[proofplan] We construct the analytic moduli map by sending $\tau \in \mathfrak H$ to the complex torus $E_\tau=\mathbb C/(\mathbb Z\tau+\mathbb Z)$ together with the cyclic subgroup generated by the class of $1/N$. The congruence condition defining $\Gamma_0(N)$ is exactly the condition that a change of lattice basis preserves this distinguished cyclic subgroup. Conversely, every pair $(E,C)$ is uniformized by a lattice and a primitive order-$N$ point, and a suitable $SL_2(\mathbb Z)$ basis change puts that point into the standard form $1/N$. Finally, the cusps are the added boundary points $\Gamma_0(N)\backslash\mathbb P^1(\mathbb Q)$, giving the compactification of the analytic quotient. [/proofplan]

[step:Construct the analytic family of elliptic curves with a cyclic subgroup]For each $\tau \in \mathfrak H$, define the lattice \begin{align*} \Lambda_\tau := \mathbb Z\tau+\mathbb Z \subset \mathbb C \end{align*} and the complex torus \begin{align*} E_\tau := \mathbb C/\Lambda_\tau. \end{align*} By the standard analytic [uniformization theorem](/theorems/3376) for complex elliptic curves applied to the rank-two lattice $\Lambda_\tau \subset \mathbb C$, the complex torus $E_\tau$ is an elliptic curve over $\mathbb C$. Define \begin{align*} P_\tau := \frac{1}{N}+\Lambda_\tau \in E_\tau. \end{align*} Since $NP_\tau=1+\Lambda_\tau=0$ in $E_\tau$, the point $P_\tau$ lies in $E_\tau[N]$. If $1\le m<N$, then \begin{align*} mP_\tau=\frac{m}{N}+\Lambda_\tau \ne 0, \end{align*} because $\frac{m}{N}\notin \mathbb Z\tau+\mathbb Z$. Hence $P_\tau$ has exact order $N$. Define \begin{align*} C_\tau := \langle P_\tau\rangle \subset E_\tau[N]. \end{align*} Then $C_\tau$ is a cyclic subgroup of $E_\tau[N]$ of order $N$. Thus we have a map \begin{align*} \Phi:\mathfrak H &\to \{\text{isomorphism classes of pairs }(E,C)\} \\ \tau &\mapsto [(E_\tau,C_\tau)]. \end{align*}[/step]

[guided]The analytic quotient $\Gamma_0(N)\backslash\mathfrak H$ should classify elliptic curves together with one cyclic subgroup of order $N$. The first task is therefore to attach such a pair to a point of the upper half-plane. For $\tau \in \mathfrak H$, define the lattice \begin{align*} \Lambda_\tau := \mathbb Z\tau+\mathbb Z \subset \mathbb C. \end{align*} The condition $\operatorname{Im}(\tau)>0$ ensures that $\tau$ and $1$ are linearly independent over $\mathbb R$, so $\Lambda_\tau$ is a rank-two lattice in $\mathbb C$. Define \begin{align*} E_\tau := \mathbb C/\Lambda_\tau. \end{align*} By the standard analytic uniformization theorem for complex elliptic curves applied to the rank-two lattice $\Lambda_\tau \subset \mathbb C$, this quotient is an elliptic curve over $\mathbb C$. Now define the point \begin{align*} P_\tau := \frac{1}{N}+\Lambda_\tau \in E_\tau. \end{align*} Multiplying by $N$ gives \begin{align*} NP_\tau=1+\Lambda_\tau=0, \end{align*} so $P_\tau$ is $N$-torsion. To see that the order is exactly $N$, let $m$ be an integer with $1\le m<N$. If $mP_\tau=0$, then $\frac{m}{N}\in\Lambda_\tau$, so \begin{align*} \frac{m}{N}=r\tau+s \end{align*} for some $r,s\in\mathbb Z$. Taking imaginary parts gives $r\operatorname{Im}(\tau)=0$, hence $r=0$. Then $\frac{m}{N}=s\in\mathbb Z$, impossible for $1\le m<N$. Therefore $P_\tau$ has exact order $N$. The subgroup \begin{align*} C_\tau := \langle P_\tau\rangle \subset E_\tau[N] \end{align*} is therefore cyclic of order $N$. This defines \begin{align*} \Phi:\mathfrak H &\to \{\text{isomorphism classes of pairs }(E,C)\} \\ \tau &\mapsto [(E_\tau,C_\tau)]. \end{align*}[/guided]

[step:Show that the construction is invariant under $\Gamma_0(N)$]Let \begin{align*} \gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N), \qquad \gamma\tau:=\frac{a\tau+b}{c\tau+d}. \end{align*} The denominator is nonzero: if $c\tau+d=0$, then either $c=0$ and $d=0$, contradicting $ad-bc=1$, or $\tau=-d/c\in\mathbb R$, contradicting $\tau\in\mathfrak H$. Thus the fractional linear expression and the map below are well-defined. Since $\det\gamma=1$, the two generators of $\Lambda_{\gamma\tau}$ satisfy \begin{align*} \Lambda_{\gamma\tau} = \mathbb Z\frac{a\tau+b}{c\tau+d}+\mathbb Z = \frac{1}{c\tau+d}\bigl(\mathbb Z(a\tau+b)+\mathbb Z(c\tau+d)\bigr) = \frac{1}{c\tau+d}\Lambda_\tau. \end{align*} Therefore multiplication by $(c\tau+d)^{-1}$ defines an isomorphism of complex tori \begin{align*} \psi_\gamma:E_\tau &\to E_{\gamma\tau} \\ z+\Lambda_\tau &\mapsto \frac{z}{c\tau+d}+\Lambda_{\gamma\tau}. \end{align*} It remains to check the subgroup. Since $c\equiv 0\pmod N$, write $c=Nc_0$ for some $c_0\in\mathbb Z$. In $E_{\gamma\tau}$, \begin{align*} \psi_\gamma(P_\tau) = \frac{1}{N(c\tau+d)}+\Lambda_{\gamma\tau}. \end{align*} Using \begin{align*} \frac{1}{c\tau+d}=a-c\gamma\tau, \end{align*} we obtain \begin{align*} \frac{1}{N(c\tau+d)} = \frac{a}{N}-c_0\gamma\tau. \end{align*} Since $c_0\gamma\tau\in\Lambda_{\gamma\tau}$, this gives \begin{align*} \psi_\gamma(P_\tau)=aP_{\gamma\tau}. \end{align*} The congruence $ad-bc=1$ and $c\equiv0\pmod N$ imply $ad\equiv1\pmod N$, so $a$ is invertible modulo $N$. Hence multiplication by $a$ preserves the cyclic subgroup generated by $P_{\gamma\tau}$: \begin{align*} \psi_\gamma(C_\tau)=C_{\gamma\tau}. \end{align*} Thus $\Phi$ is constant on $\Gamma_0(N)$-orbits and descends to a map \begin{align*} \overline{\Phi}:\Gamma_0(N)\backslash\mathfrak H \to \{\text{isomorphism classes of pairs }(E,C)\}. \end{align*}[/step]

[guided]We must show that changing $\tau$ by an element of $\Gamma_0(N)$ changes only the chosen lattice basis, not the resulting moduli pair. Let \begin{align*} \gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N), \qquad \gamma\tau:=\frac{a\tau+b}{c\tau+d}. \end{align*} The expression is defined because $c\tau+d\ne0$: otherwise either $c=d=0$, contradicting $ad-bc=1$, or $\tau=-d/c$ is real, contradicting $\tau\in\mathfrak H$. Since $\det\gamma=1$, the integer span of $a\tau+b$ and $c\tau+d$ equals the lattice $\Lambda_\tau=\mathbb Z\tau+\mathbb Z$. Therefore \begin{align*} \Lambda_{\gamma\tau} &=\mathbb