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David Roberts ♦
David Roberts ♦
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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 not 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.

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 not 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.

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.

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David Roberts ♦
David Roberts ♦
  • 35.5k
  • 11
  • 124
  • 349

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 not 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.