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Sep 29, 2013 at 6:13 comment added Marguax @Piotr: Galois cohomology at the (henselian) residue fields of the points of the Berkovich spaces provides an "obstruction" to the equality of topological and etale cohomology (say for constant coefficients), made more precise by a spectral sequence, so purely topological features such as contractibility don't have the same consequences as over $\mathbf{C}$. This is all explained very clearly in Berkovich's IHES paper. I recommend that you read it.
Sep 29, 2013 at 4:53 comment added none @Marguax: Thanks a lot for your comments. I wish I could up-vote and accept them as answers :)
Sep 29, 2013 at 4:40 comment added Piotr Achinger @Marguax: How does it relate to the fact that the Berkovich space is contractible in the case of good reduction?
Sep 29, 2013 at 3:23 comment added Marguax @LMN: Yes, but one can (and must) do so much better than that: the comparison ismorphisms are really at the level of R$f_{\ast}$'s and R$f_{!}$'s with constructible coefficients. The relativization is an essential feature of the proofs, as it is in the complex-analytic case (and Berkovich's proof, inspired by ideas of Deligne, yields a simpler proof for R$f_{\ast}$-comparison even in the complex-analytic case).
Sep 29, 2013 at 2:50 comment added LMN @Marguax: Wonderful! Just to be super careful, when you say "the same cohomology" you are talking about as Galois modules, right.
Sep 29, 2013 at 1:09 comment added Marguax The comparison theorem (analogous to Artin's over $\mathbf{C}$) shows that for coefficients in a constructible $\ell$-adic sheaf (and its analytification) one gets the same cohomology as in the algebro-geometric theory (for a separated scheme of finite type over a non-archimedean field), and likewise with proper supports. The relevance of the approach through non-archimedean geometry is when you try to compute cohomology of (or using) objects that are not analytifications of algebraic schemes.
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