Related to Proposition 2.1.1 in Diamond 's Modular Forms The following is from Diamond and Shurman's A First Course in Modular Forms book:


I can't understand not even a fraction of all of underlined statements:
1- How homeomorphism comes out?
2- How Prop. 2.1.1 is related to? 
Simple clear explanation would be much appreciated.
 A: So we are trying to make the topological space $Y(\Gamma)$ into a one dimensional complex manifold.  This means that we are trying to cover $Y(\Gamma)$ with charts, which are pairs $(V, \varphi)$ consisting of an open set $V$ of $Y(\Gamma)$ together with a homemorphism $\varphi$ of $V$ with an open subset of $\mathbb C$.
Let $\tau$ be an element of $\mathcal H$, so that $\pi(\tau)$ is an element of $Y(\Gamma)$.  In your question, we are considering the special case where the only elements of $\Gamma$ which fix $\tau$ are $\pm I$ (or just $I$ if $-I \not\in \Gamma$).  We will now construct a chart $(V,\varphi)$ in $Y(\Gamma)$ such that $\pi(\tau) \in V$.
Applying Proposition 2.1.1 (with $\tau_1 = \tau_2 = \tau$), take $U$ to be the intersection of the two open neighborhoods $U_1$ and $U_2$ of $\tau$.  Then $U$ is a neighborhood of $\tau$ with the property that if $\gamma \in \Gamma$, and $\gamma(U) \cap U \neq \emptyset$, then $\gamma(\tau) = \tau$.  
But for $\gamma \in \Gamma$, our hypothesis on $\tau$ is that $\gamma(\tau)$ is never equal to $\tau$, unless $\gamma = \pm I$, which case $\gamma$ fixes every point of $\mathcal H$.    Therefore, $\gamma(U)$ is disjoint from $U$, except if $\gamma = \pm I$.  
This implies that the restriction of $\pi$ to $U$ is a bijection of $U$ onto $\pi(U)$.  Of course this is surjective; to show that it is injective, suppose that $\pi(\tau') = \pi(\tau'')$ for $\tau', \tau'' \in U$.  Then there exists a $\gamma \in \Gamma$ such that $\gamma(\tau') = \tau''$, which implies that $\gamma(U) \cap U \neq  \emptyset$.  This then implies that $\gamma = \pm I$, hence $\tau' = \tau''$.
Now, $\pi$ is moreover a homeomorphism of $U$ onto $\pi(U)$.  This is because $\pi: \mathcal H \rightarrow Y(\Gamma)$ is an open map.  Thus $\varphi = \pi^{-1}: \pi(U) \rightarrow U$ together with $\pi(U)$ is a chart.
