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This question is somehow related to my question at https://math.stackexchange.com/questions/683915/derived-pseudo-functor and to the question here: A homotopy commutative diagram that cannot be strictified .

Consider a model category $M$. Let us assume that, for all small category $S$, we can endow the category $Fun (S, M) $ with the injective model structure.

So, considering a given small category $S$, we got the notion of a homotopy limit $ Ho(Fun (S, M))\to Ho(M) $.

My first question is: which would be the correct notion of homotopy limit $ Fun (S, Ho(M))\to Ho(M) $? The second question is: how can I study the existence of such a homotopy limit?

I know that if the diagonal functor $ \Delta : Ho (M)\to Fun (S, Ho(M)) $ has a right adjoint, then the answer is clear. But, also, I do know that this situation is rare for model categories $ M $ (considering $S$ being not discrete). However, in the general situation, I guess that the correct notion of a such homotopy limit is the following:

Consider the projections $\lambda : M\to Ho(M) $ and $ \alpha : Fun(S, M)\to Ho (Fun (S, M)) $. The functor $ Fun ( S, \lambda ) $ has a total derived functor $ i: Ho ( Fun (S, M) ) \to Fun (S , Ho (M) ) $. I would define the homotopy limit $ Fun ( S, Ho(M)) $ as being a right Kan extension of the usual homotopy limit $ Ho(Fun (S, M))\to Ho (M) $ along thar derived functor $ i $.

I guess that, one example of such a homotopy limit is the 2 (weak) - limit (homotopy limit) of a pseudo functor $ A: S\to Cat $.

But, if this is the correct notion of such a homotopy limit, my question remains: how can I study the existence of that right Kan extension? Where can I find about this kind of homotopy limit?

Thank you in advance

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  • $\begingroup$ I don't understand what you mean by homotopy limit $Fun(S,Ho(M))\to Ho(M))$. Do you mean a right adjoint of the diagonal functor ? Then this is just a limit. The homotopy category of any cofibrantly generated model category is weakly complete and weakly cocomplete, i.e. limits and colimits exist but are not unique. This is explained in M. Hovey's book Model Categories. $\endgroup$
    – Philippe Gaucher
    Commented Feb 21, 2014 at 14:04
  • $\begingroup$ @PhilippeGaucher I don't mean the right adjoint of the diagonal functor. I'm trying to deal with the case in which the right adjoint to the diagonal functor doesn't exist. And, in this case, I would define the homotopy limit as the Kan extension I defined above. In almost all the cases I know, $Ho (M) $ is not complete or cocomplete. $\endgroup$
    – Fernando
    Commented Feb 21, 2014 at 14:15
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    $\begingroup$ Answering 1), there is no such a correct notion. $\endgroup$
    – Fernando Muro
    Commented Feb 21, 2014 at 15:09
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    $\begingroup$ I mean, it would be a great surprise for me (I dare say for many) if such a good notion existed at all. I don't think it's tivial but unfeasible, but I may be wrong! $\endgroup$
    – Fernando Muro
    Commented Feb 21, 2014 at 17:29
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    $\begingroup$ I don't think it's trivial in that sense, but honestly I don't know. $\endgroup$
    – Fernando Muro
    Commented Feb 22, 2014 at 1:31

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