TITLE: Good analytic spaces over a field into locally ringed spaces is fully faithful
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: 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.