I have a Galois cover $f \colon Y \rightarrow X$, i.e $f$ is finite étale and $deg(f) = Aut_X(Y)$. The étale fundamental group $\pi_1(X,\bar{x})$ acts on the geometric fiber $Y_{\bar{x}}$, but is that also true that there is an action of the étale fundamental group $\pi_1(X,\bar{x})$ on $Y$? If yes how can I define such an action?
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1$\begingroup$ The étale fundamental group comes by definition with a canonical projection $\pi_1^{ét}(X,\bar x)\to \mathrm{Aut}(Y/X)$. In fact it is defined as the limit of those automorphism groups as $Y$ runs through the Galois covers... $\endgroup$– Denis NardinDec 11, 2019 at 17:28
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2$\begingroup$ @AnhDungLe Yes they are equivalent. I recommend studying Grothendieck Galois theory before asking questions about the topic on MO. $\endgroup$– Denis NardinDec 11, 2019 at 17:48
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1$\begingroup$ You should be heavily analogizing with algebraic topology at every step. The definitions given by Denis Nardin are equivalent because the fiber, seen as a $\pi_1$-set, has the same automorphism group as the cover does — namely, the deck group. $\endgroup$– Santana AftonDec 11, 2019 at 19:59
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1$\begingroup$ Unless I'm mistaken, the definition I gave is the one in construction 3.15 of Lenstra's Galois theory for schemes $\endgroup$– Denis NardinDec 11, 2019 at 20:48
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2$\begingroup$ The key point is that given a group $G$, acting on a set $S$ simply transitively, the automorphisms of $S$ that commute with the action of $G$ are themselves isomorphic to $G$. Here $G$ is the group of deck transformations, which acts simply transitively on $Y_{\overline{x}}$, and which the automorphisms of the fiber functor must commute with, showing that the automorphisms of the fiber functor map to the group of deck transformations. That's all that's going on. $\endgroup$– Will SawinDec 12, 2019 at 2:40
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