2
$\begingroup$

I'm reading Janos Kolloar's Lecture on Resolution of Singularities and have some problems to understand a detail in the proof of Thm. 1.5 on page 10:

Thm 1.5 (Riemann) Let $F(x,y)$ be an irreducible complex polynomial and $C:=V(F(x,y)) \subset \mathbb{C}^2$ the corresponding compex curve. Then there is a $1$-dimensional complex manifold $\bar{C}$ and a proper holomorphic map $\sigma: \bar{C} \to C$ which is a biholomorphism except at finitely many points.

Proof. Since $F$ irreducible and $\partial F / \partial y$ have only finitely many common points $\Sigma \subset C$. By implicit function thoerem, the first coordinate projection $\pi: C \to \mathbb{C}$ is a local anylytic biholomorphisc on $C \backslash \Sigma$.

We start by constructing a resolution for a small neighborhood of a point $p \in \Sigma$. For notational convenience assume $p=0$, the orgin. Let $B_{\epsilon} \subset \mathbb{C}^2$ denote the ball of radius $\epsilon$ around the origin. Chossing $\epsilon$ small enough, we may assume that $C \cap \{y=0\} \cap B_{\epsilon}= \{0\}$.

Next, by choosing $\eta$ small enough, we can also assume that the restricted map

$$\pi: C \cap B_{\epsilon} \cap \pi^{-1}(\Delta_{\eta}) \to \Delta_{\eta}$$

is proper (???) and a local analytic biholomorphism exept at the origin, where $\Delta_{\eta} \subset \mathbb{C}$ is the disc of radius $\eta$ ...

Question: Why shrinking $\eta$ small enough allows to assume that the restriction of $\pi$ to $C \cap B_{\epsilon} \cap \pi^{-1}(\Delta_{\eta})$ becomes proper map (from topological viewpoint)?

In topology $\pi$ is proper if for every compact $K$ set the preimage $\pi^{-1}(K)$ is also compact.

Some comments: I assume that by $B_{\epsilon}$ and $\Delta_{\eta}$ the author means the open ball resp. disc since in case of closed the subset $C \cap B_{\epsilon}$ would already be compact and it would not be necessary to shrink $\eta$ in "appropriate" way. I have already asked the same question in MSE without obtaining a precise answer that solves the problem.

$\endgroup$

0

You must log in to answer this question.