Differential map of a dominant morphism in char zero Let $k$ be a field of characteristic zero and $X,Y$ be integral schemes of finite type. Assume we have a dominant morphism $\pi\colon X\to Y$.
Then we know that $\pi$ is generically smooth (i.e. on an open subset of the source $X$)
Question: can we say that there exists an open subset $U\subset Y$ such that the differential map $d\pi$ is surjective at all points in $\pi^{-1}(U)$ ? (and if not in general, under which conditions on $X$ and $Y$?)
 A: I am just posting my comment as an answer.  The question has a positive answer.
Generic Smoothness.  For every dominant morphism $\pi:X\to Y$ of finite type, integral $k$-schemes,  denote by $V$ the maximal open subscheme of $X$ on which $\pi$ is smooth.  Then $V$ contains the generic point of $X$ (and hence is a dense open subset of $X$) if and only if the field extension $k(Y)\to k(X)$ is separably generated.  In particular, if $\text{char}(k)$ equals $0$, then $V$ is dense.
Proposition.  Assume that $\text{char}(k)$ equals $0$.  Let $\pi:X\to Y$ be a morphism of finite type $k$-schemes.  For every closed subset $Z\subset X$ with its reduced induced scheme structure, there exists a dense open subset $U$ of $Y$ and a partition of $Z\times_Y U$ into locally closed subsets $Z_i$ such that every $Z_i\to U$ is smooth.  
Proof. By Noetherian induction, it suffices to prove this when $Z$ is irreducible, and we may assume that the result is proved for all proper closed subsets $C$ of $Z$.  Define $W$ to be the union of $W'$, the interior of $\pi(Z)$, and $W''$, the open complement of the closure of $\pi(Z)$.  The open sets $W'$ and $W''$ are disjoint, and $W'\cup W''$ is dense in $Y$.  The assertion is vacuously true for $Z\times_Y W'' \to W''$.  Thus, it remains to find a partition as above of the open subset $Z\times_Y W'$ of $Z$; denote this open by $Z_W$.  
Since $Z_W\to W'$ is a dominant morphism of finite type, integral $k$-schemes, generic smoothness implies the density of the maximal open subscheme $V$ of $Z_W$ on which the morphism is smooth.  Define $C$ to be the closed complement of $V$ in $Z_W$.  By the induction hypothesis, there exists a dense open subset $U'$ of $W'$ such that $C\times_{W'} U'$ has a partition into locally closed subsets $C_i$ with every $C_i\to U'$ smooth.  Define $U$ to be $U'\cup W''$.  Then $Z\times_Y U'$ has a partition into the set $V\times_{W'}U'$ and the sets $C_i$, and each of these is smooth over $U'$.  Thus, the proposition is proved by Noetherian induction. QED 
Corollary. For the closed subset $Z$ equal to all of $X$, for the dense open subset $U$ in the proposition, for every point $x$ of $X\times_Y U$, the following pullback morphism of $\kappa(x)$-vector spaces is injective, $$d\pi_x^\dagger:\Omega_{Y,\pi(x)}\otimes_{\mathcal{O}_{Y,\pi(x)}} \kappa(x) \to \Omega_{X,x}\otimes_{\mathcal{O}_{X,x}} \kappa(x).$$ 
Proof.  The point $x$ is contained in one of the locally closed subsets $Z_i$.  Since $Z_i\to U$ is smooth, the following pullback morphism is injective, 
$$\Omega_{Y,\pi(x)}\otimes_{\mathcal{O}_{Y,\pi(x)}} \kappa(x) \to \Omega_{Z_i,x}\otimes_{\mathcal{O}_{Z_i,x}} \kappa(x).$$ By transitivity of differentials, this injective map factors as the composition, $$\Omega_{Y,\pi(x)}\otimes_{\mathcal{O}_{Y,\pi(x)}} \kappa(x) \xrightarrow{d\pi_x^\dagger} \Omega_{X,x}\otimes_{\mathcal{O}_{X,x}} \kappa(x)\to \Omega_{Z_i,x}\otimes_{\mathcal{O}_{Z_i,x}} \kappa(x).$$ 
Therefore also $d\pi_x^\dagger$ is injective. QED
