# Understanding definition of gerbe over a stack

I am reading Differentiable Stacks and Gerbes by Kai Behrend and Ping Xu.

They define gerbe over a stack as follows.

Let $\mathfrak{X}$ be a differentiable stack. An $\mathfrak{S}$-stack $\mathfrak{R}$ endowed with a morphism $F:\mathfrak{R}\rightarrow \mathfrak{X}$ is called a gerbe over $\mathfrak{X}$ if both $\mathfrak{R}\rightarrow \mathfrak{X}$ and $\mathfrak{R}\rightarrow \mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R}$ are epimorphisms.

I am having trouble realizing the map $\mathfrak{R}\rightarrow \mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R}$. Some people call this diagonal map with out even mentioning the map explicitly.

I will recall the definition of $2$-fibered product

Let $$\pi_X:X\rightarrow \mathcal{C}, \pi_Y:Y\rightarrow \mathcal{C},\pi_Z:Z\rightarrow \mathcal{C}$$ be categories fibered in groupoids and $$f:Y\rightarrow X, g:Z\rightarrow X$$ be morphisms of categories fibered in groupoids.

The $2$-fibered product $Z\times_XY$ is the category whose objects are given by $$(Z\times_XY)_0=\{(y,z,\alpha)\in Y_0\times Z_0 \times X_1 : \pi_Y(y)=\pi_Z(z), \alpha:f(y)\rightarrow g(z)\}.$$ $$\text{Hom}_{Z\times_X Y}((y,z,\alpha),(y',z',\alpha'))= \left\{(u:y\rightarrow y',v:z\rightarrow z') |\alpha'\circ f(u)=g(v)\circ \alpha \right\}$$

Is the map $\mathfrak{R}\rightarrow \mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R}$ is just given by

• $\mathfrak{R}_0\rightarrow (\mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R})_0$ with $y\mapsto (y,y,id:F(y)\rightarrow F(y))$.
• $\mathfrak{R}_1\rightarrow (\mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R})_1$ with $(y\xrightarrow{\tau}z)\mapsto (y\xrightarrow{\tau} z,y\xrightarrow{\tau}z)$??

I am not very sure if I understand this correctly. Can some one please confirm this? What does this map and the original map $F:\mathfrak{R}\rightarrow \mathfrak{X}$ being epimorphism signifies?

• If R→X is an epimorphism, then each fiber F_x is nonempty. Then if R→R ×_X R is an epimorphism, this means that π_0(F_x) is a point, i.e., F_x is (noncanonically) equivalent to the stack BG for some smooth group G. But a gerbe over X is precisely a fiber bundle whose typical fiber is noncanonically equivalent to BG. – Dmitri Pavlov Jul 29 '18 at 19:57
• Can you please make it as an answer.. it is little difficult to read here @DmitriPavlov – Praphulla Koushik Jul 29 '18 at 20:34
• One way of seeing this (and I apologize if this misses the OP's point or is too narrow) is that $F:R\to X$ is an epimorphism if $F$ is locally surjective on objects and $R\to R_XR$ is an epimorphism if $F$ is locally surjective on morphisms. So gerbe means: locally surjective on objects and morphisms. – inkspot Jul 30 '18 at 21:55
• @inkspot I am looking for something like that(but did not knew what to look for).. please make it as an answer... this is clearly reasonable thing one can expect.. there is a notion of gerbe on a space which is a stack that is locally non empty and locally connected.. here also it is something similar.. I will try to figure out how they are related.. you please add your comment as answer.. – Praphulla Koushik Jul 31 '18 at 1:59
• @inkspot I have added some justification for your statements.. would you like to check it... – Praphulla Koushik Aug 2 '18 at 4:36

If R→X is an epimorphism, then each fiber F_x is nonempty. Then if R→R ×_X R is an epimorphism, this means that π_0(F_x) is a point (otherwise π_0(F_x)→π_0(F_x) × π_0(F_x) cannot be an epimorphism), i.e., F_x is (noncanonically) equivalent to the stack BG for some smooth group G. But a gerbe over X is precisely a fiber bundle whose typical fiber is noncanonically equivalent to BG.

• Latex please... – Praphulla Koushik Jul 29 '18 at 20:56
• @PraphullaKoushik: Sorry, I normally do not use MathJax in my posts because I need to retain the copy-paste capability. I do not think that using underscores for subscripts presents any substantial difficulty. – Dmitri Pavlov Jul 29 '18 at 21:23
• Ok. No problem :) My first question is to check if my understanding of diagonal map is correct or not... I am not sure if I understand your notation $\pi_0(F_x)$.. What does that mean? Can you please give reference for "But a gerbe over $X$ is precisely a fiber bundle whose typical fiber is noncanonically equivalent to $BG$." – Praphulla Koushik Jul 30 '18 at 6:40
• Excuse me...... – Praphulla Koushik Jul 30 '18 at 16:54
• @PraphullaKoushik: π_0(X) is defined for any stack X by defining π_0(X)(S) to be π_0(X(S)), where (S) denotes the category/groupoid of S-points of a stack. – Dmitri Pavlov Jul 30 '18 at 16:55

I am trying to write down what does it mean to say those two maps $$\mathcal{D}\rightarrow \mathcal{C}$$ and $$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathcal{C}}\mathcal{D}$$ to be epimorphisms.

I am not very comfortable to use the definition of epimorphism as in Differentiable Stacks and Gerbes.

I use the definition of epimorphism as in Principal actions of stacky Lie groupoids. I am hoping they are equivalent.

A morphism $$F: \mathcal{X}\rightarrow \mathcal{Y}$$ of categories fibered in groupoids is said to be an epimorphism if, given a manifold $$U$$, the restriction functor $$\mathcal{X}_U\rightarrow \mathcal{Y}_U$$ is almost essentially surjective i.e., given $$y\in \mathcal{Y}_U$$, there exists an open cover $$\{U_\alpha\rightarrow U\}$$ such that, there exists $$x_\alpha\in \mathcal{X}_{U_\alpha}$$ and isomorphisms $$F(x_\alpha)\rightarrow y|_{U_\alpha}$$ in $$\mathcal{Y}_{U_{\alpha}}$$.

Let $$U$$ be a smooth manifold. As $$F:\mathfrak{R}\rightarrow \mathfrak{X}$$ is an epimorphism, the functor $$\mathfrak{R}(U)\rightarrow \mathfrak{X}(U)$$ is almost essentially surjective. For simplicity, I assume it is essentially surjetcive i.e., given $$x\in \mathfrak{X}(U)$$ there exists $$r\in \mathfrak{R}(U)$$ such that there is an isomorphism $$F(r)\rightarrow x$$.

Is this what you(@inkspot) mean when you say locally surjective on objects??

As $$G:\mathfrak{R}\rightarrow \mathfrak{R}\times _{\mathfrak{X}}\mathfrak{R}$$ is an epimorphism, the functor $$\mathfrak{R}(U)\rightarrow (\mathfrak{R}\times _{\mathfrak{X}}\mathfrak{R})(U)$$ is almost essentially surjective. For simplicity, I assume it is essentially surjective i.e., given an element $$(y,z,\alpha)\in (\mathfrak{R}\times _{\mathfrak{X}}\mathfrak{R})(U)$$ there exists $$x\in \mathfrak{R}(U)$$ and an isomorphism $$G(x)\rightarrow (y,z,\alpha)$$. Here, $$\alpha:F(y)\rightarrow F(z)$$.

Now, what does it have anything to do with "locally surjective on morphisms"??

Let $$a\rightarrow b$$ be an arrow in $$\mathfrak{X}(U)$$. As we have seen above, there is an isomorphism $$F(y)\rightarrow a$$ and an isomorphism $$F(z)\rightarrow b$$ where $$y,z\in \mathfrak{R}(U)$$.

Considering the composition $$F(y)\rightarrow a\rightarrow b\rightarrow F(z)$$ gives $$\alpha:F(y)\rightarrow F(z)$$ giving an element $$(y,z,\alpha)\in (\mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R})(U)$$.

For this, there exists $$x\in \mathfrak{R}(U)$$ and an isomorphism $$G(x)\rightarrow (y,z,\alpha)$$. As mentioned in question, $$G(x)=(x,x,id:F(x)\rightarrow F(x))$$.

Thus, we have an isomorphsim $$(y,z,\alpha:F(y)\rightarrow F(z))\rightarrow (x,x,id:F(x)\rightarrow F(x))$$. This means, there exists $$u:x\rightarrow y, v:x\rightarrow z$$ such that $$\alpha\circ F(u)=F(v)\circ id$$ i.e., $$\alpha\circ F(u)=F(v)$$.

To write down explicitly, it means the map $$a\rightarrow F(y)\rightarrow F(x)\rightarrow F(z)\rightarrow b$$ is same as that of the map $$a\rightarrow b$$ we started with i.e., $$\phi=\tau\circ F(v\circ u^{-1})\circ \tau'$$

Ignoring $$\tau, \tau'$$ this says any arrow $$a\rightarrow b$$ is of the form $$F(\eta)$$ for some arrow $$\eta$$ in $$\mathfrak{R}(U)$$.

Is this what you(@inkspot) mean when you say locally surjective on morphisms??

Overall, Is this definition of gerbe over stack mean given a manifold $$U$$ the restriction functor $$\mathcal{D}(U)\rightarrow \mathcal{C}(U)$$ has following properties

• given an object $$x\in \mathfrak{X}(U)$$ there is an open cover $$\{U_\alpha\rightarrow U\}$$ such that there exists $$y_\alpha\in \mathfrak{R}(U_\alpha)$$ and isomorphisms $$F(y_\alpha)\rightarrow x|_{U_\alpha}$$ for each index $$\alpha$$.
• given an arrow $$\phi:a\rightarrow b$$ in $$\mathfrak{X}(U)$$ there is an open cover $$\{U_\alpha\rightarrow U\}$$ such that there exists an arrow $$\tau_\alpha:p_\alpha\rightarrow q_\alpha$$ (after identifying $$F(p_\alpha)$$ with $$a_\alpha$$ and $$F(q_\alpha)$$ with $$b_\alpha$$) such that $$F(\tau_\alpha)=\phi_\alpha$$ for each index $$\alpha$$.