Instead of trying to say what flatness in analytic geometry means I'll give you some street-fighting tricks for recognizing whether a morphism of analytic spaces ( not necessarily reduced) $f:X\to Y $ is , or has a chance to be, flat.

a) A flat map is always open. . Hence, contraspositely, the embedding
$\lbrace 0\rbrace \hookrightarrow \mathbb C$ is not flat. More generally, the embedding of a closed (and not open !) subspace $X \hookrightarrow Y$ is never flat.

b) An open map need not be flat: think of the open map $ Spec(\mathbb C) \to Spec(\mathbb C[\epsilon ]) $ from the reduced point to the double point, which is open but not flat.

[It is not flat because the $\mathbb C[\epsilon]$- algebra $\mathbb C=\mathbb C[\epsilon]/(\epsilon)$ is not flat : recall that a quotient ring $A/I$ can only be flat over $A$ if $I=I^2$ and here
$I=(\epsilon) \neq I^2=(\epsilon)^2=(0)$]

An example with both spaces reduced is the normalization [see g) below] $f:X^{nor} \to X $ of the cusp $X\subset \mathbb C^2$ given by the equation $y^2=x^3$ .That normalization is a homeomorphism and so certainly open, but it is not flat : this results either from g) or from h) below.

c) Given the morphism $f:X\to Y $ , consider the following property:

$\forall x \in X, \quad dim_x(X)=dim_{f(x)} (Y) +dim_x(f^{-1}(f(x)))$ $\quad (DIM) $

We then have: $ f \; \text {flat} \Rightarrow f \;\text { satisfies } (DIM)$

For example a (non-trivial) blowup is *not* flat.

d) For a morphism $f:X\to Y $ between connected holomorphic *manifolds* we have:

$$f \text { is flat} \iff f \text { is open} \quad \iff (DIM) \;\text {holds}$$
For example a submersion is flat, since it is open.

e) Given two complex spaces $X,Y$ the projection $X\times Y\to X$ is flat ( Not trivial: recall that open doesn't imply flat!).

f) flatness is preserved by base change.

g) The normalization $f:X^{nor} \to X $ of a non-normal space is never flat.

For example if $X\subset \mathbb C^2$ is the cusp $y^2=x^3$, the normalization morphism $\mathbb C\to X: t\mapsto (t^2,t^3)$ is not flat.

h) Given a finite morphism $f:X\to Y $, each $y\in Y$ has a fiber $X(y) \subset X$ and for $x\in X$ we can define $\mu (x)= dim_{\mathbb C} (\mathbb C \otimes_{\mathcal O_{Y,y}} \mathcal O_{X,x})$ . Now for $ y\in Y$ we put $\; \nu (y)= \Sigma_{x\in X(y)} \mu(x)$ and we obtain :

$$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { locally constant}$$

[Of course for connected $Y$, locally constant = constant]

**Bibliography:**

A. Douady, *Flatness and Privilege*, L'Enseignement Mathématique, Vol.14 (1968)

G.Fischer, *Complex Analytic Geometry*, Springer LNM 538, 1976