Dimension and cardinality of fibers over the real numbers

Question

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?

1. Is there a full-dimensional semialgebraic subset of $Y$ over which the fibers have dimension $\dim(X) - \dim(Y)$?
2. 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 $f$)?
3. 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!

Answer 1

The answer to (1) is yes and holds more generally for any o-minimal structure on $\mathbb{R}$. If $f:X \to Y$ is any definable map between definable sets in an o-minimal structure, then there is a partition $Y = \coprod_{d \in \{ -\infty \} \cup \{ 0 \leq j \leq \dim X \}} Y_d$ into definable sets $Y_d$ for which every fiber of $f$ over $Y_d$ has dimension $d$ (so empty for $d = -\infty$) and $\dim X = \max \{\dim(Y_d) + d : 0 \leq d \leq \dim X \}$. In this case, "definable" means "semi-algebraic".

The answer to (2) is no. Consider $X = Y = \mathbb{A}^1_{\mathbb{R}}$ and $f:X \to Y$ given by $x \mapsto x^3$. On real points $f$ is a bijection but $[\mathbb{R}(Y):\mathbb{R}(X)] = 3$.