Discrete set of critical points of a holomorphic map. Let $U$ be an open domain in $\mathbb{C}^{n}$. Let $m\ge n$ and let $F:U\to C^{m}$ be a holomorphic map. 

What are the sufficient conditions for $F$ to have a discrete set of critical points?

One can show that if $F$ is an injection, then the set of critical points is thin, and so if $n=1$, this is sufficient. I am interested in a general case.
Thank you.
 A: The set of critical points is the set where the derivative $Df_p \colon T_p {\mathbb C}^n \to T_{f(p)}{\mathbb C}^m$ is not of maximum rank (i.e. $Df_p$ is the Jacobian matrix).  So the trick is to look at the variety described by the $m \times m$ or $n \times n$ subdeterminants of $Df_p$.  So if $m=n$, then the set of critical points is either empty or a hypersurface as there is only one function.  In other words, if $m=n$ and there are critical points, they are discrete only if $m=n=1$.  If $m < n$, the there always at least $n$ possible $m \times m$ subdeterminants, so generically the set of critical points will be discrete.  If $m > n$, then there are always at least $m$ possible $n \times n$ subdeterminants, and $m > n$, so again generically the set of critical points will be discrete.
To check, I assume you generally have to compute the Jacobian matrix and those particular determinants.  The $n=1$ case is simple because a zero set of a holomorphic function is either everything, discrete, or empty.
An example in $n=2$, $m=3$, where the mapping is 1-1, but the set of critical points is not discrete is for example $(z,w) \mapsto (z^2,z^3,w)$.  The Jacobian matrix is
$$
\begin{bmatrix}
2z & 0 \\
3z & 0 \\
0 & 1
\end{bmatrix}
$$
The three subdeterminants are $0$, $2z$, and $3z$, so their common zero set is $z=0$, which happens to be a hypersurface.
If on the other hand we perturb the thing very slightly to $(z,w) \mapsto (z^2+\epsilon zw,z^3,w)$, we find the matrix
$$
\begin{bmatrix}
2z + \epsilon w & \epsilon z \\
3z & 0 \\
0 & 1
\end{bmatrix}
$$
and the three subdeterminants are $3 \epsilon z^2$, $2z+\epsilon w$, and $3z$.  Their common zero set is $z=w=0$, a discrete set.
