# When is the homotopy category of an accessible $\infty$-category accessible?

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:

1. $$ho(\mathcal C) = \mathcal C$$;

2. $$ho(\mathcal C)$$ 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:

1. 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)?

2. Given such an example, is the functor $$\mathcal C \to ho(\mathcal C)$$ accessible (i.e. does it preserve $$\kappa$$-filtered colimits for some $$\kappa$$?)

3. 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?

4. 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.

• Give a look at Example 5.3 in Generalized Brown representability in Homotopy categories by Rosicky. The author shows that $\mathsf{Ho}(\text{SSet}_n)$ is accessible. Feb 12, 2020 at 21:58
• @IvanDiLiberti Thanks! Just to be clear, this is not the homotopy category of $n$-truncated spaces as one might naively guess (and the homotopy category of $n$-truncated spaces is not accessible for $n \geq 1$ because idempotents do not split -- the counterexample appearing in HTT 4.4.5.19, which I think is due to Freyd, uses only 1-truncated spaces). Feb 12, 2020 at 22:29
• Not directly related to the question, but a paper you might be interested, with a similar line of inquiry (for combinatoriality) is: arxiv.org/abs/1702.00240v2 Feb 15, 2020 at 15:12
• @DavidWhite Thanks! Feb 16, 2020 at 1:24
• @IvanDiLiberti Does your criterion here show, perhaps, that the derived category of a ring $R$ is not concrete (and hence not accessible) unless $R$ is a division ring? If so, that would be very interesting. Feb 16, 2020 at 1:26

Here's something we can say which addresses a large class of examples.

Claim: Let $$R$$ be a commutative ring which is not zero-dimensional (i.e. there exists $$r \in R$$ which is not a unit or a zero-divisor). Then the derived category $$ho(D(R))$$ of left $$R$$-modules is not concrete (i.e. does not admit a faithful functor to $$Set$$) and in particular is not accessible.

The proof is a straightforward generalization of Freyd's argument that the homotopy category of spaces is not concrete, and in particular is robust enough that "$$D(R)$$" could be interpreted to have various "boundedness" conditions if one likes.

Beginning of Proof of Claim: Pick $$r \in R$$ which is neither a unit nor a zero-divisor. If $$M$$ is a left $$R$$-module, define the submodule $$r^\alpha M$$ for any ordinal $$\alpha$$ inductively by $$r^0 M = M$$, $$r^{\alpha+1} M = rr^\alpha M$$, and taking intersections at limit ordinals. Note that if $$\phi: M \to N$$ is any $$R$$-module map, then $$\phi(r^\alpha M) \subseteq r^\alpha N$$ for any $$\alpha$$.

Let $$\alpha$$ be an ordinal. For each $$n \in \mathbb N$$, let $$W_\alpha^{(n)}$$ be the set of strictly increasing sequences of ordinals $$\alpha_0 < \alpha_1 < \dots < \alpha_m$$ where $$m \leq n$$ and $$\alpha_m \leq \alpha$$. Let $$F_\alpha^{(n)} = R\{W_\alpha^{(n)}\}$$ be the free left $$R$$-module on generators given by $$W_\alpha^{(n)}$$, and define $$M_\alpha^{(n)} = F_\alpha^{(n)} / K_\alpha^{(n)}$$, where $$K_\alpha^{(n)}$$ is generated by those elements of the form $$[\alpha_1,\dots,\alpha_n] - r[\alpha_0, \alpha_1, \dots, \alpha_n]$$ (when $$n = 0$$ we interpret this to mean that $$r[\alpha_0] \in K_\alpha^{(n)}$$).

Lemma: For $$n \in \mathbb N$$, the map $$M_\alpha^{(n)} \to M_\alpha^{(n+1)}$$ is injective.

Proof: This is straightforward, using the fact that $$r$$ is not a right zero-divisor.

Define $$M_\alpha = \cup_{n \in \mathbb N} M_\alpha^{(n)}$$. We have a natural filtration $$M_\alpha^{(0)} \subseteq \cdots \subseteq M_\alpha^{(n)} \subseteq \cdots M_\alpha$$, and the associated graded is naturally identified with $$\oplus_{n \in \mathbb N} \oplus^{ W_\alpha^{(n)} \setminus W_\alpha^{(n-1)}} (R/(Rr))$$.

Lemma: Pick a set of "canonical" coset representatives in $$R$$ of the nonzero elements of $$R/(Rr)$$. Then every element of $$M_\alpha$$ may be written uniquely in the form $$\sum_i r_i w_i$$ where the $$r_i$$ are canonical coset representatives and the $$w_i$$ are distinct words of some $$W_\alpha^{(n)}$$.

Proof: The existence of such a representation is basically obvious. Suppose that two such representations denote the same element of $$M_\alpha^{(n)}$$, where $$n$$ is minimal: $$\sum_i r_i w_i = \sum_j s_j v_j$$. Then they have the same image in the associated graded, and so the $$w_i$$'s of length $$n$$ match up with the $$v_j$$'s of length $$n$$; since the choice of coset representatives has been normalized, their coefficients in fact coincide. Deleting these words, we get an identification of represntations in $$M_\alpha^{(n-1)}$$, contradicting the minimality of $$n$$.

Lemma: For all ordinals $$\beta$$, $$r^\beta M_\alpha$$ consists of those elements whose canonical representation as chosen above contains only words $$[\alpha_0,\dots,\alpha_n]$$ where $$\alpha_0 \geq \beta$$.

Proof: Using the canonical representation, this is now an easy transfinite induction. (Here we use the fact that $$R$$ is commutative, though.)

Conclusion of Proof of Claim: Note that $$r^\alpha M_\alpha = R/(Rr) \neq 0$$ canonically, because $$r$$ is not a right unit. Let $$f_\alpha$$ be the map $$R/(Rr) = r^\alpha M_\alpha \subseteq M_\alpha$$. Then for $$\alpha < \beta$$, there is no factorization of $$f_\beta$$ through $$f_\alpha$$. Pick $$d \in \mathbb Z$$ such that $$(R/r)[d]$$, each $$M_\alpha[d]$$, and the fiber $$(M_\alpha/f_\alpha)[d-1]$$ of $$f_\alpha[d]$$ are all in whichever flavor of $$ho(D(R))$$ we are working with. Then the maps $$f_\alpha[d]$$ are a proper class of pairwise-inequivalent weak cokernels of their fibers in $$ho(D(R))$$. But as Freyd shows, a proper class of pairwise-inequivalent weak cokernels out of a fixed object in a pointed category imply that the category is not concrete.

So the outline, as in Freyd's proof, is to use the fact that in $$ho(D(R))$$, basically every map is a weak cokernel (in fact, all maps are if we're working in the unbounded derived category).

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.