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In stacks 0BMQ, a Galois category is defined to be a functor $F:\mathsf C\longrightarrow \mathsf{FinSet}$ such that $\mathsf C$ is finitely bicomplete, every object of $\mathsf C$ is a finite coproduct of connected objects, and $F$ is exact and conservative.

In a finitely extensive category $\mathsf C$, finite coproduct decompositions are unique up to isomorphism if they exist, so if a Galois category were finitely extensive, there would be a functorial assignment of connected components, i.e a functorial connected component decomposition $\pi_0:\mathsf C\longrightarrow \mathsf{FinSet}$. The existence of such a functor is equivalent to saying $\mathsf C\simeq \mathsf{FinFam}(\mathsf A)$ where the latter is the free finite coproduct completion of the subcategory of connected objects of $\mathsf C$. Whenever it exists, $\pi_0$ is a fibration.

Are Galois categories extensive? Is the functor $F$ just the a connected components functor (fibration)? How does this follow from the axioms?

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  • $\begingroup$ Galois category are always equivalent to finite $G$-sets for some profinite group $G$, so the answer is yes. However $F$ is the forgetful functor to sets, so $\pi_0X$ is $F(X)/Aut(F)$ $\endgroup$
    – Denis Nardin
    Jul 15, 2016 at 10:12
  • $\begingroup$ @DenisNardin I'm not familiar with all the theory, I'm trying to sort a few things out first - why does such an equivalence mean the fundamental functor is a connected components functor? $\endgroup$
    – Arrow
    Jul 15, 2016 at 10:15
  • $\begingroup$ More or less by definition of $Aut(F)$, $F$ factors through the forgetful functor $Aut(F)-Fin\to Fin$ (there is a small nuance given by the fact that $Aut(F)$ has the structure of a profinite group). However $F$ is not $\pi_0$ $\endgroup$
    – Denis Nardin
    Jul 15, 2016 at 10:17
  • $\begingroup$ @DenisNardin I don't understand what you're saying then: $\pi_0$ can't be defined no the nose, every choice of connected component decompositions gives such a functor. If $F$ is not $\pi_0$, how is it a connected component functor? $\endgroup$
    – Arrow
    Jul 15, 2016 at 10:21
  • $\begingroup$ It is not a connected component functor. $F$ is the forgetful functor from $G-Fin$ to $Fin$ that forgets the action. Think about the classical case of a Galois category, finite covering spaces. Then $F$ is the fiber above the basepoint of the base, which is not a connected component functor $\endgroup$
    – Denis Nardin
    Jul 15, 2016 at 10:25

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