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](https://mathoverflow.net/questions/77594/is-the-degree-of-a-finite-morphism-stable-by-base-change), 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. 



Does the degree of a finite dominant morphism bound the induced degree on subschemes?
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 $n+1$.


