Every $\tau$ in the U.H.P. is equivalent to *exactly one* point in the region In Ahlfors' complex analysis text, page 281 he states the following theorem:

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 region bounded by the lines $\Re z=0, \Re z=1$ and the upper half of the circle $|z-1/2|=1/2 $, $\Omega'$ is its reflection across the imaginary axis, and the closure is taken in the upper half plane.
Also, the congruence subgroup mod 2 is the group of all Möbius 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 have read the proof given in the book, and I'm stuck at the last stage, namely proving that the equivalent point is unique.
Suppose $\tau$ is in the upper half plane, and let $T',T''$ be two elements of the subgroup, such that both $\tau'=T' \tau$ and $\tau''=T'' \tau$ lie in $\overline{\Omega} \cup \Omega'$, why must $\tau'=\tau''$?
I can see that $T'^{-1} \tau'= T''^{-1} \tau''$ which proves that $\tau',\tau''$ are equivalent under the subgroup, but this is still not enough to show that they coincide.
Any help is appreciated, thanks!  
EDIT: at the end of the chapter Ahlfors writes 

The uniqueness follows readily from the fact that the $S_k$ as well as the $S_k'$ are mutually incongruent. We shall leave it to the reader to work out the details.

here $S_k,S_k'$ are some fractional linear maps which "generate" the subgroup. There is a relevant figure that I can't show here sadly...
 A: Let $G$ be the ($\pm$) modular group $SL^{\pm}(2,\mathbb Z)$ and $H$ its subgroup formed by the matrices that are $\equiv I\mod 2$. Ahlfors observes that $H$ has index $6$ in $G$, and writes down the coset representatives $S_k^{-1}$ on page 281. 
Theorem 2 says that the orbit $G \cdot \tau$ has exactly one point (call it $g\tau $) in $\overline{\Delta}\cup \Delta'$. Suppose $g\tau \in \overline{\Delta}$. Then  for every $k=1,\dots,6$ the orbit $G\cdot \tau$ has exactly one point in $S^{-1}_k(\overline{\Delta} )$, namely $S_k^{-1}g \tau $. The union of these six sets is   all the shaded parts $\overline{\Omega}\cup\Omega'$. Exactly one of the points $S_k^{-1}g \tau $ is in $H\tau$, because exactly one of the elements $S_k^{-1}g$ is in $H$. Moreover, $G\tau$ does not meet any non-shaded parts of $\overline{\Omega}\cup\Omega'$, because it does not meet $\Delta'$.
The case $g\tau \in  {\Delta'}$ is the same, except with $S_k'$ instead of $S_k$. 

Remark 1: the proof would be more transparent if we used a different fundamental domain of $H$ instead of $\overline{\Omega}\cup\Omega'$, namely the union of $S^{-1}_k(\overline{\Delta} \cup \Delta')$ over $k$.
Remark 2: The deduction of Theorem 8 from Theorem 2 can be summarized as follows: once we have a fundamental domain $D$ for group $G$, we get a fundamental domain for its subgroup $H$ by taking the union of the images of $D$ under coset representatives, one from each coset (including $H$ itself).
