# Localization, model categories, right transfer

Suppose that we have a locally presentable category $$M$$ and $$N$$ is a locally presentable full subcategory of $$M$$. Both categories are complete and cocomplete. Lets suppose that we have an adjunction $$F:M\leftrightarrow N:i$$ where $$i$$ is the embedding functor and $$(i \circ F)^{2}=i \circ F=L$$. If $$N$$ is a combinatorial model category, could we define a model structure on $$M$$ such that $$a\rightarrow b$$ is a weak equivalence (cofibration) if and only if $$F(a)\rightarrow F(b)$$ is a weak equivalence (cofibration) in $$N$$?

Edit: In case it is not true in general, we assume one more condition, there exists a combinatorial left proper model structure on $$M$$ such that $$F:M\leftrightarrow N:i$$ is a Quillen adjunction.

The statement is unfortunately not true in full generality, I will give a counter example at the end.

What I know is that:

• One in general gets a "Right semi-model structure" on $$M$$ with the properties you want (see below)

• If one assumes the localization is left exact then one gets a Quillen model structure on $$M$$ with the properties you want.

Also note that the additional assumption you propose in your edit is always satisfied: you can always put on $$M$$ the model structure where only isomorphisms are cofibrations and every morphism is a trivial fibration. Then any left adjoint functor out of $$M$$ is a left Quillen functor. But one has:

• If there is a model structure on $$M$$ such that the adjunction $$F: M \leftrightarrows N : i$$ is Quillen, and either every object in $$M$$ is cofibrant, or more generally $$F$$ sends weak equivalences to weak equivalences, then the model structure you want exists on $$M$$.

The general case:

As usual, I'm defining weak equivalences, cofibrations and trivial cofibrations as the morphisms whose image by the localizations are cofibrations/weak equivalences/trivial cofibrations. "Fibrations" and "trivial fibrations" are defined by the right lifting property against trivial cofibrations and cofibrations.

The key observation is the following:

Let $$v:X \rightarrow Y$$ be a morphism in $$M$$ that is inverted by the localization $$F:M \rightarrow N$$. Then by definition, $$v$$ is a trivial cofibration.

Consider now the relative codiagonal morphisms $$\nabla_v : Y \coprod_X Y \rightarrow Y$$

Then, as the localization preserve colimits, the image of $$\nabla_v$$ by the localization is also an isomorphism, hence $$\nabla_v$$ is also a trivial cofibration.

It follows that any fibration has the unique lifting property against all morphisms inverted by the localization. The existence of the lifting follows from the fact that $$v$$ is a trivial cofibration, the uniqueness from the fact that $$\nabla_v$$ is a trivial cofibration.

In particular the fibrant objects of $$M$$ are exactly the objects that are in (the image of) $$N$$ and fibrant as objects of $$N$$.

It immediately follows that any trivial fibration with fibrant target is in fact a weak equivalence: it is in the image of the reflection, by the observation above, and is a trivial fibration there.

This is sufficient to get a "Right semi-model structure" with the desired properties on $$M$$, i.e. something similar to a model structure but the identification of 'trivial fibration' with things that are fibrations and weak equivalences only holds for arrows with fibrant target. See for example C.Barwick "On left and right model categories and left and right Bousfield localizations".

Note: this is proved exactly as the transfer theorem mentioned by D.White: you get the existence of weak factorization system by a general theorem of left transfer of weak factorization system, we proved that trivial fibration where fibration and equivalences and the converse holds by the classical retract argument. If you are willing to slightly weaken further the definition of "right semi-model structure" by only asking the factorization to exists for arrows with fibrant target, you don't even need to use the theorem about transfer of factorization system, and you can get rid of the assumption of presentability.

The left exact case:

If one further assume that the localization $$M \rightarrow N$$ is left exact, then $$M$$ is endowed with a unique factorization system where every arrow is factored into: an arrow that is inverted in the localization, followed by an arrow which is the pullback of its image by the reflection (along the unit).

In particular, any fibration is the pullback of its image by the reflection.

Given a general arrow $$f:X \rightarrow Y$$ one factors it as $$X \rightarrow P \rightarrow Y$$ where $$P \rightarrow Y$$ is the pullback of the image of $$f$$ by the reflection. $$X \rightarrow P$$ is inverted by the localization hence is a trivial cofibration. If one factors the image of $$f$$ by the reflection $$L$$ into a (trivial) cofibration followed by a (trivial) fibrations (in $$N$$), then the pullback of this factorization to $$Y$$ is also a (trivial) cofibration followed by a (trivial) fibration.

Hence every arrow factors as a (trivial) cofibration followed by a (trivial) fibration whose image by $$L$$ is also a (trivial) fibration.

