Given a section \(\sigma:N\rightarrow P\) we produce a smooth map (trivialization) \(\Phi_P:N\times G\rightarrow P\) given by \((n,g)\mapsto \sigma(n)g\). This is smooth for obvious reasons. The map \(N\rightarrow P\) given by \(n\mapsto \sigma(n)\) is smooth so is the map \(N\times G\rightarrow P\times G\) given by \((n,g)\mapsto (\sigma(n),g)\). The multiplication map \(P\times G\rightarrow P\) given by \((p,g)\mapsto pg\) is smooth. Thus the composition \(N\times G\rightarrow P\times G\rightarrow P\) is smooth which is simply the map \(\Phi_P:N\times G\rightarrow P\) is smooth. We see that this map is a diffeomorphism. What obvious map can you think of \(P\rightarrow N\times G\)? Given \(p\in P\) we need to associate an element \((n,g)\in N\times G\). For first coordinate, obvious choice is \(\pi(p)\in N\). Remember that we are already with a guess that \(\Phi\) is a bijection and this map \(P\rightarrow N\times G\) has to be inverse of \(\Phi:N\times G\rightarrow P\). So, given \(p\in P\) we choose \(g\in G\) such that \(\Phi(\pi(p),g)=p\) i.e., \(\sigma(\pi(p)).g=p\). The point is, we can always choose such \(g\) and it is unique as action is free. See that \(\sigma(\pi(p))\in \pi^{-1}(\pi(p))\) and \(p\in \pi^{1}(p)\). So, as any two elements in fibre are related by an element in \(G\) we have \(g\in G\) such that \(\sigma(\pi(p)).g=p\). Thus, we have an obvious map \(P\rightarrow N\times G\) given by \(p\mapsto (\pi(p),g)\) where \(g\in G\) is the unique such \(g\) satisfying \(\sigma(\pi(p))g=p\). It is upto you to see that this map is a smooth map. This is smooth on first projection to \(N\) being just the map \(\pi\). It needs some work to see the projectionto \(G\) is smooth. It is by definition that this map is actually inverse of \(\Phi_P:N\times G\rightarrow P\) and thus we have a diffeomorphism. This diffeomorphism is \(G\)-equivariant if you know what it means. Thus, knowing that \(P\rightarrow N\) is a principal \(G\) bundle, a section \(\sigma:N\rightarrow P\) gives a trivialization \(N\times G\rightarrow P\). Given a trivialization \(N\times G\xrightarrow{\Phi} P\), we have a section \(\sigma:N\rightarrow P\) given by \(\sigma(n)=\Phi(n,1)\). Thus, giving a section is same thing as giving local trivialization.

Let \(G\) be a Lie group and \(\pi_P:P\rightarrow M, \pi_Q:Q\rightarrow M\) be principal \(G\) bundles. Then, any \(G\)-equivariant map \(f:P\rightarrow Q\) inducing identity on \(M\) is a diffeomorphism. The same holds when we have Lie groupoids instead of Lie groups. Let \(\mathcal{G}\) be a Lie groupoid and \(P\rightarrow M, Q\rightarrow M\) be principal \(\mathcal{G}\) bundles. Then, any \(\mathcal{G}\)-equivariant map \(f:P\rightarrow Q\) inducing identity on \(M\) is a diffeomorphism. Above result is very basic thing when defining a stack associated for a Lie groupoid \(\mathcal{G}\). Given a Lie groupoid \(\mathcal{G}\), we define a category fibered in groupoids \(B\mathcal{G}\rightarrow \text{Man}\) by associating for each manifold \(U\) a category \(B\mathcal{G}(U)\) whose objects are principal \(\mathcal{G}\) bundles whose base space is \(U\) i.e., of the form \(P\rightarrow U\) and morphism from an object \(P\rightarrow U\) to another object \(Q\rightarrow U\) is a \(\mathcal{G}\)-equivariant map \(P\rightarrow Q\) that induces \(Id:U\rightarrow U\) on base space of those principal bundles. Thus, to say \(B\mathcal{G}(U)\) is a Lie groupoid, we need to prove that every arrow \((P\rightarrow U)\rightarrow (Q\rightarrow U)\) is an isomorphism which is what we are trying to prove. Let us see the proof for the case of Lie groups. See the set up as following diagram. Let \(p,p'\in P\) are such that \(f(p)=f(p')\), thus, \(\pi_Q(f(p))=\pi_Q(f(p'))\). As \(\pi_Q\circ f=\pi_P\), we have \(\pi_P(p)=\pi_P(p')\) i.e., there exists \(g\in G\) such that \(p'=p.g\). Thus, \(f(p')=f(pg)\). As \(f\) is \(G\)-equivariant, we have \(f(pg)=f(p)g\). Thus, we have \(f(p')=f(p)g\). As the action of \(G\) on \(Q\) is free, \(f(p')=f(p),f(p')=f(p)g\) implies \(g=1\). Thus, \(p'=p\). So, \(f\) is one to one mapping. Let \(q\in Q\). We have \(\pi_Q(q)\in M\). As \(\pi_P\) is surjective, there exists \(p\in P\) such that \(\pi_P(p)=\pi_Q(q)\). As \(\pi_Q\circ f=\pi_P\), we have \(\pi_Q(f(p))=\pi_P(p)=\pi_Q(q)\). As \(\pi_Q(f(p))=\pi_Q(q)\), there exists \(g\in G\) such that \(f(p)g=q\). As \(f\) is \(G\)-equivariant, we have \(f(p)g=f(pg)\). Thus, we have \(q=f(pg)\) which implies that \(f\) is an onto mapping. Suppose that \(\pi_P:P\rightarrow M\) is trivial \(G\) bundle, not for simplicity but because every principal \(G\) bundle is locally trivial and diffeomorphism is something that needs to be checked locally. As \(\pi_P:P\rightarrow M\) is trivial, it has a global section for \(\pi_P\) i.e., a smooth map \(\sigma:M\rightarrow P\) such that \(\pi_P\circ \sigma=1\). This gives a trivialization \(M\times G\xrightarrow{\Phi} P\) i.e., an isomorphism. Consider the cimposition \(f\circ \pi:M\rightarrow Q\). This is again a smooth map such that \(\pi_Q\circ (f\circ \sigma)=(\pi_Q\circ f)\circ \sigma=\pi_P\circ \sigma=1\) ...

Let \(F:\mathcal{I}\rightarrow \mathcal{C}\) is a functor. This is also called as diagram indexed by \(\mathcal{I}\). By the Limit of this diagram, we mean an object (universal) \(L\) of \(\mathcal{C}\) and a collection of arrows (universal again) \(\pi_i:L\rightarrow F(i)\) such that, for each arrow \(m:i\rightarrow j\) in \(\mathcal{I}\) the following diagram is commutative. This is usually denoted by \(\varprojlim_{\mathcal{I}}F(i)\) or simply by \(\varprojlim_{\mathcal{I}}F\). Fixing an object \(X\) in \(\mathcal{C}\), I want to prove that \(\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i))) =\text{Hom}_{\mathcal{C}}(X,\varprojlim_{\mathcal{I}}F(i))\) ...

A morphism of stacks \(F:\mathcal{D}\rightarrow \mathcal{C}\) is said to be a gerbe over stack if following two conditions hold : Given a manifold \(U\) and an object \(\xi\in \mathcal{C}(U)\), there exists a covering \(\{U_i\rightarrow U\}\) (depending on the Grothendieck topology that we have fixed on the category \(Man\) of manifolds) and objects \(x_i\in \mathcal{D}(U_i)\) with an isomorphism \(F(x_i)\rightarrow \xi|_{U_i}\) for each \(i\). Given a manifold \(U\) and an arrow \(\xi\rightarrow \eta\) in \(\mathcal{C}(U)\), there exists a covering \(\{U_i\rightarrow U\}\) (depending on the Grothendieck topology that we have fixed on the category \(Man\) of manifolds) and arrows \(x_i\rightarrow y_i\) in \(\mathcal{D}(U_i)\) such that

Given a morphism of Lie groups \(\theta:G\rightarrow H\) and a principal \(G\) bundle \(\pi:P\rightarrow M\) there are (at least) two ways to assign a principal \(H\) bundle. See that the morphism of Lie groups \(\theta:G\rightarrow H\) gives an action of \(G\) on \(H\) by \(g.h=\theta(g).h\). Given an action of \(G\) on manifold (Lie group in this case) \(H\) there is an associated fibre bundle \(P\times_G H\rightarrow M\) with fibre \(H\). This gives a principal \(H\) bundle. For principal bundle \(\pi:P\rightarrow M\), we can find an open cover \(\{U_\alpha\}\) of \(M\) and (transition) maps \(g_\alpha g_\beta:U_{\alpha\beta}\rightarrow G\) satifsying the cocycle condition \(g_{\alpha\beta}g_{\beta\gamma}=g_{\alpha\gamma}\) on \(U_\alpha\cap U_\beta\cap U_\gamma\). Then the compositions \(\tau_{\alpha\beta}=\theta\circ g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G\rightarrow H\) also satifies the cocycle condition \(\tau_{\alpha\beta}\tau_{\beta\gamma}=\tau_{\alpha\gamma}\) on \(U_\alpha\cap U_\beta\cap U_\gamma\). One can then produce a principal \(H\) bundle over \(M\) given this open cover \(\{U_\alpha\}\) of \(M\) and smooth maps \(\tau_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow H\) satisfying the cocycle condition. This gives a principal \(H\) bundle. It is a good exercise (that I have not tried) to check that principal \(H\) bundles obtained from above two methods are (naturally) isomorphic i.e., one and the same. Given a Lie group \(G\), let \(BG\) denote the category of principal \(G\) bundles. Objects are principal \(G\) bundles and morphisms are \(G\)-equivariant morphisms. Given a morphism of Lie groups \(\theta:G\rightarrow H\), above construction gives a functor (at the level of objects) \(B\theta:BG\rightarrow BH\). It is not difficult to see that, a \(G\)-equivarint map induce a \(H\)-equivariant map. This gives a functor \(BG\rightarrow BH\).

