Given a model category $\mathcal{M}$, Goerss and Hopkins constructed a subcategory (see Structured Ring Spectra, p. 160) $\mathbf{E}$ of $\mathcal{M}$ such that:

  • If $X\in\mathbf{E}$ and $Y$ is weakly equivalent to $X$, then $Y\in\mathbf{E}$.

  • A morphism $f$ in $\mathcal{M}$ is in $\mathbf{E}$ if and only if it is a weak equivalence.

If $\mathbf{E}$ is the category whose objects are $\mathcal{M}$ and morphisms are the weak equivalences, then $\mathbf{E}$ is written $\mathbf{E}(\mathcal{M})$. My question is as follows: is it possible to recover the model category $\mathcal{M}$ from $\mathbf{E}(\mathcal{M})$? Is there any nontrivial model structure on $\mathcal{M}$ that is useful/interesting?


migrated from mathoverflow.net Feb 3 '15 at 23:52

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    $\begingroup$ What do you mean by your last sentence? M comes with a model structure... In general I think not, since you can consider a model structure on a complete/cocomplete category where isomorphisms are the weak equivalences, and so E is just the core (assuming we take all the objects of M). There's no way to recover a category from its core. $\endgroup$ – David Roberts Feb 3 '15 at 4:51
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    $\begingroup$ Part of the point of model categories is to endow a category with weak equivalences with enough structure to make localization by those weak equivalences manageable. So there's something nontrivial going on, and it is not hard to write down categories with weak equivalences with horrible localizations (which therefore don't extend to model categories). $\endgroup$ – Achim Krause Feb 3 '15 at 7:30
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    $\begingroup$ Supposing that you remember at least the underlying category $\mathcal M$, which is a must, the answer is even no, since there are well known different model structures in the same category with the same weak equivalences (e.g. chain complexes + quasi-isomorphisms). (Co)fibrations are just a way of constructing resolutions in order to compute derived functors, and we know that this can be done in several different ways, in general. $\endgroup$ – Fernando Muro Feb 3 '15 at 7:32
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    $\begingroup$ I'm voting to close this question as off-topic because Fernando's comment already provides an answer. $\endgroup$ – David White Feb 3 '15 at 13:46
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    $\begingroup$ I wouldn't say the question is stupid, just not for this forum since the answer is well documented in standard books on model categories (e.g. Hovey's). It would be a very good question for the math stack exchange, and explaining explicit examples would be pertinent there. $\endgroup$ – Fernando Muro Feb 3 '15 at 15:49

A long comment: When you have a model category $\mathcal{M}$ in particular you have an enriched $Ho(sSet)$-module structure on $Ho(\mathcal{M})$ (in the sense of Hovey) i.e. the homotopy category $Ho(\mathcal{M})$ is tensored and cotensored over $Ho(sSet)$ in a compatible way. Let me use the following analogy: suppose that $R$ is a ring and $M$ is an abelian group and imagine you have an action of some elements of $R$ on some elements of $M$ and you are asking the question wether this action is a restriction of honest $R$-module structure on $M$. So your question (if I understand it) should be formulated as follows:

Suppose that we have a (co)complete category $\mathcal{M}$ and a subcategory $ \mathbf{E}(\mathcal{M})$ such that the category $\mathcal{M}[\mathbf{E}(\mathcal{M})^{-1}] $ exists , does it come from a model structure on $\mathcal{M}$ such that $Ho(\mathcal{M})\simeq \mathcal{M}[\mathbf{E}(\mathcal{M})^{-1}]$?


1) In order to have a model structure on $\mathcal{M}$ whit a subcategory $ \mathbf{E}(\mathcal{M})$ of weak equivalences such that $\mathcal{M}[\mathbf{E}(\mathcal{M})^{-1}] $ exists a necessary condition is to have an enriched $Ho(sSet)$-module structure on $\mathcal{M}[\mathbf{E}(\mathcal{M})^{-1}]$.

2) The Homotopy category $Ho(\mathcal{M})$ depends on the class of weak equivalences but the additional structure of enriched $Ho(sSet)$-module structure on $Ho(\mathcal{M})$ is given by the class of cofibrations (fibrations).


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