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Apr 15, 2023 at 9:28 vote accept Spai
Apr 13, 2023 at 9:04 answer added Konrad Waldorf timeline score: 8
Apr 12, 2023 at 20:24 answer added Dmitri Pavlov timeline score: 4
Apr 12, 2023 at 12:20 history edited LSpice CC BY-SA 4.0
Consistent TeX
Apr 12, 2023 at 11:52 answer added Tim Porter timeline score: 6
Apr 12, 2023 at 7:56 comment added მამუკა ჯიბლაძე Furthermore these are not just any fibrations but the ones satisfying stack conditions, maybe these also have to be taken into account in determining Aut, I am not sure.
Apr 12, 2023 at 7:55 comment added მამუკა ჯიბლაძე Thanks, this is clearer to me now. Still, there is, I think, a subtlety there. On categorifying, properties turn into structures. So when you say $\operatorname{Aut}(\mathfrak P)$ is taken over Man, this actually means that objects of this category are pairs $(A,\alpha)$ where $A$ is an autoequivalence of $\mathfrak P$ and $\alpha$ is an isomorphism between $\pi_{\mathfrak P}$ and $\pi_{\mathfrak P}\circ A$, where $\pi_{\mathfrak P}$ is the fibration of $\mathfrak P$. Morphisms in Aut($\mathfrak P$) must respect those $\alpha$s.
Apr 12, 2023 at 7:36 history edited Spai CC BY-SA 4.0
added more details to the question after the suggestion given below.
Apr 12, 2023 at 7:34 comment added Spai The action is defined as a monoidal functor from the 2-group G to Aut($\mathfrak{P}$) where the automorphisms are over Man. So there is compatibility with that structure. Even the equivalence of categories is over Man. Let me add that clearly in the post. Let me think about the last point.
Apr 12, 2023 at 5:44 comment added მამუკა ჯიბლაძე Both $\mathfrak P$ and $\mathfrak X$ are not mere categories, they are both fibred over your site. Don't you need some condition about compatibility of the action with this structure? Also, you should somehow ensure that $\mathfrak X$ is actually equivalent, via $\pi$, to the quotient of $\mathfrak P$ by the action of $G$.
Apr 12, 2023 at 4:59 history asked Spai CC BY-SA 4.0