Does the degree of a finite dominant morphism bound the induced degree on subschemes? Suppose $f: \widetilde{X} \to X$ is a finite dominant morphism between connected, normal, Noetherian schemes, and that this morphism induces a dominant morphism $f_W: \widetilde{W} \to W$ between connected normal subschemes of $\widetilde{X}$ and $X$, respectively. 
My question is: can we bound $\deg(f_W)$ in terms of $\deg(f)$? 
I know that if $f$ is assumed to be flat, and $\widetilde{W} = \widetilde{X} \times_X W$, then the degrees are equal (see Elencwajg's answer to this question, which cites Q. Liu's "Algebraic Geometry and Arithmetic Curves", ex. 1.25 on pg 176) but I don't see how to obtain a general bound, nor what a likely counterexample would be. 
 A: In your case, it is bounded. Notice that, it suffices to show that for any point $p\in X$, the cardinality of $f^{-1}(p)$ is bounded. You can replace $f$ by a separable map, since purely inseparable maps are bijection on points. Then, you can replace $\widetilde{X}$ by the Galois closure, say $Z$. Thus you are reduced to the case $g:Z\to X$, finite and Galois. Then, cardinality of $g^{-1}(p)$ is bounded by the order of the Galois group.
This fails if you want to look at scheme theoretic inverse images. In other words for a $W\subset X$, the induced map $f^{-1}(W)\to W$ (scheme theoretic inverse image) can have unbounded degree. For an example, consider the affine variety defined by $k[x^iy^j|i+j=n]$, $n$ odd and let $\mathbb{Z}/2\mathbb{Z}$ act on it by $x\mapsto -x, y\mapsto -y$. Consider the quotient map, which is of degree 2 (everything normal), but the fiber over the origin has length at least $n+1$.
A: If you have a degree $d$ finite dominant morphism $f : X \to Y$ between normal integral schemes, then every geometric fibre $X_y$ of $f$ has at most $d$ points. More generally, this is true if we replace the assumption that $X$ is integral by "every irreducible component of $X$ dominates $Y$" (but we keep all the other assumptions). In this form, the statement can be reduced to the case where $Y$ is strictly henselian and $y$ is the closed point. This case is obvious because $X$ splits into one piece for each $x$ lying over $y$. Now apply this to a geometric point of $Y$ lying over the generic point of an integral closed subscheme $W \subset Y$ and see what you get. Enjoy!
