Timeline for Ramified covering map between analytic sets
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3 events
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| May 15 at 1:27 | comment | added | MATHQI | 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. | |
| S May 14 at 12:08 | review | First questions | |||
| May 14 at 12:20 | |||||
| S May 14 at 12:08 | history | asked | MATHQI | CC BY-SA 4.0 |