A genus-two compact surface gives a cocompact Fuchsian group

The affine curve and its local parameters. Let $F(x,y):=y^2-P(x)$ and let $(x_0,y_0)\in X_0$. The gradient $(-P^{\mathrm{\prime }}(x),2y)$ does not vanish at $(x_0,y_0)$: if $y_0\mathrm{\ne }0$ this is read off the second component, while if $y_0=0$ then $P(x_0)=0$, so $x_0=a_j$ for a unique $j$ and $-P^{\mathrm{\prime }}(a_j)=-{\prod }_{k\mathrm{\ne }j}(a_j-a_k)\mathrm{\ne }0$ because the six roots are distinct. By [F2] the curve is therefore locally a graph over a coordinate: where $y_0\mathrm{\ne }0$ there are discs $A\ni x_0$, $B\ni y_0$ and a holomorphic $\eta :A\to B$ with $X_0\cap (A\times B)=\{(x,\eta (x)):x\in A\}$, and at $(a_j,0)$ there are discs $A_j\ni a_j$, $B_j\ni 0$ and a holomorphic ${\psi }_j:B_j\to A_j$ with $X_0\cap (A_j\times B_j)=\{({\psi }_j(y),y):y\in B_j\}$. The maps $x\mapsto (x,\eta (x))$ and $y\mapsto ({\psi }_j(y),y)$ are homeomorphisms onto their images, with inverses given by the holomorphic coordinate functions $x$ and $y$, so the two families of charts are compatible in both directions [F1]. Moreover $u_j$ is holomorphic near $a_j$ with $u_j(a_j)=P^{\mathrm{\prime }}(a_j)\mathrm{\ne }0$, so from $y^2=P({\psi }_j(y))=({\psi }_j(y)-a_j)u_j({\psi }_j(y))$ one obtains ${\psi }_j(y)-a_j=y^2{w}_j(y)$ with $w_j:=1\mathrm{/}(u_j\circ {\psi }_j)$ holomorphic near $0$ and $w_j(0)=1\mathrm{/}{P}^{\mathrm{\prime }}(a_j)\mathrm{\ne }0$. Hence the projection ${\pi }_0(x,y)=x$ has, in the chart with parameter $y$ at $(a_j,0)$ and the centred target chart $z\mapsto a_j+z$, the expression $y\mapsto y^2{w}_j(y)$, of order $2$ at $0$, while at a point with $y\mathrm{\ne }0$ its expression in the chart with parameter $x$ is $z\mapsto z$.

The complement of the six roots is path connected. Let $V:=\mathbb{C}\setminus \{a_1,\dots ,a_6\}$ and fix $R>\max \{1\mathrm{/}\epsilon ,1+{\max }_j\mathrm{\mid }{a}_j\mathrm{\mid }\}$. For $x\in V$ with $\mathrm{\mid }x\mathrm{\mid }\le R$, choose a direction $\theta$ different from the at most six directions toward the roots. The segment of length $3R$ from $x$ in that direction avoids the roots and ends in $E:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }>R\}$, since its endpoint has modulus at least $3R-\mathrm{\mid }x\mathrm{\mid }\ge 2R>R$. If $\mathrm{\mid }x\mathrm{\mid }>R$, then $x\in E$ already. The set $E$ is path connected by [F9] and is contained in $V$, so any two points of $V$ can be joined in $V$ by paths through $E$. Hence $V$ is path connected and connected.

