[proofplan] We attach to a point $z \in \mathbb{H}$ the elliptic curve $E_z = \mathbb{C}/(\mathbb{Z}z+\mathbb{Z})$ together with the cyclic subgroup generated by the class of $1/N$. The first task is to check that this construction is invariant under the action of $\Gamma_0(N)$, so it descends to a map from $\Gamma_0(N)\backslash \mathbb{H}$. Conversely, starting from a pair $(E,C)$, we choose an oriented basis of the period lattice adapted to the cyclic subgroup $C$ and recover a point $z \in \mathbb{H}$. The adapted-basis ambiguity is exactly the action of $\Gamma_0(N)$, which gives both well-definedness and bijectivity. [/proofplan] [step:Attach a cyclic level subgroup to each point of the upper half-plane] For $z \in \mathbb{H}$, define the lattice \begin{align*} \Lambda_z := \mathbb{Z}z+\mathbb{Z} \subset \mathbb{C} \end{align*} and the complex elliptic curve \begin{align*} E_z := \mathbb{C}/\Lambda_z. \end{align*} Let \begin{align*} \pi_z: \mathbb{C} &\to E_z \\ w &\mapsto w+\Lambda_z \end{align*} denote the quotient map. Define \begin{align*} C_z := \langle \pi_z(1/N)\rangle \le E_z. \end{align*} Since $N\pi_z(1/N)=\pi_z(1)=0$ and no smaller positive integer $m<N$ satisfies $m/N \in \Lambda_z$, the element $\pi_z(1/N)$ has exact order $N$. Hence $C_z$ is a cyclic subgroup of $E_z[N]$ of order $N$. [guided] Fix $z \in \mathbb{H}$. The analytic uniformisation of the elliptic curve attached to $z$ uses the lattice \begin{align*} \Lambda_z := \mathbb{Z}z+\mathbb{Z} \subset \mathbb{C}. \end{align*} Because $\operatorname{Im}(z)>0$, the two complex numbers $z$ and $1$ are linearly independent over $\mathbb{R}$, so $\Lambda_z$ is a rank-two lattice in $\mathbb{C}$. We define \begin{align*} E_z := \mathbb{C}/\Lambda_z \end{align*} and let \begin{align*} \pi_z: \mathbb{C} &\to E_z \\ w &\mapsto w+\Lambda_z \end{align*} be the quotient homomorphism. The desired level structure should be a cyclic subgroup of order $N$. The standard choice is the subgroup generated by the class of $1/N$: \begin{align*} C_z := \langle \pi_z(1/N)\rangle \le E_z. \end{align*} We verify its order. First, \begin{align*} N\pi_z(1/N)=\pi_z(1)=0 \end{align*} because $1 \in \Lambda_z$. If $1 \le m < N$, then $m\pi_z(1/N)=0$ would mean $m/N \in \Lambda_z$. Since $m/N$ is real, an equality \begin{align*} m/N = a z + b \end{align*} with $a,b \in \mathbb{Z}$ forces $a=0$ by comparing imaginary parts. Then $m/N=b \in \mathbb{Z}$, which is impossible for $1 \le m < N$. Thus $\pi_z(1/N)$ has exact order $N$, and $C_z$ is a cyclic subgroup of $E_z[N]$ of order $N$. [/guided] [/step] [step:Check invariance under the action of $\Gamma_0(N)$] Let \begin{align*} \gamma := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N) \end{align*} and set $z' := \gamma z = \frac{az+b}{cz+d}$. Multiplication by $cz+d$ gives a complex-[linear map](/page/Linear%20Map) \begin{align*} m_\gamma: \mathbb{C} &\to \mathbb{C} \\ w &\mapsto (cz+d)w. \end{align*} It carries $\Lambda_{z'}$ onto $\Lambda_z$, since \begin{align*} (cz+d)z' &= az+b,\\ (cz+d) &= cz+d. \end{align*} Therefore $m_\gamma$ induces an isomorphism of complex elliptic curves \begin{align*} \bar m_\gamma: E_{z'} &\to E_z \\ w+\Lambda_{z'} &\mapsto (cz+d)w+\Lambda_z. \end{align*} Since $c \equiv 0 \pmod N$, write $c=Nr$ with $r \in \mathbb{Z}$. Then \begin{align*} \bar m_\gamma(\pi_{z'}(1/N)) = \pi_z((cz+d)/N) = \pi_z(rz+d/N) = d\,\pi_z(1/N). \end{align*} The