TITLE: When is the homotopy category of an accessible $\infty$-category accessible?
QUESTION [9 upvotes]: Let $\mathcal C$ be an accessible $\infty$-category, and let $ho(\mathcal C)$ be its homotopy category. I can think of two "trivial" reasons for $ho(\mathcal C)$ to be accessible:

$ho(\mathcal C) = \mathcal C$;
$ho(\mathcal C)$ is small with split idempotents.

Otherwise, I am aware of very few examples where $ho(\mathcal C)$ is accessible. Indeed, given that $ho(Spaces)$ is very far from accessible, I think I should expect that it is quite rare for $ho(\mathcal C)$ to be accessible.
However, I know of one interesting class of examples where $ho(\mathcal C)$ is "nontrivially" accessible. Let $k$ be a field.

Write $D(k)$ for "derived $\infty$-category of $k$", i.e. the category of chain complexes over $k$ localized ($\infty$-categorically) at the quasi-isomorphisms. This is a presentable $\infty$-category.
Then $ho(D(k))$ is the usual derived category of $k$, which is equivalent to the usual 1-category of graded $k$-vector spaces (though of course the triangulated structure is different), and so is obviously accessible (locally presentable, in fact).

This one class of examples has me second-guessing my expectation that taking homotopy categories rarely preserves accessibility.
Questions:

What are some other examples of accessible $\infty$-categories $\mathcal C$ such that $ho(\mathcal C)$ is also accessible (which do not satisfy (i) or (ii) above)?
Given such an example, is the functor $\mathcal C \to ho(\mathcal C)$ accessible (i.e. does it preserve $\kappa$-filtered colimits for some $\kappa$?)
Can we (partially) characterize this condition, giving necessary and / or sufficient conditions for $ho(\mathcal C)$ and $\mathcal C \to ho(\mathcal C)$ to be accessible?
Is there anything to be said in particular about the case where $\mathcal C$ is stable and presentable, or even more particularly the case where $\mathcal C$ is the derived $\infty$-category of a ring?

More broadly, at the moment it seems very mysterious to me that $ho(D(k))$ comes out to be accessible. I'd appreciate any perspective which makes this fact look less mysterious.

REPLY [2 votes]: Here's a further generalization:

Claim: Let $\mathcal T$ be a triangulated category and with a $t$-structure such that $\mathcal T^\heartsuit$ is has coproducts, which are exact. Suppose there exists $M \in \mathcal T^\heartsuit$ and a monic endomorphism $M \overset r \rightarrowtail M$ in $\mathcal T^\heartsuit$ which is not an isomorphism, and such that $Hom(M,M/r)$ is an $r$-torsion $End(M)$-module. Then $\mathcal T$ is not concrete, and in particular not accessible.

Proof: For most of the proof, we work in $\mathcal T^\heartsuit$. Similar to before, for every ordinal $\alpha$, we define $W_\alpha$ to be the set of finite subsets of $\alpha+1$, and set $M_\alpha$ to be the cokernel of the map $\oplus^{W_\alpha} M \xrightarrow{1-s} \oplus^{W_\alpha} M$, where $s$ carries the $[\alpha_0<\dots< \alpha_n]$ -copy of $M$ to the $[\alpha_1<\dots<\alpha_n]$ copy of $M$ via the map $r$. A straightforward transfinite induction shows that the $[\alpha_0<\dots<\alpha_n]$'th structure map $M \to M_\alpha$ is in $r^{\alpha_0} Hom(M,M_\alpha)$, and in particular the $[\alpha]$th structure map is in $r^\alpha Hom(M,M_\alpha)$.
It is straightforward to see that the natural map $M_\alpha \to M_{\alpha+1}$ actually splits, and its cokernel $Q$ sits in an exact sequence $M/r \to Q \to M_\alpha \to 0$, where the map to $M_\alpha$ is obtained by deleting the last  letter of each word (which is always $\alpha$ here), and the copy of $M/r$ corresponds to the generator $[\alpha]$ (moreover, the $\alpha$th structure map $M \to M_\alpha$ is nonzero and factors through $M/r$). We claim now that $r^{\alpha+1}Hom(M,M_\alpha) = 0$. This is proved by induction on $\alpha$ using the following
Lemma: Let $M$ be an object of an abelian category and $r: M \to M$ a map. If $0 \to A \to B \to C$ is exact, and if $Hom(M,C)$ is $r$-torsion, then any map $\phi \in r^{\alpha+1} Hom(M,B)$ factors (necessarily uniquely) through $A$, and the factored map lies in $r^\alpha Hom(M,A)$. Dually, if $A \to B \to C \to 0$ is exact and $Hom(M,A)$ is $r$-torsion, then if $r^\alpha Hom(M,C) = 0$ it follows that $r^\alpha Hom(M,B) = 0$, for $\alpha \geq 1$.
We apply the lemma to the exact sequences $0 \to M_\alpha \to M_{\alpha+1} \to Q \to 0$ and $M/r \to Q \to M_\alpha \to 0$ at the successor steps of a transfinite induction to conclude that indeed $r^{\alpha+1} Hom(M,M_\alpha) = 0$.
Now we conclude as before: the $\alpha$th structure map $M \to M_\alpha$ doesn't factor through the $\beta$th structure map $M \to M_\beta$ for $\beta > \alpha$ because then it would lie in $r^\beta Hom(M,M_\alpha) = 0$, and it is in fact nonzero. Since $M\to M_\alpha$ are weak cokernels in $\mathcal T$, they form a proper class of pairwise-inequivalent weak cokernels out of $M$, so that $\mathcal T$ is not concrete.