Is it possible to define a Berkovich analytic space via its functor of points?

Let $k$ be a complete non-Archimedean field, possibly the trivial one. I am tempted to define a Berkovich analytic space $X$ over $k$ as a functor $$X:(k-\mathbf{Aff})^{op}\longrightarrow\mathbf{Sets}$$ from the category of $k$-affinoid spaces into the category of sets that fulfills the following two conditions:

(1) The functor $X$ is a sheaf in the $G$-topology on $(k-\mathbf{Aff})^{op}$.

(2) Locally in the $G$-topology, the functor $X$ is isomorphic to the functor of points associated to an affinoid algebra.

I realize that this is probably not quite the right definition, since it fails to take into account the net-structure of an affinoid atlas of a Berkovich analytic space. (see Berkovich, Étale Cohomology for Non-Archimedean Analytic Spaces, Section 1.2)

Is it possible to add an additional piece of data to this definition to get a definition that is equivalent to Berkovich's?

  • $\begingroup$ The answer to your last question is trivially yes -- given any two definitions A and B, you can always "modify" A in such a way that you get B (first erase A, then insert B instead) ... . $\endgroup$ – Stefan Kohl Nov 15 '13 at 16:48
  • $\begingroup$ Thanks for pointing this out. I edited the question to avoid such a trivial answer. $\endgroup$ – Martin Ulirsch Nov 15 '13 at 17:38
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    $\begingroup$ I am not sure that I understand what you mean by "locally in the G-topology". Could you be please be more precise? $\endgroup$ – Jérôme Poineau Nov 25 '13 at 14:06
  • $\begingroup$ @JérômePoineau: The functor $X$ should have a finite covering by $G$-open subfunctors that are isomorphic to the functors of points associated to affinoid spaces. Hereby a subfunctor $U$ of a $X$ is said to be $G$-open, if for all morphisms $X'\rightarrow X$ from the functor of points $X'$ associated to an affinoid space to $X$ the base change $U'\rightarrow X'$ is induced by the embedding of a $G$-open subset, i.e. a compact analytic domain of $X'$. $\endgroup$ – Martin Ulirsch Nov 25 '13 at 18:24
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    $\begingroup$ In case you did not see it yet, there should be what you want in the preprint arxiv.org/abs/1312.0338 . $\endgroup$ – Jérôme Poineau Dec 3 '13 at 12:15

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