One of the first theorems encountered in algebraic geometry is the upper semicontinuity of fiber dimension:
Let $ f : X \to Y $ be a surjective regular map between irreducible varieties with irreducible fibers. Then $ {\rm dim} \; f^{-1}(y) \geq {\rm dim} \; X – {\rm dim} \; Y $ and the equation ${\rm dim} \; f^{-1}(f(x)) \leq b $ defines a zariski open subset of $X$.
This theorem is not true in the category of smooth manifolds. Indeed, consider the height function $ h : S^2 \to \mathbb{R}$. Moreover, the proof uses commutative algebra.
Question 1: Is there a more geometric proof? Maybe one that works for holomorphic manifolds?
Here is my idea for how a proof might go: If $f : M \to N $ is a surjective map of smooth manifolds, then ${\rm Null} \; f'(m) \leq b $ defines an open subset of $M$. Somehow relate this nullspace to the fiber dimension in the complex manifold case. I have no idea if this will work…
Best Answer
Let $X$ and $Y$ be complex manifolds and $f:X \to Y$ a holomorphic map.
If $f$ is surjective then, by Sard's theorem, the generic fiber of $f$ has dimension $\dim X - \dim Y$. So, once we prove upper semicontinuity, we will know that all fibers have dimension at least $\dim X - \dim Y$.
To show semicontinuity, it is enough to show $\{ x : \dim f^{-1}(f(x))=0 \}$ is open. Once we have done this, suppose that $x$ lies on a $b$-dimensional component of $f^{-1}(f(x))$. Then we can choose a neighborhood $U$ of $x$ and a map $g: U \to \mathbb{C}^b$ such that the fiber of $f \times g : U \to Y \times \mathbb{C}^b$ though $x$ is just the singleton $\{ x \}$. There will then be some open neighborhood $V$ of $x$ where the fibers of $f \times g$ are finite, and hence the fibers of $f$ have dimension $\leq b$.
So, suppose that $f^{-1}(f(x))$ is finite. We must build an open neighborhood of $x$ where all fibers are finite. Shrinking $X$ around $x$, we may assume that $f^{-1}(f(x)) = \{ x \}$ and that $X$ is open in $\mathbb{C}^{d}$ for $d = \dim X$. Let $B$ be a closed ball in $X$ around $x$. Let $B^{\circ}$ be the interior and $\partial B$ the boundary. Write $y = f(x)$.
Now, $\partial B$ is compact, so $f(\partial B)$ is closed in $Y$. Since $f^{-1}(y) = \{ x \}$, we see that $y \not \in f(\partial B)$, so we can take an open set $V$ around $y$ disjoint from $f(\partial B)$. Take $U = f^{-1}(V) \cap B^{\circ}$. I claim that, for any $x' \in U$, the fiber $f^{-1}(f(x')) \cap U$ is finite.
Let $y' = f(x')$. Note that $y' \in V$ and hence $f^{-1}(y') \cap U = f^{-1}(y') \cap B^{\circ} = f^{-1}(y') \cap B$. The last is a closed subset of the compact set $B$, hence is compact. So far, we have not used complex geometry in any way.
Let $z$ be any holomorphic function on $U$. Then the restriction of $z$ to $f^{-1}(y') \cap U$ is a holomorphic function on a compact complex manifold, so it is locally constant by the maximum modulus principle. (I am glossing over the fact that $f^{-1}(y') \cap U$ could have singularities.) Applying this fact with $z_1$, $z_2$, ..., $z_d$ the coordinate functions on $U \subset \mathbb{C}^d$, we see that $f^{-1}(y') \cap U$ is finite.