congruence condition $c \equiv 0 \pmod N$ and the identity $ad-bc=1$ imply $\gcd(d,N)=1$. Hence multiplication by $d$ permutes the cyclic group $C_z$, and $\bar m_\gamma(C_{z'})=C_z$. Thus $(E_{z'},C_{z'}) \cong (E_z,C_z)$. [guided] We must show that the pair constructed from $z$ depends only on the $\Gamma_0(N)$-orbit of $z$. Let \begin{align*} \gamma := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N), \qquad z' := \gamma z = \frac{az+b}{cz+d}. \end{align*} The standard relation between the two lattices is given by multiplication by $cz+d$. Define \begin{align*} m_\gamma: \mathbb{C} &\to \mathbb{C} \\ w &\mapsto (cz+d)w. \end{align*} We verify that $m_\gamma(\Lambda_{z'})=\Lambda_z$. The lattice $\Lambda_{z'}$ is generated by $z'$ and $1$, and \begin{align*} (cz+d)z' &= az+b \in \Lambda_z,\\ (cz+d)\cdot 1 &= cz+d \in \Lambda_z. \end{align*} Thus $m_\gamma(\Lambda_{z'}) \subseteq \Lambda_z$. Because $\gamma \in SL_2(\mathbb{Z})$, the integer matrix \begin{align*} \begin{pmatrix} a & c \\ b & d \end{pmatrix} \end{align*} has determinant $ad-bc=1$, so the two elements $az+b$ and $cz+d$ form another $\mathbb{Z}$-basis of $\Lambda_z$. Hence $m_\gamma(\Lambda_{z'})=\Lambda_z$. Therefore $m_\gamma$ descends to an isomorphism \begin{align*} \bar m_\gamma: E_{z'} &\to E_z \\ w+\Lambda_{z'} &\mapsto (cz+d)w+\Lambda_z. \end{align*} We now check that this isomorphism respects the cyclic subgroup. Since $\gamma \in \Gamma_0(N)$, there is $r \in \mathbb{Z}$ with $c=Nr$. Then \begin{align*} \bar m_\gamma(\pi_{z'}(1/N)) &= \pi_z((cz+d)/N)\\ &= \pi_z(rz+d/N)\\ &= d\,\pi_z(1/N), \end{align*} because $rz \in \Lambda_z$. Finally, $ad-bc=1$ and $N \mid c$ imply $ad \equiv 1 \pmod N$, so $d$ is invertible modulo $N$. Multiplication by $d$ therefore sends the cyclic group generated by $\pi_z(1/N)$ onto itself. Hence $\bar m_\gamma(C_{z'})=C_z$, proving that the construction is constant on $\Gamma_0(N)$-orbits. [/guided] [/step] [step:Descend the construction to a map from $Y_0(N)(\mathbb{C})$] By the preceding step, the assignment \begin{align*} z \mapsto [(E_z,C_z)] \end{align*} is constant on $\Gamma_0(N)$-orbits. Hence it defines a map \begin{align*} \Phi: \Gamma_0(N)\backslash \mathbb{H} &\to \{ \text{isomorphism classes of pairs }(E,C)\} \\ \Gamma_0(N)z &\mapsto [(E_z,C_z)]. \end{align*} Under the analytic identification $Y_0(N)(\mathbb{C}) \cong \Gamma_0(N)\backslash \mathbb{H}$, this is the required candidate moduli map. [/step] [step:Put an arbitrary cyclic subgroup into standard position] Let $(E,C)$ be a pair as in the statement. Choose a lattice $\Lambda \subset \mathbb{C}$ and an isomorphism of complex elliptic curves \begin{align*} \alpha: \mathbb{C}/\Lambda &\to E. \end{align*} Let \begin{align*} \pi_\Lambda: \mathbb{C} &\to \mathbb{C}/\Lambda \\ w &\mapsto w+\Lambda \end{align*} be the quotient map, and define \begin{align*} \widetilde C := \alpha^{-1}(C) \le \mathbb{C}/\Lambda. \end{align*} Since $\widetilde C$ is cyclic of order $N$, the preimage \begin{align*} \Lambda' := \pi_\Lambda^{-1}(\widetilde C) \end{align*} is a lattice containing $\Lambda$ with quotient $\Lambda'/\Lambda \cong \widetilde C \cong \mathbb{Z}/N\mathbb{Z}$. By the classification of sublattices of a rank-two free abelian group with cyclic quotient, there is an oriented $\mathbb{Z}$-basis $(\omega_1,\omega_2)$ of $\Lambda$ such that \begin{align*} \Lambda' = \mathbb{Z}(\omega_1/N)+\mathbb{Z}\omega_2. \end{align*} Replacing $(\omega_1,\omega_2)$ by $(-\omega_1,-\omega_2)$ if necessary, assume \begin{align*} z := \omega_2/\omega_1 \in \mathbb{H}. \end{align*} Multiplication by $\omega_1$ induces an isomorphism \begin{align*} \beta_z: E_z &\to \mathbb{C}/\Lambda \\ w+\Lambda_z &\mapsto \omega_1 w+\Lambda. \end{align*} Moreover, \begin{align*} \beta_z(C_z) = \langle \pi_\Lambda(\omega_1/N)\rangle = \Lambda'/\Lambda = \widetilde C. \end{align*} Thus $(E,C)$ is isomorphic to $(E_z,C_z)$, so $\Phi$ is surjective. [guided] We now start with an arbitrary pair $(E,C)$ and show that it comes from some point of the upper half-plane. Choose a lattice $\Lambda \subset \mathbb{C}$ and an isomorphism \begin{align*} \alpha: \mathbb{C}/\Lambda &\to E. \end{align*} Let \begin{align*} \pi_\Lambda: \mathbb{C} &\to \mathbb{C}/\Lambda \\ w &\mapsto w+\Lambda \end{align*} be the quotient map. Pull the subgroup $C$ back to \begin{align*} \widetilde C := \alpha^{-1}(C) \le \mathbb{C}/\Lambda. \end{align*} This subgroup is again cyclic of order $N$. The subgroup $\widetilde C$ can be encoded as a larger lattice. Define \begin{align*} \Lambda' := \pi_\Lambda^{-1}(\widetilde C). \end{align*} Then $\Lambda \subset \Lambda'$, and the quotient $\Lambda'/\Lambda$ is exactly $\widetilde C$. Hence \begin{align*} \Lambda'/\Lambda \cong \mathbb{Z}/N\mathbb{Z}. \end{align*} The classification of sublattices of a rank-two free abelian group with cyclic quotient says that there is an oriented basis $(\omega_1,\omega_2)$ of $\Lambda$ for which \begin{align*} \Lambda' = \mathbb{Z}(\omega_1/N)+\mathbb{Z}\omega_2. \end{align*} This is the algebraic meaning of putting the subgroup into standard position: the chosen generator of the cyclic quotient is represented by $\omega_1/N$. The ratio $\omega_2/\omega_1$ lies either in $\mathbb{H}$ or in the lower half-plane, depending on the orientation convention. Replacing both basis vectors by their negatives does not change the lattice or the displayed formula for $\Lambda'$, and we choose the sign so that \begin{align*} z := \omega_2/\omega_1 \in \mathbb{H}. \end{align*} Then $\Lambda=\omega_1(\mathbb{Z}+\mathbb{Z}z)=\omega_1\Lambda_z$. Multiplication by $\omega_1$ therefore induces an isomorphism \begin{align*} \beta_z: E_z &\to \mathbb{C}/\Lambda \\ w+\Lambda_z &\mapsto \omega_1 w+\Lambda. \end{align*} It sends the standard cyclic subgroup to the pulled-back subgroup because \begin{align*} \beta_z(C_z) = \langle \pi_\Lambda(\omega_1/N)\rangle = \Lambda'/\Lambda = \widetilde C. \end{align*} Composing $\beta_z$ with $\alpha$ gives an isomorphism $(E_z,C_z)\cong(E,C)$. Thus every pair occurs in the image of $\Phi$. [/guided] [/step] [step:Show that two adapted bases differ by an element of $\Gamma_0(N)$] Suppose $z,z' \in \mathbb{H}$ and $(E_z,C_z)\cong(E_{z'},C_{z'})$. Choose an isomorphism \begin{align*} f: E_z &\to E_{z'} \end{align*} with $f(C_z)=C_{z'}$. Any complex-linear isomorphism of complex tori lifts to multiplication by some $\lambda \in \mathbb{C}^{\times}$ satisfying \begin{align*} \lambda \Lambda_z=\Lambda_{z'}. \end{align*} Thus there exist integers $a,b,c,d \in \mathbb{Z}$ such that \begin{align*} \lambda z &= az'+b,\\ \lambda &= cz'+d. \end{align*} The induced matrix has determinant $ad-bc=1$ after choosing the orientation compatible with $z,z' \in \mathbb{H}$. Solving for $z'$ gives \begin{align*} z' = \frac{dz-b}{-cz+a}. \end{align*} The condition $f(C_z)=C_{z'}$ says that $\lambda/N+\Lambda_{z'}$ lies in the subgroup generated by $1/N+\Lambda_{z'}$. Hence there is $k \in \mathbb{Z}$ such that \begin{align*} \frac{\lambda}{N}-\frac{k}{N}\in \Lambda_{z'}. \end{align*} Using $\lambda=cz'+d$, this becomes \begin{align*} \frac{cz'+d-k}{N}\in \mathbb{Z}z'+\mathbb{Z}. \end{align*} Comparing coefficients in the $\mathbb{Z}$-basis $(z',1)$ of $\Lambda_{z'}$ gives $N \mid c$. Therefore the matrix \begin{align*} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \end{align*} belongs to $\Gamma_0(N)$ and sends $z$ to $z'$. Thus $z$ and $z'$ define the same point of $\Gamma_0(N)\backslash \mathbb{H}$, so $\Phi$ is injective. [guided] It remains to prove that the only ambiguity in the construction is the action of $\Gamma_0(N)$. Suppose \begin{align*} f: E_z \to E_{z'} \end{align*} is an isomorphism of elliptic curves with $f(C_z)=C_{z'}$. A holomorphic group isomorphism between complex tori lifts to a complex-linear map of universal covers. Thus there exists $\lambda \in \mathbb{C}^{\times}$ such that multiplication by $\lambda$ maps $\Lambda_z$ isomorphically onto $\Lambda_{z'}$: \begin{align*} \lambda \Lambda_z=\Lambda_{z'}. \end{align*} Since $\Lambda_z$ has basis $(z,1)$ and $\Lambda_{z'}$ has basis $(z',1)$, there are integers $a,b,c,d \in \mathbb{Z}$ with \begin{align*} \lambda z &= az'+b,\\ \lambda &= cz'+d. \end{align*} The corresponding change-of-basis matrix has determinant $\pm 1$. Because both ordered bases define the positive orientation of the complex plane through points of $\mathbb{H}$, the determinant is $1$, so \begin{align*} ad-bc=1. \end{align*} Solving the two displayed equations for $z'$ gives \begin{align*} z'=\frac{dz-b}{-cz+a}. \end{align*} Thus $z'$ is obtained from $z$ by an element of $SL_2(\mathbb{Z})$; we still need to prove the extra congruence condition defining $\Gamma_0(N)$. The congruence comes exactly from preservation of the subgroup. Since $f(C_z)=C_{z'}$, the image of the generator $1/N+\Lambda_z$ lies in $C_{z'}$. Under the lift, this image is \begin{align*} \lambda/N+\Lambda_{z'}. \end{align*} Therefore there is $k \in \mathbb{Z}$ such that \begin{align*} \frac{\lambda}{N}-\frac{k}{N}\in \Lambda_{z'}. \end{align*} Using $\lambda=cz'+d$, this becomes \begin{align*} \frac{cz'+d-k}{N}\in \mathbb{Z}z'+\mathbb{Z}. \end{align*} Because $(z',1)$ is a $\mathbb{Z}$-basis of $\Lambda_{z'}$, membership in $\Lambda_{z'}$ forces both coefficients to be integers. In particular, \begin{align*} c/N \in \mathbb{Z}. \end{align*} Hence $N \mid c$. Consequently \begin{align*} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \in \Gamma_0(N), \end{align*} and the formula above says that this element sends $z$ to $z'$. Thus $z$ and $z'$ represent the same point of $\Gamma_0(N)\backslash\mathbb{H}$. This proves injectivity of $\Phi$. [/guided] [/step] [step:Conclude the moduli interpretation] The map \begin{align*} \Phi: \Gamma_0(N)\backslash \mathbb{H} \to \{ \text{isomorphism classes of pairs }(E,C)\} \end{align*} is well-defined by $\Gamma_0(N)$-invariance, surjective by adapted bases, and injective because two adapted bases differ by an element of $\Gamma_0(N)$. Therefore $\Phi$ is a bijection. Using the analytic identification $Y_0(N)(\mathbb{C}) \cong \Gamma_0(N)\backslash \mathbb{H}$ gives the claimed natural bijection between $Y_0(N)(\mathbb{C})$ and isomorphism classes of pairs $(E,C)$. [/step]