Z\frac{a\tau+b}{c\tau+d}+\mathbb Z \\ &=\frac{1}{c\tau+d}\bigl(\mathbb Z(a\tau+b)+\mathbb Z(c\tau+d)\bigr) \\ &=\frac{1}{c\tau+d}\Lambda_\tau. \end{align*} It follows that multiplication by $(c\tau+d)^{-1}$ carries $\Lambda_\tau$ onto $\Lambda_{\gamma\tau}$ and hence defines an isomorphism of complex tori \begin{align*} \psi_\gamma:E_\tau &\to E_{\gamma\tau} \\ z+\Lambda_\tau &\mapsto \frac{z}{c\tau+d}+\Lambda_{\gamma\tau}. \end{align*} It remains to check that this isomorphism preserves the chosen cyclic subgroup. Since $\gamma\in\Gamma_0(N)$, we have $c\equiv0\pmod N$, so write $c=Nc_0$ with $c_0\in\mathbb Z$. In $E_{\gamma\tau}$, \begin{align*} \psi_\gamma(P_\tau)=\frac{1}{N(c\tau+d)}+\Lambda_{\gamma\tau}. \end{align*} The identity \begin{align*} \frac{1}{c\tau+d}=a-c\gamma\tau \end{align*} gives \begin{align*} \frac{1}{N(c\tau+d)}=\frac{a}{N}-c_0\gamma\tau. \end{align*} Because $c_0\gamma\tau\in\Lambda_{\gamma\tau}$, this says \begin{align*} \psi_\gamma(P_\tau)=aP_{\gamma\tau}. \end{align*} Finally, $ad-bc=1$ and $c\equiv0\pmod N$ imply $ad\equiv1\pmod N$, so $a$ is invertible modulo $N$. Multiplication by such an $a$ permutes the generators of the cyclic group $\langle P_{\gamma\tau}\rangle$, hence \begin{align*} \psi_\gamma(C_\tau)=C_{\gamma\tau}. \end{align*} Thus $\Phi$ is constant on $\Gamma_0(N)$-orbits and descends to \begin{align*} \overline{\Phi}:\Gamma_0(N)\backslash\mathfrak H \to \{\text{isomorphism classes of pairs }(E,C)\}. \end{align*}[/guided]

[step:Put every cyclic subgroup into standard lattice form]Let $(E,C)$ be a pair consisting of a complex elliptic curve $E$ and a cyclic subgroup $C\subset E[N]$ of order $N$. By the standard analytic uniformization theorem for complex elliptic curves applied to the elliptic curve $E$ over $\mathbb C$, there is a rank-two lattice $\Lambda\subset\mathbb C$ and an analytic isomorphism \begin{align*} E \cong \mathbb C/\Lambda. \end{align*} Choose an oriented $\mathbb Z$-basis $(\omega_1,\omega_2)$ of $\Lambda$, meaning that $\operatorname{Im}(\omega_1/\omega_2)>0$. Then \begin{align*} \tau:=\frac{\omega_1}{\omega_2}\in\mathfrak H, \end{align*} and multiplication by $\omega_2^{-1}$ identifies $\mathbb C/\Lambda$ with $E_\tau$. The $N$-torsion subgroup is \begin{align*} (\mathbb C/\Lambda)[N]=\frac{1}{N}\Lambda/\Lambda. \end{align*} Choose a generator $P\in C$. Then \begin{align*} P=\frac{r\omega_1+s\omega_2}{N}+\Lambda \end{align*} for some integers $r,s\in\mathbb Z$. Since $P$ has exact order $N$, the class $(r,s)\in(\mathbb Z/N\mathbb Z)^2$ is primitive, meaning that it has exact order $N$. [claim:Lift a primitive residue vector to a unimodular row] Let $(r,s)\in(\mathbb Z/N\mathbb Z)^2$ have exact order $N$. Then there are integers $\gamma,\delta\in\mathbb Z$ with $(\gamma,\delta)\equiv(r,s)\pmod N$ and $\gcd(\gamma,\delta)=1$. [/claim] [proof] Choose integer representatives $r,s\in\mathbb Z$. Exact order $N$ is equivalent to $\gcd(r,s,N)=1$. We seek $\gamma=r+Nk$ and $\delta=s$ for a