Every Riemann surface is a quotient of a simply connected model

The cover, its deck group and the model. By [F5] the holomorphic universal cover $p:\tilde{X}\to X$ exists, $\tilde{X}$ is a simply connected Riemann surface, the deck transformations of $p$ are biholomorphic, and there is a biholomorphism $\phi :\tilde{X}\to M$ onto exactly one model $M\in \{\widehat{\mathbb{C}},\mathbb{C},\mathbb{D}\}$. The deck group $\mathrm{Deck}(p)$ consists of the homeomorphisms $h$ of $\tilde{X}$ with $p\circ h=p$ [F2].

Small invariant-free neighbourhoods. We record the following consequence of free proper discontinuity, used twice below. Let a group $G$ act on a locally compact Hausdorff space $Y$ by homeomorphisms, freely and properly discontinuously. Then every $y\in Y$ has an open neighbourhood $V$ with $gV\cap V=\mathrm{\varnothing }$ for every $g\mathrm{\ne }e$. Indeed, by local compactness choose a compact neighbourhood $K$ of $y$; by proper discontinuity the set $F:=\{g\in G:gK\cap K\mathrm{\ne }\mathrm{\varnothing }\}$ is finite and contains $e$; for each $g\in F\setminus \{e\}$ freeness gives $g\cdot y\mathrm{\ne }y$, and since $Y$ is Hausdorff there are disjoint open sets $U_g\ni g\cdot y$ and $W_g\ni y$. Then $V:=\mathrm{int}(K)\cap {\bigcap }_{g\in F\setminus \{e\}}(W_g\cap g^{-1}{U}_g)$ is a finite intersection of open neighbourhoods of $y$ [F8], hence an open neighbourhood of $y$, and for $g\in F\setminus \{e\}$ one has $gV\subseteq U_g$ and $V\subseteq W_g$, so $gV\cap V=\mathrm{\varnothing }$, while for $g\mathrm{\notin }F$ one has $V\subseteq K$ and $gV\subseteq gK$, so $gV\cap V=\mathrm{\varnothing }$.

The deck group acts freely and simply transitively on fibres. By [F3] the deck group of the covering $p$, whose total space $\tilde{X}$ is connected, acts freely on $\tilde{X}$. By [F4] it acts transitively, hence simply transitively, on every fibre of $p$: this uses that $X$ is path-connected, locally path-connected and semilocally simply connected, which holds because $X$ is a connected locally Euclidean space [F8].

The deck group acts properly discontinuously on $\tilde{X}$. Let $K\subseteq \tilde{X}$ be compact. Consider all pairs consisting of an evenly covered coordinate-disc neighbourhood $U$ and a smaller open neighbourhood $W$ with compact closure $\bar{W}\subseteq U$; local coordinate discs supply such pairs at every point of $p(K)$. Compactness of $p(K)$ gives finitely many such $W_i$ covering it, with ${\bar{W}}_i\subseteq U_i$. For each $i$, the set $K\cap p^{-1}({\bar{W}}_i)$ is compact: it is closed in $K$. The sheets over $U_i$ form an open cover of this set, so only finitely many of those sheets meet it; call their family ${\mathcal{S}}_i$. If $hK\cap K\mathrm{\ne }\mathrm{\varnothing }$, write $h(x)=y$ with $x,y\in K$, choose $i$ with $p(x)=p(y)\in W_i$, and let $V,V^{\mathrm{\prime }}\in {\mathcal{S}}_i$ be the sheets containing $x,y$. Since $h$ preserves $p$, it maps the connected sheet $V$ onto the sheet $V^{\mathrm{\prime }}$ over the same $U_i$. Two deck transformations mapping $V$ to $V^{\mathrm{\prime }}$ agree at the point of $V$ above any prescribed point of $U_i$, so they agree everywhere by [F3]. Thus at most ${\sum }_i\mathrm{\mid }{\mathcal{S}}_i{\mathrm{\mid }}^2$ deck transformations have $hK\cap K\mathrm{\ne }\mathrm{\varnothing }$, proving proper discontinuity.

