Let $X$ be a quasiprojective variety over a $p$adic field. Denote by $X^{an}$ its rigid analytification. Does $\pi_1^{et}(X)=\pi_1^{et}(X^{an})$?
1 Answer
Yes, if $X$ is proper (by GAGA). Otherwise, this is false for covers of degree divisible by $p$. See Example 51 in Ducros's survey "Étale Cohomology of Schemes and Analytic Spaces", though unfortunately he does not give a reference.
For an example of this failure in equal characteristic $p$, see section 7.4 in de Jong–van der Put "Etale cohomology of rigid analytic spaces" (Doc. Math. 1995). They construct an example of a $\mathbf{Z}/p$covering of $\mathbf{A}^{1, \rm an}$ which is not the analytification of a finite etale covering of the line.
In mixed characteristic, almost the same argument should work if we replace ArtinSchreier coverings with Kummer coverings of degree $p$ (i.e. replace equations $T^pT=f$ with $T^p=f$), but I didn't check the details.

4$\begingroup$ Piotr, Thanks! I agree with you about the proper and equal characteristic case. In mixed characteristic case, I found the answer right now. The result is also ture (of course, here the fundamental group means finite etale ones). see Theorem 3.1 in scribd.com/document/14391207/… $\endgroup$– YangJan 20 at 15:09

1$\begingroup$ Oops! Indeed, my answer was too hasty. I will leave it here for now and edit or delete later. $\endgroup$ Jan 20 at 18:36