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 $[X/G]$.

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.

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.

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