[proofplan] We first use that $\Gamma_0(N)$ has finite index in $SL_2(\mathbb{Z})$, so a finite union of translates of the standard modular fundamental domain controls the quotient. Away from the rational boundary points this gives the usual orbifold quotient of $\mathcal{H}$, whose elliptic stabilisers are finite and hence give legitimate Riemann-surface charts after quotienting by finite rotation groups. The only noncompact parts are finitely many cuspidal ends, and each cusp has a finite width $w$ so that the local parameter $\exp(2\pi i z/w)$ identifies a punctured cusp neighbourhood with a punctured disc. Filling in these finitely many punctures gives compactness, while connectedness follows from the connectedness of $\mathcal{H}$ and the fact that adjoining cusp points preserves connectedness. [/proofplan]

[step:Reduce the quotient to finitely many translates of the standard fundamental domain]Let \begin{align*} \mathcal{F} := \left\{z \in \mathcal{H} : |\operatorname{Re}(z)| \leq \frac{1}{2},\ |z| \geq 1\right\} \end{align*} denote the closed standard fundamental domain for the action of $SL_2(\mathbb{Z})$ on $\mathcal{H}$. Let $r \in \mathbb{N}$ be the finite index \begin{align*} r := [SL_2(\mathbb{Z}) : \Gamma_0(N)]. \end{align*} Choose matrices $\gamma_1,\dots,\gamma_r \in SL_2(\mathbb{Z})$ whose images are representatives for the left cosets $\Gamma_0(N)\backslash SL_2(\mathbb{Z})$. Since $\Gamma_0(N)$ has finite index in $SL_2(\mathbb{Z})$ and $\mathcal{F}$ is a fundamental domain for $SL_2(\mathbb{Z})$, every point of $\mathcal{H}$ is $\Gamma_0(N)$-equivalent to a point of the finite union \begin{align*} \mathcal{D} := \bigcup_{j=1}^{r} \gamma_j \mathcal{F}. \end{align*} Indeed, for $z \in \mathcal{H}$, choose $A \in SL_2(\mathbb{Z})$ and $u \in \mathcal{F}$ such that $z = A u$. Since the $\gamma_j$ are left-coset representatives, there are $\Gamma \in \Gamma_0(N)$ and $j \in \{1,\dots,r\}$ such that $A = \Gamma \gamma_j$. Hence $z = \Gamma(\gamma_j u)$, so the image of $\mathcal{D}$ covers $\Gamma_0(N)\backslash \mathcal{H}$. The finiteness of $r$ follows directly from the reduction map \begin{align*} \rho_N: SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N\mathbb{Z}), \end{align*} because $\Gamma_0(N)$ contains the inverse image under $\rho_N$ of the finite subgroup of matrices whose lower-left entry is $0$ modulo $N$. Thus $\Gamma_0(N)$ has finite index in $SL_2(\mathbb{Z})$.[/step]

[guided]The first point is that although $\Gamma_0(N)$ is smaller than $SL_2(\mathbb{Z})$, it is only smaller by finitely many cosets. We make this finite reduction explicit. Define \begin{align*} \mathcal{F} := \left\{z \in \mathcal{H} : |\operatorname{Re}(z)| \leq \frac{1}{2},\ |z| \geq 1\right\}, \end{align*} the usual closed fundamental domain for $SL_2(\mathbb{Z})$ acting on $\mathcal{H}$. We need $\Gamma_0(N)$ to have finite index in $SL_2(\mathbb{Z})$. Define the reduction map \begin{align*} \rho_N: SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N\mathbb{Z}) \end{align*} by reducing each matrix entry modulo $N$. The group $SL_2(\mathbb{Z}/N\mathbb{Z})$ is finite. The subgroup $\Gamma_0(N)$ contains the inverse image of the finite subgroup consisting of matrices whose lower-left entry is $0$ modulo $N$, so $\Gamma_0(N)$ has finite index. Let \begin{align*} r := [SL_2(\mathbb{Z}) : \Gamma_0(N)] \end{align*} and choose representatives $\gamma_1,\dots,\gamma_r \in SL_2(\mathbb{Z})$ for the left cosets $\Gamma_0(N)\backslash SL_2(\mathbb{Z})$. Now define the finite union \begin{align*} \mathcal{D} := \bigcup_{j=1}^{r} \gamma_j \mathcal{F}. \end{align*} We verify that its image covers $\Gamma_0(N)\backslash \mathcal{H}$. Given $z \in \mathcal{H}$, the standard fundamental-domain property for $SL_2(\mathbb{Z})$ gives $A \in SL_2(\mathbb{Z})$ and $u \in \mathcal{F}$ with $z = A u$. Since the $\gamma_j$ represent the left cosets, there are $\Gamma \in \Gamma_0(N)$ and $j \in \{1,\dots,r\}$ such that $A = \Gamma \gamma_j$. Therefore \begin{align*} z = \Gamma(\gamma_j u), \end{align*} so $z$ and $\gamma_j u \in \mathcal{D}$ define the same point of $\Gamma_0(N)\backslash \mathcal{H}$. Thus a finite union of standard translates controls the entire quotient.[/guided]

[step:Give the interior quotient its Riemann-surface structure]Let \begin{align*} Y_0(N) := \Gamma_0(N)\backslash \mathcal{H} \end{align*} be the noncompact modular curve. We verify the hypotheses of the quotient-surface theorem. The group $\Gamma_0(N)$ is a subgroup of $SL_2(\mathbb{Z})$, hence its image in $PSL_2(\mathbb{R})$ is a discrete Fuchsian group. By the [proper discontinuity of Fuchsian group actions](/page/Fuchsian%20Group), a discrete subgroup of $PSL_2(\mathbb{R})$ acts properly discontinuously on $\mathcal{H}$ by Möbius transformations. For each $z \in \mathcal{H}$, define the stabiliser subgroup \begin{align*} \Gamma_{0}(N)_z := \{\Gamma \in \Gamma_0(N) : \Gamma z = z\}. \end{align*} The image of $\Gamma_{0}(N)_z$ in $PSL_2(\mathbb{R})$ is a discrete subgroup of the compact rotation group fixing $z$; a discrete subset of a compact group is finite. The kernel of $SL_2(\mathbb{R}) \to PSL_2(\mathbb{R})$ is $\{I,-I\}$, so $\Gamma_{0}(N)_z$ itself is finite. The action is by holomorphic automorphisms because each matrix \begin{align*} \Gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N) \end{align*} acts by $z \mapsto (az+b)/(cz+d)$, whose denominator is nonzero on $\mathcal{H}$. Therefore the [quotient-surface theorem for properly discontinuous holomorphic actions with finite stabilisers](/page/Riemann%20Surface%20Quotient) applies and gives $Y_0(N)$ the structure of a Riemann surface, with the quotient map \begin{align*} \pi: \mathcal{H} \to Y_0(N) \end{align*} holomorphic. Connectedness follows at this stage because $\mathcal{H}$ is connected, $\pi$ is continuous for the [quotient topology](/page/Quotient%20Topology), and $\pi$ is surjective; the image of a connected topological space under a continuous map is connected.[/step]

[guided]We now put the complex structure on the interior quotient \begin{align*} Y_0(N) := \Gamma_0(N)\backslash \mathcal{H}. \end{align*} Each element \begin{align*} \Gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N) \end{align*} acts by the holomorphic automorphism \begin{align*} z \mapsto \frac{az+b}{cz+d} \end{align*} of $\mathcal{H}$. The action is properly discontinuous because $\Gamma_0(N)$ is a discrete subgroup of $SL_2(\mathbb{R})$ acting on $\mathcal{H}$ by Möbius transformations. For each $z \in \mathcal{H}$, the stabiliser \begin{align*} \Gamma_{0}(N)_z := \{\Gamma \in \Gamma_0(N) : \Gamma z = z\} \end{align*} is finite; geometrically these are the elliptic stabilisers, and algebraically they are finite cyclic subgroups after passing to $PSL_2(\mathbb{R})$. We now verify the hypotheses of the [quotient-surface theorem for properly discontinuous holomorphic actions with finite stabilisers](/page/Riemann%20Surface%20Quotient). First, $\Gamma_0(N)$ maps to a discrete subgroup of $PSL_2(\mathbb{R})$ because it is contained in $SL_2(\mathbb{Z})$, whose entries form a discrete subset of $\mathbb{R}^4$. The [proper discontinuity of Fuchsian group actions](/page/Fuchsian%20Group) then gives proper discontinuity of the action on $\mathcal{H}$. Second, the action is holomorphic because every group element acts by a Möbius transformation. Third, stabilisers are finite: the stabiliser of a point $z \in \mathcal{H}$ in $PSL_2(\mathbb{R})$ is conjugate to $SO(2)$, hence compact, and its intersection with the discrete image of $\Gamma_0(N)$ is finite. Lifting back from $PSL_2(\mathbb{R})$ to $SL_2(\mathbb{R})$ adds at most the two central matrices $I$ and $-I$, so $\Gamma_0(N)_z$ is finite. All hypotheses of the quotient theorem are therefore satisfied. Applying it gives a Riemann-surface structure on \begin{align*} Y_0(N) = \Gamma_0(N)\backslash \mathcal{H} \end{align*} and makes the quotient map \begin{align*} \pi: \mathcal{H} \to Y_0(N) \end{align*} holomorphic. Finally, $Y_0(N)$ is connected. The upper half-plane $\mathcal{H}$ is connected, the quotient map $\pi$ is continuous by the quotient topology, and $\pi$ is surjective by definition. Hence $Y_0(N)$ is the continuous image of a connected space, so it is connected.[/guided]

[step:Identify the finitely many cuspidal ends] Let \begin{align*} \mathbb{P}^1(\mathbb{Q}) := \mathbb{Q} \cup \{\infty\} \end{align*} be the rational boundary of $\mathcal{H}$. The set of cusps for $\Gamma_0(N)$ is \begin{align*} C_0(N) := \Gamma_0(N)\backslash \mathbb{P}^1(\mathbb{Q}). \end{align*} This set is finite. Indeed, $SL_2(\mathbb{Z})$ acts transitively on $\mathbb{P}^1(\mathbb{Q})$, and $\Gamma_0(N)$ has finite index in $SL_2(\mathbb{Z})$, so the number of $\Gamma_0(N)$-orbits in $\mathbb{P}^1(\mathbb{Q})$ is at most $[SL_2(\mathbb{Z}) : \Gamma_0(N)]$. Choose representatives \begin{align*} \alpha_1,\dots,\alpha_m \in \mathbb{P}^1(\mathbb{Q}) \end{align*} for the finitely many cusps, where \begin{align*} m := |C_0(N)|. \end{align*} For each $k \in \{1,\dots,m\}$, choose a scaling matrix $\sigma_k \in SL_2(\mathbb{Z})$ satisfying \begin{align*} \sigma_k(\infty) = \alpha_k. \end{align*} The cusp width $w_k \in \mathbb{N}$ at $\alpha_k$ is the positive generator of the subgroup \begin{align*} \{n \in \mathbb{Z} : \sigma_k \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \sigma_k^{-1} \in \Gamma_0(N)\} \subset \mathbb{Z}. \end{align*} This subgroup is nonzero because $\Gamma_0(N)$ has finite index in $SL_2(\mathbb{Z})$, and hence $w_k$ is finite. [/step]

