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$\def\D{\mathcal{D}} \def\ind{\operatorname{Ind}} \def\K{\mathcal{K}} \def\A{\mathcal{A}}$Inside [GW, Remark F.168, p. 794], we find:

[Let $\K$ be a category and let $\K_S$ be its localization with respect to a right multiplicative system $S$.] If $\iota_\K:\K\to\ind(\K)$ is the fully faithful canonical functor and $Q:\K\to\K_S$ is the localization functor, one obtains a diagram of functors $$ \begin{matrix} \K&\xrightarrow{Q}&\K_S\\ \iota_\K&\searrow&\downarrow&\alpha\\ &&\ind(\K) \end{matrix} $$ which is not commutative. But there exists a natural morphism $$ \tag{F.42.1}\label{nat} \iota_\K\to\alpha\circ Q. $$ Let us suppose that $\K$ is a triangulated subcategory of $K(\A)$ for an abelian category $\A$ and that $\K_S$ is the localization by the system $S$ of quasi-isomorphisms in $\K$. Then $\ind(\K)$ is also triangulated, all functors above are triangulated and the morphism \eqref{nat} is a morphism of triangulated functors.

Here $\alpha:\K_S\to\ind(\K)$ is the canonical embedding of the localization of a category into its ind-completion, see e.g. [KS, Proposition 7.4.1].

My questions are:

(Q1). Given a triangulated category $\K$, is there some canonical triangulated structure on its ind-completion $\ind(\K)$ turning $\K\to\ind(\K)$ into a triangulated functor?

(Q2). If the answer is to (Q1) is 'no' in general but 'yes' if $\K$ is a full subcategory of $K(\A)$, where $\A$ is abelian (as [GW] claims in the quote above), do you know any reference for this? And same for [GW]'s claims that $\iota_K$ and $\alpha$ are triangulated functors and \eqref{nat} is a morphism of triangulated functors. In [KS] there's nothing on these issues.

References

[GW]. Görtz, Wedhorn, Algebraic Geometry II

[KS]. Kashiwara, Schapira, Categories and Sheaves

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    $\begingroup$ One of the main reasons to work with higher enhancements of triangulated categories (such as differential graded categories, model categories, or any flavour of $\infty$-categories) is that 'basic' category theory constructions behave better. Taking Ind-categories is one example of this (others include taking $\operatorname{Fun}(\mathcal I,\mathcal K)$ from some small category $\mathcal I$). The usual derived category is a homotopy category of one of these enhancements, and taking homotopy category does not commute with most basic operations in (higher) category theory. $\endgroup$ Commented 12 hours ago

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The category $\mathcal{F}$ of finite spectra is self-dual under Spanier-Whitehead duality, so $\operatorname{Pro}(\mathcal{F})\simeq\operatorname{Ind}(\mathcal{F})$. The $\operatorname{Pro}$ category is equivalent to the category of homology theories, or equivalently spectra mod phantoms, by Proposition 4.19 of Morava K-theories and localisation. Of course the category of spectra is triangulated, but I am fairly sure that the quotient spectra/phantoms is not, although I do not immediately see a proof. Although the cited proof is written for the category of finite spectra, it will apply much more generally, although some kind of countability hypothesis may be required (as in Theorem 4.1.5 of Axiomatic stable homotopy).

On the other hand, if you interpret everything in Lurie's framework, I think it is true that the $\operatorname{Ind}$ completion of any stable $\infty$-category is a stable $\infty$-category, and that is in some sense the right version of your question.

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  • $\begingroup$ I think I don't understand anything from this answer xD. I don't know what spectra are (my topology knowledge is not great), and I don't know anything about $\infty$-categories either. $\endgroup$ Commented 15 hours ago
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    $\begingroup$ Let me reinforce Neil's answer here with another example. If you ind complete the small stable module category of a finite group as a triangulated category, you get the big stable module category modulo phantoms, which is (probably) not triangulated, and not really what you want. However, if you regard it as a stable infinity category and ind complete, you get the big stable module category that you wanted, as a stable infinity category. An intermediate solution is to use differential graded categories, and this works well, but comes with its own problems. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ Now I am confused: Isn't the homotopy category of a stable $\infty$-category triangulated? I guess, the Ind-completion does not commute with taking the homotopy category? $\endgroup$ Commented 13 hours ago
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    $\begingroup$ @UlrichPennig Yes it is; no it doesn't. $\endgroup$ Commented 13 hours ago

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