Open immersion of affinoid adic spaces If $R$ and $S$ are complete Huber rings with $\varphi: R \to S$ a continuous map, then is it true in general that if $\mathrm{Spa}(S, S^\circ) \to \mathrm{Spa}(R, R^\circ)$ is an open immersion of adic spaces (here $S^\circ$ and $R^\circ$ are the power-bounded subrings) then $\mathrm{Spec}(S) \to \mathrm{Spec}(R)$ is injective?
For example, this is true if $R$ and $S$ both have the discrete topology, because if $\frak p$ and $\frak q$ are two prime ideals in $S$ which are equal after restricting to $R$ then $(\frak p, |\cdot|_{\rm triv})$ and $(\frak q, |\cdot|_{\rm triv})$ (trivial valuations), which are both points in $\mathrm{Spa}(S,S)$, restrict to the trivial valuation on $R/\varphi^{-1}(\frak p)$.
But I'm not sure how generally to expect that this is true. If it makes it easier, we can assume that $R$ and $S$ are Tate, so the adic spaces are analytic.
 A: This is not correct in general.
There are in fact two examples in Bosch's Lectures on Formal and Rigid Geometry p.61-63. Let me sketch the first one.
While it uses rigid-analytic spaces, it can be easily transferred to adic spaces: Weierstraß subdomains can be seen as special cases of rational subdomains (in adic spaces).
Take $(R,R^\circ)=(K,\mathcal{O}_K)$ a non-archimedean field and let $D=Spa(R,R^\circ )$ be the unit disk. Choosing a $c\in K$ such that $0 < |c| < 1$, we can look at the subspace
$$ U=\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}.$$
Then $U=Spa(S,S^\circ)$ and $S\cong K\langle \frac{T(T-1)}{c}\rangle$, which in turn is isomorphic to
$$ S_0\times S_1:=K\langle\frac{T}{c}\rangle\times K\langle\frac{T-1}{c}\rangle.$$
One way to see this is that as rigid-analytic spaces we have
$$\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}=\{x\in D \mid |T(x)|\leq |c|\}\coprod \{x\in D \mid |(T(x)-1)|\leq |c|\}. $$
But now
$$ Spec(S)\cong Spec(S_1)\coprod Spec(S_2)\to Spec(R)$$
is not injective: Both the generic point of $S_1$ and $S_2$ map to the generic point of $R$.
