[proofplan] We realize each point $z \in \mathbb{H}$ as the elliptic curve $\mathbb{C}/(\mathbb{Z}z+\mathbb{Z})$ with the marked $N$-torsion point represented by $1/N$. The key computation is that a change of lattice basis by a matrix in $SL_2(\mathbb{Z})$ preserves this marked point exactly when the matrix lies in $\Gamma_1(N)$. Surjectivity follows by choosing a lattice basis adapted to a given torsion point of exact order $N$, and injectivity follows by lifting an isomorphism of complex tori to multiplication by a nonzero complex number and reading off the induced basis change. [/proofplan] [step:Verify that the standard marked point has exact order $N$] Fix $z \in \mathbb{H}$. The lattice $\Lambda_z := \mathbb{Z}z+\mathbb{Z}$ is a rank-two discrete subgroup of $\mathbb{C}$, and $E_z := \mathbb{C}/\Lambda_z$ is the associated complex elliptic curve. Define the marked point $P_z \in E_z$ by \begin{align*} P_z := \frac{1}{N}+\Lambda_z. \end{align*} Since \begin{align*} N P_z = 1+\Lambda_z = 0 \in E_z, \end{align*} we have $P_z \in E_z[N]$. If $1 \leq k \leq N$ and $kP_z = 0$, then $k/N \in \Lambda_z$. Thus there exist $m,n \in \mathbb{Z}$ such that \begin{align*} \frac{k}{N} = mz+n. \end{align*} Because $\operatorname{Im}(z)>0$, the complex number $z$ is not real, so equality with the real number $k/N$ forces $m=0$. Hence $k/N=n \in \mathbb{Z}$, and therefore $N \mid k$. For $1 \leq k \leq N$, this gives $k=N$. Thus $P_z$ has exact order $N$. [/step] [step:Compute precisely when a modular change of basis preserves the marked point] Let \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) \end{align*} and define $\gamma z \in \mathbb{H}$ by \begin{align*} \gamma z := \frac{az+b}{cz+d}. \end{align*} Since $\operatorname{Im}(z)>0$ and $\gamma \in SL_2(\mathbb{Z})$, one has $\gamma z \in \mathbb{H}$. Define the complex-[linear map](/page/Linear%20Map) \begin{align*} \varphi_\gamma: \mathbb{C} &\to \mathbb{C} \\ u &\mapsto (cz+d)u. \end{align*} The identities \begin{align*} (cz+d)(\gamma z) &= az+b, \\ (cz+d)\cdot 1 &= cz+d \end{align*} show that \begin{align*} (cz+d)\Lambda_{\gamma z} &= \mathbb{Z}(az+b)+\mathbb{Z}(cz+d) \\ &= \mathbb{Z}z+\mathbb{Z} \\ &= \Lambda_z, \end{align*} because the integer matrix with columns $(a,b)$ and $(c,d)$ has determinant $ad-bc=1$. Hence $\varphi_\gamma$ descends to an isomorphism of complex elliptic curves \begin{align*} \overline{\varphi}_\gamma: E_{\gamma z} &\to E_z \\ u+\Lambda_{\gamma z} &\mapsto (cz+d)u+\Lambda_z. \end{align*} The image of the standard marked point is \begin{align*} \overline{\varphi}_\gamma(P_{\gamma z}) &= \frac{cz+d}{N}+\Lambda_z \\ &= \frac{c}{N}z+\frac{d}{N}+\Lambda_z. \end{align*} Therefore $\overline{\varphi}_\gamma(P_{\gamma z})=P_z$ if and only if \begin{align*} \frac{c}{N}z+\frac{d-1}{N}\in \Lambda_z. \end{align*} Since $z$ and $1$ are $\mathbb{R}$-linearly independent, this condition is equivalent to \begin{align*} c \equiv 0 \pmod N, \qquad d \equiv 1 \pmod