Completion at infinity. Put $t:=1\mathrm{/}x$ and $v:=y\mathrm{/}{x}^3$, so that on $X_0$ with $x\mathrm{\ne }0$ the equation $y^2=P(x)$ reads $v^2=P(x)\mathrm{/}{x}^6=Q(t)={\prod }_{j=1}^6(1-a_{j}t)$, and $Q(0)=1$. Choose $\epsilon >0$ with $Q(t)\mathrm{\ne }0$ for $\mathrm{\mid }t\mathrm{\mid }<\epsilon$; by [F3] there is a holomorphic $L$ on that disc with $\exp L=Q$, and $\rho :=\exp (L\mathrm{/}2)$ satisfies ${\rho }^2=Q$ and $\rho (0)\mathrm{\ne }0$. Add two points ${\mathrm{\infty }}_{+}$ and ${\mathrm{\infty }}_{-}$ to $X_0$ and declare $t\mapsto (t,+\rho (t)),t\mapsto (t,-\rho (t))(\mathrm{\mid }t\mathrm{\mid }<\epsilon ),$ to be charts at them. For $t\mathrm{\ne }0$ the corresponding point of the first chart is $(x,y)=(1\mathrm{/}t,\rho (t)t^{-3})$, and $y^2=t^{-6}\rho (t{)}^2=t^{-6}Q(t)=P(1\mathrm{/}t)=P(x)$, so it lies in $X_0$; the two charts are glued to the affine charts of step 1.1 by the transition maps $t=1\mathrm{/}x$, $v=y\mathrm{/}{x}^3$ and their inverses $x=1\mathrm{/}t$, $y=v\mathrm{/}{t}^3$, holomorphic on $t\mathrm{\ne }0$. Thus $X:=X_0\cup \{{\mathrm{\infty }}_{+},{\mathrm{\infty }}_{-}\}$ carries the atlas of steps 1.1 and 2.1, and towards the chart ${\varphi }_{\mathrm{\infty }}$ of [F4] the projection $\pi$ has at ${\mathrm{\infty }}_{\pm }$ the expression $t\mapsto t$. Moreover for every $R>1\mathrm{/}\epsilon$ the points of $X$ with $\mathrm{\mid }x\mathrm{\mid }>R$, equivalently with $t=1\mathrm{/}x$ satisfying $0<\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}R$, are exactly the points of the two chart images with $t\mathrm{\ne }0$, and their second coordinates are $\pm x^3\rho (1\mathrm{/}x)$.

The punctured affine curve is a connected two-sheeted cover of $V$. Put $Z:=\{(x,y)\in X_0:x\in V\}$. For $x\in V$ one has $P(x)\mathrm{\ne }0$, so by [F3] there are a disc $D(x,r)\subseteq V$ and a holomorphic square root ${\rho }_x$ of $P$ on it; the two maps $z\mapsto (z,\pm {\rho }_x(z))$ are local inverses of ${\pi }_0$, and they exhibit ${\pi }_0^{-1}(D(x,r))$ as the disjoint union of two open sets each mapped homeomorphically onto $D(x,r)$. Hence ${\pi }_0:Z\to V$ is a covering of degree $2$ [F16], so it is continuous and open, and $Z$ is Hausdorff and second countable as a subspace of ${\mathbb{C}}^2$. Suppose $Z$ were disconnected, say $Z=Z_1\bigsqcup Z_2$ with the $Z_i$ nonempty, open and closed. Over each small evenly covered disc, each of the two connected sheets lies wholly in one of the clopen $Z_i$, so the number of sheet points in $Z_i$ is locally constant on $V$. Thus the images ${\pi }_0(Z_i)$ are nonempty, open and closed in the connected space $V$ (step 1.2), hence equal to $V$; and since each fibre of ${\pi }_0$ consists of exactly two points, one over each $Z_i$, the restriction ${\pi }_0{\mathrm{\mid }}_{Z_1}$ is a bijection onto $V$ with local continuous inverse, hence a homeomorphism. Its inverse provides a continuous $s:V\to \mathbb{C}$ with $s(x{)}^2=P(x)$ for all $x\in V$. Fix $0<{\epsilon }_1<{\min }_{k\mathrm{\ne }1}\mathrm{\mid }{a}_1-a_k\mathrm{\mid }$ and, by [F3], a holomorphic $\rho$ on $D(a_1,{\epsilon }_1)$ with ${\rho }^2=u_1$; there $u_1$ is holomorphic and nowhere zero. Define $c(\theta ):=a_1+({\epsilon }_1\mathrm{/}2)e^{i\theta }$, $w(\theta ):=({\epsilon }_1\mathrm{/}2{)}^{1\mathrm{/}2}{e}^{i\theta \mathrm{/}2}$, so $w(\theta {)}^2=c(\theta )-a_1$, and $g(\theta ):=s(c(\theta ))\mathrm{/}(w(\theta )\rho (c(\theta )))$ for $\theta \in [0,2\pi ]$. Then $g(\theta {)}^2=\frac{P(c(\theta ))}{(c(\theta )-a_1)u_1(c(\theta ))}=1,$ so $g$ maps the connected interval $[0,2\pi ]$ continuously into $\{1,-1\}$ [F8]; a continuous image of a connected set is connected while $\{1,-1\}$ is disconnected, so $g$ is constant [F8]. But $s(c(2\pi ))=s(c(0))$, $\rho (c(2\pi ))=\rho (c(0))$ and $w(2\pi )=-w(0)$, so $g(2\pi )=-g(0)\mathrm{\ne }g(0)$, a contradiction. Hence $Z$ is connected.