[step:Use the cusp parameter to fill each punctured end]Fix $k \in \{1,\dots,m\}$. Let $w_k$ be the cusp width at $\alpha_k$, and define the holomorphic map \begin{align*} q_k: \{z \in \mathcal{H} : \operatorname{Im}(z) > R\} &\to \{q \in \mathbb{C} : 0 < |q| < e^{-2\pi R/w_k}\} \\ z &\mapsto \exp\left(\frac{2\pi i z}{w_k}\right), \end{align*} where $R > 0$ is chosen sufficiently large that the corresponding horoball neighbourhoods for the finitely many cusps are disjoint modulo $\Gamma_0(N)$. The definition of $w_k$ gives invariance under the translation $z \mapsto z+w_k$. We also need injectivity modulo the full group, not only invariance under this one translation. By the [standard cusp-neighbourhood theorem for finite-index subgroups of $SL_2(\mathbb{Z})$](/page/Modular%20Curve%20Cusp), after increasing $R$ if necessary, the only elements of $\Gamma_0(N)$ identifying two points of \begin{align*} \sigma_k\{z \in \mathcal{H} : \operatorname{Im}(z) > R\} \end{align*} are the elements of the cusp stabiliser \begin{align*} \Gamma_0(N)_{\alpha_k} := \{\Gamma \in \Gamma_0(N) : \Gamma \alpha_k = \alpha_k\}. \end{align*} Conjugating by $\sigma_k$ identifies this stabiliser, modulo the central sign, with the translation subgroup generated by $z \mapsto z+w_k$. Hence two points in the chosen horoball have the same image in $\Gamma_0(N)\backslash\mathcal{H}$ exactly when their $q_k$-values agree. Therefore $q_k$ descends to a biholomorphic coordinate from the punctured cusp neighbourhood \begin{align*} \Gamma_0(N)\backslash \sigma_k\{z \in \mathcal{H} : \operatorname{Im}(z) > R\} \end{align*} onto the punctured disc \begin{align*} \{q \in \mathbb{C} : 0 < |q| < e^{-2\pi R/w_k}\}. \end{align*} Adjoin one point $c_k$ corresponding to $q=0$. This turns the punctured cusp neighbourhood into the full disc \begin{align*} \{q \in \mathbb{C} : |q| < e^{-2\pi R/w_k}\}. \end{align*} Doing this for all $k \in \{1,\dots,m\}$ defines \begin{align*} X_0(N) := Y_0(N) \cup \{c_1,\dots,c_m\}. \end{align*} The topology on $X_0(N)$ is generated by the open sets of $Y_0(N)$ together with the full cusp discs in the $q_k$-coordinates. These $q_k$-coordinates are complex charts at the added cusp points. On any overlap with an interior quotient chart $\varphi: U \to V \subset \mathbb{C}$, the transition function has the form $\varphi \circ \pi \circ \sigma_k \circ q_k^{-1}$ on a punctured disc or punctured annulus. It is holomorphic because it is a composition of the inverse exponential branch, a Möbius transformation, the quotient map, and the interior chart. After shrinking the cusp disc, its image lies in the bounded coordinate disc $V$. Hence the [removable singularity theorem](/page/Removable%20Singularity%20Theorem) extends the transition holomorphically across $q=0$ whenever the overlap accumulates at the cusp; on annular overlaps away from $q=0$ no extension is needed. Thus the cusp charts are compatible with the interior quotient atlas.[/step]

[guided]Fix a cusp representative $\alpha_k$ and a scaling matrix $\sigma_k \in SL_2(\mathbb{Z})$ satisfying \begin{align*} \sigma_k(\infty) = \alpha_k. \end{align*} The role of $\sigma_k$ is to move the chosen cusp to $\infty$, where the stabiliser is described by translations. The cusp width $w_k \in \mathbb{N}$ was defined by \begin{align*} \{n \in \mathbb{Z} : \sigma_k \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \sigma_k^{-1} \in \Gamma_0(N)\} = w_k\mathbb{Z}. \end{align*} Thus the translation $z \mapsto z+w_k$ is exactly the primitive parabolic identification at the cusp. Choose $R > 0$ sufficiently large so that the horoball neighbourhoods \begin{align*} \sigma_k\{z \in \mathcal{H} : \operatorname{Im}(z) > R\} \end{align*} for the finitely many cusp representatives are pairwise disjoint after passing to the quotient by $\Gamma_0(N)$. This is possible because there are only finitely many cusp classes and the action is properly discontinuous away from the boundary. Define the cusp parameter \begin{align*} q_k: \{z \in \mathcal{H} : \operatorname{Im}(z) > R\} &\to \{q \in \mathbb{C} : 0 < |q| < e^{-2\pi R/w_k}\} \\ z &\mapsto \exp\left(\frac{2\pi i z}{w_k}\right). \end{align*} This map is invariant under $z \mapsto z+w_k$, because \begin{align*} \exp\left(\frac{2\pi i (z+w_k)}{w_k}\right) = \exp\left(\frac{2\pi i z}{w_k}\right)\exp(2\pi i) = \exp\left(\frac{2\pi i z}{w_k}\right). \end{align*} It is also the standard universal covering map from the upper half-strip modulo $w_k\mathbb{Z}$ to a punctured disc. However, to get a coordinate on the quotient by all of $\Gamma_0(N)$, we must know that no additional group elements identify points inside this small horoball. The [standard cusp-neighbourhood theorem for finite-index subgroups of $SL_2(\mathbb{Z})$](/page/Modular%20Curve%20Cusp) gives exactly this after increasing $R$: if two points of \begin{align*} \sigma_k\{z \in \mathcal{H} : \operatorname{Im}(z) > R\} \end{align*} are equivalent under $\Gamma_0(N)$, then the identifying element lies in the cusp stabiliser \begin{align*} \Gamma_0(N)_{\alpha_k} := \{\Gamma \in \Gamma_0(N) : \Gamma \alpha_k = \alpha_k\}. \end{align*} Conjugating this stabiliser by $\sigma_k^{-1}$ gives, up to the central sign, the translation group generated by \begin{align*} z \mapsto z+w_k. \end{align*} Thus $q_k$ is not merely invariant: it separates the equivalence classes in the punctured cusp neighbourhood. Hence it descends to a biholomorphic coordinate from \begin{align*} \Gamma_0(N)\backslash \sigma_k\{z \in \mathcal{H} : \operatorname{Im}(z) > R\} \end{align*} onto \begin{align*} \{q \in \mathbb{C} : 0 < |q| < e^{-2\pi R/w_k}\}. \end{align*} We now add one point $c_k$ corresponding to $q=0$. In the $q_k$-coordinate, the punctured disc becomes the full disc \begin{align*} \{q \in \mathbb{C} : |q| < e^{-2\pi R/w_k}\}. \end{align*} Repeating this construction for all cusp representatives defines \begin{align*} X_0(N) := Y_0(N) \cup \{c_1,\dots,c_m\}. \end{align*} The topology on $X_0(N)$ is the topology generated by the original open sets of $Y_0(N)$ and these full cusp discs. The charts at the newly added points are the $q_k$-charts. It remains to justify compatibility with the existing complex structure. Let $\varphi: U \to V \subset \mathbb{C}$ be an interior quotient chart with $U \subset Y_0(N)$ and $V$ chosen to be a bounded coordinate disc. On a punctured overlap with the cusp chart, the transition map is \begin{align*} \varphi \circ \pi \circ \sigma_k \circ q_k^{-1}. \end{align*} Here $q_k^{-1}$ denotes a local branch of the logarithm inverse to $q_k$, $\sigma_k$ acts by a Möbius transformation, $\pi$ is holomorphic on the interior quotient, and $\varphi$ is holomorphic by definition of the interior atlas. Therefore the transition map is holomorphic on the punctured overlap. If the overlap accumulates at $q=0$, then after shrinking the cusp disc its image lies in the bounded coordinate disc $V$, so the [removable singularity theorem](/page/Removable%20Singularity%20Theorem) extends the transition holomorphically across $q=0$. If the overlap is an annulus bounded away from $q=0$, the transition is already holomorphic on its whole domain. Thus the cusp charts are compatible with the interior atlas, and the filled cusp is a genuine point of a Riemann surface.[/guided]

