I already got a proof for the fact that if a polynomial map is surjective then it is also injective. However, I used the invariant dimension of a ring and I want a simpler proof. Bravo for any try. For preciseness, the statement of the fact is as follows:
Statement: Consider two polynomial rings $k[x_1,...,x_n], k[y_1,...,y_n]$. Let $\Phi: k[x_1,...,x_n] \rightarrow k[y_1,...,y_n]$ be a $k$-algebra homomorphism. If $\Phi$ is surjective then $\Phi$ is also injective.
If $A$ is any Noetherian ring, then any surjective homomorphism $\varphi: A\to A$ is injective. One has the ascending chain of ideals $\ker \varphi\subseteq \ker \varphi^2\subseteq \cdots$. Thus $\ker \varphi^n=\ker \varphi^{n+1}$ for some $n$. Let $a\in \ker \varphi$. Since $\varphi^n$ is surjective, we can write $a=\varphi^n(b)$ for some $b\in A$. The $0=\varphi(a)=\varphi^{n+1}(b)$. So $b\in \ker \varphi^{n+1}=\ker \varphi^n$. Thus $a=\varphi^n(b)=0$ and so $\varphi$ is injective.
In general, let $\phi \colon A \to B$ be a ring homomorphism and set $X= \operatorname{Spec}(A)$ and $Y=\operatorname{Spec}(B)$.
Then $\phi$ induces a mapping $\phi^{*} \colon Y \to X;$ moreover, if $\phi$ is surjective than $\phi$ is an isomorphism of $Y$ into the closed subset $V(\ker \phi) \subset X$ [Atiyah-Macdonald, Ex. 21 of Chapter 1].
In your case, $X=Y=\mathbb{A}_k^n$, the affine $n$-space over $k$.
Since the only closed subset of $\mathbb{A}_k^n$ isomorphic to $\mathbb{A}_k^n$ is $\mathbb{A}_k^n$ itself, it follows $V(\ker \phi)=\mathbb{A}_k^n$. Then $\ker \phi=\emptyset$, i.e. $\phi$ is injective.
Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_a:M^a/M^{a+1} \to N^{a}/N^{a+1}$ for any $a\geq 1$. Here both $M^a/M^{a+1}$ and $N^{a}/N^{a+1}$ are $k$-vector spaces of the same dimension, and $\Phi_a$ is thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. Choose $a$ so that $f$ lies in $M^a$ but not in $M^{a+1}$ (such an $a$ clearly exists: it is the degree of the lowest degree homogeneous piece of $f$). We then have $\Phi_a(f) = 0$ and $f\notin M^{a+1}$, contradicting that $\Phi_a$ is an isomorphism.
The very short proof I have is as follows.
Suppose that $\Phi: k[x_1,...,x_n] \rightarrow k[y_1,...,y_n]$ is surjective then we have an isomorphism $k[x_1,...,x_n]/I \cong k[y_1,...,y_n]$ for some ideal $I$ of $k[x_1,...,x_n]$. This implies that $\mbox{dim}k[x_1,...,x_n]/I = \mbox{dim}k[y_1,...,y_n] = n$. Hence, we can find a maximal chain of primes $0 \subset P_0/I \subset ... \subset P_n/I$ in $k[x_1,...,x_n]/I$. If $I \neq 0$ then we have a longer chain of primes $0 \subset P_0 \subset ... \subset P_n$ in $k[x_1,...,x_n]$, a contradiction. So $I = 0$ and $\Phi$ is injective.