The completed space is Hausdorff and second countable, and $\pi$ is holomorphic for the atlas. Distinct points of $X_0$ are separated by the Hausdorff topology of ${\mathbb{C}}^2$, the affine charts being restrictions of the coordinate projections; the two points ${\mathrm{\infty }}_{+}$ and ${\mathrm{\infty }}_{-}$ are separated because their chart values at $t=0$ are $\rho (0)$ and $-\rho (0)$, which are distinct and give disjoint chart images. A point $(x_0,y_0)\in X_0$ with $R>\max \{\mathrm{\mid }{x}_0\mathrm{\mid },1\mathrm{/}\epsilon \}$ and an infinity point are separated by the open sets $\{\mathrm{\mid }x\mathrm{\mid }<R\}\cap X_0$ and the image under the relevant chart of the disc $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}R$, which is disjoint from the first by step 2.1. Hence $X$ is Hausdorff. A countable base of the topology of $X$ is obtained from a countable base of the open subspace $X_0$, which is second countable as a subspace of ${\mathbb{C}}^2$, together with the images under the two chart maps of a countable base of the disc $\mathrm{\mid }t\mathrm{\mid }<\epsilon$; these sets are open and every open subset of $X$ is the union of its intersections with the three open pieces $X_0$ and the two chart images, so $X$ is second countable. The chart expressions of $\pi$ are holomorphic: $z\mapsto z$ and $y\mapsto {\psi }_j(y)$ on the affine charts of step 1.1 and $t\mapsto t$ on the two charts at infinity of step 2.1. Consequently, once $X$ is known to be connected, it is a Riemann surface with this atlas and $\pi$ is a nonconstant holomorphic map of Riemann surfaces [F1].

The completed space is compact. Fix $R>1\mathrm{/}\epsilon$. The set $K:=\{(x,y)\in X_0:\mathrm{\mid }x\mathrm{\mid }\le R\}$ is the intersection of the closed set $X_0$ with the closed cylinder $\{\mathrm{\mid }x\mathrm{\mid }\le R\}$ in ${\mathbb{C}}^2$, hence closed; on it $\mathrm{\mid }y{\mathrm{\mid }}^2=\mathrm{\mid }P(x)\mathrm{\mid }\le {\prod }_j(R+\mathrm{\mid }{a}_j\mathrm{\mid })$, so it is bounded in ${\mathbb{C}}^2$, hence compact [F7]. The image of the closed disc $\mathrm{\mid }t\mathrm{\mid }\le 1\mathrm{/}R$ under each of the two charts at infinity is compact, being a continuous image of a compact set [F7], and it contains the corresponding point ${\mathrm{\infty }}_{\pm }$; by step 2.1 every point of $X$ with $\mathrm{\mid }x\mathrm{\mid }>R$ lies in one of these two images. Therefore $X=K\cup (\text{image of the }+\text{-chart})\cup (\text{image of the }-\text{-chart})$ is a finite union of compact subsets, hence compact.

The completed space is connected. By step 2.2 the set $Z$ is connected and contained in $X_0$. Every point of $X_0\setminus Z$, namely each $(a_j,0)$, is a limit point of $Z$: in the chart $y\mapsto ({\psi }_j(y),y)$ of step 1.1 the points with $0<\mathrm{\mid }y\mathrm{\mid }<\delta$ have $x={\psi }_j(y)=a_j+y^2{w}_j(y)\mathrm{\ne }{a}_j$, so they lie in $Z$, and they tend to $(a_j,0)$ as $y\to 0$. Hence $X_0$ lies between the connected set $Z$ and its closure, so $X_0$ is connected [F8]. Likewise each ${\mathrm{\infty }}_{\pm }$ is a limit point of $X_0$: the points of its chart with $0<\mathrm{\mid }t\mathrm{\mid }<\delta$ belong to $X_0$ by step 2.1 and tend to ${\mathrm{\infty }}_{\pm }$ as $t\to 0$; hence $X$ lies between the connected set $X_0$ and its closure, so $X$ is connected [F8]. By step 3.1, $X$ is a Riemann surface.

