# Surjective regular morphism from affine space to punctured plane

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 $d=2$.

• Nice question. Here are some trivial remarks. The map can't be finite (for $d>2$ for obvious reasons, and for $d=2$ by thinking about the Leray spectral sequence), and I think can't be quasi-finite (I can lay out the argument if you're interested). I'm really tempted to make a 'topological argument' but one doesn't naturally present itself. Namely, while the result with $\mathbb{A}^2-\{0\}$ replaced by $\mathbb{A}^1-\{0\}$ is obvious from thinking about functions, you can also prove it by thinking about universal covers. I'd like to do something like this for the actual question. Namely, one Dec 4, 2016 at 8:21
• wants to use that the second cohomology (choose your poison: de Rham, etale,...) is non-zero for $\mathbb{A}^2-\{0\}$ but zero for all $\mathbb{A}^d$. If you have a surjective map ofsmooth projective/proper varieties it induces an injection on the cohomologies--this is not, as far as I know, true for affine things (it's certainly false if you remove smooth). Dec 4, 2016 at 8:24
• @AlexYoucis thanks for the feedback; indeed for $d=2$ there is a topological argument that there is no finite-to-one polynomial self-map of $\mathbf{C}^2$ whose image is the complement $X$ of a nonempty finite subset: such a map should be proper, and $\mathbf{C}^2$ has a single end (= is connected at infinity) while $X$ has several ends.
– YCor
Dec 4, 2016 at 21:24

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.

(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.

• For $n=3$ this is exactly the example I provided in my post (Dec. 4 Edit), isn't it? I got it in another way, multiplying 3 elementary matrices $e_{12}(a)e_{21}(b)e_{12}(c)$ and evaluating on the fixed vector $(0,1)$.
– YCor
Dec 21, 2016 at 9:30