This post is based on (wanted to write after reading) Lie Groupoids and Differentiable stacks by Matias L. del Hoyo. I would suggest this for any one who wants to know about Lie groupoids and Differentiable stacks. This is well written. By a manifold, we always mean a smooth manifold. A Groupoid is a category where every arrow is invertible. A Lie groupoid is a groupoid with additional smooth structures on object set/morphism set and maps between them. Definition : A Lie groupoid consists of a manifold \(\mathcal{G}_0\) of objects, a manifold \(\mathcal{G}_1\) of arrows and following maps : \(s:\mathcal{G}_1\rightarrow \mathcal{G}_0\), a submersion, called the source map. \(t:\mathcal{G}_1\rightarrow \mathcal{G}_0\), a submersion, called the target map. \(m:\mathcal{G}_1\times_{s,\mathcal{G}_0,t}\mathcal{G}_1\rightarrow \mathcal{G}_1\), a smooth map, called the multiplication map. \(u:\mathcal{G}_0\rightarrow \mathcal{G}_1\), a smooth map, called the unit map. \(i:\mathcal{G}_1\rightarrow \mathcal{G}_1\), a smooth map, called the inverse map. with some compatibility conditions. We denote this Lie groupoid by \(\mathcal{G}_1\rightrightarrows \mathcal{G}_0\). Definition : Let \(\mathcal{G}_1\rightrightarrows \mathcal{G}_0\) be a Lie groupoid and \(x\in \mathcal{G}_0\). The set \(s^{-1}(x)=:G(x,-)\) is called the \(s\)-fibre of \(x\) . The set \(t^{-1}(x)=:G(x,-)\) is called the \(s\)-fibre of \(x\) The set \(s^{-1}(x)\cap t^{-1}(x)=:G_x\) is called the Isotropy group of \(x\). The set \(t(s^{-1}(x))=\{y:x\rightarrow y\in \mathcal{G}_1\}=:O_x\) is called the orbit of \(x\). Proposition : Given a Lie groupoid \(\mathcal{G}_1\rightrightarrows \mathcal{G}_0\) and \(x,y\in \mathcal{G}_0\), the subset \(G(y,x)\subseteq G\) is an embedded submanifold. In particular, \(G_x\) is a Lie group. the subset \(O_x\) is a (may not be embedded) submanifold in a canonical way. By a morphism of Lie groupoids \(\phi: (\mathcal{G}_1\rightrightarrows \mathcal{G}_0)\rightarrow (\mathcal{H}_1\rightrightarrows \mathcal{H}_0)\) we mean a pair of smooth maps \(\phi^{ar}:\mathcal{G}_1\rightarrow \mathcal{H}_1\) and \(\phi^{ob}:\mathcal{G}_0\rightarrow \mathcal{H}_0\) compatible with structure maps \(s,t,m,u,i\). We write \(\phi\) for both \(\phi^{ar}\) and \(\phi^{ob}\).

