# Miranda Pag. 66 Plugging Holes

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?

• 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. Sep 13, 2013 at 15:16
• 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. Sep 16, 2013 at 21:13

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.