The ramification points of $\pi$ are the six branch points. At a point $(x_0,y_0)\in X_0$ with $y_0\mathrm{\ne }0$ the chart of step 1.1 has local parameter $x$ and the chart expression of $\pi$ towards ${\varphi }_0$ is $z\mapsto z$, so the ramification index is $1$ there, by the description of the index as an order [F10]. Over each $a_j$ the only point of $X$ is $(a_j,0)$, because $y^2=P(a_j)=0$ forces $y=0$; in the chart with local parameter $y$ and the centred target chart at $a_j$ the expression of $\pi$ is $y\mapsto y^2{w}_j(y)$ with $w_j(0)\mathrm{\ne }0$ (step 1.1), whose order at $0$ is $2$, so $e_{(a_j,0)}(\pi )=2$ [F10]. At ${\mathrm{\infty }}_{\pm }$ the chart expression towards ${\varphi }_{\mathrm{\infty }}$ is $t\mapsto t$ (step 2.1), so the index is $1$ and $\mathrm{\infty }$ is not a branch value. Hence the branch values of $\pi$ are exactly the six distinct numbers $a_1,\dots ,a_6$, each with exactly one preimage, of index $2$, and every other value has all its preimages of index $1$.

$\pi$ is proper of degree two. By steps 3.1 and 4.1 the space $X$ is a Riemann surface and $\pi :X\to \widehat{\mathbb{C}}$ is nonconstant holomorphic. For compact $K\subseteq \widehat{\mathbb{C}}$ the preimage ${\pi }^{-1}(K)$ is closed in $X$, because $\pi$ is continuous and $K$ is closed in the Hausdorff space $\widehat{\mathbb{C}}$; being a closed subset of the compact space $X$ (step 3.2), it is compact [F7]. So $\pi$ is proper, and the degree theorem [F11] applies. For $b\in \mathbb{C}\setminus \{a_1,\dots ,a_6\}$ the fibre is ${\pi }^{-1}(b)=\{(b,\sqrt{P(b)}),(b,-\sqrt{P(b)})\}$, two distinct points, each of index $1$ by step 5.1, so $d=\mathrm{deg}\pi =2$.

The genus is two. By steps 3.2, 4.1 and 6.1 the map $\pi$ is a nonconstant holomorphic map of degree $2$ between compact connected Riemann surfaces, so Riemann-Hurwitz [F12] gives $2g(X)-2=2(2g(\widehat{\mathbb{C}})-2)+{\sum }_{x\in X}(e_x(\pi )-1)\mathrm{.}$ By step 5.1 the ramification points are exactly the six points $(a_j,0)$, each with index $2$, while all other points, namely the affine points with $y\mathrm{\ne }0$ and the two points ${\mathrm{\infty }}_{\pm }$, have index $1$; hence the sum equals $6$. The sphere $\widehat{\mathbb{C}}$ is compact, being homeomorphic to the closed bounded subset $S^2$ of ${\mathbb{R}}^3$ [F5, F7], and connected [F6]; and $g(\widehat{\mathbb{C}})=0$, because $\widehat{\mathbb{C}}\cong S^2={\mathrm{\#}}_0{T}^2$ and the genus is the unique handle number [F13]. Therefore $2g(X)-2=2(0-2)+6=2$, that is $g(X)=2$.

