Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Thanks.

share|cite|improve this question
3  
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
2  
...or you can just say everything using the language of $\infty$-categories. – Aaron Mazel-Gee Nov 3 '12 at 5:22

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.

share|cite|improve this answer

Although most model structures appearing "in practice" do not have a reflective subcategory of fibrant objects (unless every object is fibrant), the following result of A. Salch provides many examples that do.

Theorem. Let $\mathcal{C}$ be a complete and cocomplete category, and let $\mathcal{A}$ be a full subcategory which is reflective and replete (i.e., closed under isomorphisms). Then there exists a model structure on $\mathcal{C}$ such that $\mathcal{A}$ consists of the fibrant objects.

See Corollary 6.3 here:

http://arxiv.org/abs/1501.00508v1

The model structure is obtained as a (left) Bousfield localization of the discrete model structure on $\mathcal{C}$. The weak equivalences are the maps that become isomorphisms upon applying the reflector $\mathcal{C} \to \mathcal{A}$.

share|cite|improve this answer

The 1-categorical question is well-defined, and its answer is no, in general. I will provide examples below.

Lemma 1. Let $\eta \colon X \to F$ be a map from $X$ to a fibrant object, initial among such maps. Then $\eta$ is a trivial cofibration. In particular, $F$ is a fibrant replacement of $X$.

Proof. $\require{AMScd}$ Factor $\eta$ as $\eta = pi$, where $i \colon X \to Z$ is a trivial cofibration and $p \colon Z \to F$ is a fibration. In particular, $Z$ is fibrant. By the universal property of $\eta$, there exists a (unique) map $\phi \colon F \to Z$ satisfying $i = \phi \eta$. The equality $p \phi \eta = p i = \eta$ then implies $p \phi = 1_F$, again by the universal property of $\eta$. The commutative diagram \begin{CD} X @= X @= X\\@V{\eta}VV @V{i}VV @V{\eta}VV\\ F @>{\phi}>> Z @>{p}>> F \end{CD} exhibits $\eta$ as a retract of $i$, hence a trivial cofibration. $\blacksquare$

Next, I will describe a model category in which the full subcategory of cofibrant objects is not coreflective. Taking the opposite model category, this provides a negative answer to the original question.

Let $R$ be a ring and consider $\mathrm{Ch}_{\geq 0}(R)$ the category of non-negatively graded chain complexes of (left) $R$-modules, equipped with the projective model structure. More precisely, the weak equivalences are quasi-isomorphisms, the fibrations are epimorphisms in positive degrees, and the cofibrations are degreewise monomorphisms with projective cokernel. In particular, a complex is cofibrant if and only if it is degreewise projective.

Observation 2. Let $M$ be an $R$-module, viewed as a complex concentrated in degree $0$. Then a chain map $f \colon C \to M$ is zero if and only if it induces the zero map on homology.

Proposition 3. Let $\epsilon \colon P \to M$ be a map from a cofibrant object to $M$, terminal among such maps. Then the following holds.

  1. Let $Q$ be a cofibrant chain complex and $g \colon Q \to P$ a chain map satisfying $\epsilon g = 0 \colon Q \to M$. Then $g$ is the zero map.
  2. The chain complex $P$ has trivial differential. In particular, its homology $H_n P = P_n$ is projective for every $n \geq 0$.
  3. $M$ is projective.

Proof. (1) follows from the universal property of $\epsilon$ and the equality $\epsilon g = 0 = \epsilon 0$. For (2), consider the chain complex $P_n[n-1]$ concentrated in degree $n-1$. The differential $d_n \colon P_n \to P_{n-1}$ defines a chain map $d_n \colon P_n[n-1] \to P$ which is null-homotopic. By Observation 2, we have $\epsilon d_n = 0$, and thus $d_n=0$. Now (3) follows from the isomorphism $H_0 P \cong M$, using Lemma 1. $\blacksquare$

Corollary 4. The full subcategory of cofibrant objects in $\mathrm{Ch}_{\geq 0}(R)$ is coreflective if and only if every $R$-module is projective (in which case every object of $\mathrm{Ch}_{\geq 0}(R)$ is cofibrant).

share|cite|improve this answer

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.

share|cite|improve this answer
2  
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.