Quotients by free proper actions are coverings. We record the general statement needed here for $M$ and again at the uniqueness stage below. Let $Y$ be a Riemann surface and let $K$ be a group of homeomorphisms of $Y$ acting freely and properly discontinuously (for instance a group of holomorphic automorphisms); let $q:Y\to Y\mathrm{/}K$ be the quotient map to the orbit space with the quotient topology. Then $q$ is a covering map. Indeed, by step 1.2 applied to $Y$ and $K$, each $y\in Y$ has an open neighbourhood $V$ with $gV\cap V=\mathrm{\varnothing }$ for all $g\mathrm{\ne }e$. Then $q^{-1}(q(V))={\bigcup }_{g\in K}gV$ is a disjoint union of open sets (if $gV\cap hV\mathrm{\ne }\mathrm{\varnothing }$ then $V\cap g^{-1}hV\mathrm{\ne }\mathrm{\varnothing }$, so $g^{-1}h=e$ and $g=h$), and $q$ restricted to each $gV$ is injective: if $q(ga)=q(gb)$ with $a,b\in V$, then $gb\in K\cdot ga$ and hence $b=g^{-1}hga\in g^{-1}hgV$ for some $h\in K$, so $V\cap g^{-1}hgV\mathrm{\ne }\mathrm{\varnothing }$, which forces $g^{-1}hg=e$, that is $h=e$ and $a=b$. $q$ is open because $q^{-1}(q(O))={\bigcup }_{g}gO$ is open for open $O$; hence $q(V)$ is open and each restriction $q{\mathrm{\mid }}_{gV}:gV\to q(V)$ is a continuous open bijection, thus a homeomorphism, so $q(V)$ is evenly covered and $q$ is a covering map [F2, F7, F8].

Topology of the general holomorphic quotient. In step 2.3 assume now that $K$ acts by biholomorphisms, and put $Z:=Y\mathrm{/}K$. The continuous image $Z$ of connected $Y$ is connected. Since $q$ is open, the images of a countable open base of $Y$ form a countable open base of $Z$. To prove Hausdorffness, take distinct orbits represented by $y,z$. Choose compact neighbourhoods $C_y,C_z$ and put $C:=C_y\cup C_z$, compact by [F8]. The set $F:=\{g:g{C}_y\cap C_z\mathrm{\ne }\mathrm{\varnothing }\}$ is finite by proper discontinuity on $C$. For every $g\in F$, $gy\mathrm{\ne }z$; choose disjoint open sets $A_g\ni gy$, $B_g\ni z$. Put $V:=\mathrm{int}{C}_y\cap {\bigcap }_{g\in F}{g}^{-1}{A}_g$ and $W:=\mathrm{int}{C}_z\cap {\bigcap }_{g\in F}{B}_g$. These are open neighbourhoods of $y,z$, and $gV\cap W=\mathrm{\varnothing }$ for $g\in F$ by construction and for $g\mathrm{\notin }F$ by the definition of $F$. Thus $q(V)$ and $q(W)$ are disjoint open neighbourhoods of the two orbits. So $Z$ is Hausdorff.

The conjugate group on the model. Put $G:=\phi \mathrm{Deck}(p){\phi }^{-1}=\{\phi \circ h\circ {\phi }^{-1}:h\in \mathrm{Deck}(p)\}$; this is a group of holomorphic automorphisms of $M$, since $\phi$ and $h$ are biholomorphic [F1, F5]. The action of $G$ on $M$ by evaluation is free: if $\phi h{\phi }^{-1}(m)=m$, then $h({\phi }^{-1}(m))={\phi }^{-1}(m)$ and $h=e$ by the freeness of step 2.1. It is properly discontinuous: for compact $K\subseteq M$ the set ${\phi }^{-1}(K)$ is compact, being a continuous image under the homeomorphism ${\phi }^{-1}$, and $\{g\in G:gK\cap K\mathrm{\ne }\mathrm{\varnothing }\}=\phi \{h\in \mathrm{Deck}(p):h{\phi }^{-1}(K)\cap {\phi }^{-1}(K)\mathrm{\ne }\mathrm{\varnothing }\}{\phi }^{-1}$ is a bijective image of a finite set by step 2.2.

