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Let $\mathcal{G}=(A\rightrightarrows X)$ be a groupoid. Here $X={\rm Ob}(\mathcal{G})$, $A={\rm Ar}(\mathcal{G})$, and we have 5 maps: $s,t\colon A\to X$ (the source and the target, surjective), $m\colon A\times_X A\to A$ (multiplication of composable arrows), ${\rm id}\colon X\to A$ ($x\mapsto{\rm id}_x$, injective), and $i\colon A\to A$ ($a\mapsto a^{-1}$), satisfying the usual axioms. I say that my groupoid is connected if for any two objects $x,y\in X$ there exists an arrow $a\colon x\to y$.

Assume that a finite group $\Gamma$ acts on $\mathcal{G}$, i.e., it acts on $X$ and $A$ such all the 5 maps are $\Gamma$-equivariant. We say that $\mathcal{G}$ is a $\Gamma$-groupoid.

Now I want to construct a fibered category (gerbe) $\mathbb{G}$ over the category (site) of finite $\Gamma$-sets, starting from a connected $\Gamma$-groupoid $\mathcal{G}$. In other words, for any finite $\Gamma$-set $S$, I want to construct a groupoid $\mathbb{G}(S)$, and for a morphism $S\to T$ of finite $\Gamma$-sets, I want to define a restriction functor $\mathbb{G}(T)\to \mathbb{G}(S)$. How can I do that? I could not find this in Giraud's book.

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If you don't give more precision on what you want, it seems to me that the simplest things work. (taking $\mathbb{G}(S)$ to be the groupoid whose object are $\Gamma$-equivariant map from $S$ to $X$ and whose arrow are $\Gamma$-equivariant map from $S$ to $A$, the five strucural map being simply composition, and the functoriality being also given by composition... But this is generally not what we want to do. –  Simon Henry Apr 14 '13 at 14:34
    
@Simon: What you propose looks fine. Why is it not what we want to do?! –  Mikhail Borovoi Apr 14 '13 at 15:54
    
@Simon: I want to construct a gerbe starting from a connected $\Gamma$-groupoid, and to describe the cohomology class of this gerbe in terms of my $\Gamma$-groupoid, thereby explicitly relating the paper of Springer on non-abelian $H^2$ in Galois cohomology with the book by Giraud. –  Mikhail Borovoi Apr 14 '13 at 15:56
    
One problem with this construction is that : start with X being any $\Gamma$-Set, you can construct a groupoid with $A = X \times X$, this groupoid is suppose to be seen as a trivial one, hence we expect the associated stacks to be also trivial. Be if $X$ doesn't have any $\Gamma$ fixed point this is not the case (for example, $\mathcal{G}({*})$ won't have any point). –  Simon Henry Apr 14 '13 at 21:24
    
What's the topology on the category of finite $\Gamma$-sets? Is it just surjections? –  David Roberts Apr 15 '13 at 6:19

2 Answers 2

up vote 3 down vote accepted

The thing that Simon gives in his first comment is just a prestack, it needs to be stackified. You can do this by letting $\mathbb{G}(S)$ be the groupoid of principal $\mathcal{G}$-bundles in $\Gamma Set$, in other words, $\Gamma$-equivariant $\mathcal{G}$-bundles in Set.

In more detail, I'm assuming you are considering the topology on $\Gamma Set$ to be the one where a covering family is a surjective map. Then we define a principal $\mathcal{G}$-bundle over a $\Gamma$-set $S$ to be a surjective map of $\Gamma$-sets (all maps from now on will be $\Gamma$-equivariant), $\pi\colon P\to S$, a map $b\colon P\to X$ and an action map $a\colon P\times_{X,s} A \to P$ such that $\pi(a(p,f)) = \pi(p)$ (the action is fibrewise); in addition, we demand that the map $P\times_X A \to P\times_S P$ is an isomorphism.

Additionally, we demand there is a cover $j\colon T\to S$ and a map $\sigma\colon T \to P$ such that the pulled back bundle $T\times_S P \to T$ is isomorphic to the pullback of the canonical $\mathcal{G}$-bundle $ t\colon A \to X$ (the action is by composition in the groupoid) along $b\circ \sigma\colon T \to P \to X$

The groupoid $\mathbb{G}(S)$ is that of principal $\Gamma$-bundles with bundle maps between them (they preserve fibres and actions, including the data of the map to $X$). The functor $\mathbb{G}(S) \to \mathbb{G}(T)$ is given by pullback.

