Suppose that we have a $\lambda$-combinatorial model category $M$ (for some cardinal $\lambda$) such that any $\lambda$-filtered colimit of fibrant-cofibrant objects is always fibrant. My question is the following: It is true that any fibrant object in $M$ is a filtered colimit of fibrant-cofibrant objects ?

Edit: After reading the comments, I have realized that I have to add some reasonable condition. So here is the additional condition. Let $R:M\rightarrow M$ be a functorial fibrant replacement i.e. for any $x\in M$ we have a functorial factorization $x\rightarrow R(x)\rightarrow 1$ there the first map is a trivial cofibration and the second map fibration (1 is terminal object).

So here is my new question: It is true that for any object $x\in M$, $R(x)$ is a $\lambda$-filtered colimit of fibrant-cofibrant objects ?

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    $\begingroup$ We can just take the set of generating cofibrations to be empty, so that every object is fibrant while only the initial object is cofibrant. You would need to add some kind of condition to ensure that you have a sufficient supply of cofibrant objects in the first place. $\endgroup$ – Reid Barton Dec 18 '18 at 16:57
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    $\begingroup$ Also, doesn't $\lambda$-combinatorial include the condition that the generating acyclic cofibrations are maps between $\lambda$-compact objects, so that any $\lambda$-filtered colimit of fibrant objects is automatically fibrant? So your condition is redundant. $\endgroup$ – Reid Barton Dec 18 '18 at 16:58
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    $\begingroup$ @ABC wait -- why is a $\lambda$-filtered colimit of fibrant-cofibrant objecs fibrant-cofibrant? It's not necessarily the case that cofibrant objects are closed under $\lambda$-filtered colimits for any $\lambda$. $\endgroup$ – Tim Campion Dec 18 '18 at 19:41
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    $\begingroup$ @ABC That's not necessarily true. I don't quite have a counterexample, but the question of whether free abelian groups are closed under $\lambda$-filtered colimits for some $\lambda$ is independent of set theory. I think (but I'm not sure) this implies that the question "Are the cofibrant objects in the projective model structure on chain complexes closed under sufficiently-filtered colimits?" is independent of set theory. $\endgroup$ – Tim Campion Dec 18 '18 at 19:50
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    $\begingroup$ The example in my first comment is still a counterexample to the amended question. $\endgroup$ – Reid Barton Dec 19 '18 at 5:07

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