Hyperbolic type and the uniformization. By step 7.1 the surface $X$ is a compact Riemann surface of genus $2$, so the compact-genus corollary [F14], whose choice hypothesis is covered by [A1], gives that $X$ has hyperbolic universal-covering type: its holomorphic universal cover $p:\tilde{X}\to X$ satisfies $\tilde{X}\cong \mathbb{D}$ [F14], and fixing a biholomorphism $\psi :\tilde{X}\to \mathbb{D}$ gives a uniformization $(p,\psi )$ [F14]. By [F17] the conjugate ${\gamma }_h:=\psi \circ h\circ {\psi }^{-1}$ is an automorphism of $\mathbb{D}$ for every $h\in \mathrm{Deck}(p)$, and $\mathrm{\Gamma }:=\psi \mathrm{Deck}(p){\psi }^{-1}=\{{\gamma }_h:h\in \mathrm{Deck}(p)\}$ is a subgroup of $\mathrm{Aut}(\mathbb{D})$, isomorphic to $\mathrm{Deck}(p)$.

The transported covering, its deck group, and freeness. Define $\mathrm{\Psi }:=p\circ {\psi }^{-1}:\mathbb{D}\to X$. A homeomorphism of the total space carries evenly covered neighbourhoods to evenly covered neighbourhoods, so $\mathrm{\Psi }$ is a covering map, with the same evenly covered sets as $p$ [F16]. A homeomorphism $h$ of $\mathbb{D}$ satisfies $\mathrm{\Psi }\circ h=\mathrm{\Psi }$ exactly when $p\circ {\psi }^{-1}h\psi =p$, that is exactly when ${\psi }^{-1}h\psi \in \mathrm{Deck}(p)$, that is exactly when $h\in \mathrm{\Gamma }$; hence $\mathrm{Deck}(\mathrm{\Psi })=\mathrm{\Gamma }$. Since $\mathbb{D}$ is connected, deck transformations of $\mathrm{\Psi }$ agreeing at one point are equal, so $\mathrm{Deck}(\mathrm{\Psi })=\mathrm{\Gamma }$ acts freely on $\mathbb{D}$ [F16].

The action of $\mathrm{\Gamma }$ is properly discontinuous. Let $K\subseteq \mathbb{D}$ be compact. Use the family of all evenly covered coordinate-disc neighbourhoods $U$ and smaller open neighbourhoods $W$ whose compact closures lie in $U$. Finitely many $W_i$ cover $\mathrm{\Psi }(K)$. For each $i$, the set $K\cap {\mathrm{\Psi }}^{-1}({\bar{W}}_i)$ is closed in $K$, hence compact; the sheets over $U_i$ cover it, so only finitely many sheets meet it. Denote these by $S_i$. If $\gamma K\cap K\mathrm{\ne }\mathrm{\varnothing }$, write $\gamma x=y$ with $x,y\in K$ and choose $i$ with $\mathrm{\Psi }(x)=\mathrm{\Psi }(y)\in W_i$. The sheets $V,V^{\mathrm{\prime }}\in S_i$ containing $x,y$ are over the same $U_i$, and $\gamma$ maps $V$ onto $V^{\mathrm{\prime }}$ because it is a deck transformation. Two deck transformations mapping $V$ onto $V^{\mathrm{\prime }}$ agree at the unique point of $V$ above any fixed base point, hence agree everywhere by [F16]. Therefore at most ${\sum }_i\mathrm{\mid }{S}_i{\mathrm{\mid }}^2$ elements of $\mathrm{\Gamma }$ move $K$ to meet itself, so the action is properly discontinuous [F15].

$\mathrm{\Gamma }$ is discrete. Fix $z_0\in \mathbb{D}$. For each $\gamma \in \mathrm{\Gamma }$ choose an evenly covered neighbourhood of $\mathrm{\Psi }(z_0)$ and its sheet $V$ containing $\gamma (z_0)$. The compact-open set $S(\{z_0\},V)$ is a neighbourhood of $\gamma$. If ${\gamma }^{\mathrm{\prime }}\in \mathrm{\Gamma }$ also lies in it, then ${\gamma }^{\mathrm{\prime }}(z_0)$ and $\gamma (z_0)$ lie in the same sheet and fibre, so injectivity on the sheet makes these values equal. Deck rigidity [F16] gives ${\gamma }^{\mathrm{\prime }}=\gamma$. Every element of $\mathrm{\Gamma }$ is therefore isolated in the compact-open topology.

