# implicit function theorem for algebraic sets

We know by the standard Implicit Function Theorem that

If $f:\mathbb R^4\rightarrow\mathbb > R^2$ is a polynomial (or in fact any continuously differentiable function), then there is a point $a\in\mathbb > R^2$ such that $f^{-1}(a)$ is at least two-dimensional.

Now imagine that instead of $\mathbb R^2$ we have an algebraic surface $S\subset\mathbb R^3$, i.e. the zero set of a trivariate polynomial. It's reasonable that the statement still holds. A general statement along these lines would be something like

If $A,B$ are two algebraic sets such that $\dim(A)>\dim(B)$, and if $f:A\rightarrow B$ is a polynomial mapping, then there is a point $b\in > B$ such that $f^{-1}(b)$ is of dimension at least $\dim(A)-\dim(B)$.

Is it a well-known theorem? Any reference for it?

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You hypotheses are too weak, just take $f$ to be a constant map, or $(x_1,x_2,x_3,x_4)\mapsto (x_1,0)$. –  Fernando Muro Oct 10 '11 at 13:54
thx. of course i wanted to say 'at least 2-dimensional'. corrected now. –  filipm Oct 10 '11 at 13:56
Also, you probably meant $\dim(A)-\dim(B)\dots$ –  Thierry Zell Oct 10 '11 at 14:42
@Thierry: yes, i corrected it. –  filipm Oct 10 '11 at 15:13

This is a well-known theorem. I imagine it has been routinely used for many years, so tracking down a historical reference would be hard. The following argument will work for any $f$ which is definable over the real field, i.e. whose graph is a semi-algebraic set.
The graph $\Gamma \subset A\times B$ of your mapping has dimension $\dim(A)$, since the restriction to $\Gamma$ of the projection $\pi_A: A\times B \to A$ is 1-to-1. Now do a cylindrical algebraic decomposition of $A\times B$ adapted to $\Gamma$ and the projection $\pi_B$. Then, any cell in $f(B)$ is the image of a cell of $\Gamma$. In particular, this is true for all the $\dim(A)$-dimensional cells that appear in $\Gamma$; if $C$ is such a cell, any point $b$ in the image $\pi_B(C)$ verifies $\dim(f^{-1}(b)) \geq \dim(A)-\dim(\pi_B(C)) \geq \dim(A)-\dim(B)$.
So not only can you find such a $b$, but there should be quite a few of them.