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I apologize in advance if my question is utterly stupid, but I can't resist asking it. So...

Is it true that in a model category ( - for example $\mbox{Set}_\Delta$ with the Joyal model structure - ) the full subcategory of fibrant objects is a reflective subcategory? More concretely, is the fibrant replacement functor a left adjoint to the inclusion functor?


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Hmmm. This seems unlikely. Adjointness would say that, for each object $X$, there exists a unique (up to isomorphism) fibrant object $L X$ equipped with a universal morphism $X \to L X$ such that all morphisms $X \to A$ with $A$ fibrant factor through $X \to L X$ uniquely. But fibrant replacements are only unique up to weak equivalence, not isomorphism. – Zhen Lin Sep 28 '12 at 1:55
@ZhenLin This sounds like minimal fibrations. – Baby Dragon Sep 28 '12 at 1:57
...or you can just say everything using the language of $\infty$-categories. – Aaron Mazel-Gee Nov 3 '12 at 5:22

In the case that all isomorphisms in the model category are fibration, cofibration, and weak equivalences, (trivial cofibrations, fibrations) and (fibrations, trivial cofibrations) form factorization systems in the sense of Borceux:

A factorization system in a category $\mathbf B$ is defined as a pair $(E, F)$ of classes of arrows in $\mathbf B$ such that

  1. every isomorphism belongs to both E and M
  2. both E and M are closed under composition
  3. $\forall e \in E \forall m \in M. e \perp m$ (i.e. e has the LLP wrt m)
  4. every arrow f in B can be factored as f = me with e in E and m in M

Then the following theorem in Borceux's Handbook of Categorical Algebra, volume 1, page 211, gives an affirmative answer:

Suppose $\mathbf B$ is a category with a terminal object 1 and a factorization system (E, F). For each object $B \in \mathbf B$, factor the unique arrow $B \xrightarrow ! 1$ into $B \xrightarrow {e_B} r(B) \xrightarrow {f_B} 1$. Then $r$ is a reflection of $\mathbf B$ into the full subcategory spanned by all $r(B)$, and each $e_B$ is the unit of the reflection.

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The proposition you are citing refers to a (proper) factorization system, whereas in a model category, we "only" have a weak factorization system. You can read about these two notions in the Appendices C and D of Joyal's The Theory of Quasi-Categories and its Applications. – Daniel Gerigk Oct 23 '15 at 22:45
@user43687 Hovey demands a model category to include two functorial factorizations $(\alpha,\beta)$ and $(\gamma,\delta)$, but he does not demand those functorial factorizations to be strong, or am I misunderstanding you? (Also, note that Hovey had to correct his definition of "functorial factorization": Errata) – Daniel Gerigk Feb 7 at 20:58
@user43687 What makes you think that a functorial factorization must necessarily be strict? For example, the functorial factorizations (à la Hovey) of the classical Quillen model structure on simplicial sets aren't strict, I think. – Daniel Gerigk Feb 8 at 4:43
@DanielGerigk I was thinking that having a functor which did the factorization forced uniqueness of the factorization. Of course this is not true as the functor doesn't have anything to do with the subcategories that do the factorizing. – user43687 Feb 8 at 5:44

I don't think this question is really meaningful at the level of 1-categories, but assuming that what your after is the model category theoretic version of reflecting into a reflexive $\infty$-subcategory, then what your looking for is the fibrant replacements arising from (left) Bousfield localizations of simplicial model categories. More precisely, suppose $C$ is a simplicial model category and let $C_{loc}$ denote a left Bousfield localization (if one exists, e.g. $C$ is combinatorial). Then by the basic properties of the localization, the identity functor $$id:C_{loc}\to C$$ preserves fibrations and acyclic fibrations, and the identity functor $$id:C\to C_{loc}$$ preserves cofibrations and acyclic cofibrations. Passing to fibrant/cofibrant objects gives an inclusion $$i:C_{loc}^{o}\to C^o.$$ In general, the fibrant replacement functor in the local model structure will not be a strict left adjoint. However, it will be a left $\infty$-adjoint. In fact, it's pretty straightforward to show (using the basic properites of left Bousfield localization) that for locally fibrant $Y$, we have a natural equivalence: $$\mathrm{Map}(R(X),Y)\simeq \mathrm{Map}(X,Y)\;$$
at the level of mapping spaces.

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