Is the ind-completion of a triangulated category triangulated?

Question

$\def\D{\mathcal{D}} \def\ind{\operatorname{Ind}} \def\K{\mathcal{K}} \def\A{\mathcal{A}}$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]. Namely, $\alpha$ is the left Kan extension of $\iota_\K$ along $Q$ [KS, Proposition 7.3.3].

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.

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Context: I was trying to find a proof of the fact that derived triangulated functors are triangulated in terms of ind-completions. Despite the negative answer to (Q1) that Neil Strickland provided, I found a workaround for the proof that I explain here.

References

[GW]. Görtz, Wedhorn, Algebraic Geometry II

[KS]. Kashiwara, Schapira, Categories and Sheaves

Answer 1

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.

Answer 2

Let $\mathscr{T}$ be a triangulated category. A cohomologycal functor is a functor $\mathscr{T}^{\operatorname{op}}\to\operatorname{Ab}$ taking exact triangles to exact sequences. Let $\operatorname{Coh}\mathscr{T}$ denote the category of cohomological functors. Let $\operatorname{Ind}\mathscr{T}$ be the ind-completion of $\mathscr{T}$. The filtered colimit of cohomological functors is cohomological, since filtered commits are exact in the category of abelian groups. Hence, the functor $\mathscr{T}\to\operatorname{Coh}\mathscr{T}\colon X\mapsto \mathscr{T}(-,X)$ extends to $\operatorname{Ind}\mathscr{T}\to\operatorname{Coh}\mathscr{T}$. The latter is known to be an equivalence. Therefore, in the first question (Q1) you wonder whether $\operatorname{Coh}\mathscr{T}$ is triangulated in such a way that the previous $\mathscr{T}\to\operatorname{Coh}\mathscr{T}$ is exact.

The answer to (Q1) is expected to be negative in almost all cases, although it is true in some. In order to give you examples, and to answer also (Q2), let me concentrate in case $\mathscr{T}=D^c(R)$ is the derived category of bounded complexes of projective modules over a ring $R$, a.k.a. perfect complexes, or compact objects.

The answer to (Q1) is positive if $R=k$ is a field because in that case $\mathscr{T}=D^c(k)$ is the category of finite-dimensional graded vector spaces and $\operatorname{Ind}\mathscr{T}\simeq\operatorname{Coh}\mathscr{T}$ is the category of all graded vector spaces, which is equivalent to the full derived category $D(k)$. The equivalence is compatible with the triangulated inclusion $D^c(k)\subset D(k)$. Similarly when $R$ is Von Neumann regular.

Now, let $R$ be a ring with a non-projective flat module $F$ (e.g. $R=\mathbb Z$ and $F=\mathbb Q$). Choose an epimorphism from a projective $R$-module $P\twoheadrightarrow F$. Let $\mathscr{S}=D(R)$ be the full derived category. The induced morphism of cohomological functors $\mathscr{S}(-,P)\twoheadrightarrow\mathscr{S}(-,F)$ restricted to $\mathscr{T}\subset\mathscr{S}$ is known to be an epimorphism in $\operatorname{Coh}\mathscr{T}$ (see the reference below). Therefore, if $\mathscr{S}$ were triangulated it would split. In particular, putting $R$ in the first variable we would obtain a splitting for $P\twoheadrightarrow F$, contradicting the fact that $F$ is not projective.

Christensen, J.Daniel. «Ideals in Triangulated Categories: Phantoms, Ghosts and Skeleta». Advances in Mathematics 136, n.º 2 (junio de 1998): 284-339. https://doi.org/10.1006/aima.1998.1735.