Let $(k,|\cdot|)$ be an algebraically closed field, complete wrt a (multiplicative) norm as in the framework of the Berkovich's analytic geometry. Given a commutative Banach $k$-algebra $\mathcal{A}\neq 0$, let $X=\mathcal{M}(\mathcal{A})$ its spectrum and let $0\neq f\in \mathcal{A}$.


For every point $P=||\cdot||\in X$, the kernel $\mathfrak{B}\_P$ of $P$ is a closed prime ideal of $\mathcal{A}$. Then $P$ induces a seminorm on the integral domain $\mathcal{A}/\mathfrak{B}\_P$ and therefre in its fraction field $F$. The completion of $F$ wrt the norm induced by $P$ is denoted $\mathcal{H}(P)$. It is a normed field extension of $k$. The image of $f$ under the above sequence of transformations is an element of $\mathcal{H}(P)$ which is denoted $f(P)$.


My point is: the function induced by $f$ on $X$ take values at different field extensions of $k$ (unless, for example, if $k=\mathbb{C}$ with the euclidean norm). $$f:X\rightarrow\bigcup\_{P\in X}\mathcal{H}(P),$$ via $P\mapsto f(P).$


Does anyone know if there is a nice way to give a topology to the above disjoint union (which can be identified with $\mathbb{C}$ in the "classical" case...) so that we have properties like path-connectedness, Hausdorff, etc as we do have in the aforementioned particular case?



The target of a regular function in Non-archimedean analytic geometry
If you want to think of $f$ in a way not too far from complex intuition, you should rather consider the induced morphism $\varphi$ from $X$ to the Berkovich affine line. If $x\in X$ then $\varphi(x)$ is the semi-norm $P\mapsto |P(f)(x)|$. 


