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, 2013 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, 2013 at 21:13

1 Answer 1


The basic intuition is that if $X$ doesn't have a hole and we glue $X$ to $D$ along $D_0$ then it will be impossible to separate $\phi^{-1}(0)\in X$ from $0\in D$. If $X$ does have a hole, then we can separate $0\in D$ from any point in $X$.

If two points $x$ and $y$ both fall in either $X$ or $D$, they can easily be separated. This leaves the case where $x\in X\backslash\{\phi^{-1}(D_0)\}$ and $y=0\in D.$ If $x\not\in \overline{\phi^{-1}(D_0)}$ they can again be separated. That leaves $x\in \overline{\phi^{-1}(D_0)}.$ Then $\phi(x)\in D_1$ by the hole chart assumption. But since $D_1\subseteq V,$ we can separate $\phi(x)$ and $0$ in $V$ with open sets we will call $W_1$ and $W_2,$ respectively. We pick $W_2$ so that $W_2\subseteq D$. Then $\phi^{-1}(W_1)$ is a neighborhood of $x.$ Because the embedding maps $X\rightarrow X\sqcup_{\phi} D$ and $D\rightarrow X \sqcup_{\phi} D$ are open, the images of $\phi^{-1}(W_1)$ and $W_2$ are open in the adjunction and contain (the images of) $x$ and $y$. Also, because $W_1$ and $W_2$ are disjoint, an in particular don't overlap anywhere in $D_0,$ the images of $\phi^{-1}(W_1)$ and $W_2$ are disjoint in the adjunction.


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