Given a manifold \(M\) we have the concept of open cover of \(M\). We usually write an open cover of a manifold \(M\) as a collection of open subsets \(\{U_i\}\) (such that \(\bigcup U_i=M\)). In this note we see an open cover of \(M\) as a collection of maps (inclusions) \(\{U_i\rightarrow M\}\). Some properties of "open cover" are. (Pull back exists and gives an open cover) Suppose \(\{U_i\rightarrow M\}\) is an open cover for \(M\) and \(\pi:V\rightarrow M\) is a smooth map. Then, \(\{\pi^{-1}(U_i) \rightarrow V\}\) is a cover for \(V\). (Diffeomorphisms gives open cover) For any manifold \(M\), \(M\) itself is considered as an open cover \(\{M\rightarrow M\}\). More generally, for any diffeomorphism \(M'\rightarrow M\), \(\{M'\rightarrow M\}\) is considered as an open cover. (Open cover of open cover is an open cover) Let \(\{U_\alpha\rightarrow U\}\) be an open cover for \(U\) i.e., \(\bigcup_{\alpha} U_\alpha=U\). Suppose \(\{V_{\alpha\beta}\rightarrow U_\alpha\}\) is an open cover for \(U_\alpha\) for each \(\alpha\) i.e., \(\bigcup_{\beta}V_{\alpha\beta}=U_\alpha\). Then, \(\bigcup_{\alpha\beta}V_{\alpha\beta}=U\) i.e., \(\{V_{\alpha\beta}\rightarrow U\}\) is an open cover for \(U\). For a category \(\mathcal{C}\) and an object \(U\) of \(\mathcal{C}\), a collection of arrows \(\{U_i\rightarrow U\}\) is said to be a cover for \(U\). Definition : Let \(\mathcal{C}\) be a category. A Grothendieck topology on \(\mathcal{C}\) is given by a collection of covers \(\mathcal{W}=\{\{U_i\rightarrow U\}: U\in \mathcal{C}_0\}\) satisfying following conditions. (Pullbacks exists and gives a cover) Suppose \(\{U_i\rightarrow U\}\in \mathcal{W}\) and \(\pi:V\rightarrow U\) be an arrow. Then, the pull back \(U_i\times_UV\) exists (as an object in \(\mathcal{C}\)) and \(\{U_i\times_UV \rightarrow V\}\) is a cover for \(V\). (Isomorphisms gives an open cover) Suppose \(V\in \mathcal{C}_0\) and \(V\rightarrow U\) is an isomorphism in \(\mathcal{C}\) then, \(\{V\rightarrow U\}\in \mathcal{W}\). (cover of a cover is a cover) Suppose \(\{U_\alpha\rightarrow U\}\in \mathcal{W}\) and \(\{U_{\alpha\beta}\rightarrow U_\alpha\}\in \mathcal{W}\) for each \(\alpha\). Then, the collection of compositions \(\{U_{\alpha\beta}\rightarrow U_\alpha\rightarrow U\}\in \mathcal{W}\). To talk about a stack over category \(\mathcal{C}\) we fix a Grothendieck topology \(\mathcal{W}\) on \(\mathcal{C}\). When we say cover, we mean it belongs to \(\mathcal{W}\). Let \(\mathcal{D}\) be a category fibered in groupoids over \(\mathcal{C}\) i.e., we have a functor \(F:\mathcal{D}\rightarrow \mathcal{C}\) satisfying some conditions. Given an object \(U\) of \(\mathcal{C}\) we have what is called fibre of \(U\) in \(\mathcal{D}\) usually denoted by \(\mathcal{D}(U)\). Given an object \(U\) of \(\mathcal{C}\) and a cover \(\{U_i\rightarrow U\}\) (i.e., it belongs to \(\mathcal{W}\)) we have what is called descent category associated to the cover \(\{U_i\rightarrow U\}\), usually denoted by \(\mathcal{D}(\{U_i\rightarrow U\})\). There is an obvious functor \(\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_i\rightarrow U\})\). Definition : Let \(\mathcal{C}\) be a category with Grothendieck topology \(\mathcal{W}\). A category fibered in groupoids \(\mathcal{D}\rightarrow \mathcal{C}\) is said to be a stack over \(\mathcal{C}\) if, for every object \(U\) of \(\mathcal{C}\) and every cover \(\{U_i\rightarrow U\}\), the functor \(\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_i\rightarrow U\})\) is an equivalence of categories. The fibre categroy \(\mathcal{D}(U)\) is a category whose objects are that of \(\mathcal{D}\) which map to \(U\) under \(F\) i.e., \(\mathcal{D}(U)_0=\{V\in \mathcal{D}_0:F(V)=U\}\). ...

