@DanielLitt Yes, you are of coure right — $K$ was meant to be algebraically closed of characteristic $0$. Thanks for pointing that out!

As its well-known that $\pi_1(\mathbb{A}^n_K)=0$ for $K$ algebraically closed of characteristic $p$, we deduce that if $K$ is further a non-archimedean field (e.g., $\mathbb{C}_p$) then $\pi_1^\mathrm{et}(\mathbb{A}^{n,\mathrm{an}}_K)=0$.

This is not whata the OP is asking it seems, but for anyone else reading this, let me respond to @R.vanDobbendeBruyn implicit question. If $K$ is a non-archimedean field of characteristic $p\geqslant 0$ then Lütkebohmert proved (and I believe also independently proven by Gabber) that the the prime-to-$p$ finite étale covers of $X$ and $X^\mathrm{an}$ are the same, for a locally if finite type $K$-scheme $X$, i.e., the Riemann existence theorem holds. So, when $p=0$ this just says the fundamental groups of $X$ and $X^\mathrm{an}$ are the same.

Considering the Levi decomposition reduces you to a Levi. But, the Levis of symplectic groups are themselves symplectic groups times general linear groups. But, both symplectic groups and general linear groups are special (see en.m.wikipedia.org/wiki/Special_group_(algebraic_group_theor‌​y) ).

The $L$-group is not functorial in the contravariant way one might expect (e.g., think about the inclusion of a maximal torus), but it is functorial in homomorphisms with normal image. Again, I learned this from Kottwitz. Both of these results are in his paper “Stable Trace Formula: Cuspidal Tempered Terms”.

Do you care about the cohomology of $p$-adic groups? Kottwitz proves (as a sort of generalization of Tate—Nakayama) that for a reductive group $G$ over a $p$-adic local field $F$, one has that $H^1(F,G)$ (i.e., the Galois cohomology of $G$) is computed as the Pontryagin dual of $\pi_0(Z(\widehat{G})^\Gamma)$, where $\Gamma$ is the Galois group of $F$.

I guess $\mathbb{Z}_p$ is a bad example of because of Lang’s theorem ,but hopefully you get my point.

Can you clarify the claim “… hence for all linear algebraic group schemes, over a Dedekind domain…”? This seems not obvious to me. For instance, what if $R=\mathbb{Z}_p$ and $G$ is a parahoric group scheme, what is the argument?

So in short: they are different points of view, but some care must be taken in the change of perspectives, similar to 'classical varieties' vs. 'schemes' change of view.

in the adic world. So, you can work in this Berkovich world and if you have something like an open/closed decomposition of topoi, things are fine, but you have to work a priori harder because this missing point contains information. But, in the end, there is no actual contradiction because this open/closed decomposition for the pseudo-adic space $Z$ should roughly capture the same information as the open/closed decomposition of topoi in this Berkovich world.

@AlexeyDo Everything changes but only superficially. Like, it's analogous to saying that for a finite type K-algebra R, that two subsets $S,T\subseteq \mathrm{Spec}(R)$ can have the property that $S\cap \mathrm{MaxSpec}(R)=T\cap\mathrm{MaxSpec}(R)$ while also $S\ne T$. Like in this MaxSpec world you are missing points, and so set-theoretic equality can be misleading (this is an issue of constructability of $S,T$ , but this is also what's happening secretly in this rigid world). In particular, the complement in the Berkovich world misses a higher rank point in the complement

@AlexeyDo I’m not sure what the other person said. Maybe if you could be more specific about what they mean, I could say, but I don’t really see how that would be possible. It’s possible what they mean is that the complement in the Berkovich space is OPEN not closed. Namely, in the example with the close disk above, at the level of the berkovich spaces this is a closed embedding and so its complement is an open, not a closed. Is this related to what you mean?

@AlexeyDo Hey. Can you say what you mean precisely? Do you mean is there an adic space and a map $Z’\to X$ which has image $Z$? I think my answer shows that is impossible. It is possible for $Z$ to be a pseudo-adic space as I said, but this is not giving $Z$ the structure of an adic space, but is the next best thing (although it depends on the tuple $(X,Z)$, although the etale topos is more flexible).

But, to say that the rigid geometry done in Tate language is obsolete is like saying any algebraic geometry done before schemes is useless. Some people might say this, but those people are frankly not very well educated.