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I do not understand the line: "isomorphic to $T\times_X S$ for some map $T\to X$". What is the map $S\to X$ in the fibered product? Is it a typo? –  Mikhail Borovoi Apr 15 '13 at 10:31
    
Yes, it's a typo. I'm going to improve the post.. –  David Roberts Apr 15 '13 at 10:45
    
Probably when you write "such that $\pi(a(p,f))=\pi(a)$", you mean $\pi(a(p,f))=\pi(p)$. When you write "in addition, we demand that the map $\dots$ is an isomorphism", you mean a bijection. Is this correct? –  Mikhail Borovoi Apr 15 '13 at 11:49
    
Whoops, you're right. And isomorphisms are bijections in this case. –  David Roberts Apr 15 '13 at 22:51
    
@DavidRoberts: Thank you for your detailed answer. This is exactly the kind of answer that I wanted to get! –  Mikhail Borovoi Apr 16 '13 at 17:24

David Roberts already gave a nice answer, which you accepted. However, let me give you another perspective. Your groupoid $\mathcal{G}$ has an action of $\Gamma,$ i.e. is a groupoid object in $Set^{B\Gamma}\cong \Gamma-Set$ (where $B\Gamma$ is $\Gamma$ viewed as a one object category). So $$\mathcal{G} \in Gpd\left(Set^{B\Gamma}\right) \cong Gpd^{B\Gamma}.$$ I.e., your groupoid $\mathcal{G}$ is the same data as a funtor $\mathbb{G}:B\Gamma \to Gpd$. Explicitly, it sends the one object to the underlying groupoid in sets of $\Gamma,$ and $g \in \Gamma$ gets sent to the functor $\mathcal{G} \to \mathcal{G}$ induced by the action.

Now, you wanted to arrive at a gerbe over the large site $\Gamma-Set$ with surjections as covers. Well, this is the canonical topology on the topos $\Gamma-Set$, and sheaves over $\Gamma-Set$ with respect to this topology are equivalent to $\Gamma-Set$ itself, so, stacks over this site, are equivalent to the bicategory of (weak) functors $B\Gamma \to Gpd$, of which $\mathbb{G}$ is an example. So it corresponds canonically to a stack on $\Gamma-Set$. Explicitly, given a $\Gamma$-set $X,$ $X$ may viewed as a functor $X:B\Gamma \to Set$ and hence also as a functor $X^{id}:\Gamma \to Gpd,$ where we view a set as a groupoid with all identity arrows. Then the stack on $\Gamma-Set$ that $\mathbb{G}$ corresponds to, sends $X$ to the groupoid of maps $Hom\left(X^{id},\mathbb{G}\right).$ Provided that $\pi_0\mathcal{G}=*$ this stack is a gerbe.

If you prefer to get this stack as a fibered category, there is another approach. Consider the Grothendieck construction of $\mathbb{G}$ $$\pi:\int_{B\Gamma} \mathbb{G} \to B\Gamma.$$ It can be canonically identified with the action groupoid $\Gamma \ltimes \mathcal{G} \to B\Gamma.$ (See e.g. the "generalized action groupoid" construction in http://arxiv.org/abs/1011.6070, or see the homework assignment I gave my topos theory class: http://people.mpim-bonn.mpg.de/carchedi/HW1.pdf). Anyway, this is a fibered category over $\Gamma$ describing the gerbe associated to $\mathcal{G}.$ The fibered category over $\Gamma-Set$ it corresponds to is "sheaves over $\int_{B\Gamma} \mathbb{G}$ with the induced Grothendieck topology" which is easily seen to be the same as $Set^{\Gamma \ltimes \mathcal{G}},$ (since the topology becomes the canonical topology again) which becomes a fibered category over $Set^{\Gamma}$ via the functor $$\pi_!:Set^{\Gamma \ltimes \mathcal{G}} \to Set^{\Gamma},$$ where $\pi_!$ is left adjoint to the restriction functor.

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Nice! [padding characters] –  David Roberts Apr 16 '13 at 4:46
    
@DavidCarchedi: Thank you! However, I am not quite familiar with your notations $Set^{B\Gamma}$, $Gpd(Set^{B\Gamma})$ and $Gpd^{B\Gamma}$. Your links did not help me. Could you please explain me these notations? Also, what is a weak functor? Where can I read about all this stuff? –  Mikhail Borovoi Apr 16 '13 at 17:37
    
@Mikhail: $Set^{B\Gamma}$ is the category of functors from the one-object groupoid with arrows $\Gamma$ to $Set$; it is equivalent to the category of $\Gamma$-sets. $Gpd(Set^{B\Gamma})$ is the 2-category of groupoid objects in this category and $Gpd^{B\Gamma}$ is the 2-category of (I guess strict) functors $B\Gamma \to Gpd$. When David says weak functor, it's the same thing as a pseudofunctor. –  David Roberts Apr 17 '13 at 0:57
    
@DavidRoberts: Thank you! –  Mikhail Borovoi Apr 17 '13 at 20:44

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