About fundamental domain of SL(2,Z)

Question

I am trying to understand the proof for fundamental domain of $SL(2,\mathbb{Z})$ using the lecture notes http://www.ltcc.ac.uk/media/london-taught-course-centre/documents/Lecture-3-Notes.pdf . The relevant part is attached as an image below.

My question is regarding Theorem 1 statement (a).

Here $\mathcal{U}$ is the upper half of complex plane, and $\Gamma_0$ is $SL(2,\mathbb{Z})$.

Also $\mathcal{D}=\{ z: \vert z \vert \geq 1, \vert\mathfrak{R}(z)\vert \leq 1/2\}$ is the fundamental domain of $SL(2,\mathbb{Z})$.

The proof for Theorem 1 statement (a) can be summarized as follows

1. Any $z$ can be mapped to $\gamma(z)$ where $\gamma\in SL(2,\mathbb{Z})$ such that $\mathfrak{I}(\gamma(z))$ is the maximum. This map can be shown to map from $\mathcal{U}\rightarrow \{ \vert z \vert \geq 1 \}$ (this is the contradiction part mentioned towards the end)
2. Any $z$ can be translated to the set $\{\vert\mathfrak{R}(z)\vert \leq 1/2 \}$

So we have proved esitence of two elements in $SL(2,\mathbb{Z})$ where one maps any $z\in \mathcal{U}$ to $\{ \vert z \vert \geq 1 \}$ and another maps any $z\in \mathcal{U}$ to $\{\vert\mathfrak{R}(z)\vert \leq 1/2 \}$

How does existence of these two elements imply that there is an element that maps any $z\in \mathcal{U}$ to $\mathcal{D} = \{ \vert z \vert \geq 1 \} \bigcap \{\vert\mathfrak{R}(z)\vert \leq 1/2 \}$ ?

I see the same proof technique is used in

• https://www.math.mcgill.ca/darmon/courses/01-02/af/nicole.pdf [Proposition 2.6]
• https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,Z).pdf [Theorem 1.1]
• https://metaphor.ethz.ch/x/2019/fs/401-4110-19L/sc/modulargroup.pdf [Theorem 1]
• https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2013-14/07_fund_dmn.pdf [Claim 2.0.2]

Answer 1

Send an arbitrary element $w$ to $w' = \gamma(w)$ with maximal imaginary part over all $\gamma \in \mathrm{SL}_2(\mathbb{Z})$, so that $|w'| \geq 1$. If the translate(s) $w''$ of $w'$ to $|\Re(z)| \leq 1/2$ still has absolute value $\geq 1$, then we are done. Otherwise, $w''$ lies in the unit disk, so $\Im(-1/w'') > \Im(w'') = \Im(w')$. But $-1/w''$ lies in the orbit of the original $w$, so this contradicts the assumption on $w'$.