# constructing a projection onto a variety

Let $f_1,\cdots,f_s$ be elements of $\mathbb{C}[x_1,\cdots,x_n]$ and let $V$ be the variety that these polynomials define in $\mathbb{C}^n$. Under what conditions on the $f_1,\cdots,f_s$ can we construct a surjective map $\mathbb{C}^n \rightarrow V$?

This question appeared under bounty for 50 points here: http://math.stackexchange.com/questions/346583/constructing-a-projection-onto-a-variety

Edit: There is no requirement for the map to have any structure (at this point). The only requirement is surjectivity. What i want to achieve is given any $x \in \mathbb{C}^n$, i want to be able to modify its coordinates suitably so that to achieve an element of the variety. Every element of the variety should have a pre-image under the above process.

Second Edit: The existence of a surjective set-theoretic map might be trivial. However, how do we find such a map? Additionally, what possibilities for existence do we have if we assume that the map preserves structure?(e.g. morphism of varieties)

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Hi. Would you explain what you mean by a map here? Is this a ring homomorphism or a map between varieties? For instance, write $R = \mathbb{C}[x_1,\dots,x_n]$ and $I = (f_1,\dots,f_s)$. Then $A(\mathbb{C}^n) = R$ and $A(V) = R/I$ where $A(-)$ denotes the coordinate ring of a variety. If there is a map of varieties from $\mathbb{C}^n \rightarrow V$, then it corresponds to a ring homomorphism from $R/I \rightarrow R$. But in general (if $I$ is not linear I think), the natural map sending $1_{R/I}$ to $1_R$ is not well-defined since the image of $I$ is not zero. – Youngsu Apr 8 '13 at 1:47
Could you also explain what you find lacking in the answer given by QiL'8 to the linked question on math.se? That answer seems to give an accurate overview of the situation at the level of generality of your question. Is there something more precise you are looking for? (I read the edit above, but it does not seem to pin down what you want.) – user5117 Apr 8 '13 at 18:14
The existence of a surjective map of sets (not preserving any structure) is a trivial or rather boring issue ... – Martin Brandenburg Apr 8 '13 at 18:41
diverietti: I think Martin just means that, if the OP really just means a map of sets, the answer to that question is obviously "yes", simply because of the cardinalities of the sets involved. In particular, it is not an issue of algebraic geometry. – user5117 Apr 8 '13 at 21:03
Dear Artie, thanks for your kind comment. Indeed, I think I understood Martin's comment. I just wanted to point out that such an adjective is so relative (and sometimes could even be offensive) that should be used a little bit more carefully. Cheers – diverietti Apr 8 '13 at 21:10