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In a combinatorial model category, every $\lambda$-filtered colimit is a homotopy colimit for $\lambda$ regular big enough. So for $\lambda$ regular big enough, every $\lambda$-filtered colimit of a diagram of cofibrant replacements of a given object $X$ is a cofibrant replacement of $X$. Does the contrary hold, i.e. is the full subcategory of cofibrant replacements of a given object accessible ?

EDIT : the class of cofibrations is accessible so for $\lambda$ regular big enough, a $\lambda$-filtered colimit of cofibrations $\varnothing \to X_i$ is a cofibration ; the class of weak equivalences is accessible so for $\lambda$- regular big enough, a $\lambda$-filtered colimit of weak equivalences $X_i\to X$ is a weak equivalence ; so for $\lambda$ regular big enough, a $\lambda$-filtered colimit of cofibrant replacements of $X$ is a cofibrant replacement of $X$.

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  • $\begingroup$ I don't see why such a colimit of cofibrant objects should be cofibrant, although I have the feeling that it is true. $\endgroup$
    – Fernando Muro
    Commented Apr 29, 2013 at 9:12
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    $\begingroup$ I added the reason in my question. $\endgroup$
    – Philippe Gaucher
    Commented Apr 29, 2013 at 10:39
  • $\begingroup$ I think the answer is yes: you should be able to describe the category of cofibrant replacements as a homotopy limit of categories you already know are accessible... If I have time I'll try to think of the exact limit you want. (Probably you need that cofibrant objects form an accessible subcategory, and the category of weak equivalences are accessible, and undercategories of accessible categories are accessible, etc.) $\endgroup$
    – Dylan Wilson
    Commented Apr 29, 2013 at 15:09
  • $\begingroup$ @PhilippeGaucher I don't believe the class of cofibrations, or even cofibrant objects, is always accessible -- or at least if it is, some set-theoretical hypothesis is needed. For, following Rosicky, there is a combinatorial model structure on $Ab$ with cofibrations generated by $0 \to \mathbb Z$ and weak equivalences the isomorphisms. The cofibrant objects are the free abelian groups. But under $V=L$, the free abelian groups are not closed under $\lambda$-filtered colimits for any $\lambda$, and in particular they aren't accessible. $\endgroup$
    – Tim Campion
    Commented Feb 4, 2020 at 20:25

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(sorry I have troubles with comments, I post here even if it is not an answer) I have a new information. In On a fat small object argument, it is proved that in a λ-combinatorial model category, every cofibrant object is a λ-filtered colimit of λ-presentable cofibrant objects, which is close to what I wanted.

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(Edited later: The answer is probably no in general given the discussion and references in the comments...)

Here is a reduction to a more basic question...

First note it suffices to show that the subcategory of cofibrant objects is accessible. Indeed, if we have this then we can prove your result by forming the homotopy pullback of accessible categories and accessible functors:

$$W \times_{\mathcal{C}} \mathcal{C}^{[1]}\times_{\mathcal{C}} \mathcal{C}_{X/}\times_{\mathcal{C}} \mathcal{C}^{cof}$$

This is the category you're interested in, and replacing $W$ by $F \cap W$ we get the other possible category where objects are $(X, Y \rightarrow X$) with the morphism a trivial fibration.

So we just need that the category of cofibrant objects is accessible... and I don't actually see how to prove this at the moment!

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    $\begingroup$ Sorry, I do mean the one that takes the composite. The proof I gave interprets "cofibrant replacement" as a pair (X,f:Y→X) where f is an acyclic fibration and Y is cofibrant. This is the category I described. There is a similar, easier argument which shows the category you're interested in is accessible. Here is a proof: The category of weak equivalences is accessible, so the category whose objects are weak equivalences with domain $X$ is accessible (by a similar argument). My above proof can be used to show that the category of cofibrant objects is accessible, so using the obvious homotopy $\endgroup$
    – Dylan Wilson
    Commented Apr 30, 2013 at 15:06
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    $\begingroup$ I cannot edit my comment so I rephrase my question. Take a "very good" cofibrant replacement $f.g$ of $X$ ($f$ cofibration and $g$ trivial fibration). Why does it come from a factorization by a functor $T$ constructed using the small object argument ? $\endgroup$
    – Philippe Gaucher
    Commented Apr 30, 2013 at 16:01
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    $\begingroup$ Good point... I've edited the 'answer' and I guess it's not much of one any more... I'll have to think about this some more. The rest would follow if I knew that the category of cofibrations were accessible, but I don't yet see why this is true. You mention this in your question- do you have a reference? $\endgroup$
    – Dylan Wilson
    Commented Apr 30, 2013 at 16:49
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    $\begingroup$ I believed that "On combinatorial model categories" by J. Rosicky was the answer (math.muni.cz/~rosicky/papers/comb2.pdf), and page 8 (top of the page), one can read that $cof(S)$ is not always accessible, $S$ being a set ! On the contrary, $inj(S)$ is always accessible (Proposition 3.3 of the same paper) as a small injectivity class... $\endgroup$
    – Philippe Gaucher
    Commented Apr 30, 2013 at 17:13
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    $\begingroup$ It is yes if the category of cofibrations is accessible... I have no more information. $\endgroup$
    – Philippe Gaucher
    Commented Apr 30, 2013 at 19:00

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