N. \end{align*} Together with $ad-bc=1$, these congruences imply $a \equiv 1 \pmod N$. Thus $\overline{\varphi}_\gamma(P_{\gamma z})=P_z$ exactly for $\gamma \in \Gamma_1(N)$. It follows that if $z$ and $z'$ lie in the same $\Gamma_1(N)$-orbit, then $(E_z,P_z)$ and $(E_{z'},P_{z'})$ are isomorphic. Hence the displayed map from $\Gamma_1(N)\backslash \mathbb{H}$ to isomorphism classes is well-defined. [/step] [step:Choose a lattice basis adapted to any point of exact order $N$] Let $(E,P)$ be a pair consisting of a complex elliptic curve $E$ and a point $P \in E[N]$ of exact order $N$. By the definition of a complex elliptic curve as a one-dimensional complex torus, there is a lattice $\Lambda \subset \mathbb{C}$ such that $E \cong \mathbb{C}/\Lambda$. Choose a $\mathbb{Z}$-basis $(\omega_1,\omega_2)$ of $\Lambda$. Since $P \in E[N]$, there exists $\lambda \in \Lambda$ such that \begin{align*} P = \frac{\lambda}{N}+\Lambda. \end{align*} Write $\lambda = r\omega_1+s\omega_2$ with $r,s \in \mathbb{Z}$. The exact order of $P$ is $N$, so the class of $(r,s)$ in $(\mathbb{Z}/N\mathbb{Z})^2$ has exact order $N$. Equivalently, \begin{align*} \gcd(r,s,N)=1. \end{align*} Choose integers $u,v$ such that \begin{align*} u \equiv r \pmod N, \qquad v \equiv s \pmod N, \qquad \gcd(u,v)=1. \end{align*} We justify the choice explicitly. If $r=0$, then $\gcd(s,N)=1$, so we may take $u=N$ and $v=s$. If $r \neq 0$, take $u=r$ and choose an integer $t$ so that $v:=s+tN$ is not divisible by any prime divisor of $r$. For a prime $p \mid r$ with $p \mid N$, the condition $\gcd(r,s,N)=1$ gives $p \nmid s$, hence $p \nmid s+tN$ for every $t$. For a prime $p \mid r$ with $p \nmid N$, exactly one residue class of $t$ modulo $p$ makes $s+tN \equiv 0 \pmod p$; avoiding these finitely many forbidden residue classes gives $p \nmid v$ for every prime $p \mid r$. Therefore $\gcd(u,v)=\gcd(r,v)=1$, and the congruences modulo $N$ hold by construction. Define \begin{align*} \lambda_2 := u\omega_1+v\omega_2 \in \Lambda. \end{align*} Then $\lambda_2 \equiv \lambda \pmod{N\Lambda}$, so \begin{align*} P = \frac{\lambda_2}{N}+\Lambda. \end{align*} Because $\gcd(u,v)=1$, Bezout's identity gives integers $x,y \in \mathbb{Z}$ such that \begin{align*} xv-yu=1. \end{align*} Define \begin{align*} \lambda_1 := x\omega_1+y\omega_2 \in \Lambda. \end{align*} The change-of-basis matrix from $(\omega_1,\omega_2)$ to $(\lambda_1,\lambda_2)$ has determinant $xv-yu=1$, so $(\lambda_1,\lambda_2)$ is a $\mathbb{Z}$-basis of $\Lambda$. If necessary, replace $\lambda_1$ by $-\lambda_1$; this keeps $(\lambda_1,\lambda_2)$ a lattice basis and makes \begin{align*} z := \frac{\lambda_1}{\lambda_2} \end{align*} belong to $\mathbb{H}$. Multiplication by $\lambda_2^{-1}$ gives an isomorphism \begin{align*} \psi: \mathbb{C}/\Lambda &\to \mathbb{C}/\Lambda_z \\ u+\Lambda &\mapsto \frac{u}{\lambda_2}+\Lambda_z, \end{align*} and under this isomorphism \begin{align*} \psi(P) = \frac{1}{N}+\Lambda_z = P_z. \end{align*} Thus every isomorphism class of pairs $(E,P)$ occurs in the image. [/step] [step:Show that isomorphic marked elliptic curves give the same $\Gamma_1(N)$-orbit] Suppose $z,w \in \mathbb{H}$ and suppose there is an isomorphism of marked elliptic curves \begin{align*} \Psi: E_z &\to E_w \end{align*} such that $\Psi(P_z)=P_w$. We use the standard lifting property for homomorphisms of complex tori: a holomorphic group homomorphism $\mathbb{C}/\Lambda_z \to \mathbb{C}/\Lambda_w$ lifts through the universal covering map $\mathbb{C} \to \mathbb{C}/\Lambda_z$ to a complex-linear map $\mathbb{C} \to \mathbb{C}$ carrying $\Lambda_z$ into $\Lambda_w$. Since $\Psi$ is an isomorphism, the lifted linear map is multiplication by some $\alpha \in \mathbb{C}^{\times}$ and carries $\Lambda_z$ bijectively onto $\Lambda_w$. Thus $\Psi$ lifts to \begin{align*} \widetilde{\Psi}: \mathbb{C} &\to \mathbb{C} \\ u &\mapsto \alpha u \end{align*} with $\alpha\Lambda_z=\Lambda_w$. Because $\alpha z$ and $\alpha$ form a $\mathbb{Z}$-basis of $\Lambda_w=\mathbb{Z}w+\mathbb{Z}$, there are integers $a,b,c,d \in \mathbb{Z}$ such that \begin{align*} \alpha z &= aw+b, \\ \alpha &= cw+d. \end{align*} Multiplication by the nonzero complex number $\alpha$ preserves orientation on the real plane $\mathbb{C}$, and the ordered bases $(z,1)$ and $(w,1)$ have the same orientation because $\operatorname{Im}(z)>0$ and $\operatorname{Im}(w)>0$. Therefore \begin{align*} ad-bc=1, \end{align*} so \begin{align*} \gamma := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} Dividing the identity $\alpha z=aw+b$ by $\alpha=cw+d$ gives \begin{align*} z = \frac{aw+b}{cw+d} = \gamma w. \end{align*} The condition $\Psi(P_z)=P_w$ means \begin{align*} \frac{\alpha}{N}+\Lambda_w = \frac{1}{N}+\Lambda_w. \end{align*} Equivalently, \begin{align*} \frac{\alpha-1}{N}\in \Lambda_w. \end{align*} Since $\alpha=cw+d$, this is \begin{align*} \frac{c}{N}w+\frac{d-1}{N}\in \mathbb{Z}w+\mathbb{Z}. \end{align*} The numbers $w$ and $1$ are $\mathbb{R}$-linearly independent, so \begin{align*} c \equiv 0 \pmod N, \qquad d \equiv 1 \pmod N. \end{align*} Using $ad-bc=1$, we also obtain $a \equiv 1 \pmod N$. Hence $\gamma \in \Gamma_1(N)$, and since $z=\gamma w$, the points $z$ and $w$ determine the same element of $\Gamma_1(N)\backslash\mathbb{H}$. This proves injectivity. [/step] [step:Conclude the moduli interpretation] The construction $z \mapsto (E_z,P_z)$ is well-defined on $\Gamma_1(N)$-orbits, every pair $(E,P)$ with $P$ of exact order $N$ is isomorphic to some $(E_z,P_z)$, and two points $z,w \in \mathbb{H}$ give isomorphic marked elliptic curves exactly when they lie in the same $\Gamma_1(N)$-orbit. Therefore the map \begin{align*} \Gamma_1(N)\backslash \mathbb{H} &\longrightarrow \left\{ \text{isomorphism classes of pairs } (E,P) \right\} \\ [z] &\longmapsto [E_z,P_z] \end{align*} is a bijection. This is precisely the claimed moduli interpretation of $Y_1(N)(\mathbb{C})$. [/step]