Question. Let $(\mathcal B,\mathcal N)$ be an analytic (animated associative) ring, $\mathcal A$ be a condensed (animated associative) ring and $f\colon\mathcal A\to\mathcal B$ a surjective map of condensed rings, i.e., it induces a surjection on $\pi_0$.
- Do we always have an analytic structure $\mathcal M$ on $\mathcal A$ such that $\mathcal N$ coincides with the induced analytic structure given by $\mathcal M_B[S]:=\mathcal B[S]\otimes_{\mathcal A}^L(\mathcal A,\mathcal M)$ for all extremally disconnected sets $S$?
- Is there always a final analytic structure $\mathcal M$ on $\mathcal A$ satisfying the condition above?
First, by the "topological invariance of analytic structure" in Lectures on Analytic Geometry (by the way, this does not seem to be a good terminology because "topological" is more referring to topological spaces than homotopy types), if I am not mistaken, we could assume that both $\mathcal A$ and $\mathcal B$ are static, and thus $\mathcal A\to\mathcal B$ is simply a surjective map of static condensed rings. However, this does not imply that $\mathcal M[S]$ is static for all extremally disconnected sets $S$.
I checked this for the simplest case that $\mathcal A=A$ is a discrete (commutative) ring, $(\mathcal B,\mathcal N)$ is given by a discrete Huber pair $(B,B^+)$ and $f\colon A\to B$ is a surjective map of discrete rings.
In this case, seemingly we could take $A^+:=f^{-1}(B^+)\subseteq A$. Given any monic polynomial $P(T)\in A^+[T]$ and any $x\in A$ such that $P(x)=0$, we have $(f^*P)(f(x))=0$ where $f^*\colon A^+[T]\to B^+[T]$ is the induced map. Since $B^+\subseteq B$ is integrally closed, $f(x)\in B^+$, which implies that $x\in A^+$. In this case, the map $f\colon(A,A^+)\to(B,B^+)$ is a map of discrete Huber rings therefore also of analytic rings. It then follows from the surjectivity that the solidification $(B,B^+)_\blacksquare$ coincides with the induced analytic structure.
Example. When $A\to B$ is given by $\mathbb Q[T]\to\mathbb Q,T\mapsto0$ and $B^+=\mathbb Z$, we have $A^+=\mathbb Z+T\cdot\mathbb Q[T]$.
Moreover, $(A,A^+)$ is obviously the final discrete Huber pair satisfying the condition. It seems reasonable to guess that this is also final as an analytic ring.
Update. The guess about the finality seems to be incorrect: consider the trivial case that $B$ is the zero ring, then $A^+=A$ but the final analytic structure on $A$ seems to be zero.
Update for motivation: $\DeclareMathOperator\AnSpec{AnSpec}$This is part of effort to understand what is the correct "closed embedding" of analytic rings / analytic spectra. It seems reasonable to assume that "closed subset" should have the induced analytic structure (it should be "partially proper"). When $\mathcal A\to\mathcal B$ is surjective, $\AnSpec(\mathcal B,\mathcal N)\to\AnSpec(\mathcal B)\to\AnSpec(\mathcal A)$ could be understood as a composition of embeddings of which the second is closed, then the analytic ring $(\mathcal A,\mathcal M)$ that I seek looks like another factorization $\AnSpec(\mathcal B,\mathcal N)\to\AnSpec(\mathcal A,\mathcal M)\to\AnSpec(\mathcal A)$ of this embedding of which the first embedding is "closed", i.e., "switching" the closedness to the first. Compare with the "immersions" (or locally closed embeddings) in algebraic geometry.
