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Suppose $F:\mathcal{X}\rightarrow\mathcal{Y}$ is gerbe over stack and $p:X\rightarrow \mathcal{X}$ is an atlas $\mathcal{X}$.

Does this imply $F\circ p:X\rightarrow \mathcal{Y}$ is an atlas for $\mathcal{Y}$?

Atlas of a stack is as in Understanding the definition of atlas of a stack

Gerbe over a stack is as in Understanding definition of gerbe over a stack

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  • $\begingroup$ I am ok with yes/no reply for now... I have tried something to say the result is positive.. $\endgroup$
    – Praphulla Koushik
    Commented Aug 16, 2018 at 18:41
  • $\begingroup$ When you say that $p:X\to\mathcal{X}$ is an "atlas", does this mean that $p:X\to \mathcal{X}$ is a smooth surjective morphism and $X$ is a scheme? (I assume yes.) If $F:\mathcal{X}\to \mathcal{Y}$ is a gerbe, then it is surjective. Is your gerbe smooth? If so, then the composition $F\circ p$ is a smooth surjective morphism, and thus an "atlas". $\endgroup$
    – Ariyan Javanpeykar
    Commented Aug 16, 2018 at 18:44
  • $\begingroup$ @AriyanJavanpeykar it seems to be too straight forward in case of algebraic stacks.. it is not so straightforward in geometric stack set up.. if you are aware of what a geometric stack is, I want to explain little more, otherwise I don’t want to bother you.. $\endgroup$
    – Praphulla Koushik
    Commented Aug 16, 2018 at 18:49
  • $\begingroup$ I only vaguely know what a geometric stack is. But, in that context, the "only" condition to verify is that $X\to \mathcal{Y}$ is representable, right? But isn't this automatic if the diagonal of $\mathcal{Y}$ is representable? Anyway, this is as far as I can help you probably. Sorry. $\endgroup$
    – Ariyan Javanpeykar
    Commented Aug 16, 2018 at 18:52
  • $\begingroup$ @AriyanJavanpeykar Ok.. please let me know if you have seen following definition somewhere... I am just trying to imitate the definition of geometric stack.. An atlas is a map of stacks $X\rightarrow \mathcal{X}$ such that the diagonal $X\times_{\mathcal{X}}X$ is a scheme and some thing extra(I do not know how to imitate that condition here)... $\endgroup$
    – Praphulla Koushik
    Commented Aug 16, 2018 at 18:58

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