# Counter-example to the existence of left Bousfield localization of combinatorial model category

Is there any known example of a combinatorial model category $$C$$ together with a set of map $$S$$ such that the left Bousefield localization of $$C$$ at $$S$$ does not exists ?

It is well known to exists when $$C$$ is left proper, and it seems that it also always exists as a left semi-model structure, but I don't known if there is any concrete example where it is known to not be a Quillen model structure.

PS: I technically already asked this question a year ago but it was mixed with other related questions and this part was not answered, so I thought it was best to ask it again as a separate question.

• Is there a standard example where $S$ is a class? – Tim Campion Mar 13 at 21:18
• @TimCampion : good question. The left Bousfield localization at $S$ of a category where only iso are weak equivalence exists if and only if the subcategory of objects orthogonal to $S$ is reflective. it seems to me that this is not always the case when $S$ is a class, and that there might be known counter example to this ? but I haven't really thought about it. – Simon Henry Mar 13 at 21:32
• Ah yes -- the statement that every orthogonality class is reflective is equivalent to weak Vopenka's principle -- this is 6.24 and 6.25 in Adamek and Rosicky. Example 6.25 is an example of an orthogonal subcategory in a locally presentable category which is not reflective (under the negation of weak Vopenka's principle), which I suppose answers your question in a rather artificial way. – Tim Campion Mar 13 at 21:59
• I see that Casacuberta and Chorny showed all Bousfield localizations exist in a left proper, combinatorial, simplicial model category. Is the "simplicial" condition removed somewhere? – Tim Campion Mar 13 at 22:03
• @TimCampion : I thought it was in Hirschhorn's book, but he does it for "cellular" model categories instead of combinatorial. I just found the statement in Barwick "On left and right model categories and left and right bousfield localizations" as theorem 4.7. He attributed the results to J.Smith. I thought it was more "well known" than that. But maybe I missed a more classical reference. – Simon Henry Mar 13 at 22:21

Here, we know that a counterexample must fail to be left proper, so start with a diagram$$\require{AMScd}$$ $$\begin{CD} a @>\sim>> b\\ @VVV @VVV\\ c @>>> d \end{CD}$$ in which $$a \to b$$ is a weak equivalence, $$a \to c$$ is a cofibration, but $$c \to d$$ is not a weak equivalence. Then $$a \to c$$ also cannot be a weak equivalence (otherwise $$b \to d$$ would be one too). Since $$a \to c$$ and $$c \to d$$ are not weak equivalences, they must be both cofibrations and fibrations and therefore the same is true of $$a \to d$$. Then $$a \to d$$ cannot be a weak equivalence (or it would be an isomorphism), so $$b \to d$$ is also not a weak equivalence, and therefore is a fibration too. In summary, all the maps are fibrations and $$a \to c$$, $$b \to d$$, $$c \to d$$ are cofibrations while $$a \to b$$ is a weak equivalence. One can check that this does in fact yield a model category structure (probably the easiest way is to verify that the (acyclic) cofibrations/fibrations are closed under composition and pushout/pullback, and that the factorization axioms hold).
Now, let's try to form the left Bousfield localization at the map $$a \to c$$, which is already a cofibration between cofibrant objects. All objects are fibrant in the original structure, and the local objects are the ones which have the same maps from $$a$$ and from $$c$$, which are the objects $$c$$ and $$d$$. The map $$c \to d$$ was not a weak equivalence originally, so it has to still not be one in the localization. However, making $$a \to c$$ a weak equivalence also makes $$b \to d$$ a weak equivalence because it is the pushout of the acyclic cofibration $$a \to c$$, which contradicts two-out-of-three.