suitable $k\in\mathbb Z$. For each prime $p$ dividing $s$, if $p\mid N$, then $p\nmid r$ and therefore $p\nmid r+Nk$ for every $k$. If $p\nmid N$, then the condition $p\mid r+Nk$ excludes exactly one residue class of $k$ modulo $p$. Since only finitely many primes divide $s$, the [Chinese remainder theorem](/theorems/734) gives an integer $k$ avoiding all excluded residue classes. For this $k$, no prime divisor of $s$ divides $r+Nk$, hence $\gcd(r+Nk,s)=1$. Set $\gamma:=r+Nk$ and $\delta:=s$. [/proof] By Bezout's identity applied to $\gcd(\gamma,\delta)=1$, choose $\alpha,\beta\in\mathbb Z$ with \begin{align*} \alpha\delta-\beta\gamma=1. \end{align*} Define \begin{align*} A= \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in SL_2(\mathbb Z). \end{align*} For the new oriented basis \begin{align*} \omega_1'&:=\alpha\omega_1+\beta\omega_2,\\ \omega_2'&:=\gamma\omega_1+\delta\omega_2, \end{align*} the congruences $\gamma\equiv r\pmod N$ and $\delta\equiv s\pmod N$ give \begin{align*} \omega_2'\equiv r\omega_1+s\omega_2 \pmod{N\Lambda}. \end{align*} Replacing the oriented basis by $(\omega_1',\omega_2')$, the subgroup $C$ is generated by \begin{align*} \frac{\omega_2'}{N}+\Lambda. \end{align*} Therefore, after scaling by $(\omega_2')^{-1}$, the pair $(E,C)$ is isomorphic to $(E_{\tau'},C_{\tau'})$ for \begin{align*} \tau':=\frac{\omega_1'}{\omega_2'}\in\mathfrak H. \end{align*} Thus $\overline{\Phi}$ is surjective.[/step]

[guided]We now prove that every moduli point comes from some $\tau\in\mathfrak H$. Let $(E,C)$ be a pair with $E$ an elliptic curve over $\mathbb C$ and $C\subset E[N]$ cyclic of order $N$. By the standard analytic uniformization theorem for complex elliptic curves, there exists a rank-two lattice $\Lambda\subset\mathbb C$ and an analytic isomorphism \begin{align*} E\cong\mathbb C/\Lambda. \end{align*} Choose an oriented $\mathbb Z$-basis $(\omega_1,\omega_2)$ of $\Lambda$, so \begin{align*} \operatorname{Im}\left(\frac{\omega_1}{\omega_2}\right)>0. \end{align*} Then \begin{align*} \tau:=\frac{\omega_1}{\omega_2}\in\mathfrak H, \end{align*} and scaling by $\omega_2^{-1}$ identifies the quotient $\mathbb C/\Lambda$ with $\mathbb C/(\mathbb Z\tau+\mathbb Z)$. The $N$-torsion points of $\mathbb C/\Lambda$ are exactly \begin{align*} (\mathbb C/\Lambda)[N]=\frac{1}{N}\Lambda/\Lambda. \end{align*} Indeed, a class $z+\Lambda$ is killed by $N$ precisely when $Nz\in\Lambda$, which is equivalent to $z\in\frac{1}{N}\Lambda$. Since $C$ is cyclic of order $N$, choose a generator $P\in C$. There are integers $r,s\in\mathbb Z$ such that \begin{align*} P=\frac{r\omega_1+s\omega_2}{N}+\Lambda. \end{align*} The point $P$ has exact order $N$, so the residue vector $(r,s)\in(\mathbb Z/N\mathbb Z)^2$ is primitive. We now prove the arithmetic input needed to change basis. Since $P$ has exact order $N$, the residue vector $(r,s)\in(\mathbb Z/N\mathbb Z)^2$ has exact order $N$, equivalently $\gcd(r,s,N)=1$ for integer representatives $r,s\in\mathbb Z$. We need a lift $(\gamma,\delta)\in\mathbb Z^2$ congruent to $(r,s)$ modulo $N$ and satisfying $\gcd(\gamma,\delta)=1$, because such a pair can be completed to a unimodular matrix. Take $\delta:=s$ and seek $\gamma=r+Nk$. For each prime $p$ dividing $s$, there are two cases. If $p\mid N$, then $p\nmid r$ because $\gcd(r,s,N)=1$, so $p\nmid r+Nk$ for every $k$. If $p\nmid N$, then $p\mid r+Nk$ excludes exactly one residue class of $k$ modulo $p$. Avoiding the finitely many excluded residue classes, using the Chinese [remainder theorem](/theorems/1707), gives $k\in\mathbb Z$ such that no prime divisor of $s$ divides $r+Nk$. Hence $\gcd(r+Nk,s)=1$. Set \begin{align*} \gamma:=r+Nk, \qquad \delta:=s. \end{align*} Then $(\gamma,\delta)\equiv(r,s)\pmod N$ and $\gcd(\gamma,\delta)=1$. By Bezout's identity, choose $\alpha,\beta\in\mathbb Z$ satisfying \begin{align*} \alpha\delta-\beta\gamma=1. \end{align*} Thus \begin{align*} A= \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in SL_2(\mathbb Z). \end{align*} For the new oriented lattice basis \begin{align*} \omega_1'&:=\alpha\omega_1+\beta\omega_2,\\ \omega_2'&:=\gamma\omega_1+\delta\omega_2, \end{align*} the congruences $\gamma\equiv r\pmod N$ and $\delta\equiv s\pmod N$ give \begin{align*} \omega_2'\equiv r\omega_1+s\omega_2 \pmod{N\Lambda}. \end{align*} This congruence means exactly that \begin{align*} \frac{\omega_2'}{N}+\Lambda = \frac{r\omega_1+s\omega_2}{N}+\Lambda = P. \end{align*} Thus, in the new basis, the cyclic subgroup $C$ is generated by the standard point $\omega_2'/N+\Lambda$. Finally scale by $(\omega_2')^{-1}$. The lattice becomes \begin{align*} \mathbb Z\frac{\omega_1'}{\omega_2'}+\mathbb Z, \end{align*} and the generator $\omega_2'/N+\Lambda$ becomes $1/N+\Lambda_{\tau'}$, where \begin{align*} \tau':=\frac{\omega_1'}{\omega_2'}\in\mathfrak H. \end{align*} Hence $(E,C)\cong(E_{\tau'},C_{\tau'})$, proving surjectivity.[/guided]

[step:Identify when two standard pairs are isomorphic]Suppose $\tau,\tau'\in\mathfrak H$ and \begin{align*} (E_\tau,C_\tau)\cong(E_{\tau'},C_{\tau'}). \end{align*} An isomorphism of elliptic curves preserves the identity element, so translations are excluded. By the standard lifting theorem for holomorphic homomorphisms of complex tori, the induced isomorphism of complex tori lifts to a complex-[linear map](/page/Linear%20Map) $\mathbb C\to\mathbb C$ fixing $0$. Every complex-linear automorphism of the one-dimensional complex [vector space](/page/Vector%20Space) $\mathbb C$ is multiplication by some $\lambda\in\mathbb C^\times$, and the lifted map carries one lattice onto the other, so \begin{align*} \lambda\Lambda_\tau=\Lambda_{\tau'}. \end{align*} Thus there exists a matrix \begin{align*} \gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb Z) \end{align*} such that \begin{align*} \tau'=\frac{a\tau+b}{c\tau+d}. \end{align*} It remains to impose preservation of the cyclic subgroup. Since the isomorphism sends $C_\tau$ to $C_{\tau'}$, the image of $P_\tau$ is a generator of $C_{\tau'}$. Therefore