11
$\begingroup$

The category of fibrations in a combinatorial model category is accessible, accessibly embedded in the arrow category. How about the cofibrations?

More generally, let $C$ be a locally presentable category and let $(L,R)$ be a weak factorization system on $C$. If $(L,R)$ is cofibrantly-generated (i.e. there is a set $I \subseteq L$ such that $R$ consists precisely of those morphisms with the right lifting property with respect to $I$), then $R$, considered as a full subcategory of $C^\to$, is accessible and accessibly embedded.

Question 1: Suppose that $(L,R)$ is cofibrantly-generated. Is $L$ accessible and accessibly embedded (as a full subcategory of $C^\to$)?

Question 2: Conversely, if $L$ is accessible and accessibly embedded, then is $(L,R)$ cofibrantly-generated?

Question 3: Similar to the above two, but use the notion of "small-generated" coming from Garner's small object argument (where $I$ can be a category rather than a set).

The proof that $R$ is accessible and accessibly embedded is not completely straightforward: it relies on the fact that the small object argument provides a functorial factorization system which preserves $\lambda$-filtered colimits and $\lambda$-presentable objects for some $\lambda$ to exhibit every $R$-morphism as a retract of a colimit of $\lambda$-presentable $R$-morphisms and to see that fibrations are closed under $\lambda$-filtered colimits.

The fact that $L$ is closed under transfinite composition sounds tantalizingly close to saying that it is closed under filtered colimits, but I'm not sure the latter is actually true.

Motivation: If the answer to both questions is yes, then it becomes very easy to prove Jeff Smith's theorem since an intersection of accessible, accessibly-embedded, replete subcategories is accessible and accessibly-embedded.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Do you know if "L accessible and accessibly embedded" follow from having an accessible factorization system (I.e there is a functorial choice of factorization, given by an accessible functor). That's sound reasonable and if it is true the answer are yes and no: cofibrantly generated wfs are accessible because of the factorization given by (Garner's version of) the small object argument, and I know some exemple of accessible wfs that are not cofibrantly generated. $\endgroup$
    – Simon Henry
    Commented Jun 29, 2018 at 14:44
  • $\begingroup$ Read: "Cofibrantly generated wfs in presentable categories are..." $\endgroup$
    – Simon Henry
    Commented Jun 29, 2018 at 14:50
  • 1
    $\begingroup$ @SimonHenry I see. I misread Theorem 4.3 in Rosicky's Accessible model categories, eliding the difference between small-generated and cofibrantly-generated. I would really like to believe that "L accessible and accessibly-embedded" is equivalent having an accessible wfs, but I don't know if either implication holds. I suppose Rosicky's Theorem 5.3 is a version of Jeff Smith's theorem along the lines I'm suggesting, although as he remarks, it's not clear if it's optimal. $\endgroup$
    – Tim Campion
    Commented Jun 29, 2018 at 14:54
  • 1
    $\begingroup$ Take a look at Lemma 2.11 in arxiv.org/abs/1802.09889. This doesn't answer your question, but is relevant. $\endgroup$
    – Karol Szumiło
    Commented Jun 30, 2018 at 5:26
  • $\begingroup$ @TimCampion: So, is it correct to say (based on my understanding of the answers given below) that we still do not know any examples of combinatorial model categories where the class of cofibrations is not accessible? $\endgroup$
    – Dmitri Pavlov
    Commented Jul 30, 2021 at 20:51

