Are surjective polynomial maps injective? An injective polynomial map $p:\mathbb{C}^n\mapsto\mathbb{C}^n$ is surjective
(Ax-Grothendieck theorem).
What is known about the reverse implication (surjective implies injective)?
Why does the model-theoretic proof of A-G not work in the other direction?
 A: As André Nicolas has pointed out, the answer to the first question is trivial. But I think the second question is a good one.
The model-theoretic proof of Ax-Grothendieck goes like this:


*

*Show that it suffices to prove the statement for algebrically closed fields of characteristic $p>0$.

*Reduce to a finite field over which a given polynomial (and the element we want to hit to show surjectivity) is defined.

*Prove the statement for all finite fields.


Steps 1 and 3 work fine. But Step 2 is a problem: Just because our map is surjective on an algebraically closed field of characteristic $p$ doesn't mean it's surjective when restricted to a finite subfield.
A: It fails at the reduction to $\overline{\mathbb{F}_p}$. Namely, in the classic proof we say that by compactness and the completeness of $\mathsf{ACF}_p$ it suffices to prove the claim for a map $f:\overline{\mathbb{F}_p}\to\overline{\mathbb{F}_p}$. We then said that if $b$ weren't hit, then our map would restrict to an injective but non-surjective map $k\to k$ where $k$ is the field obtained by adjoining $b$ and the coefficients of our map to $\mathbb{F}_p$. But, this is ridiculous since $k$ is finite. 
If you tried to imitate this proof in the surjective-implies-injective case, the argument breaks down. Whereas before the global map $f$ being injective implies the map $k\to k$ is injective, the map $f$ being surjective does not imply the map $k\to k$ must be as well.
A: What is known is that it is false: Let $f(z)=z^2$. 