Charts on the general holomorphic quotient. For every chart $\psi :U\to \mathbb{C}$ of $Y$ and every coordinate-disc restriction $V\subseteq U$ satisfying $gV\cap V=\mathrm{\varnothing }$ for $g\mathrm{\ne }e$, give $q(V)$ the chart $\psi \circ (q{\mathrm{\mid }}_V{)}^{-1}$. Such $V$ exist about every point by step 1.2, and $q{\mathrm{\mid }}_V$ is a homeomorphism by step 2.3. On an overlap of two quotient charts, write $s_V=(q{\mathrm{\mid }}_V{)}^{-1}$ and $s_W=(q{\mathrm{\mid }}_W{)}^{-1}$. At a point $a$ of the overlap there is $g\in K$ with $s_W(a)=g{s}_V(a)$. On the open neighbourhood where $s_V$ takes values in $V\cap g^{-1}W$, one has $s_W=g\circ s_V$, because $q{\mathrm{\mid }}_W$ is injective. The transition is therefore locally ${\psi }_W\circ g\circ {\psi }_V^{-1}$, holomorphic by [F1]. These charts, with step 3.1, make $Z$ a Riemann surface, and $q:Y\to Z$ a local biholomorphism: on $V$ its quotient-coordinate expression is ${\psi }_V$. This structure is uniquely determined by requiring $q$ to be a local biholomorphism, since then every inverse local section and every displayed quotient chart is holomorphic.

The transported covering and its deck group. Define $\tilde{p}:=p\circ {\phi }^{-1}:M\to X$. Then $\tilde{p}$ is a covering map: it is continuous, it is surjective because $p$ is [F2], and if $U\subseteq X$ is evenly covered for $p$ with sheets $V_j$, then the sets $\phi (V_j)$ are pairwise disjoint open subsets of $M$ covering ${\tilde{p}}^{-1}(U)$, and $\tilde{p}$ restricted to each $\phi (V_j)$ is the composite of the homeomorphism ${\phi }^{-1}{\mathrm{\mid }}_{\phi (V_j)}$ with the homeomorphism $p{\mathrm{\mid }}_{V_j}$, hence a homeomorphism onto $U$; so $U$ is evenly covered for $\tilde{p}$ [F2, F8]. Its deck group is exactly $G$: a homeomorphism $h$ of $M$ satisfies $\tilde{p}\circ h=\tilde{p}$ if and only if $p\circ {\phi }^{-1}h=p\circ {\phi }^{-1}$, that is ${\phi }^{-1}h\phi \in \mathrm{Deck}(p)$, which is exactly $h\in G$.

The quotient map is a covering with deck group $G$. Applying step 2.3 to $M$ and $G$ (which acts freely and properly discontinuously by step 3.2, by holomorphic automorphisms) gives that the canonical projection $q:M\to M\mathrm{/}G$ onto the orbit space is a covering map; in particular $M\mathrm{/}G$ is a topological space with the quotient topology of $q$ [F7, step 3.2, step 2.3].

The induced map $\bar{p}:M\mathrm{/}G\to X$. Define $\bar{p}(q(y)):=\tilde{p}(y)$ for $y\in M$. This is well defined: if $q(y^{\mathrm{\prime }})=q(y)$ then $y^{\mathrm{\prime }}=g\cdot y=\phi h{\phi }^{-1}(y)$ for some $h\in \mathrm{Deck}(p)$ (step 3.2), and $\tilde{p}(y^{\mathrm{\prime }})=p({\phi }^{-1}\phi h{\phi }^{-1}(y))=p(h({\phi }^{-1}(y)))=p({\phi }^{-1}(y))=\tilde{p}(y)$ because $p\circ h=p$ [F2, step 4.2].

$\bar{p}$ is continuous. Let $W\subseteq X$ be open. By the definition of the quotient topology, ${\bar{p}}^{-1}(W)$ is open in $M\mathrm{/}G$ if and only if $q^{-1}({\bar{p}}^{-1}(W))={\tilde{p}}^{-1}(W)$ is open in $M$ [F7]; this holds because $\tilde{p}$ is continuous (it is a covering map, step 4.2).

$\bar{p}$ is a bijection. It is surjective because $\tilde{p}$ is surjective [F2, step 4.2]. It is injective: if $\tilde{p}(y)=\tilde{p}(y^{\mathrm{\prime }})$, then ${\phi }^{-1}(y)$ and ${\phi }^{-1}(y^{\mathrm{\prime }})$ lie in the same fibre of $p$, so by the simple transitivity of step 2.1 there is $h\in \mathrm{Deck}(p)$ with ${\phi }^{-1}(y^{\mathrm{\prime }})=h({\phi }^{-1}(y))$; then $y^{\mathrm{\prime }}=\phi h{\phi }^{-1}(y)=g\cdot y$ lies in the orbit of $y$, so $q(y^{\mathrm{\prime }})=q(y)$.

