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By https://arxiv.org/abs/1406.4419 (The fundamental groupoid as a terminal costack, Ilia Pirashvili), we know that for a topological space $X$, the $2$-functor $$\text{Top}(X)\rightarrow \text{Gpd}, \quad (U\rightarrow X)\mapsto \Pi_1(U)$$ is a $2$-cosheaf, in fact the terminal one. In particular, it follows that $$\text{Top}(X)^{op}\rightarrow \text{Gpd}, \quad (U\rightarrow X)\mapsto \text{Hom}(\Pi_1(U),\mathcal{G})$$ is a $2$-sheaf, where $\mathcal{G}$ is a groupoid. However, if we assume $X$ is a manifold (probably locally simply connected does the job, but I want to be cautious), then for any point $x\in X$, open simply connected sets $U\ni x$ are final in all open neighborhoods of $x$. Hence the stalk of this $2$-sheaf is $$\text{colim}_{U\ni x}\text{Hom}(\Pi_1(U),\mathcal{G})\cong \text{colim}_{U\ni x,\text{ simp. conn}}\text{Hom}(\Pi_1(U),\mathcal{G})\cong \text{colim}_{U\ni x,\text{ simp. conn}}\text{Hom}(\{x\},\mathcal{G})$$ where I denote by $\{x\}$ the groupoid whose objects are $\{x\}$ and has no automorphism (I think this is called the trivial groupoid). If we assume that $\mathcal{G}$ is a group $G$, then we get that the stalk consits only of the trival morphism.

However, having trivial stalks means that $\text{Hom}(\Pi_1(-),G)$ is trivial $2$-sheaf. But surely this can not be true, as that would mean $$\text{Hom}(\Pi_1(X),\text{Gl}_n(\mathbb{C})$$ is trivial, which is not true. Hence there is a gap/flaw in my reasoning, but I don't see it.

Basically, as $\Pi_1(-)$ is the terminal $2$-cosheaf, $\text{Hom}(\Pi_1(-),\mathcal{G})$ is the initial $2$-sheaf, but that would mean that there are almost no representations of the fundamental group, which seems wrong to me.

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  • $\begingroup$ Do you mean $\hom$ with values in $Gpd$ or with values in $Set$ ? (note, by the way, that you never use, in your "proof", that $\Pi_1(-)$ is the terminal $2$-cosheaf) $\endgroup$ Commented Sep 23, 2020 at 19:40
  • $\begingroup$ I mean $\text{hom}$ in $Gpd$ to get the sheaf property. $\endgroup$ Commented Sep 23, 2020 at 19:45
  • $\begingroup$ But then $\hom(\{x\},\mathcal G)\cong \mathcal G$. So you probably want $\hom$ to take values in $Set$, so the thing you're after is the fact that $\Pi_1(-)$ is a $1$-cosheaf (which is also claimed in the paper) $\endgroup$ Commented Sep 23, 2020 at 19:52
  • $\begingroup$ Shouldn't $\{x\}$ be the initial groupoid? So in partiuclar $\text{hom}(\{x\},\mathcal{G})$ would be a singleton as a set. $\endgroup$ Commented Sep 23, 2020 at 20:01
  • $\begingroup$ So you're looking at $\hom$ valued in $Set$, which is what I first asked (it's not the initial groupoid, it's the terminal one -in particular $\hom(\{x\}, \mathcal G) \cong Ob(\mathcal G)$ in general, although that doesn't change your question when $\mathcal G= BG$ has only one object) $\endgroup$ Commented Sep 23, 2020 at 20:03

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First of all, note that you haven't used the fact that $\Pi_1(-)$ was the terminal $2$-cosheaf, just that it was a ($2$-)cosheaf.

Then, as I pointed out in the comments, there's a question of whether you're considering the usual $\hom$ functor into $Set$, or the internal $\hom$ with values in $Gpd$.

  • If you're considering the usual, external $\hom$, then the question is more about whether $\Pi_1(-)$ is a cosheaf than a $2$-cosheaf. But note that if you're considering $\hom$ into $Set$, then $\hom$ is not equivalence invariant. In particular, $\hom(\Pi_1(U),\mathcal G) \not\cong \hom(\{x\},\mathcal G)$ even when $U$ is simply-connected. For instance if you take for $\mathcal G$ an indiscrete groupoid on a set $S$ of objects, the former is $\hom_{Set}(U,S)$, while the latter is $\hom_{Set}(\{x\},S)$. So your stalk computation doesn't work anymore.

(to perhaps get a better grasp on this, compare $\hom(2^{in},BG)$ and $\hom(\{x\},BG)$ where $2^{in}$ is the indiscrete groupoid on $2$ objects, and $BG$ the one-object groupoid associated to the group $G$)

  • If you're considering the internal $\hom$, and looking at stacks/$2$-sheaves (so that you do have invariance under equivalence), then $\hom(\{x\},\mathcal G)\cong \mathcal G$, so it's definitely nontrivial.

(moreover, note that in this context of stacks, or $2$-sheaves, you need to be more careful with colimits: this would probably be a $2$-colimit, so you'd need to check that the subcategory of simply-connected neighbourhoods of $x$ is final in the $2$-categorical sense in the category of neighbourhoods. This is a rather stronger condition than just finality in the $1$-categorical sense, and here I'm not sure whether it would be satisfied in general - I'm definitely not an expert on stacks and $2$-categories so I can't say much more)

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    $\begingroup$ Directed sets have the wonderful property that cofinal = homotopy cofinal, so it isn't a problem. $\endgroup$
    – Zhen Lin
    Commented Sep 23, 2020 at 22:17
  • $\begingroup$ I'm still confused by your statement that $hom_{gpd}(\{x\},\mathcal{G})\cong \mathcal{G}$. To define a functor $\{x\}\rightarrow \mathcal{G}$ corresponds to a choice of $\mathcal{G}$, so therefore shouldn't $hom_{gpd}(\{x\},\mathcal{G})\cong \text{Ob}(\mathcal{G})$? $\endgroup$ Commented Sep 23, 2020 at 23:59
  • $\begingroup$ You should read more carefully : there are two flavours of $\hom$, one which is set-valued (for this one, you are correct, and I explain in the first bullet point why the argument doesn't work for this one), and one which is groupoid-valued : this is a different functor $\endgroup$ Commented Sep 24, 2020 at 7:09
  • $\begingroup$ @ZhenLin : the poset of simply-connected neighbourhoods if $x$ isn't directed in general, is it ? Ah maybe up to a cofinal subset, in manifolds ? $\endgroup$ Commented Sep 24, 2020 at 7:36
  • $\begingroup$ The poset of open neighbourhoods, ordered by reverse inclusion, is directed. The simply connected open neighbourhoods are cofinal in this poset, for a manifold or more generally a locally simply connected space. $\endgroup$
    – Zhen Lin
    Commented Sep 24, 2020 at 7:55

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