In Ahlfors complex analysis book, 2nd edition Pg. 273-274 Chapter 7 Section 3.5 we have the theorem
$\textbf{Theorem 8.}$ Every point $\tau$ in the upper half plane is equivalent under the congruence subgroup mod $2$ to exactly one point in $\overline{\Omega}\cup \Omega'$.
Here $\Omega$ is the open region bounded by the lines $\text{Re}(\tau)=0$, $\text{Re}(\tau)=1$ and $|\tau-1/2|=1/2$. The region $\Omega'$ is the reflection in the imaginary axis of $\Omega$. Closure is taken in the upper half plane and so $\{0,1\}\not\in \overline{\Omega}$.
The congruence subgroup mod $2$ is the group of mobius transformations $ \frac{a \tau+b}{c \tau+d}$ with integer entries, determinant $\pm 1$, where $a \equiv d \equiv1$ (mod 2) and $b \equiv c \equiv 0$ (mod 2).
I understand the proof and how he uses Theorem $2$ (quite long to type but can if needed) to map the fundamental region of the modular group $G$ to $\overline{\Omega}\cup\overline{\Omega'}$.
Therefore he ends up proving that $\tau$ is equivalent under the congruence subgroup mod $2$ to a point in the region $\overline{\Omega}\cup\overline{\Omega'}$.
He then states, which is the part I don't understand:
Thus there is always a $T\tau$ in $\overline{\Omega}\cup\overline{\Omega'}$, and a $\textit{trivial consideration}$ shows that it can be chosen in $\overline{\Omega}\cup\Omega'$.
What is this trivial consideration? How is this obvious? Thanks any help much appreciated.
Note that $\partial \overline{\Omega'}$ consists of the lines $\Re(\tau)=-1$, $\Re(\tau)=0$ and $|\tau+1/2|=1/2$. (Taking the intersection of everything with the upper half plane of course).
The line $\Re(\tau)=0$ lies in $\overline\Omega$.
The map $z \mapsto \frac{1z+2}{0z+1}$ lies in $G$ and sends the line $\Re(\tau)=-1$ to the line $\Re(\tau)=1$ which lies in $\overline\Omega$.
The map $z\mapsto \frac{z}{2z+1}$ sends $1/2(-1+e^{i\varphi})$ to $1/2(-e^{-i \varphi}+1)$, ie the circle in $\overline \Omega$.