Let $K$ be a local field, with a completed algebraic closure $\hat{\bar{K}}$ and $A$ be an affinoid $K$-algebra. Is it then true, that for $f_0, \ldots ,f_n \in A$, the map $\phi:$ Sp$A \rightarrow$ $\mathbb{P}_K^{n,rig}, x \mapsto [f_0(x):\ldots:f_n(x)]$ defines a morphism of rigid spaces?
I know by a lecture a similar result for varieties, but the proof I have is not applicable in this case.
EDIT: To clarify what it means. For $f \in A$ and $x \in$ Max$A$, $f(x)$ is the residue class in $A/x$, which can be embedded in $\bar{K}$, so $f(x) \in \bar{K}$, is determined up to conjugation. Points in $\mathbb{P}_K^{n,rig}$ are also meant as equivalence classes of conjugated elements of $\mathbb{P}^{n,rig}(\hat{\bar{K}})$ (see https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf, page 8)
I thought what you need is
- $\phi$ is continuous with respect to the Grothendieck topology of rigid spaces
- for all admissible open $U \subset \mathbb{P}_K^{n,rig}$ and for all $f \in \mathcal{O}_{\mathbb{P}_K^{n,rig}}(U)$, we have that $f\circ \phi_{|\phi^{-1}(U)} \in \mathcal{O}_{\text{Sp}A}(\phi^{-1}(U))$, i.e we have a map $\phi^*_U:\mathcal{O}_{\mathbb{P}_K^{n,rig}}(U) \rightarrow \mathcal{O}_{\text{Sp}A}(\phi^{-1}(U))$ compatible with restrictions
As rigid geometry is relatively new for me, both points, if true, are at the moment not obvious for me and I don't know what a good way could be to verify this.