$\mathrm{\Gamma }$ is torsion-free. Let $\gamma \in \mathrm{\Gamma }$ with ${\gamma }^m=\mathrm{i}\mathrm{d}$ for some $m\ge 1$; we show $\gamma =\mathrm{i}\mathrm{d}$. By [F19] there are $a\in \mathbb{D}$ and $\theta \in \mathbb{R}$ with $\gamma =e^{i\theta }{\phi }_a$, where ${\phi }_a(z)=(a-z)\mathrm{/}(1-\bar{a}z)$; written as a quotient of linear polynomials this exhibits $\gamma$ as a Mobius transformation [F20]. Assume $\gamma \mathrm{\ne }\mathrm{i}\mathrm{d}$; the classification [F20] gives two alternatives. In the parabolic alternative $\gamma$ is conjugate to $z\mapsto z+1$, so ${\gamma }^m$ is conjugate to $z\mapsto z+m\mathrm{\ne }\mathrm{i}\mathrm{d}$, contradicting ${\gamma }^m=\mathrm{i}\mathrm{d}$. In the other alternative $\gamma$ has two fixed points and is conjugate to $z\mapsto \lambda z$ with $\lambda \in {\mathbb{C}}^{\times }\setminus \{1\}$; then ${\gamma }^m=\mathrm{i}\mathrm{d}$ forces ${\lambda }^m=1$, so $\mathrm{\mid }\lambda \mathrm{\mid }=1$, $\lambda \mathrm{\ne }1$, and the invariant $\tau (\gamma )=\mathrm{tr}(A{)}^2\mathrm{/}\det A$ of any representing matrix $A$ equals $\lambda +2+{\lambda }^{-1}=2+2\mathrm{Re}\lambda$, which lies in $[0,4)$ [F20]. The matrix $A=\left(\begin{array}{cc}-e^{i\theta }& e^{i\theta }a\\ -\bar{a}& 1\end{array}\right)$ represents $\gamma$, and $\mathrm{tr}A=1-e^{i\theta }$, $\det A=-e^{i\theta }(1-r^2)$ with $r:=\mathrm{\mid }a\mathrm{\mid }<1$, so $\tau (\gamma )=\frac{(1-e^{i\theta }{)}^2}{-e^{i\theta }(1-r^2)}=\frac{4{\sin }^2(\theta \mathrm{/}2)}{1-r^2}\mathrm{.}$ Hence $4{\sin }^2(\theta \mathrm{/}2)<4(1-r^2)$, that is $r^2<{\cos }^2(\theta \mathrm{/}2)$; writing $c:=\cos (\theta \mathrm{/}2)$ we have $r<\mathrm{\mid }c\mathrm{\mid }$. If $a=0$ then $\gamma (z)=-e^{i\theta }z$ fixes $0\in \mathbb{D}$, contradicting the freeness of the action of $\mathrm{\Gamma }$ (step 9.1); so $a\mathrm{\ne }0$ and $r>0$. The fixed points of $\gamma$ solve $\gamma (z)=z$, that is $\bar{a}{z}^2-(1+e^{i\theta })z+e^{i\theta }a=0$; substituting $z=e^{i\theta \mathrm{/}2}w$ and dividing by $e^{i\theta }$ turns this into $\bar{a}{w}^2-2cw+a=0$, and multiplying by $a$ gives $r^2{w}^2-2acw+a^2=0$, whose roots are $w=a(c\pm s)\mathrm{/}{r}^2$ with $s:=(c^2-r^2{)}^{1\mathrm{/}2}>0$ real. Hence $z_{\pm }=e^{i\theta \mathrm{/}2}a(c\pm s)\mathrm{/}{r}^2$ are the two fixed points of $\gamma$, and $\mathrm{\mid }{z}_{\pm }\mathrm{\mid }=\mathrm{\mid }c\pm s\mathrm{\mid }\mathrm{/}r$. Since $c^2-s^2=r^2>0$, one has $s<\mathrm{\mid }c\mathrm{\mid }$ and the two numbers $c\pm s$ have the same sign. Thus $\min \{\mathrm{\mid }{z}_{+}\mathrm{\mid },\mathrm{\mid }{z}_{-}\mathrm{\mid }\}=(\mathrm{\mid }c\mathrm{\mid }-s)\mathrm{/}r$. Moreover $\mathrm{\mid }c{\mathrm{\mid }}^2-s^2=r^2$, so $(\mathrm{\mid }c\mathrm{\mid }-s)\mathrm{/}r=r\mathrm{/}(\mathrm{\mid }c\mathrm{\mid }+s)<1$. Hence one of the fixed points lies in $\mathbb{D}$, contradicting freeness (step 9.1).

