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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.

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    $\begingroup$ Is there a standard example where $S$ is a class? $\endgroup$ – Tim Campion Mar 13 at 21:18
  • $\begingroup$ @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. $\endgroup$ – Simon Henry Mar 13 at 21:32
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    $\begingroup$ 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. $\endgroup$ – Tim Campion Mar 13 at 21:59
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    $\begingroup$ 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? $\endgroup$ – Tim Campion Mar 13 at 22:03
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    $\begingroup$ @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. $\endgroup$ – Simon Henry Mar 13 at 22:21
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A surprisingly effective way to construct counterexamples in model category theory is to just write down all the objects and morphisms involved and try to give the resulting (finite!) diagram the structure of a model category.

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.

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    $\begingroup$ Woa ! This is a very nice example ! I was going to say that it is not combinatorial... but it actually is $\endgroup$ – Simon Henry Mar 13 at 22:48

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