there is an integer $u\in\mathbb Z$ with $\gcd(u,N)=1$ such that \begin{align*} \lambda\left(\frac{1}{N}\right)+\Lambda_{\tau'} = \frac{u}{N}+\Lambda_{\tau'}. \end{align*} Equivalently, \begin{align*} \lambda-u\in N\Lambda_{\tau'}. \end{align*} Writing the lattice-basis relation in the form $\lambda=(c\tau+d)^{-1}$ and using \begin{align*} \frac{1}{c\tau+d}=a-c\tau', \end{align*} we get \begin{align*} a-c\tau'-u\in N(\mathbb Z\tau'+\mathbb Z). \end{align*} Comparing the coefficient of $\tau'$ in the basis $(\tau',1)$ gives \begin{align*} c\equiv0\pmod N. \end{align*} Hence $\gamma\in\Gamma_0(N)$, so $\tau$ and $\tau'$ determine the same point of $\Gamma_0(N)\backslash\mathfrak H$. Together with the invariance already proved, this shows that \begin{align*} \overline{\Phi}:\Gamma_0(N)\backslash\mathfrak H \to \{\text{isomorphism classes of pairs }(E,C)\} \end{align*} is bijective.[/step]

[guided]We now prove injectivity of the descended map. Suppose $\tau,\tau'\in\mathfrak H$ and \begin{align*} (E_\tau,C_\tau)\cong(E_{\tau'},C_{\tau'}). \end{align*} An isomorphism of elliptic curves preserves the identity element, so it is not followed by an arbitrary translation. By the standard lifting theorem for holomorphic homomorphisms of complex tori, the induced torus isomorphism lifts to a complex-linear map $\mathbb C\to\mathbb C$ fixing $0$. Since $\mathbb C$ is one-dimensional over itself, this lift is multiplication by some $\lambda\in\mathbb C^\times$, and compatibility with the quotient means \begin{align*} \lambda\Lambda_\tau=\Lambda_{\tau'}. \end{align*} Choosing the ordered bases $(\tau,1)$ and $(\tau',1)$ of these two lattices gives a matrix \begin{align*} \gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb Z) \end{align*} such that \begin{align*} \tau'=\frac{a\tau+b}{c\tau+d}. \end{align*} The remaining question is which matrices preserve the cyclic subgroup generated by the class of $1/N$. Since the isomorphism sends $C_\tau$ to $C_{\tau'}$, the image of the generator $P_\tau$ must be another generator of $C_{\tau'}$. Thus there is an integer $u\in\mathbb Z$ with $\gcd(u,N)=1$ such that \begin{align*} \lambda\left(\frac{1}{N}\right)+\Lambda_{\tau'}= \frac{u}{N}+\Lambda_{\tau'}. \end{align*} Equivalently, \begin{align*} \lambda-u\in N\Lambda_{\tau'}. \end{align*} For the matrix above, the lattice-basis relation gives $\lambda=(c\tau+d)^{-1}$, and the identity \begin{align*} \frac{1}{c\tau+d}=a-c\tau' \end{align*} turns the subgroup-preservation condition into \begin{align*} a-c\tau'-u\in N(\mathbb Z\tau'+\mathbb Z). \end{align*} In the basis $(\tau',1)$ of $\Lambda_{\tau'}$, the coefficient of $\tau'$ on the left is $-c$. Membership in $N\Lambda_{\tau'}$ therefore forces \begin{align*} c\equiv0\pmod N. \end{align*} Hence $\gamma\in\Gamma_0(N)$, so $\tau$ and $\tau'$ are the same point of $\Gamma_0(N)\backslash\mathfrak H$. Together with the invariance step, this proves that $\overline{\Phi}$ is bijective.[/guided]

