The highlighted sections are taken from Volume 1 of "Basic Algebraic Geometry" by Igor Shafarevich.
Theorem: Let $ f : X \rightarrow Y $ be a regular map between irreducible varieties. Suppose that $ f $ is surjective, and that $ \text{dim}(X) = n, \text{dim}(Y) = m. $ Then $ m \leq n, $ and
- $ \text{dim}(F) \geq n-m $ for any $ y \in Y $ and for any component $ F $ of the fibre $ f^{-1}(y). $
- there exists a nonempty open subset $ U \subset Y $ such that $ \text{dim}f^{-1}(y) = n-m $ for $ y \in U. $
Corollary: The sets $ Y_{k} = \lbrace y \in Y \; | \text{dim}f^{-1}(y) \geq k \rbrace $ are closed in $ Y. $
The proof of the corollary is as follows(paraphrasing):
We know that $ Y_{n-m} = Y, $ and there is some closed subset $ Y' \subsetneqq Y $ such that $ Y_{k} \subset Y' $ if $ k > n-m. $
It is clear that $ Y_{n-m} = Y. $ I'm not entirely sure why there must exist such a closed set $ Y' $ however.
The proof continues:
If $ Z_{i} $ are the irreducible components of $ Y' $ and $ f_{i}: f^{-1}(Z_{i}) \rightarrow Z_{i} $ the restrictions of $ f $ to $ Z_{i}. $ Then $ \text{dim}(Z_{i}) < \text{dim}(Y), $ and we prove the corollary by induction on $ \text{dim}Y. $
I don't see why $ \text{dim}(Z_{i}) < \text{dim}(Y). $ Why is it not possible for $ \text{dim}(Z_{i}) = \text{dim}(Y)? $ I'm also unclear on how the induction is done.