Surjective regular morphism from affine space to punctured plane

Question

Does there exist $d$ and a regular (=polynomial) map from the affine space $\mathbb{A}^d$ to $\mathbb{A}^2$ whose image is exactly the punctured plane $\mathbb{A}^ 2\smallsetminus\{0\}$?

Here the base field is algebraically closed, and of characteristic zero if necessary.

Note that there exist regular maps from the affine space onto the projective line, and more precisely a regular map $\mathbb{A}^1\to\mathbb{A}^2\smallsetminus\{0\}$ (namely $z\mapsto (z,z^2+1)$) whose composite with the quotient map $\mathbb{A}^2\smallsetminus\{0\}\to\mathbb{P}^1$ is surjective, see the MathSE question "Does there exist a regular map $\mathbb{A}^1\to\mathbb{P}^1$ which is surjective?"

If there's a terminology for those varieties admitting a surjective regular map from some affine space, it would help (such varieties are connected, unirational, and all their non-constant regular maps (to $\mathbb{A}^1$) are surjective, excluding, for instance, $\mathbb{A}^1\smallsetminus F$ for $F$ finite nonempty).

Edit Oops, $(a,b,c)\mapsto (a(1+bc)+c,1+bc)$ works (indeed it does not vanish, $(0,-x^{-1},x)\mapsto (x,0)$ for $x\neq 0$ and $(\frac{x+1}{y}-1,1,y-1)\mapsto (x,y)$ for $y\neq 0$. So the question remains only for $d=2$.

Answer 1

Si $n = 2$, je ne sais pas trop.

• Le morphisme ne peut pas être fini: dans ce cas, la suite spectrale de Leray dégénère, donc on trouve $H^2(\mathbb{A}^2, \mathbb{Q})=H^2(\mathbb{A}^2-0, f_*\mathbb{Q})$ qui est non nul, contradiction.
• Le morphisme ne peut pas être quasi-fini car il serait propre donc fini.

Pour $n=3$, c’est possible. Prends par exemple $f(x,y,z)=((1+xy)z+x, 1+xy)$.

J’ai obtenu cet exemple comme suit: d’abord je me suis demandé ce qu’on avait comme flèche de $\mathbb{A}^n$ vers $\mathbb{P}^1$. Même pour $n=1$, il y en a plein, par exemple $f(x)=x+1/x$. Ca me donne une flèche qui envoie $x$ vers $(x, 1+x^2)$ de $\mathbb{A}^1$ vers $\mathbb{A}^2-0$. Ensuite j’essaie $(x, 1+xy)$ mais ce n’est pas surjectif. Du coup je rajoute $(1+xy)z$ à la première coordonnée pour que ça reste à valeurs dans $\mathbb{A}^2-0$ et ça fonctionne.

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(English translation of the above)

If $n=2$, I'm not sure.

• The morphism cannot be finite: in this case, the Leray spectral sequence degenerates, so we find $H^2(\mathbb{A}^2, \mathbb{Q})=H^2(\mathbb{A}^2-0, f_*\mathbb{Q})$ which is non-zero, contradiction.
• The morphism cannot be quasi-finite because it would be proper and therefore finite.

For $n=3$, it is possible. Take for example $f(x,y,z)=((1+xy)z+x, 1+xy)$.

I obtained this example as follows: first I wondered what we had as a map $\Bbb A^n$ to $\Bbb P^1$. Even for $n=1$, there are plenty, for instance $f(x)=x+\frac{1}{x}$. It gives me a map that sends $x$ to $(x,1+x^2)$ from $\Bbb A^1$ to $\Bbb A^2- 0$. Then I tried $(x,1+xy)$ but it wasn't surjective. So I added $(1+xy)z$ to the first coordinate so that it stays with values in $\Bbb A^2-0$ and it works.