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Let $\pi _1:SS\to Grpd$ denote the fundamental groupoid functor, from simplicial sets to groupoids, and let $N:Grpd\to SS$ denote the nerve functor. Then $\pi _1$ is left adjoint to $N.$

On simplicial sets we consider the usual model category and on groupoids the model category structure inherited by the one given by Thomason. It is known that $\pi _1$ preserves weak equivalences, cofibrations, fibrations and colimits, and that $N$ preserves weak equivalences, fibrations and limits. From these facts one can easily deduce that $\pi _1$ preserves homotopy colimits and $N$ preserves homotopy limits. I wonder whether the following questions have positive answers:

  1. Does $N$ preserve homotopy colimits?
  2. Does $\pi _1$ preserve homotopy limits?
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  • $\begingroup$ The nerve functor preserves coproducts. By a result of Dwyer and Kan, the nerve functor preserves certain pushouts. Explicitly, for $G,H,K$ groupoids and for $f:G\to H$ and $g:G\to K$ groupoid homomorphisms, if (1) these groupoids have the same set of objects ($G_0=H_0=K_0$) and (2) both $f_0:G_0\to H_0$ and $g_0:G_0\to K_0$ is the identity morphism on objects, then $N(H\coprod_G K) \simeq NH\coprod_{NG} NK$. $\endgroup$
    – user2529
    Jul 13, 2014 at 15:29
  • $\begingroup$ The fundamental group preserves products. $\endgroup$
    – user2529
    Jul 13, 2014 at 15:29
  • $\begingroup$ Thanks for your answer, @colin. So if I understood well, as a consequence we have that the nerve functor preserves (homotopy) pushouts of groups, right? What is the reference for Dwyer and Kan's result? $\endgroup$
    – user55956
    Jul 13, 2014 at 18:32
  • $\begingroup$ @ColinTan Yes, but not all pushouts are homotopy pushouts. $\endgroup$
    – Zhen Lin
    Jul 13, 2014 at 19:39

3 Answers 3

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Just for fun, here's a purely abstract way of seeing neither of these can happen:

The functors you wrote down both preserve weak equivalences, so are morphisms of relative categories, and induce adjunctions of the associated $\left(\infty,1\right)$-categories, which are $\infty$-groupoids and $1$-groupoids (which is actually a $(2,1)$-category) respectively. The adjunction is precisely the one exhibiting $1$-groupoids as a reflective subcategory of $\infty$-groupoids. The full and faithful inclusion is modeled by $N$. The $\left(\infty,1\right)$-category of $\infty$-groupoids is freely generated by the (any) contractible $\infty$-groupoid (i.e. the one point space), (i.e. it's the free colimit cocompletion of the terminal category). Since the $\left(2,1\right)$-category of groupoids is cocomplete, and contains the terminal groupoid, if the inclusion preserved colimits, it would have to be essentially surjective:

If $X$ is any $\infty$-groupoid, you can write $X$ as the colimit of the terminal functor $X \to \infty\mbox{-}\mathbf{Gpd}$ (the constant functor with value the point) (this is basically just the Yoneda Lemma for $\infty$-categories), and this functor (obviously) factors through the inclusion $\mathbf{Gpd} \hookrightarrow \infty\mbox{-}\mathbf{Gpd}$.

If the left adjoint $\tau_1$ of this inclusion (which is modeled by $\Pi_1$) preserved even finite limits, then this would exhibit the $(2,1)$-category of groupoids as a left-exact localization of $\infty\mbox{-}\mathbf{Gpd}=\mathbf{Psh}_\infty\left(*\right),$ i.e. it would exhibit $\mathbf{Gpd}$ as an $\infty$-topos. This means there would exist a unique Grothendieck topology $J$ on the terminal category such that $\mathbf{Gpd}$ sat somewhere between $\mathbf{Sh}_\infty\left(*,J\right)$ and its hypercompletion $\mathbf{HSh}_\infty\left(*,J\right).$ But, there is a unique Grothendieck topology on the terminal category (the trivial one!), so we'd have $$\mathbf{HSh}_\infty\left(*,J\right)=\mathbf{Sh}_\infty\left(*,J\right)=\mathbf{Psh}_\infty\left(*\right).$$ So we'd have to have $\mathbf{Gpd}\simeq \mathbf{Psh}_\infty\left(*\right)=\infty\mbox{-}\mathbf{Gpd}$ which is clearly nonsense.

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    $\begingroup$ For another concrete example: let $S^1$ be the 1-groupoid which is the fundamental groupoid of the circle and form the homotopy pushout where both maps are $S^1 \to *$. If you form this pushout in 1-groupoids you get something equivalent to the terminal groupoid; if you do it in SS you get $S^2$. $\endgroup$ Jul 14, 2014 at 8:36
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The fundamental groupoid does not preserve homotopy limits. The simplest examples can be obtained from the path-space sequence $\Omega X\to \operatorname{pt}\to X$ which shows that $\Omega X$ is the homotopy limit of $\operatorname{pt}\to X\leftarrow \operatorname{pt}$. If $\pi_1$ preserves homotopy limits, then the fundamental groupoid of $\Omega X$ should be contractible for any simply-connected space $X$. But any abelian group can be realized as $\pi_2(X)$ for $X$ simply-connected, so the fundamental groupoid can not preserve homotopy limits.

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  • $\begingroup$ That's right, thank you! If we restrict to aspherical spaces, would we reach to the same conclusion? $\endgroup$
    – user55956
    Jul 13, 2014 at 18:45
  • $\begingroup$ The fundamental groupoid should preserve homotopy limits for aspherical spaces, I guess for the same reason as in the comment to Zhen Lin's answer. $\endgroup$ Jul 13, 2014 at 21:02
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The nerve functor does not preserve homotopy colimits. Indeed, take any simplicial set $X$ with non-trivial $\pi_n$ ($n > 1$) and consider $X$ as a simplicial diagram of sets. In $\mathbf{sSet}$, its homotopy colimit will be $X$ (up to weak equivalence, by the Bousfield–Kan theorem), but the nerve of any 1-groupoid has trivial $\pi_n$ ($n > 1$).

The fundamental groupoid functor does not preserve homotopy limits. As Matthias Wendt explained, the loop space can be computed by a homotopy limit, so if we choose a simplicial set $X$ with trivial $\pi_0$ and $\pi_1$ but non-trivial $\pi_2$, then the homotopy pullback diagram $$\begin{array}{ccc} \Omega X & \rightarrow & \Delta^0 \\ \downarrow & & \downarrow \\ \Delta^0 & \rightarrow & X \end{array}$$ cannot be preserved by the fundamental groupoid functor.

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  • $\begingroup$ Thank you! And if we restric to aspherical spaces? $\endgroup$
    – user55956
    Jul 13, 2014 at 18:49
  • $\begingroup$ I suppose you mean spaces with trivial $\pi_n$ for $n > 1$? Then yes, because 1-groupoids are a reflective $(\infty, 1)$-subcategory of $\infty$-groupoids. $\endgroup$
    – Zhen Lin
    Jul 13, 2014 at 19:44

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