Exactly one model. Suppose that $X$ is biholomorphic to $N\mathrm{/}H$ for another model $N\in \{\widehat{\mathbb{C}},\mathbb{C},\mathbb{D}\}$ and a group $H$ of holomorphic automorphisms of $N$ acting freely and properly discontinuously. The argument of step 2.3 applies verbatim to $Y:=N$ and $K:=H$, and uses only that $N$ is a Riemann surface and $H$ acts freely and properly discontinuously, so the quotient map $N\to N\mathrm{/}H$ is a covering map; steps 3.1 and 4.1 give $N\mathrm{/}H$ its quotient Riemann-surface structure and make this projection a local biholomorphism. Compose it with the supposed biholomorphism $N\mathrm{/}H\to X$. The total space $N$ is simply connected [F11], so $N\to N\mathrm{/}H$ is a universal covering of $N\mathrm{/}H\cong X$, as is $p:\tilde{X}\to X$; by the uniqueness of universal covers over the common base $X$, after basepoints are fixed there is a homeomorphism $\phi :N\to \tilde{X}$ over $X$ [F10], and $\phi$ is biholomorphic: both projections $N\to X$ and $p:\tilde{X}\to X$ are holomorphic coverings, the latter by [F5] and the former because step 4.1 applied to $N$ and $H$ makes $N\to N\mathrm{/}H$ a holomorphic covering while $N\mathrm{/}H\to X$ is a biholomorphism, hence both projections are local biholomorphisms [F9]; so on an evenly covered open set $W\subseteq X$ the map $\phi$ is a composite of local inverses of local biholomorphisms, hence holomorphic, and the same argument applied to ${\phi }^{-1}$ gives the reverse. Hence $N$ is biholomorphic to $\tilde{X}$, and $\tilde{X}$ is biholomorphic to $M$ by step 1.1; since no two of the models are biholomorphic [F11], $N=M$. Therefore the model occurring in the statement is unique: $X$ is a quotient of exactly one of the three models.

$\bar{p}$ is an open map and hence a homeomorphism. The covering map $\tilde{p}$ of step 4.2 is open: if $O\subseteq M$ is open and $x\in \tilde{p}(O)$, choose an evenly covered neighbourhood $U$ of $x$ and a sheet $V$ of ${\tilde{p}}^{-1}(U)$ meeting $O$; then $V\cap O$ is open and $\tilde{p}(V\cap O)$ is open in $X$ because $\tilde{p}{\mathrm{\mid }}_V$ is a homeomorphism, and it contains $x$, so $\tilde{p}(O)={\bigcup }_{x\in \tilde{p}(O)}\tilde{p}(V_x\cap O)$ is open [F2, F7, F8]. Now let $O\subseteq M\mathrm{/}G$ be open; then $q^{-1}(O)$ is open in $M$ by the quotient topology, and $\bar{p}(O)=\bar{p}(q(q^{-1}(O)))=\tilde{p}(q^{-1}(O))$ is open in $X$ because $q$ is surjective. Hence the continuous bijection $\bar{p}$ is an open map, so it is a homeomorphism $M\mathrm{/}G\to X$ [F7].

Complex structure on the quotient and existence. Apply steps 3.1 and 4.1 to $Y=M$ and $K=G$. The quotient $M\mathrm{/}G$ is a Riemann surface with its quotient atlas, and $q:M\to M\mathrm{/}G$ is a holomorphic local biholomorphism. The map $\tilde{p}=p\circ {\phi }^{-1}$ is also a holomorphic local biholomorphism by [F5], [F9]. Since $\bar{p}\circ q=\tilde{p}$, on a sufficiently small sheet $V$ the homeomorphism $\bar{p}$ of step 6.1 is $\tilde{p}{\mathrm{\mid }}_V\circ (q{\mathrm{\mid }}_V{)}^{-1}$, a local biholomorphism. Thus $\bar{p}$ and its inverse are holomorphic, so $X$ is biholomorphic to the quotient $M\mathrm{/}G$. This proves existence for the model of step 1.1.

Conclusion and choice accounting. Steps 7.1 and 5.4 prove the statement: $X$ is biholomorphic to $M\mathrm{/}G$ for the model $M$ of its universal-covering type and the group $G$ of holomorphic automorphisms of $M$ acting freely and properly discontinuously, and the model is exactly one of the three. The Axiom of Choice [A1] is used exactly through the Countable Choice consumed by the lifted holomorphic structure [F5] and through the uniformization theorem inside the definition of the type [F5]; the remaining selections are finite (finitely many evenly covered sets and sheets in step 2.2, finitely many group elements in step 1.2).