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) $a_1 a_2=1$.
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Could you explain how PSL(2,Z) acts on lattices? – Moishe Kohan Sep 23 '13 at 20:23
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Does my answer here (http://math.stackexchange.com/questions/491437/isomorphism-between-torus/491450#491450) answer your question? – Alex Youcis Sep 23 '13 at 20:30
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Yes, all is clear now! – Gabriel Soranzo Sep 24 '13 at 16:25