Here, I will add links to WordPress pages where I have written something about Stacks. Papers I am reading are Differentiable Stacks and Gerbes by Kai Behrend and Ping Xu. Orbifolds as Stacks by Eugene Lerman. Non abelian Differentiable Gerbes by Camille, Stienon and Ping Xu. -- --  

Definition : A stack \(\mathcal{D}\rightarrow \text{Man}\) is differentiable if there exists a manifold \(X\) with an atlas \(p:\underline{X}\rightarrow \mathcal{D}\) i.e., \(p\) is representable surjective submersion. We see a criterion for a map \(p:\underline{X}\rightarrow \mathcal{D}\) to be an atlas. By \(p:\underline{X}\rightarrow \mathcal{D}\) to be representable surjective submersion, we mean given a map of stacks \(\underline{Y}\rightarrow \mathcal{D}\) the fibered product \(\underline{X}\times_{\mathcal{D}}\underline{Y}\) is representable by a manifold and that the map of manifolds \(\underline{X}\times_{\mathcal{D}}\underline{Y}\rightarrow \underline{Y}\) is a surjective submersion. As \(\underline{X}\times_{\mathcal{D}}\underline{Y}\) is representable by a manifold for any map of stacks \(\underline{Y}\rightarrow \mathcal{D}\), in particular, taking \(\underline{Y}\rightarrow \mathcal{D}\) to be the same map \(\underline{X}\rightarrow \mathcal{D}\) we see that, in particular \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold. Remark : If \(p:\underline{X}\rightarrow \mathcal{D}\) is an atlas for \(\mathcal{D}\) then \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold. As \(\underline{X}\times_{\mathcal{D}}\underline{Y}\rightarrow \underline{Y}\) is a submersion for any map of stacks \(\underline{Y}\rightarrow \mathcal{D}\), in particular, taking \(\underline{Y}\rightarrow \mathcal{D}\) to be the same map \(\underline{X}\rightarrow \mathcal{D}\) we see that, projecion map \(\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) is a submersion. It is not relevant which projection is it as both maps are same. So, both projection maps \(pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) and \(pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) are submersions. Remark : If \(p:\underline{X}\rightarrow \mathcal{D}\) is an atlas for stack \(\mathcal{D}\) then projection maps \(pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) and \(pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) are submersions. As any representable surjective submersion is an epimorphism we have following remark. Remark : If \(p:\underline{X}\rightarrow \mathcal{D}\) is an atlas for \(\mathcal{D}\) then \(p:\underline{X}\rightarrow \mathcal{D}\) is an epimorphism. Combining all these remarks we have following remark. If \(p:\underline{X}\rightarrow \mathcal{D}\) is an atlas for \(\mathcal{D}\) then, \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold and projection maps \(pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) and \(pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) are submersions and \(p:\underline{X}\rightarrow \mathcal{D}\) is an epimorphism. It turns out that converse of above remark is true. Proposition : Let \(p:\underline{X}\rightarrow \mathcal{D}\) is a morphism of stacks such that \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold and projection maps \(pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) and \(pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) are submersions and that \(p:\underline{X}\rightarrow \mathcal{D}\) is an epimorphism. Then, Then, \(p:\underline{X}\rightarrow \mathcal{D}\) is a representable surjective submersion i.e., an atlas for \(\mathcal{D}\). Before we give proof of this, we recall a result. Lemma : Let \(\mathcal{D}\rightarrow\mathcal{C}\) be a morphism of stacks. Suppose \(U\) be a manifold and \(\underline{U}\rightarrow \mathcal{C}\) is an epimorphism of stacks such that fiber product \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\) is represented by a manifold and the map of manifolds \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U\) is a submersion. Then, \(\mathcal{D}\rightarrow \mathcal{C}\) is a representable submersion. To prove \(\underline{X}\rightarrow \mathcal{D}\) is a representable submersion, consider an epimorphism of stacks, namely \(\underline{X}\rightarrow \mathcal{D}\) (it is given to be an epimorphism, condition \(2\) above). See that the fibre product \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold (it is given in condition \(1\) above) and that the projection map \(\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow X\) is a submersion (it is in condition \(1\) above). Thus, by above lemma, \(p:\underline{X}\rightarrow \mathcal{D}\) is a representable submersion. Note that, a representable submersion that is an epimorphism is a representable surjective submersion. Thus, \(p:\underline{X}\rightarrow \mathcal{D}\) is an atlas for \(\mathcal{D}\). So, we have the following result. Proposition : Let \(p:\underline{X}\rightarrow \mathcal{D}\) is a morphism of stacks such that \(\underline{X}\times_{\mathcal{D}}\underline{X}\) is representable by a manifold and projection maps \(pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) and \(pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}\) are submersions and that \(p:\underline{X}\rightarrow \mathcal{D}\) is an epimorphism. Then, Then, \(p:\underline{X}\rightarrow \mathcal{D}\) is a representable surjective submersion i.e., an atlas for \(\mathcal{D}\).