Let $X$ be a rigid analytic space over a non-Archimedean field $k$. If $U_1,\ldots,U_n\subseteq X$ are affinoid opens, then it's usually not clear whether or not the admissible open $U=U_1\cup\cdots\cup U_n$ is affinoid. But we have (due to the sheaf axioms, and the fact that the $U_i$ constitute an admissible covering of $U$) the canonical injection (arising from the restrictions $\mathscr{O}_X(U)\to\mathscr{O}_X(U_i)$)

$$\mathscr{O}_X(U)\hookrightarrow\prod_{i=1}^n\mathscr{O}_X(U_i)\quad (*)$$

The target is a product of affinoid $k$-algebras, and in particular is a Banach algebra over $k$. One can endow $\mathscr{O}_X(U)$ (or $\mathscr{O}_X(V)$ for any admissible open $V$ of $X$) with a locally convex topology via the $k$-algebra isomorphism (again due to the sheaf axioms) $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$, where $W$ runs over the affinoid opens of $X$ contained in $U$, by pulling back the projective limit topology on the target (using the canonical $k$-Banach topologies on each $\mathscr{O}_X(W)$). For this topology, the map (*) is certainly continuous. My question is as follows:

**Is the map (*) a topological embedding?**

This would follow if the sheaf of rings of a rigid space were in fact a sheaf of topological rings, but I don't think this is required in the definition of a $G$-ringed space in e.g. BGR like it is for adic spaces and formal schemes. I feel like this is the case for $X$ affinoid though...at least, I think this is true if it is the case that $k$-algebra maps between $k$-affinoid algebras, which are always continuous, are in fact strict (I don't know if this is true, though it's true for maps of finite modules over Noetherian $k$-Banach algebras, which affinoid algebras are by definition, but the base Tate algebra depends on the given affinoid algebra).

A positive answer to my question would imply that the locally convex topology of $\mathscr{O}_X(U)$, while possibly not that of an affinoid $k$-algebra, is at least defined by a norm (the one got by restricting the max norm for a choice of norms on each $\mathscr{O}_X(U_i)$ along (*)). This consequence is what I really want. (I guess if we really have a sheaf of topological rings, then the map would be a closed topological embedding, since it's the first arrow in an equalizer sequence of Hausdorff topological rings, but I don't actually need this.)

EDIT: As user grghxy points out, I should assume that $X$ is quasi-separated to ensure that my set $U$ is in fact admissible open with admissible covering given by the $U_i$.

quasi-compact$U_i \cap U_j$. You can check via the Banach open mapping theorem that the natural Banach space structure on the equalizer (a closed subspace of $\prod O(U_i)$) is the intrinsic topology you are considering. $\endgroup$ – grghxy Jun 28 '15 at 18:54definethe topology on $O(U)$ to be that transported from the equalizer, and directly prove that this is independent of all choices. One has no need for any kind of open mapping theorem beyond the setting of Banach spaces. $\endgroup$ – grghxy Jun 29 '15 at 0:07