$\mathbb{D}\mathrm{/}\mathrm{\Gamma }$ is homeomorphic to $X$ and compact. Let $q:\mathbb{D}\to \mathbb{D}\mathrm{/}\mathrm{\Gamma }$ be the quotient map of the action of $\mathrm{\Gamma }$ and give $\mathbb{D}\mathrm{/}\mathrm{\Gamma }$ the quotient topology [F22]. For ${\gamma }_h\in \mathrm{\Gamma }$ one has $\mathrm{\Psi }({\gamma }_{h}z)=p({\psi }^{-1}\psi h{\psi }^{-1}z)=p(h{\psi }^{-1}z)=p({\psi }^{-1}z)=\mathrm{\Psi }(z)$, so $\mathrm{\Psi }$ is $\mathrm{\Gamma }$-invariant and induces a map $\mathrm{\Phi }:\mathbb{D}\mathrm{/}\mathrm{\Gamma }\to X$ with $\mathrm{\Phi }\circ q=\mathrm{\Psi }$; by the characteristic property of the quotient topology $\mathrm{\Phi }$ is continuous [F22]. It is surjective because $\mathrm{\Psi }$ is. It is injective: if $\mathrm{\Psi }(z)=\mathrm{\Psi }(z^{\mathrm{\prime }})$, then ${\psi }^{-1}z$ and ${\psi }^{-1}{z}^{\mathrm{\prime }}$ lie in one fibre of $p$, and the deck group of the universal cover acts transitively on each fibre [F18], so ${\psi }^{-1}{z}^{\mathrm{\prime }}=h{\psi }^{-1}z$ for some $h\in \mathrm{Deck}(p)$ and hence $z^{\mathrm{\prime }}={\gamma }_{h}z$, that is $q(z^{\mathrm{\prime }})=q(z)$. The map $\mathrm{\Psi }$ is open: if $O\subseteq \mathbb{D}$ is open and $w=\mathrm{\Psi }(u)\in \mathrm{\Psi }(O)$, choose an evenly covered $U\ni w$ with sheet $V\ni u$; then $V\cap O$ is open and $\mathrm{\Psi }(V\cap O)$ is open in $U$, hence in $X$, and contains $w$. Consequently for every open $O\subseteq \mathbb{D}\mathrm{/}\mathrm{\Gamma }$ the set $\mathrm{\Phi }(O)=\mathrm{\Psi }(q^{-1}(O))$ is open in $X$ [F22], since $q$ is surjective; so the continuous bijection $\mathrm{\Phi }$ is a homeomorphism [F22]. Therefore $\mathbb{D}\mathrm{/}\mathrm{\Gamma }\cong X$ is compact by step 3.2, that is, $\mathrm{\Gamma }$ is cocompact.

Conclusion. Steps 3.1 to 4.1 exhibit $X$ as a compact connected Riemann surface (steps 3.2 and 4.1). Step 6.1 shows that the projection $\pi :X\to \widehat{\mathbb{C}}$ is a proper holomorphic map of degree $2$; step 5.1 identifies its branch values as the six numbers $a_1,\dots ,a_6$, each with a single point of index $2$ and with $\mathrm{\infty }$ unramified; and step 7.1 computes $g(X)=2$. By step 8.1 the surface $X$ has hyperbolic universal-covering type, with uniformization $\mathrm{\Psi }:\mathbb{D}\to X$ whose deck group is $\mathrm{\Gamma }\le \mathrm{Aut}(\mathbb{D})$ (step 9.1). The group $\mathrm{\Gamma }$ acts freely (step 9.1) and properly discontinuously (step 10.1), so it is Fuchsian in the sense of the Example; it is torsion-free (step 10.3) and discrete for the compact-open topology (step 10.2); and $\mathbb{D}\mathrm{/}\mathrm{\Gamma }\cong X$ is compact (step 10.4), so the quotient is cocompact. This proves all the assertions of the Example.