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I am exploring a bit the world of groupoids. What I have in mind is that infinity groupoids correspond to spaces. So my first question is the following:

  1. Consider the model category $\infty-Grpd$ of infinity groupoids, meant as the full subcategory of Kan complexes in $sSet$, with Quillen model structure. On the other side, consider the model category $Top$ with the Quillen model structure.

Is it true that the two model categories are Quillen equivalent under Sing and geometric realization?

  1. In case of 1, the Quillen equivalence restricts to 1-connected groupoids and yields 1-connected spaces. This is kind of tautological because I define homotopy groups of a grpd as the homotopy groups of its geometric realization. The real question is: does the functors (Nerve, homotopy category) induce a Quillen equivalence between classical groupoids (with the model structure induced by the classical one on categories) and 1-connected infinity groupoids ?

It would help maybe if you could provide references, many thanks!

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    $\begingroup$ Kan complexes do not have finite (co)limits, so cannot have a model structure. Simplicial sets do have a model structure, whose fibrant objects are Kan complexes. Furthermore, |−| and Sing form a Quillen equivalence. This answers (1). $\endgroup$
    – Dmitri Pavlov
    Commented Jan 14, 2020 at 15:49
  • $\begingroup$ If 1-connected really means 1-truncated (e.g., 2-coskeletal simplicial sets), then indeed the fundamental groupoid functor and the nerve functor induce a Quillen equivalence between 2-coskeletal simplicial sets and groupoids. This answers (2). $\endgroup$
    – Dmitri Pavlov
    Commented Jan 14, 2020 at 15:50
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    $\begingroup$ I am curious though as to what source on model categories does not mention the Quillen equivalence between sSet and Top. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 14, 2020 at 15:53
  • $\begingroup$ Yes,Indeed your first comment answers me. I knew about the classical Quillen equivalence, and it seemed straightforward to me that it restricts to Kan complexes (as sing has image in Kan complexes ). But as you pointed out the problem is that they do not have colimits. 1 connected here means that the homotopy groups of the geometric realization are trivial for n greater than 2. Not sure if the 2-coskeletization of a 1-comnected sset is weakly equivalent to himself, which would conclude. $\endgroup$
    – Andrea Marino
    Commented Jan 14, 2020 at 16:41
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    $\begingroup$ @AndreaMarino That's what's commonly known as 1-truncated, not 1-connected (that'd be $\pi_0=\pi_1=\ast$) $\endgroup$
    – Denis Nardin
    Commented Jan 14, 2020 at 16:45

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