[step:Show the filled quotient is compact]Let \begin{align*} \mathcal{F}_R := \mathcal{F} \cap \{z \in \mathcal{H} : \operatorname{Im}(z) \leq R\} \end{align*} be the truncated standard fundamental domain, and define \begin{align*} \mathcal{K}_R := \bigcup_{j=1}^{r} \gamma_j \mathcal{F}_R. \end{align*} The set $\mathcal{F}_R$ is closed and bounded in $\mathbb{C}$, so $\mathcal{K}_R$ is compact because it is a finite union of continuous images of the compact set $\mathcal{F}_R$ under Möbius transformations whose denominators do not vanish on $\mathcal{H}$. It remains to check that the omitted parts of $\mathcal{D}$ are contained in the cusp neighbourhoods already filled in the previous step. For each $j \in \{1,\dots,r\}$, the point $\gamma_j(\infty) \in \mathbb{P}^1(\mathbb{Q})$ is equivalent under $\Gamma_0(N)$ to exactly one chosen cusp representative $\alpha_{k(j)}$. Choose $\Delta_j \in \Gamma_0(N)$ with \begin{align*} \Delta_j \gamma_j(\infty) = \alpha_{k(j)}. \end{align*} Then \begin{align*} \tau_j := \sigma_{k(j)}^{-1}\Delta_j\gamma_j \in SL_2(\mathbb{Z}) \end{align*} fixes $\infty$, hence $\tau_j = \pm\begin{pmatrix}1 & n_j \\ 0 & 1\end{pmatrix}$ for some $n_j \in \mathbb{Z}$. Therefore, for the standard tail $\mathcal{F} \cap \{\operatorname{Im}(z)>R\}$, \begin{align*} \gamma_j\bigl(\mathcal{F} \cap \{\operatorname{Im}(z)>R\}\bigr) = \Delta_j^{-1}\sigma_{k(j)}\tau_j\bigl(\mathcal{F} \cap \{\operatorname{Im}(z)>R\}\bigr) \subset \Delta_j^{-1}\sigma_{k(j)}\{z \in \mathcal{H}: \operatorname{Im}(z)>R\}. \end{align*} Since $\Delta_j \in \Gamma_0(N)$, the image of this set in $\Gamma_0(N)\backslash\mathcal{H}$ is contained in the cusp neighbourhood associated to $\alpha_{k(j)}$. Thus the image of $\mathcal{K}_R$ covers the complement of the finitely many open cusp discs \begin{align*} \{q \in \mathbb{C} : |q| < e^{-2\pi R/w_k}\}, \qquad k=1,\dots,m. \end{align*} The image of $\mathcal{K}_R$ in $X_0(N)$ is compact because it is the continuous image of a compact set. Each closed cusp disc \begin{align*} \{q \in \mathbb{C} : |q| \leq e^{-2\pi R/w_k}\} \end{align*} is compact in $\mathbb{C}$. Therefore $X_0(N)$ is covered by finitely many compact sets: the compact image of $\mathcal{K}_R$ together with the finitely many closed cusp discs. Hence $X_0(N)$ is compact.[/step]

[guided]The point requiring care is that the unbounded part of the finite union \begin{align*} \mathcal{D}=\bigcup_{j=1}^r \gamma_j\mathcal{F} \end{align*} does not necessarily lie inside the particular geometric horoballs $\sigma_k\{\operatorname{Im}(z)>R\}$ before passing to the quotient. We therefore truncate the standard fundamental domain first and then identify each omitted tail with one of the chosen cusp neighbourhoods modulo $\Gamma_0(N)$. Define \begin{align*} \mathcal{F}_R := \mathcal{F} \cap \{z \in \mathcal{H} : \operatorname{Im}(z) \leq R\}, \qquad \mathcal{K}_R := \bigcup_{j=1}^{r} \gamma_j \mathcal{F}_R. \end{align*} The set $\mathcal{F}_R$ is closed and bounded in $\mathbb{C}$, hence compact by the [Heine-Borel theorem](/theorems/315). Each matrix $\gamma_j$ acts on $\mathcal{H}$ by a Möbius transformation, and its denominator cannot vanish at a point of $\mathcal{H}$ because such a zero would be real. Hence $\gamma_j$ is continuous on $\mathcal{F}_R$, so $\gamma_j\mathcal{F}_R$ is compact. Since $\mathcal{K}_R$ is a finite union of these compact sets, $\mathcal{K}_R$ is compact. Now consider one omitted tail \begin{align*} \gamma_j\bigl(\mathcal{F} \cap \{\operatorname{Im}(z)>R\}\bigr). \end{align*} The endpoint of this tail on the rational boundary is $\gamma_j(\infty)$. By construction of the cusp representatives, there is a unique index $k(j) \in \{1,\dots,m\}$ and an element $\Delta_j \in \Gamma_0(N)$ such that \begin{align*} \Delta_j\gamma_j(\infty)=\alpha_{k(j)}. \end{align*} Set \begin{align*} \tau_j := \sigma_{k(j)}^{-1}\Delta_j\gamma_j. \end{align*} Then $\tau_j \in SL_2(\mathbb{Z})$ and $\tau_j(\infty)=\infty$. The stabiliser of $\infty$ in $SL_2(\mathbb{Z})$ consists exactly of the matrices $\pm\begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$ with $n \in \mathbb{Z}$, so \begin{align*} \tau_j = \pm\begin{pmatrix}1 & n_j \\ 0 & 1\end{pmatrix} \end{align*} for some $n_j \in \mathbb{Z}$. Such a matrix preserves imaginary parts. Therefore \begin{align*} \gamma_j\bigl(\mathcal{F} \cap \{\operatorname{Im}(z)>R\}\bigr) = \Delta_j^{-1}\sigma_{k(j)}\tau_j\bigl(\mathcal{F} \cap \{\operatorname{Im}(z)>R\}\bigr) \subset \Delta_j^{-1}\sigma_{k(j)}\{z \in \mathcal{H}: \operatorname{Im}(z)>R\}. \end{align*} Because $\Delta_j \in \Gamma_0(N)$ represents the identity operation after passing to the quotient $\Gamma_0(N)\backslash\mathcal{H}$, this tail maps into the cusp neighbourhood associated to $\alpha_{k(j)}$. Thus every unbounded part removed from $\mathcal{D}$ is accounted for by one of the finitely many filled cusp discs. The quotient image of $\mathcal{K}_R$ is compact because continuous images of compact spaces are compact. The filled cusp neighbourhood for $\alpha_k$ is represented by the closed disc \begin{align*} \{q \in \mathbb{C} : |q| \leq e^{-2\pi R/w_k}\}, \end{align*} which is compact in $\mathbb{C}$. We have covered $X_0(N)$ by the compact image of $\mathcal{K}_R$ and the finitely many compact closed cusp discs. A finite union of compact subsets is compact, so $X_0(N)$ is compact.[/guided]

[step:Conclude connectedness after adjoining the cusp points] The space $Y_0(N)=\Gamma_0(N)\backslash\mathcal{H}$ is connected by the earlier quotient argument. The compactification $X_0(N)$ is obtained from $Y_0(N)$ by adjoining finitely many cusp points, each of which lies in the closure of a punctured cusp neighbourhood contained in $Y_0(N)$. Thus $Y_0(N)$ is dense in $X_0(N)$. Since the closure of a connected subset is connected, $X_0(N)=\overline{Y_0(N)}$ is connected. Combining this connectedness with the Riemann-surface structure constructed by the interior quotient charts and the cusp $q$-charts, and with compactness from the previous step, proves that $X_0(N)$ is a compact connected Riemann surface. [/step]