lately I have been interested in functional analysis, especially with a view towards its applications in the world of (complex) analytic geometry. I have been using R. Taylor's book Several complex variables with connections to algebraic geometry and Lie groups (as it is suggested in this MO question) and J.P. Demailly excellent book [D]: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf.
However, my poor background in functional analysis doesn't allow to me to answer to a very natural question I have: in [D], definition 5.32 the notion of transverse nuclear $A$-modules over a nuclear Fréchet algebra $A$ is introduced. In the subsequent Proposition 5.33, it is shown (in particular) that if $X$ is a (finite dimensional) Stein space and $\mathcal F$ is a coherent sheaf over $X$, then for every pair of open Stein subspaces $U' \subset U \Subset X$, $\mathcal F(U)$ and $\mathcal O_X(U')$ are transverse over $\mathcal O_X(U)$.
So far so good, but the proof leaves me astonished: I would expect that $\mathcal O_X(U')$ is "analytically flat" over $\mathcal O_X(U)$ (in the sense that the functor $\mathrm{Tôr}^{\mathcal O_X(U)}_q(\mathcal O(U'), -)$ is identically zero. The reason I would expect that is simply that $U'\to U$ is an open immersion in analytic geometry, and I would expect it to be flat "in the correct sense" (of course it is flat stalkwise, but I would expect more).
Can someone explain to me if the intuition coming to me from Algebraic Geometry leads me to some mistake or if it is possible to prove the above statement?