Cofibrant replacements of a given object in a combinatorial model category 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 $X$.
 A: (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.
A: (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!