The usual retract lemma shows that the image of every (trivial) fibration by $$L$$ is again a (trivial) fibration.

In particular any trivial fibration is a fibration and any fibration that is an equivalence is a trivial fibration.

Hence one also gets a model structure.

Note that in this case one has constructed the factorization in $$M$$ explicitly from the factorization in $$N$$, so one does not need the presentability assumption at all.

The case of a "cofibrant" model structure on $$M$$:

Here I'm assuming there is a first model structure on $$M$$ such that the adjunction $$M \leftrightarrows N$$ is Quillen, and either that every object is cofibrant or that $$F$$ send equivalences to equivalences. As left Quillen functors send weak equivalences between cofibrant objects to weak equivalences it is enough to work under the second assumption.

Here every cofibration in $$M$$ is a cofibration for the transferred model structure. It follows that every trivial fibration of the transferred model structure is a trivial fibration, hence a weak equivalence, for the model structure of $$M$$. In particular, its image by $$F$$ is a weak equivalence, hence it is a weak equivalence for the transferred model structure, and hence one can conclude by the transfer theorem.

The counter-example:

Let $$G^+$$ be the category of oriented graphs where every vertices has a unique self loop. I claim that there is a model structure on $$G^+$$ whose equivalences are the bijection on $$\pi_0$$ and the cofibrations are the morphisms which are monic on vertices.

Let $$G$$ be the category of all oriented graphs. $$G^+$$ is a reflective full subcategory of $$G$$. The cofibrations (resp. weak equivalences) in $$G$$ for the transferred model structure are still the monomorphisms on vertices (resp. the bijections on $$\pi_0$$).

Now, the map from the two point graph (with no arrows) to the one point graph (with no arrows) has the right lifting property against all cofibrations, but is not an equivalence.

• This is a fantastic answer! Also, I love the counterexample. I will add it to my mental list of things that are semi-model categories but provably not model categories. A few such examples have appeared in my papers, but this one is so clean and elegant! In this example, what are the fibrant objects? I would tend to assume the one point graph would be fibrant (by way of analogy to many other examples). But, if so, then wouldn't the last line imply it's not even a right semi-model structure? Mar 7, 2019 at 18:24
• Thanks ! The fibrant objects are the graphs where there is at most one arrow from $x$ to $y$ for each $x,y$, and the existance of an arrow from $x$ to $y$ is an equivalence relation. The one point graph is hence not fibrant, but the one point graph with a loop is (well it is terminal so this not very surprising). Examples of combinatorial right semi-model structure that are provably not Quillen model structure are actually a lot more common than for their "left" cousins. I guess this is because trivial fibrations are in general easier to understand than trivial cofibration. (...) Mar 7, 2019 at 19:15
• I work out some examples in my paper Arxiv 1807.02650, including a variant of this model structure on graph with cofibrant being the mono and fibrant object the "setoids" and a semi-simplicial version of Kan-Quillen model structures. Mar 7, 2019 at 19:18
• Very nice! As far as I understand, the left exactness is sufficient condition, do you think that if $F$ is cartesian it is also a sufficient condition ?
– num
Mar 7, 2019 at 19:58
• @num : If by cartesian you just mean preserve products, then I don't think so, but I don't think I know a counter examples. Mar 7, 2019 at 20:10

The question of when $$i$$ induces a model structure from $$N$$ to $$M$$ has been well-studied in recent years (it is called a left-induced model structure). A necessary and sufficient condition was given in this paper, which I also summarize here. This condition is summarized very clearly in this recent paper, where it is called the left acyclicity condition.

Let $$\mathcal{C}$$ (resp. $$\mathcal{W}$$) be the classes of maps $$f$$ in $$M$$ such that $$i^{-1}(f)$$ is a cofibration (resp. weak equivalence) in $$N$$. The left acyclicity condition states that all maps with the right lifting property with respect to $$i^{-1}\mathcal{C}$$ must be weak equivalences (i.e. must lie in $$i^{-1}\mathcal{W}$$). This condition is difficult to verify. A number of examples appear in the linked papers above, and I verified several more in this joint paper with Donald Yau. You are asking if the condition comes for free when $$i\circ F$$ is a localization. This is not immediately obvious to me, and I'm in the middle of a heavy teaching week, so might not have time to think about it till the weekend. I wanted to share the equivalent formulation in case it helps in the meanwhile.

• Interesting, I took a look to the references, It seems to me that the condition (in the cited paper) does not come for free when $i\circ F$ is a localization. I think my question has a negative answer in general... Maybe I'm wrong.
– num
Mar 6, 2019 at 21:52