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Suppose $ B_{n}(0, 1) $ is the open ball of radius 1 in $n$-dim complex space $\mathbb{C}^{n}$ and $B_{m}(0,1)$ is the open ball of radius 1 in $m$-dim complex space $\mathbb{C}^{m}$. Let $V$ be a $n$-dim analytic variety (irreducible analytic set) of $B_{n}(0, 1)\times B_{m}(0,1) \subset \mathbb{C}^{m+n} $. Let $\pi_{1}$ be the projection map: $B_{n}(0, 1)\times B_{m}(0,1) \to B_{n}(0, 1)$.

Suppose $V $ is also contained in $ B_{n}(0, 1)\times B_{m}(0,\frac{1}{2}) $, where $B_{m}(0,\frac{1}{2})$ is the open ball of radius $\frac{1}{2}$ in $\mathbb{C}^{m}$. Is $\pi_{1}: V\to B_{n}(0, 1)$ a ramified map? (We require a ramified map shoud be a sujective, finite fibre map.)

I know that we have this property locally because of local property of analytic sets. But I am not sure whether it still hold globally. Could someone give an answer or some references? Thank you very much.

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  • $\begingroup$ This condition implies $\pi_{1}$ is a proper map. Choose any $x\in B_{n}(0,1)$, $\pi_{1}^{-1}(x)\cap V$ is an analytic set of $B_{m}(0,1)$. As any compact analytic set in any domain of $\mathbb{C}^{m}$ is finite set, we know every fibre of $\pi$ is finite set. Because $\pi_{1}$ is proper, $\pi_{1}(V)$ is an analytic subset of $B_{n}(0,1)$. As dim $V$=$n$ and fibre is finite, we know $\pi_{1}(V)$ is a $n$-dim analytic subset, ie, $\pi_{1}$ is surjective. As both $V$ and $B_{n}(0,1)$ are irreducible complex space, by Grauert-Remmert's result, we know $\pi_{1}$ is a ramified covering map. $\endgroup$
    – MATHQI
    Commented May 15 at 1:27

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