I have a finite étale morphism $ f \colon Y \rightarrow X$ of degree $d$. Here we have a result which basically says finite étale morphism is like finite covering map in topology, i.e. each point $x \in X$ has an étale neighbourhood $U$, whose preimage $U \times_X Y$ is a disjoint union of (d??) copies of $U$. In topology the fundamental group acts by permutation of these copies, so the étale fundamental group should do the same. Is that true? Because the étale fundamental group seems to be defined as a group acting on the geometric fiber and not the usual fiber. Is there any problem one should be careful of?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ As mentioned in your later question, the étale fundamental group acts on $Y$ by automorphisms of the étale morphism. In order for this action on $Y$ to give an automorphism of the cover, should fibers be preserved? $\endgroup$– Santana AftonCommented Dec 12, 2019 at 1:39
-
$\begingroup$ Rather than being careful, maybe try using the definition of the geometric fiber and trying to prove, or disprove, that in this case the geometric fiber of $Y$ over $x$ is in bijection with the usual fiber of $U \times_X Y$ over a point in $U$. If you do one of those, you don't need to be careful of anything! $\endgroup$– Will SawinCommented Dec 12, 2019 at 2:46
Add a comment
|