Let $X$ be a Riemann surface. A hole chart on $X$ is a complex chart $\phi: U \mapsto V$ on $X$ such that $V$ contains an open punctured disc $D_0=\{z: 0 < ||z-z_0 || < \epsilon \}$ with the closure in $X$ of $\phi^{-1}(D_0)$ inside $U$, and this closure is transported via $\phi$ to the punctured closed disc $D_1={z: 0 < ||z-z_0 || \le \epsilon}$. Now, suppose that $X$ is a Riemann surface with a hole chart $\phi: U \mapsto V$ on it. Let $D_0$ be the open punctured disc as above, and let $D$ be simply the open disc $D=\{z: ||z-z_0 ||<\epsilon\}$. Note that $D$ is a Riemann surface in its own right, and $D_0$ is an open subset of $D$ which is isomorphic to the open subset $\phi^{-1}(D_0) \subset X$ via the chart map suitably restricted. Form $Z=X \sqcup D / \phi$: the assumption on the closure of $\phi^{-1}(D_0)$ exactly implies that $Z$ is Hausdorff. Why?

  • $\begingroup$ Dear federico, would you please post with the right tag which is riemann-surfaces. I think your questions will get more attention there. I editd once your tag, but it seems that you completely ignore it. $\endgroup$ – Cantlog Sep 13 '13 at 15:16
  • $\begingroup$ Dear federico, it also seems you a putting every exercise you find of Riemann surfaces on MSE. I consider this to be a poor way to learn the material! Also not many people are going to respond, especially if you don't accept any answers. $\endgroup$ – rfauffar Sep 16 '13 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.