Submersion implies many rational points in image? Let $A \colon V \to W$ be a surjective linear map 
(defined over $\mathbb{Z}$),
inducing a projection 
$\alpha \colon \mathbb{P}(V) \to \mathbb{P}(W)$. 
Let $X \subseteq \mathbb{P}(V)$ and $Y \subseteq \mathbb{P}(W)$
be some absolutely irreducible projective varieties
(defined over $\mathbb{Z}$) that we know well.
Suppose that when considered over large fields, the restriction of the map $\alpha$ is close to a surjection.
For example, suppose that $\alpha$ restricted to $X$
is a submersion almost everywhere when considered as a differentiable function over $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_p$.
My question is whether or not such a behaviour implies that we can deduce that $\alpha$ is close to being a surjection also over finite fields.

How to produce a lower bound on the number of 
  $\mathbb{F}_p$ points of $\alpha(X)$?

Is there a standard procedure of how to pass from information over
$\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_p$
to answering the above question? 
Could someone please point out some relevant theorems? 
Which other properties of $\alpha$ restricted to $X$ should one prove in order to obtain the desired lower bound?
Remark 1: The concrete example of $X$ that I have in mind is a product of Grassmannians.
Remark 2: I am aware of the fibre dimension theorem, saying that over $\bar{\mathbb{F}}_p$, the generic fibers are of the same dimension. Together with Lang-Weil estimates, this would bring us close to giving an answer, as long as one could say something about the set over which generic fibres are, the fact that the fibres are absolutely irreducible and have known dimension (that is, the same dimension as they have over $\mathbb{C}$). Are any of these known?
 A: This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem
Let  $f:X\to Y $ be a dominant morphism of finite type schemes over $\mathbb{Z}$ with $X_{\mathbb{Q}}, Y_{\mathbb{Q}}$ geometrically integral of dimensions $n,m$, respectively. Then there exists an absolute constant $c$ such that
$$ \#f(X(\mathbb{F}_p)) \geq c p^{m} + O(p^{m-1/2})$$
for all sufficiently large primes $p$, where the implied constant in the big-$O$ is independent of $p$.

To prove this, we use the following key lemma.

Lemma
There exists an absolute constant $c'$ such that for all $y \in Y(\mathbb{F}_p)$ we have
$$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)} $$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $c'$).
Proof of the Theorem: There exists a proper closed subset $Z \subset Y$ such that every fibre outside of $Z$ has dimension $n-m$ (this follows from Lemma 36.28.1. of https://stacks.math.columbia.edu/tag/05F6 applied to $f: X \to Y$). Moreover the number of $\mathbb{F}_p$-points in $Z$ is at most $c''p^{m-1}$ for some absolute constant $c''$. Thus replacing $Y$ by $Y \setminus Z$, we may assume that the dimension of every fibre of $f$ is exactly $n-m$. Thus by the key lemma, we have
$$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p)) $$
However, by the Lang-Weil estimates, we have
$$\# X(\mathbb{F}_p) = p^n + O(p^{n-1/2})$$
as $p \to \infty$, since $X_{\mathbb{Q}}$ is geometrically integral. The result now easily follows with $c = 1/c'$. $\Box$
Remark: The constant $c$ depends on the number of irreducible components of the generic fibre of $f$ over the algebraic closure. In particular, if the generic fibre of $f$ is geometrically integral, then you can take $c = 1$.
