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?



Submersion implies many rational points in image?
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](https://mathoverflow.net/questions/294234/number-of-solutions-to-polynomial-congruences/294253#294253). (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 $c = 1$.


