Consider a central extension of Lie groups $1\rightarrow S^1\rightarrow \hat{G}\xrightarrow{\pi} G\rightarrow 1$.


I understand that this mean $\pi:\hat{G}\rightarrow G$ is a surjective homomorphism of Lie groups (not sure if this has to be submersion) and that $S^1\subseteq Z(\hat{G})$. There is a local section $\sigma:U\rightarrow \hat{G}$ such that $\pi\circ \sigma=1\_U$ where $U$ is an open nbd of $1\in G$. Correct me if I am missing some conditions.


Let $X$ be a manifold with an action of $G$ on it. Then we have the notion of [quotient stack](https://mathoverflow.net/questions/319038/motivation-for-definition-of-quotient-stack) $[X/G]$. 


There is an action of $\hat{G}$ on $X$ given by $(\hat{g},x)\mapsto \pi(\hat{g})\cdot x$.


We can then consider the quotient stack $[X/\hat{G}]$. 


Given a manifold $Y$, objects of $[X/G](Y)$ are pairs $(P\rightarrow Y,P\rightarrow X)$ where $P\rightarrow Y$ is a principal $G$ bundle and $P\rightarrow X$ is a $G$-equivariant space (see that $G$ acts on $P$ and $X$).



> 
> "As locally any map $T\rightarrow G$ can be lifter to $\tilde{G}$", the map of stacks $[X/\hat{G}]\rightarrow [X/G]$ is a gerbe over stack.
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I see that, locally any map $\theta: T\rightarrow G$ can be lifted to $\hat{G}$. As there is a section $\sigma:U\rightarrow \hat{G}$, we can consider $\theta^{-1}(U)\xrightarrow{\theta} U\xrightarrow{\sigma} \hat{G}$ and $\pi\circ (\sigma\circ \theta)=\theta$. Thus, any map $\theta:T\rightarrow G$ can be locally lifted to $\hat{G}$. But, I am not able to see why this imply $[X/\tilde{G}]\rightarrow [X/G]$ is a gerbe over stack. 


Any comments are welcome.



Central extension gives a gerbe over stack
As discussed in the comments, $[X/\hat{G}] \to [X/G]$ is the pullback of $[pt/\hat{G}] \to [pt/G]$ along the canonical map $[X/G] \to [pt/G]$, so it suffices to show that $[pt/\hat{G}] \to [pt/G]$ is a gerbe. Since every principal $G$-bundle is locally trivial, it can be locally lifted to a principal $\hat{G}$-bundle, which is another way of saying that $[pt/\hat{G}] \to [pt/G]$ is an epimorphism of stacks. Using the fact that $\hat{G}\to G$ is surjective, then there is an equivalence of stacks 
$$
[pt/\hat{G}\times\_G\hat{G}] \stackrel{\simeq}{\to} [pt/\hat{G}]\times\_{[pt/G]}[pt/\hat{G}]
$$
Thus the diagonal $[pt/\hat{G}] \to [pt/\hat{G}]\times\_{[pt/G]}[pt/\hat{G}]$ is equivalent to $[pt/\hat{G}] \to [pt/\hat{G}\times\_G\hat{G}]$, induced by the diagonal homomorphism $\hat{G} \to \hat{G}\times\_G\hat{G}$. Such a map of stacks is an epimorphism by the same argument as before, and so we are done.


