Complexes Torus and $\mathrm{PSL}_2(\mathbb{Z})$

Question

I want to prove that if $\omega_1\equiv\omega_2$ modulo $\mathrm{PSL}_2(\mathbb{Z})$ then $X(\omega_1)\simeq X(\omega_2)$ where $X(\omega)=\mathbb{C}/(\mathbb{Z}+\omega\mathbb{Z})$. I see that if $\omega_1\equiv\omega_2$ I have $\mathbb{Z}+\omega_2\mathbb{Z}\subseteq a_1 (\mathbb{Z}+\omega_1\mathbb{Z})$ and $\mathbb{Z}+\omega_1\mathbb{Z}\subseteq a_2 (\mathbb{Z}+\omega_2\mathbb{Z})$ where $a_1=\frac{1}{c\omega_1+d}$ and $a_2=\frac{1}{c'\omega_2+d'}$. I know that this induce morphismes $X(\omega_1)\to X(\omega_2)$ and $X(\omega_2)\to X(\omega_1)$ but I don't see why these morphismes are inverses of each others because we don't have (it's seems) $a_1 a_2=1$.