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Let $H$ be a set with a binary operation $\cdot _H$ on it. To show that it is a group, one has to show that $\cdot _H$ is associative, find an identity element in $H$, and so forth; it might take some work. However, if we knew in advance that $H$ is a sub-set of some group $G$, with $\cdot _H$ induced from the multiplication of $G$, then it would be much easier to check that $H$ is indeed a group.

In a similar manner, suppose that $\mathcal N$ is a subcategory of a model category $\mathcal M$, and consider $\mathcal N$ with the structure induced from $\mathcal M$. Is there any condition that allows one to check easily if $\mathcal N$ with the induced structure is indeed a model category?


As a motivating example, consider the following. Let $\kappa$ be some infinite cardinal. Denote by $\mathcal S$ the full subcategory category of $sSet$, of simplicial sets which have at most $\kappa$ many nondegenerate simplicies, all of which are elements in some fixed large enough "universe" set $\mathcal U$. $\mathcal U$ be chosen such that $\mathcal S$ contains also the generating cofibrations $\partial\Delta\left[n\right]\hookrightarrow\Delta\left[n\right]$ and the generating trivial cofibrations $\Lambda^{k}\left[n\right]\hookrightarrow\Delta\left[n\right]$ of $sSet$.

$\mathcal S$ is a small subcategory of $sSet$, hence cannot have all small (co)limits (because it's not a preorder). However, if one is willing to compromise the standard limits axiom with some weaker version of it, then perhaps $\mathcal S$ could be given a model category structure, one that is induced from the model structure on $sSet$? In particular, it would seem very appealing if $\mathcal S$ could be cofibrantly generated, with the same sets of generating (trivial) cofibrations.

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Sure, the transfer theorem, it involves the existence of a left adjoint to the inclusion, $\mathcal M$ being cofibrantly generated, and some other hypothesis. See any modern text on model categories (Hirschhorn, Hovey...) – Fernando Muro Oct 4 '12 at 9:45
In a little more detail, the requirement that the inclusion is a Quillen functor is analogous to saying that $\cdot_H$ is induced from the multiplication in $G$, except that a Quillen functor is more complicated and involves extra structure. – David Roberts Oct 4 '12 at 10:52
Yes, I see. Thank you. However, the solution that the two of you offer requires the existence of a reflector. What if there is none? – Shlomi A Oct 4 '12 at 19:40
That's, I think, the best answer to your question with the currently available techniques. If your problem doesn't fit into the hypothesis of the transfer theorem you will probably have to develop new techniques. It would help that you motivated your question with the real example you're thinking of. – Fernando Muro Oct 5 '12 at 6:45
@Fernando, I've added my motivating example. Thanks. Following the techniques offered, it does not seem reasonable to me that the subcategroy $\mathcal S$ would be a reflective one. – Shlomi A Oct 5 '12 at 10:54

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