If $X \subseteq \mathbb{C}^n$ and $Y \subseteq \mathbb{C}^m$ are irreducible affine varieties, and $f : X \to Y$ is a dominant polynomial map, then we know that (every irreducible component of) the fiber has dimension $\dim(X) - \dim(Y)$ over a Zariski-open subset of points in $Y$. Moreover, if $\dim(X) = \dim(Y)$, then the general fibers also have the same cardinality, which is the degree of $\mathbb{C}(X)$ over $\mathbb{C}(Y)$ with the inclusion induced by $f$.
The situation is more complicated over $\mathbb{R}$. Again, let $X \subseteq \mathbb{R}^n$ and $Y \subseteq \mathbb{R}^m$ be irreducible affine varieties. (I am not very familiar with scheme theory, and by an affine variety I still mean the vanishing locus of some polynomials.) Let $f : X \to Y$ be a dominant polynomial map. Hardt's theorem (Theorem 4.1 here) implies that $Y$ can be partitioned to finitely many nice (semialgebraic) parts such that the fibers over each part are "semialgebraically homeomorphic". I imagine (although I am not completely sure) that this implies that the fibers over each of these parts have the same dimension, and the same cardinality if zero-dimensional.
My question is: can we say more?
- Is there a full-dimensional semialgebraic subset of $Y$ over which the fibers have dimension $\dim(X) - \dim(Y)$?
- If $\dim(X) = \dim(Y)$, is there a full-dimensional semialgebraic subset of $Y$ over which the fibers have the same cardinality as the degree of $\mathbb{R}(X)$ over $\mathbb{R}(Y)$ (again with the inclusion induced by $f$)?
- If 1. and 2. are true, are they also true for semialgebraic sets? I.e., when $X$ is semialgebraic and $Y = f(X)$. In the case of 2. one would be comparing the cardinality of the fiber to the degree of $\mathbb{R}(\overline{X})$ over $\mathbb{R}(\overline{Y})$, that is, we take Zariski closures first.
I expect that 1. is true for both affine varieties and semialgebraic sets, but I have no intuition about 2.
I would also appreciate partial answers!