2 Answers 2

Reset to default
10
$\begingroup$

A cofibrantly generated $(L,R)$ does not need to have $L$ accessible, see Example 3.5 in my paper "On combinatorial model categories." Also, $L$ accessible does not imply that $(L,R)$ is cofibrantly generated, even accessible. Take regular monos in Boolean algebras. This $L$ is accessible but $(L,R)$ cannot be accessible because regular injectives are complete Boolean algebras which are not accessible.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks! The second example in Example 3.5 -- the independence of the accessibility of free abelian groups -- is even one that I knew at some point, and was trying to remember -- I looked for it in your On projectivity in locally presentable categories but didn't find it there. Apparently the original reference is 5.5.1 in Makkai and Pare. $\endgroup$
    – Tim Campion
    Commented Jun 30, 2018 at 15:00
  • $\begingroup$ Just for completeness, let me record the first example in Example 3.5. The split monos in the category of posets form the cofibrant closure of the split monos between finite posets, but they are not closed under $\lambda$-filtered colimits for any $\lambda$, and so not accessibly embedded. There is no dependence on set theory in this example. $\endgroup$
    – Tim Campion
    Commented Jul 1, 2018 at 16:14
  • $\begingroup$ I'm suddenly doubtful. Let $S$ be a set, regarded as a discrete poset, let $S_1$ be $S$ with a top element added, and let $S_2 = S_1 \cup_S S_1$. Then the inclusion $S_1 \to S_2$ is a split mono. But if $S$ is infinite, I think that $S_1 \to S_2$ is not cellular in split monos between finite posets (and I believe the cellular maps in split monos between finite posets are closed under retracts). $\endgroup$
    – Tim Campion
    Commented Jul 1, 2018 at 18:39
  • $\begingroup$ After all, transfinite composition doesn't buy us anything here, since we're only adding one element. So we would have to exhibit $S_1 \to S_2$ as the pushout of a split mono between finite posets. But by construction, any pushout of a split mono between finite posets which factors $S_1 \to S_2$ can only make finitely many of the elements of $S$ lie below the new top element, so we cannot achieve $S_1 \to S_2$. $\endgroup$
    – Tim Campion
    Commented Jul 1, 2018 at 18:47
2
$\begingroup$

Here's an elaboration on the example in Professor Rosický's paper. I'll make it community-wiki.

Let $Pos$ be the category of posets, and let $L$ be the class of split monomorphisms in $Pos$. Let $L_\omega$ be the set of split monomorphisms between finite posets.

Claim 1: $L$ is the cofibrant closure of $L_\omega$.

Proof: One can check that in any category the class of split monomorphisms is closed under coproduct, cobase-change, transfinite composition, and retracts. Conversely, if $P \to Q$ is a split mono, one can add the elements of $Q$ one at a time in a chain, so we may assume without loss of generality that $Q$ has only one element $q$ which is not in $P$. Now we may express $P \to Q$ as the colimit of a chain, each link of which adds one relation $p \leq q$ or $q \leq p$ for some $p \in P$. Each of these links is a pushout by a split mono between 2-element posets. I'm not sure how do do this!

Claim 2: $L$ is not closed in $Pos^{\to}$ under $\lambda$-filtered colimits for any $\lambda$.

*Proof:** The closure of $L$ under $\lambda$-filtered colimits consists of the $\lambda$-pure monomorphisms in $Pos$. So we just need an example of a $\lambda$-pure monomorphism which doesn't split, for each regular cardinal $\lambda$. The inclusion $\lambda \to \lambda+1$ fits the bill -- see Example 2.28(3) in Adamek and Rosicky's Locally Presentable and Accessible Categories.

Thus $L$ is cofibrantly generated, but not accessibly embedded.

In the other direction, I don't know a source for Professor Rosický's claim that regular monos in Boolean algebras are a counterexample. But I'm pretty sure that in any locally presentable category, both (epi, strong mono) and (strong epi, mono) are accessible orthogonal factorization systems. And Example 4.4(2) in the same book says that complete Boolean algebras are the injective objects in the category of distributive lattices, citing

Banaschewski, B. and G. Bruns (1967): Categorical characterization of MacNellie completion. Arch. Math. 18, 369-377.

I think it's well-known that complete Boolean algebras don't form an accessible category. To show this it suffices to construct a Boolean algebra of cardinality $\kappa$ which is $\kappa$-complete but not $\kappa^+$ complete, for arbitrarily large $\kappa$. The set of $<\kappa$-sized subsets of a set of size $\kappa$ works (where $\kappa$ is regular).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Tim, you are right. Split monos in posets are not cofibrantly generated. The resulting weak factorization system is only accessible. This follows from 1.6 in my joint paper with Adámek, Herrlich and Tholen "Weak factorization systems and topological functors". So, we are missing an example in ZFC. Concerning regular monos in Boolean algebras, you gave a proof. This and other examples can be found in my recent paper On the uniqueness of cellular injectives. $\endgroup$
    – Jiří Rosický
    Commented Jul 2, 2018 at 7:38
  • $\begingroup$ @JiříRosický Thanks, that's helpful. I see that 1.6 in "Weak factorization systems and topological functors" implies that split monos in posets are the left half of a weak factorization system, but why is it accessible? $\endgroup$
    – Tim Campion
    Commented Jul 2, 2018 at 12:57
  • 2
    $\begingroup$ The factorization of $f:A\to B$ is $A\to A\times B\to B$ and $colim (A_i\times B_i)\cong (colim A_i)\times (colim B_i)$. $\endgroup$
    – Jiří Rosický
    Commented Jul 2, 2018 at 13:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .