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
$\alpha$.