Are maps corresponding to affinoid subdomains flat in the Banach sense?

Question

$\newcommand{\Sp}{\mathrm{Sp}}\newcommand{\abs}[1]{\lvert #1\rvert}\newcommand{\comptensor}{\mathbin{\hat{\otimes}}}$ Let $k$ be a complete non-archimedian field and let $X = \Sp(B)$ be a $k$-affinoid space. Let $V = \Sp(B') \subseteq X$ be an affinoid subdomain. It is well-known that the corresponding map $B \to B'$ is a flat ring homomorphism; see e.g. Cor. 7.3.2/6 in Bosch-Güntzer-Remmert.

Let us say that $B'$ is Banach-flat over $B$ if whenever $M \to N \to P$ is an admissible exact sequence of Banach $B$-modules then the completed tensor product sequence $M \comptensor_B B' \to N \comptensor_B B' \to P \comptensor_B B'$ is also admissible exact.

(A map $f \colon M \to N$ of Banach $B$-modules is called admissible if there is a constant $C>0$ such that any $n \in f(M)$ there is a preimage $m \in M$ such that $f(m) = n$ and $\abs{m} \le C \abs{n}$. By Banach's open mapping theorem, this condition is equivalent to $f(M)$ being closed in $N$. Generally exact sequences of Banach modules can only be expected to behave well, if all mappings are admissible.)

Is it true that if $V = \Sp(B') \subseteq X = \Sp(B)$ is an affinoid subdomain, then $B'$ is a flat Banach algebra over $B$?

Note that for a functor being exact in the Banach sense, it suffices to consider short exact sequences $0 \to M \to N \to P \to 0$. Taking the completed tensor products is always Banach-right exact (because being admissible right exact is equivalent to being a cokernel diagram and because completed tensor product is left adjoint to Banach-Hom), so it suffices here to show that tensoring with $B'$ preserves admissible injective maps of Banach modules.

One can show that fixing $M$, the association $V = \Sp(B') \mapsto M \comptensor_B B'$ is a sheaf on $X$ (more precisely, its Čech complex is admissible exact). In particular, $M \comptensor B' \to \prod_i M \comptensor B_i'$ is admissible injective, if $\Sp(B') = \bigcup_i \Sp(B_i')$. By the theorem of Gerritzen and Grauert we can therefore assume that $V = X(f_1/f, \dots, f_r/f)$ is a rational subdomain of $X$, for which we have a rather explicit description of the algebra $B'$. But so far I had no success.

Answer 1

This is not true. Assume that $X$ is the closed unit disc (given by $|T| \le 1$, with algebra $B$) and $V$ is a smaller disc (given by $|T| \le r$ for some $r \in (0,1)$, with algebra $B_V$). Consider the annulus $W$ defined by $|T|=1$ with algebra $B_W$. Then the restriction map $B \to B_W$ is injective and admissible (because function on the disc reach their maximum on the boundary).

If we do the completed tensor product with $B_V$, we get the map $B_V \to B_W \hat{\otimes}_B B_V$. But $B_W \hat{\otimes}_B B_V$ is the algebra of $V\cap W = \emptyset$, that is to say 0, so the map is not injective.