map of tangent spaces surjective on an open set Let $\phi: \mathbb{A}^n \rightarrow \mathbb{A}^m$ be a morphism of affine spaces given by $\phi(\alpha) = (P_1(\alpha),\dots,P_m(\alpha))$, where $P_1,\dots,P_m$ are in $k[x_1,\dots,x_n]$, the polynomial ring of regular functions of $\mathbb{A}^n$. Suppose that $k$ is of characteristic zero. Suppose that the restriction $\phi|_{X}$ of $\phi$ on some affine variety $X \subset \mathbb{A}^n$, takes $X$ to an affine variety $Y \subset \mathbb{A}^m$. 
For $\alpha \in \mathbb{A}^n$, $\phi$ induces a map of tangent spaces
$(d \phi)_{\alpha}: T_{\alpha}(\mathbb{A}^n) \rightarrow T_{\phi(\alpha)}(\mathbb{A}^m)$, where $T_{\alpha}(\mathbb{A}^n) \cong k^n$ and 
$T_{\alpha}(\mathbb{A}^m) \cong k^m$. The map $(d \phi)_{\alpha}$ is given by the Jacobian matrix $J|_{\alpha}$ of $\phi$ evaluated at $\alpha$, with $J$ defined as defined by $J_{ij} = \partial P_i / \partial x_j$. Notice that $J$ is an $m \times n$ matrix. 
The restriction of $(d \phi)_{\alpha}$ on $T_{\alpha}(X)$ gives a map of tangent spaces
$(d \phi)_{\alpha}|_{T_{\alpha}(X)}: T_{\alpha}(X) \rightarrow T_{\phi(\alpha)}(Y)$. 
Question: How do we see that there is an open set $U \subset X$, such that for every $\alpha \in U$, $(d \phi)_{\alpha}|_{T_{\alpha}(X)}$ is surjective? What is the geometric interpretation and connection of this result with the generic fiber dimension theorem, which says that for a generic point $\alpha \in X$ we have that $\dim \phi^{-1}(\phi(\alpha)) = \dim X - \dim Y$? 
 A: Here is an answer by "elementary" arguments that i put together, which i find very insightful (thanks to @Mohan for providing some brief but essential comments).
By the fiber dimension theorem, we have $\dim X \ge \dim Y$ and for every $\alpha \in X$ we have $\dim \phi^{-1}(\phi(\alpha)) \ge \dim X - \dim Y$. In addition, there is an open set $V_1 \subset Y$ such that
for every $\alpha \in \phi^{-1}(V_1)$ we have that $\dim \phi^{-1}(\phi(\alpha)) = \dim X - \dim Y$. Since the singular locus of $Y$ is a proper closed subset, let $V_2$ be the open set in $Y$ such that $Y$ is nonsingular on $V_2$. Similarly, let $U_1$ be the open set of $X$ on which $X$ is nonsingular. Now, let $U_2 = U_1 \cap \phi^{-1}(V_1 \cap V_2)$. The open set $U_2$ is non-empty because $X,Y$ are irreducible. We will see that for every $\alpha \in U^*=U_2 \cap U_3$, the map of tangent spaces $(d\phi)_{\alpha}: T(X)_{\alpha} \rightarrow T(Y)_{\phi(\alpha)}$ is surjective, where $U_3$ is going to be some other open set, to be specified a little later.
First, observe that for any $\alpha \in X$, the fiber $\phi^{-1}(\phi(\alpha))$ is a subvariety of $X$ given by the same equations as $X$ together with the additional equations $0=\phi(\alpha)_1 - P_1= \cdots = \phi(\alpha)_m-P_m$, where $\phi(\alpha) = (\phi(\alpha)_1,\dots,\phi(\alpha)_m)$. The sequence of morphisms $\phi^{-1}(\phi(\alpha)) \hookrightarrow X \stackrel{\phi}{\rightarrow} Y$ induces a sequence of maps of tangent spaces $T(\phi^{-1}(\phi(\alpha)))_{\alpha} \hookrightarrow T(X)_{\alpha} \rightarrow T(Y)_{\phi(\alpha)}$, where the first arrow is an inclusion and the second arrow is given by the $m \times n$ Jacobian matrix $J|_{\alpha}$ evaluated at $\alpha$, where $J_{ij} = \partial P_i / \partial x_j, \, i=1,\dots,m, \, j=1,\dots,n$. Now, by definition of $T(\phi^{-1}(\phi(\alpha)))_{\alpha}$, every point $\xi=(\xi_1,\dots,\xi_n) \in T(\phi^{-1}(\phi(\alpha)))_{\alpha}$ must satisfy on top of the equations of $T(X)_{\alpha}$ the equations
$\sum_{j=1}^n \partial P_i/\partial x_j|_{\alpha} \xi_j = 0, \, i=1,\dots,m$. This says that $T(\phi^{-1}(\phi(\alpha)))_{\alpha}$ is a subspace of the right nullspace of $J|_{\alpha}$, the latter to be denoted as $N(J|_{\alpha})$. 
By our choice of $U_2$, any $\alpha \in U_2$ is a nonsingular point of $X$, and so $\dim T(X)_{\alpha} = \dim X$. Similarly, $\phi(\alpha)$ is a nonsingular point of $Y$ and so $\dim T(Y)_{\phi(\alpha)} = \dim Y$. Thus, $J|_{\alpha}$ is a linear transformation from a vector space of dimension $\dim X$ to a vector space of dimension $\dim Y$. Since $\dim Y \le \dim X$, the rank of $J|_{\alpha}$ can be at most $\dim Y$. Now, the dominance of $\phi$ implies
that at least $\dim Y$ polynomials among $P_1,\dots,P_m$ are algebraically independent over $k$ (this comes from the fact that the $k$-algebra homorphism $\phi^*:A(Y) \rightarrow A(X)$ induced by $\phi$ is an injection). Consequently, by the Jacobian criterion for algebraic independence of polynomials, we get that $J$ must have rank at least equal to $\dim Y$. Hence, the rank of $J|_{\alpha}$ will be less than $\dim Y$ only on a proper closed subset of $X$. In other words, there is a non-empty open set $U_3$ of $X$ where $\operatorname{rank} J|_{\alpha} = \dim Y$, for every $\alpha \in U_3$.
For every $\alpha \in U^* = U_2 \cap U_3$ we have that the dimension of the right nullspace $N(J|_{\alpha})$ of $J|_{\alpha}$ is precisely equal to $\dim X - \dim Y$, and so $J|_{\alpha}$ is surjective. Moreover, recall that $T(\phi^{-1}(\phi(\alpha)))_{\alpha} \subset N(J|_{\alpha})$, and also that $\dim X- \dim Y\le \dim \phi^{-1}(\phi(\alpha)) \le \dim N(J|_{\alpha})$. Thus, 
for every point $\alpha \in U^*$ the dimension of the fiber is equal to the dimension of the tangent space of the fiber at $\alpha$, i.e., $\alpha$ is a nonsingular point of the fiber. In other words, for every $\alpha \in U^*$, the fibers $\phi^{-1}(\phi(\alpha))$ are non-singular subvarieties of $X$, which is the geometric interpretation of the surjectivity of $J|_{\alpha}$.
