Let $k$ be a field which is complete with respect to a non-trivial non-archimedean rank-1 valuation, and let $X$ be scheme which is locally of finite type over $k$. In section of 3.5 of Berkovich's book, he defines the analytification of $X$ to be the closed $k$-analytic space $X^{\textrm{an}}$ representing the following functor: let $\textrm{An}_k$ be the category of good analytic spaces over $k$, and let $F_X \colon \textrm{An}_k \to (\textrm{Sets})$ be the functor which sends a good analytic space $Z$ over $k$ to the set of morphisms $Z \to X$ of locally $k$-ringed spaces.

In order for Berkovich's proof to work, it seems that we need the functor from $\textrm{An}_k$ to the category of locally ringed spaces to be fully faithful; that is, every morphism of locally ringed spaces between two good analytic spaces must in fact be a morphism of analytic spaces over $k$. This fact is stated without proof on page 16 of Berkovich's Trieste notes, but how could one prove this?

However, there is also Remark 2.1.13 of Berkovich's book, which gives an example of a morphism (of locally ringed spaces) between $k$-affinoid spaces which is not a morphism of $k$-affinoid spaces. As $k$-affinoid spaces are examples of good analytic spaces, this seems to contradict the above assertion of fully faithfulness (granted, this particular example is over a trivially-valued field). This issue also does not seem to be addressed in Berkovich's IHES paper.


1 Answer 1


First, I would like to say that I do not understand why you need the full faithfullness of the analytification functor. It seems to me that the main point is to prove that giving a morphism from an analytic space $X$ to the affine analytic space of dimension $n$ is equivalent to giving $n$ global section of $X$.

Second, as you rightly say, the functor is not fully faithful over trivially valued fields nor for non-strict spaces (see Remark 2.1.13 of Berkovich's book), so let us assume that the valuation is non-trivial and that the spaces are strict. We can reduce to the case of a morphism between strictly $k$-affinoid spaces, say $f : X \to Y$. By taking global sections, you find an induced morphism between affinoid algebras $A_Y \to A_X$ and [BGR, Theorem 6.3.1/1] tells you that such a morphism is always continuous. Arguing as for schemes, you then prove that the induced morphism $X = \mathcal{M}(A_X) \to \mathcal{M}(A_Y) = Y$ coincides with $f$ for rigid points. By density of rigid points (which holds over a non-trivially valued field), it coincides with $f$ everywhere.

  • $\begingroup$ I agree that one does not need the fully faithfullness of the analytification functor, but rather of the natural functor from good analytic spaces over $k$ into locally ringed spaces. The reason I think this is needed is exactly for what you pointed out: if $Z \in \textrm{An}_k$ is good, then $\Gamma(Z, \mathcal{O}_Z) \simeq \textrm{Hom}_{\textrm{An}_k}(Z,\mathbf{A}^1)$ but you need that $\Gamma(Z,\mathcal{O}_Z)$ is also isomorphic to the morphisms $Z \to \mathbf{A}^1$ of locally ringed spaces. This would follow from the fully faithfulness. $\endgroup$
    – msteve
    Jul 22, 2016 at 16:45
  • $\begingroup$ OK, sorry, I did not comment on that point. You should find what you need here (for instance): stacks.math.columbia.edu/tag/01I1. $\endgroup$ Jul 22, 2016 at 21:12
  • $\begingroup$ Also, it seems that $A_Y \to A_X$ is just homomorphism between two $k$-affinoid algebras; in order for it to be continuous, we would need that they are strictly $k$-affinoid, would we not? Or can one initially reduce to the strictly $k$-affinoid case somehow? $\endgroup$
    – msteve
    Jul 29, 2016 at 22:44
  • $\begingroup$ Yes, I totally overlooked that point. I think you are right and that Remark 2.1.13 of Berkovich's book shows that this is not always the case. Note that for the second example, you do not need $k$ to be trivially valued. I will edit my answer. $\endgroup$ Jul 30, 2016 at 8:06
  • $\begingroup$ So are you saying that Fact 3.2.10(i) of Berkovich's Trieste notes is wrong? (That is, it should be replaced with a functor from the category of good strictly $k$-analytic spaces.) For my question about analytification, the current answer is fine though, since analytifications are strictly $k$-analytic. $\endgroup$
    – msteve
    Jul 31, 2016 at 22:57

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