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Let $M$ and $N$ two very nice simplicial model categories and let $F:N\rightarrow M$ be a (nice) simplicial functor which induces an equivalence of homotopy categories, i.e. $Ho(F): Ho(N)\rightarrow Ho(M)$ is an equivalence of homotopy categories and it is well defined. We define the category $\pi_{0}M$ as the category having the same objects of $M$ and $Hom_{\pi_{0}M}(a,b)=\pi_{0}Map_{M}(a,b)$.

What can we say about the functor $\pi_{0}F:\pi_{0}N\rightarrow \pi_{0}M$? is it an equivalence of categories. Maybe we should assume that $F$ takes cofibrant-fibrant objects to cofibrant-fibrant objects... or something close to that.

Clarification: A simplicial functor between simplicial model categories $M$ and $N$ is a simplicial enriched functor between simplicial enriched categories.

Clarification II: A simplicial model category is a model category $M$ tensored and cotensored over the model category of simplicial sets in a way compatible with model structure of both model structure on $M$ and on simplicial sets.

Edit I think there is a counterexample: Let $M=N$ be the standard model category of simplicial sets, and let $F: sSet\rightarrow sSet$ be the functor of fibrant replacement such that it is simpllicial (I guess such functor do exist). Then $\pi_{0}F$ is not an equivalence of categories, while $Ho(F)$ is. I'm I right ?

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  • $\begingroup$ How are you defining $\mathrm{Map}$? In general you'll need to take some kind of (co)fibrant replacement of the arguments to get the correct homotopy type of the mapping space... $\endgroup$ – Denis Nardin Sep 8 '18 at 15:01
  • $\begingroup$ @DenisNardin M is a simplicial model category, $Map$ is given enrichment by definition. Do I misunderstand your question ? $\endgroup$ – Amadeus Sep 8 '18 at 15:05
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    $\begingroup$ I think the clarification for $Map$ comes from different authors using it for different things, for example, simplicial enrichment, homotopy function complexes, etc etc $\endgroup$ – Charles Sep 8 '18 at 15:24
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    $\begingroup$ It would be helpful if some context was provided as to the origin of this question. Why would we want to consider π_0(M) in the first place? This is a fairly atypical thing to do to a simplicial model category. $\endgroup$ – Dmitri Pavlov Sep 9 '18 at 14:53
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    $\begingroup$ @DmitriPavlov You are right, I should not use the term homotopy equivalences for these maps. However, what I meant by "closure by transitivity" is that when one wants to compute actual sets of homotopy classes of maps between two spaces, one needs to quotient out the set of maps by the homotopy relation. And to do so, a closure by transitivity is needed unless both spaces are bifibrant. Furthermore, this set of homotopy classes is exactly $\pi_0Map_M(a,b)$, and even though it does not correspond to the Homset in the homotopy category, there is still a well defined map $[X,Y]\to Hom_{Ho}(X,Y)$ $\endgroup$ – S. Douteau Sep 12 '18 at 8:28
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Since there is an answer to the question, I think I should write it down.

There is a simple counterexample to my question: Let $M=N=sSet$ the standard model category of simplicial sets. Let $ex^{\infty}:sSet\rightarrow sSet$ the fibrant replacement functor. It is simplicial as it was noticed in the comments. $Ho(F)$ is well defined and induce an (auto)equivalence of the homotopy category $Ho(sSet)$. On the other hand $\pi_{0}(F)$ is clearly not an autoequivalence of $\pi_{0}sSet$.

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