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When can I localise a model category by a set(or class) of morphisms, and how do I describe the localised model category ?

By 'localise' I mean to find a localisation functor $q_S : M \rightarrow M_S$ with the usual universality property: that is, $q_S(s)$ is a weak equivalence for any $s\in S$, and any functor $q:M \rightarrow M'$ with this property factors through $q_S$. Importantly, $M_S$ and $M$ may have different underlying categories and so it is not necessarily a Bousfield localisation.

The model category (defined here) I am interested in, is very degenerate: its underlying category is a partially ordered set (or class).

flag	asked Apr 27 2011 at 12:13 mmm 772●3●8

	If I'm not mistaken, it's essentially a sort of set-theoretic condition on the class of morphisms we want to localize at? That is, we basically take (homotopy) colimits over acyclics w.r.t. to our class W of morphisms, and if that class is generated under homotopy colimits by a set, then our localization exists. – Jon Beardsley Feb 10 2012 at 21:40
	I believe your universal property will force the underlying categories to be equal. I mean, you could take $M'$ to be the model structure on $M$ where the weak equivalences are the isomorphisms and all maps are both fibrations and cofibrations. – David White Nov 10 at 0:22
	If my comment above is correct, then the Bousfield localization (if it exists) will be the localization in your sense, by its well-known universal property. However, your type should always exist (because the class of $M'$ is nonempty, so you can just take the one "closest" to $M$) even if Bousfield localization does not. But because you haven't required the cofibrations in $M$ and $M_S$ to match up, I'm not sure if yours is good for anything. Still, +1 since it got me thinking. – David White Nov 10 at 0:28

“non-Bousfield” localisations of model categories

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