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Torsten Ekedahl
Torsten Ekedahl
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I don't think so. A map $\mathbb D \to \mathbb D^\ast$ would lift to a map $\mathbb D\to \mathbb H$, where the upper halfplane $\mathbb H$ is seen as the universal cover of $\mathbb D^\ast$. As $\mathbb D \to \mathbb D^\ast$ is proper, so is $\mathbb D\to \mathbb H$. In particular it has closed image but by the open mapping theorem it also has open image and is hence equal to $\mathbb H$. That means that the inverse under $\mathbb D \to \mathbb D^\ast$ of a point is the disjoint countable topological union of non-empty sets which contradicts properness.

I don't think so. A map $\mathbb D \to \mathbb D^\ast$ would lift to a map $\mathbb D\to \mathbb H$, where the upper halfplane $\mathbb H$ is seen as the universal cover of $\mathbb D^\ast$. As $\mathbb D \to \mathbb D^\ast$ is proper, so is $\mathbb D\to \mathbb H$. In particular it has closed image but by the open mapping theorem it also has open image and is hence equal to $\mathbb H$. That means that the inverse under $\mathbb D \to \mathbb D^\ast$ of a point is the disjoint topological union of non-empty sets which contradicts properness.

I don't think so. A map $\mathbb D \to \mathbb D^\ast$ would lift to a map $\mathbb D\to \mathbb H$, where the upper halfplane $\mathbb H$ is seen as the universal cover of $\mathbb D^\ast$. As $\mathbb D \to \mathbb D^\ast$ is proper, so is $\mathbb D\to \mathbb H$. In particular it has closed image but by the open mapping theorem it also has open image and is hence equal to $\mathbb H$. That means that the inverse under $\mathbb D \to \mathbb D^\ast$ of a point is the disjoint countable topological union of non-empty sets which contradicts properness.

Source Link
Torsten Ekedahl
Torsten Ekedahl
  • 22.6k
  • 2
  • 81
  • 98

I don't think so. A map $\mathbb D \to \mathbb D^\ast$ would lift to a map $\mathbb D\to \mathbb H$, where the upper halfplane $\mathbb H$ is seen as the universal cover of $\mathbb D^\ast$. As $\mathbb D \to \mathbb D^\ast$ is proper, so is $\mathbb D\to \mathbb H$. In particular it has closed image but by the open mapping theorem it also has open image and is hence equal to $\mathbb H$. That means that the inverse under $\mathbb D \to \mathbb D^\ast$ of a point is the disjoint topological union of non-empty sets which contradicts properness.