at 13:28
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    $\begingroup$ @DylanWilson The example I'm thinking of here is where C is something that is already abelian or at least additive, so I think the point is that CMG is implicitly thinking of the construction Ho Stab Ch^+(C) (identifying Ch^+(C) with sC, in which case this does have a universal property I think. $\endgroup$ Aug 9, 2020 at 14:37

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I will give a partial answer. I note that the OP has asked a LOT of questions recently (I count 12 so far in the first 9 days of August), and many of them are good questions on which much research has already been done. I would encourage the OP to slow the rate of question-asking, to spend more time reading the references that have been provided, and to think carefully in future questions to avoid easily avoided problems like those that have been raised in the comments.

Now to the answer. Morally, what the OP is suggesting is exactly the kind of thing we love to do as homotopy theorists, but the devil is in the details. Specifically, in this case, the devil is in "...and $\mathcal{C} \to \mathcal{T}$ satisfying some properties..." The issue is that there could be multiple "obvious" ways to stabilize $\mathcal{C}$, and the functor $F:\mathcal{C} \to \mathcal{T}$ has to know that you mean the one you suggested. For example, suppose $\mathcal{C}$ is the empty category. Then whatever conditions you have in mind will probably be satisfied vacuously, and you're asking for a triangulated category $Ho(Stab(C))$ that is supposed to admit a map from every triangulated category $\mathcal{T}$. That's probably not what you really meant.

That said, homotopy theorists have thought long and hard along the direction you have in mind. I recommend the following papers:

  • Dugger's Universal homotopy theories: given any small category $C$, create a "universal" model category $UC$, which is essentially the free homotopy theory generated by $C$.
  • Hovey's Spectra and symmetric spectra in general model categories: given a model category $C$ and a Quillen endofunctor $G$, create the stabilization $Sp(C,G)$ where $G$ becomes a Quillen equivalence, just like if $C = Top$, $G$ is the suspension functor, and $Sp(C,G)$ is spectra.
  • Lurie's Higher Topos Theory, which expands the work above into the realm of infinity categories, e.g., so that $Sp(C,G)$ is a stable $\infty$-category, when you start with a presentable $\infty$-category $C$. Similarly, Dugger's construction can be made to work to produce a presentable $\infty$-category $UC$.
  • Hovey's book on model categories: chapter 6 shows how to start from a pointed model category and produce a pre-triangulated category in a universal way. But note that when Hovey says "pre-triangulated" that doesn't mean the same as when other authors say "pre-triangulated." Here the structure that would need to be preserved by $F$ has to do with fiber and cofiber sequences, and you need $C$ to be pointed to define these.
  • Beligiannis and Reiten's Homological and homotopical aspects of torsion theories: Section 5 shows how, given a left, right, or pretriangulated category, there is a universal stabilization that preserves the old partial triangulated structure.
  • There are many, many other papers in this vein. For example, building on Beligiannis's paper Relative homological algebra and purity in triangulated categories (from 2000), Balmer and Stevenson just this year showed how to take a triangulated category, take a quotient (making it no longer triangulated), and then stabilize in a nice way.

The point is that you have to know what structure of $\mathcal{C}$ to you want to be preserved by $F: \mathcal{C} \to \mathcal{T}$. For example, if you want to assume that $\mathcal{T}$ can be realized by a stable $\infty$-category, then you can probably get a positive answer by making sure $F$ plays nicely with the eventual $\infty$-category structure on $sC$.

However, it's not true that every triangulated category comes from a stable $\infty$-category (Muro and others have constructed counterexamples), so the universal properties shown by Lurie do not provide an affirmative answer to your question in general. Broadly speaking, the collection of triangulated categories breaks down into two types: those that are "geometric/topological" (e.g., the homotopy category of a stable $\infty$-category) and those that are algebraic. Unless you have some way to connect the geometric-type triangulated categories to the algebraic-type (a problem that many have thought about and none have solved to my knowledge) via whatever conditions you place on $F$, then your question is unlikely to have a positive answer in the generality in which you asked it. Hope this helps!

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