Let $F=\{z\in H: |z|\geq1, |Re(z)|\leq 1/2\}$. It is a fundamental domain of the modular group $\Gamma$ acting on the upper half plane $H$. (Strictly speaking, one should take part of the boundary from $F$ to make it a fundamental domain. For example, one can use $F\_1=F-\{z\in F:Re(z)=1/2\}-\{z\in F:Re(z)>0 \text{ and }|z|=1\}$.


For a matrix $A\in SL\_2(\mathbb{R})$, it is said that $A(F)=\{A(x):x\in F\}$ gives a fundamental domain for $\Gamma$. My question is how to prove this.


To prove it, one thing we should show is: for every $s\in H$, there is a $g\in \Gamma$ such that $g(s)\in A(F)$.


This claim is used in the proof that the Hecke operators are Hermitian with respect the Petersson inner product. It seems like a trivial result, but I couldn't prove this. I think I must have missed something. Thus any comment would be appreciated.



the Action of $SL_2(\mathbb{R})$ on a fundamental domain of $\Gamma$
As Misha says, this is false as stated. To see this: Apply the dilation $A \colon z \mapsto z/2$ to $F$. Note $i$ lies in the interior of $A(F)$. Since $i$ is fixed by a torsion element of $SL(2,\mathbb{Z})$, deduce that $A(F)$ is not a fundamental domain. 


There are several different things you might be trying to ask here. I suggest you have a bit of a think, and then ask a new question. 


