There is a notion of Gerbe over a Manifold and a notion of Gerbe over a stack. Given a manifold $M$, there is a way to associate a stack $\underline{M}$ with it and this gives an embedding of category of smooth manifolds into category of Categories fibered in groupoids which can be found in Orbifolds as Stacks.

This gives a hope that notion of gerbe on a manifold $M$ has to be some special case of notion of gerbe on stack $\underline{M}$. This question is about the difficulty I faced when realize that.

I have written definition of Gerbe over manifold and Gerbe over Stack in comments (hoping the question looks smaller with out the definitions).

The problem I am looking at is :

Is a gerbe over a manifold a special case of a gerbe over a stack? In particular, if $\mathfrak{X}=\underline{M}$, does a gerbe over $\mathfrak{X}=\underline{M}$ give a gerbe on manifold $M$? If so, How? How is the main question here.

Has any one looked at this before?

My previous question says the definition of gerbe over stack mean that

Given a manifold $U$, there exists an open cover $\{U_\alpha\rightarrow U\}$ such that the restriction functor $\mathcal{D}(U)\rightarrow \mathfrak{X}(U)$ has following properties

- given an object $x\in \mathfrak{X}(U)$ there is an object $y_\alpha\in \mathcal{D}(U_\alpha)$ and an isomorphism $F(y_\alpha)\rightarrow x|_{U_\alpha}$ for each index $\alpha$.
- given an arrow $\phi:a\rightarrow b$ in $\mathfrak{X}(U)$ there exists an arrow $\tau_\alpha:y_\alpha\rightarrow z_\alpha$ in $\mathcal{D}(U_\alpha)$ such that $F(\tau_\alpha)=\phi_\alpha$ for each index $\alpha$.

I start with a stack $\mathcal{D}\rightarrow \mathfrak{X}=\underline{M}$ and construct a gerbe over manifold $M$. I need to declare what is $\mathbb{G}(U)$ for each open set $U$ of $M$. I define that to be $\mathcal{D}(U)$ i.e., $\mathbb{G}(U)=\mathcal{D}(U)$.

**motivation for declaring like that is first condition says that object set of $\mathcal{D}(U_\alpha)$ is non empty and second condition says that morphism set of $\mathcal{D}(U_\alpha)$ is non empty if I can produce some objects and morphisms in $\mathcal{C}(U_\alpha)=\underline{M}(U_\alpha)$.**

I have to check that it is locally non empty and locally connected.

I was able to see that this makes $\mathcal{G}$ to be locally non empty stack. Proof I have written in comments.

Let $y,y'\in \mathbb{G}(U)=\mathfrak{R}(U)$. Then, we have $F(y),F(y')\in \mathfrak{X}(U)=\underline{M}(U)$. Assuming there is an arrow $F(y)\rightarrow F(y')$ in $\mathfrak{X}(U)$, second condition in my answer says that there is an arrow $y\rightarrow y'$ (locally atleast) in $\mathcal{D}(U)=\mathbb{G}(U)$ that maps to the arrow $F(y)\rightarrow F(y')$. This says $\mathbb{G}(U)$ is locally connected.

Thus, we have a Gerbe on manifold $M$ given a gerbe over stack $\underline{M}$ taking something for granted.

We have $F(x),F(y)\in \mathfrak{X}(U)=\underline{M}(U)$ i.e., $F(x):U\rightarrow M$ and $F(y):U\rightarrow M$ (smooth maps). An arrow $F(x)\rightarrow F(y)$ is a smooth map $\psi:U\rightarrow U$ such that $F(y)\circ \psi=F(x)$. One can quickly realize that this is not possible always (for example when $F(x)$ is constant map $p\in M$ and $F(y)$ is constant map $q\in M$ with $p\neq q$). Thus, **this idea does not work**.

We redefine $\mathbb{G}(U)$ as a category whose objects are objects of $\mathcal{D}(U)$ for which $F(y)$ is a diffeomorphism onto $U$. We have $$\mathbb{G}(U)=\{y\in \mathcal{D}(U): F(y)\text{ is a diffeomorphism onto } U\}.$$ Clearly, the $y$ we have choosen (locally) as preimage of inclusion map $i:U\rightarrow M$ is in $\mathbb{G}(U)$. Now, for $F(y),F(y)'\in \mathcal{C}(U)$, define $\Phi=F(y')^{-1}\circ F(y)$. Then, $F(y')\circ \Phi=F(y)$ which means there is an arrow $F(y)\rightarrow F(y')$. This solves the problem. Given $y,y'\in \mathcal{D}(U)$ there is an arrow $\phi:F(y)\rightarrow F(y')$ in $\mathcal{C}(U)$. As we are in case of gerbe, there exists arrow $\tau_{\alpha}:y|_{U_\alpha}\rightarrow y'|_{U_\alpha}$ such that $F(\tau_{\alpha})=\phi|_{U_\alpha}$, which in particular say that given $y,y'\in \mathbb{G}(U)$, they are locally connected by arrows $y|_{U_\alpha}\rightarrow y'|_{U_\alpha}$ which says $\mathbb{G}$ defines a gerbe on manifold $M$.

Is my construction correct? Can it be made more natural? For me, it was most reasonable thing to do. Any comments are welcome.

given an object $x\in \mathfrak{X}(U)$ there is an object $y\in \mathcal{D}(U)$ and an isomorphism $F(y)\rightarrow x$. As $\mathfrak{X}=\underline{M}$, $\mathfrak{X}(U)$ is just the collection of smooth maps $U\rightarrow M$. There is an obvious map we can go for, i.e., the inclusion map $x=i:U\rightarrow M$. For this, there is an object $y\in \mathcal{D}(U)$ and an isomorphism $F(y)\rightarrow x$. This in particular says that $\mathcal{D}(U)\neq \emptyset$. $\endgroup$ – Praphulla Koushik Aug 3 '18 at 16:58Locally non emptyi.e., i.e., given $x\in X$ there is an object/open set $U$ containing $x$ such that $\mathbb{G}(U)$ is non empty. 2.Locally connectedi.e., given $a,b\in\mathbb{G}(U)$ and $x\in U$ there exists an open subset $V$ of $U$ containing $x$ such that $a|_V$ is isomorphic to $b|_V$. The definition is from arxiv.org/pdf/math/0212266.pdf $\endgroup$ – Praphulla Koushik Aug 3 '18 at 17:16$\mathcal{D}\rightarrow \mathfrak{X}$ is an epimorphismand 2.$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathfrak{X}}\mathcal{D}$ is an epimorphism. The definition is from arxiv.org/pdf/math/0605694.pdf $\endgroup$ – Praphulla Koushik Aug 3 '18 at 17:18