Let $f:X\to Y$ be a morphism of complex analytic spaces. Assume $f$ is flat (or, more generally, that there is a coherent sheaf on $X$ with support $X$ which is $f$-flat). Is $f$ an open map?
The rigid-analytic analogue is true (via Raynaud's formal models): see Corollary 7.2 in S. Bosch, Pure Appl. Math. Q., 5(4) :1435–1467, 2009. I don't know about the Berkovich side.
In the algebraic case it's also true (that's what led me to the question, see http://arxiv.org/abs/1010.0341). Specifically, if $K$ is an algebraically closed field with an absolute value, and $f:X\to Y$ is a universally open morphism of $K$-schemes of finite type, the the induced map on $K$-points is open (for the strong topology).
Note that in the complex analytic case, I don't know any reasonable substitute for "universally open". If I believe in the analogy, the result ($f$ is open) should be true assuming for instance that $Y$ is locally irreducible and $f$ is "equidimensional" in some sense (e.g. surjective, $X$ irreducible and the fiber dimension is constant). In this setting the case of fiber dimension 0 is known.
Best Answer
The answer is yes.
In fact, there is the following result, see Banica-Stanasila, Algebraic methods in the global theory of Complex Spaces, Theorem 2.12 p. 180.
In particular, every flat morphism is open.