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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.

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    $\begingroup$ 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. $\endgroup$ Feb 12, 2020 at 21:58
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    $\begingroup$ @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). $\endgroup$
    – Tim Campion
    Feb 12, 2020 at 22:29
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    $\begingroup$ 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 $\endgroup$ Feb 15, 2020 at 15:12
  • $\begingroup$ @DavidWhite Thanks! $\endgroup$
    – Tim Campion
    Feb 16, 2020 at 1:24
  • $\begingroup$ @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. $\endgroup$
    – Tim Campion
    Feb 16, 2020 at 1:26

2 Answers 2

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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).

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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.

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