[step:Add the cusps to obtain the compact modular curve]The group $\Gamma_0(N)$ is a congruence subgroup of $SL_2(\mathbb Z)$, so the standard analytic compactification theorem for congruence subgroup quotients applies. The theorem states that the Riemann surface $\Gamma_0(N)\backslash\mathfrak H$ admits a unique compact Riemann surface compactification whose underlying set is \begin{align*} \Gamma_0(N)\backslash\bigl(\mathfrak H\cup\mathbb P^1(\mathbb Q)\bigr), \end{align*} with the complex-analytic structure determined as follows: if $\rho\in\mathbb P^1(\mathbb Q)$ is a cusp and $\sigma\in SL_2(\mathbb Z)$ satisfies $\sigma\infty=\rho$, then after conjugating the stabilizer of $\rho$ to translations, a local parameter is $q_\rho=\exp(2\pi i\,\sigma^{-1}z/w_\rho)$, where $w_\rho\in\mathbb N$ is the width of the cusp. The added points \begin{align*} \Gamma_0(N)\backslash\mathbb P^1(\mathbb Q) \end{align*} are, by definition in that theorem, the cusps. We take this compact Riemann surface as the complex analytic modular curve $X_0(N)(\mathbb C)$. The previous steps identify the open part \begin{align*} \Gamma_0(N)\backslash\mathfrak H \end{align*} with the moduli set of pairs $(E,C)$ consisting of an elliptic curve over $\mathbb C$ and a cyclic subgroup $C\subset E[N]$ of order $N$. Therefore the compact Riemann surface obtained by adjoining the cusps is precisely the complex analytic modular curve $X_0(N)(\mathbb C)$, and its non-cuspidal points have the asserted moduli interpretation.[/step]

[guided]The last step explains what is added to the open moduli space. Since $\Gamma_0(N)$ is defined by congruence conditions modulo $N$, it is a congruence subgroup of $SL_2(\mathbb Z)$. The standard analytic compactification theorem for congruence subgroup quotients therefore applies to the Riemann surface $\Gamma_0(N)\backslash\mathfrak H$. The theorem compactifies the quotient by adjoining the parabolic boundary orbits. Set-theoretically the compactification is \begin{align*} \Gamma_0(N)\backslash\bigl(\mathfrak H\cup\mathbb P^1(\mathbb Q)\bigr). \end{align*} The points in \begin{align*} \Gamma_0(N)\backslash\mathbb P^1(\mathbb Q) \end{align*} are called cusps. The complex structure near such a cusp is described by a $q$-parameter: if $\rho\in\mathbb P^1(\mathbb Q)$ and $\sigma\in SL_2(\mathbb Z)$ satisfies $\sigma\infty=\rho$, then after conjugating the stabilizer of $\rho$ to a translation subgroup, one may take \begin{align*} q_\rho=\exp(2\pi i\,\sigma^{-1}z/w_\rho), \end{align*} where $w_\rho\in\mathbb N$ is the cusp width. This parameter fills in the missing point by allowing $q_\rho=0$. The bijection proved above identifies the open part \begin{align*} \Gamma_0(N)\backslash\mathfrak H \end{align*} with isomorphism classes of pairs $(E,C)$, where $E$ is an elliptic curve over $\mathbb C$ and $C\subset E[N]$ is cyclic of order $N$. Thus the compact Riemann surface obtained by adding the cusp orbits is $X_0(N)(\mathbb C)$, and precisely its non-cuspidal points have the stated moduli interpretation.[/guided]