A complex torus has a lattice of parabolic deck translations

The two periods are a real basis. The map $T_0:{\mathbb{R}}^2\to \mathbb{C}$, $T_0(s,t):=s{\omega }_1+t{\omega }_2$, is real-linear and injective: $s{\omega }_1+t{\omega }_2=0$ forces $s=t=0$ by $\mathbb{R}$-linear independence. Hence the composite $\mathrm{\Phi }\circ T_0:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is an injective real-linear map of a $2$-dimensional space into itself, so it is bijective and $T_0$ is a bijection [F2, F4]. Applying the boundedness bound of [F4] to the inverse linear map $(\mathrm{\Phi }\circ T_0{)}^{-1}$ gives $K\ge 0$ with $\parallel (\mathrm{\Phi }\circ T_0{)}^{-1}v{\parallel }_2\le K\parallel v{\parallel }_2$ for all $v$; here $K>0$, since $K=0$ would make the inverse map zero, impossible for a bijection. Put $c:=1\mathrm{/}K>0$. For $\lambda =m{\omega }_1+n{\omega }_2\in \mathrm{\Lambda }$ one has $(m,n)=(\mathrm{\Phi }\circ T_0{)}^{-1}\mathrm{\Phi }(\lambda )$, hence $\max (\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n\mathrm{\mid })\le \parallel (m,n){\parallel }_2\le K\mathrm{\mid }\lambda \mathrm{\mid }$, that is $\mathrm{\mid }\lambda \mathrm{\mid }\ge c\max (\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n\mathrm{\mid })$; in particular $\mathrm{\mid }\lambda \mathrm{\mid }\ge c$ for every nonzero $\lambda \in \mathrm{\Lambda }$.

The plane is simply connected. Under the dictionary of [F2] the space $\mathbb{C}$ is the nonempty convex set ${\mathbb{R}}^2$, which is simply connected [F8]; in particular $\mathbb{C}$ is nonempty and path connected with trivial fundamental group.

Continuity and translations. The map $T_0$ is continuous: $\mathrm{\Phi }\circ T_0$ is real-linear, hence bounded and Lipschitz, hence continuous by [F4], and ${\mathrm{\Phi }}^{-1}$ is an isometry by [F2]. For each $\lambda \in \mathbb{C}$ the translation ${\tau }_{\lambda }(z):=z+\lambda$ satisfies $\mathrm{\mid }{\tau }_{\lambda }(z)-{\tau }_{\lambda }(w)\mathrm{\mid }=\mathrm{\mid }(z+\lambda )-(w+\lambda )\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$ for all $z,w$ by [F3], so ${\tau }_{\lambda }$ is an isometry of $\mathbb{C}$; it is therefore a bijection with continuous inverse ${\tau }_{-\lambda }$, that is, a homeomorphism of $\mathbb{C}$ [F4, F5].

The translation action is a covering-space action. The set $\mathrm{\Lambda }$ is a subgroup of $(\mathbb{C},+)$ [F3], and $\lambda \cdot z:=\lambda +z$ defines a left action of $\mathrm{\Lambda }$ on $\mathbb{C}$ by homeomorphisms: $(\lambda +\mu )\cdot z=z+(\lambda +\mu )=\lambda +(z+\mu )=\lambda \cdot (\mu \cdot z)$ and $0\cdot z=z$ by the field laws [F3], while each map $z\mapsto \lambda \cdot z$ is the homeomorphism ${\tau }_{\lambda }$ of step 2.1 [F5]. It is a covering-space action: fix $z\in \mathbb{C}$ and put $U:=D(z,c\mathrm{/}4)$; if $w\in (U+\lambda )\cap U$ for some nonzero $\lambda \in \mathrm{\Lambda }$, then $w=u+\lambda =u^{\mathrm{\prime }}$ for some $u,u^{\mathrm{\prime }}\in U$, so $\lambda =u^{\mathrm{\prime }}-u$ and $\mathrm{\mid }\lambda \mathrm{\mid }\le \mathrm{\mid }{u}^{\mathrm{\prime }}-z\mathrm{\mid }+\mathrm{\mid }z-u\mathrm{\mid }<c\mathrm{/}2<c$, contradicting $\mathrm{\mid }\lambda \mathrm{\mid }\ge c$ from step 1.1; hence $(U+\lambda )\cap U=\mathrm{\varnothing }$ for every nonzero $\lambda$, as required by [F5].

The quotient map is open. For open $W\subseteq \mathbb{C}$ one has $q^{-1}(q(W))={\bigcup }_{\lambda \in \mathrm{\Lambda }}(W+\lambda )$, because the classes of $q$ are the orbits $\{\lambda +z:\lambda \in \mathrm{\Lambda }\}$; each $W+\lambda ={\tau }_{\lambda }(W)$ is open by step 2.1, so $q^{-1}(q(W))$ is open in $\mathbb{C}$ and $q(W)$ is open in $T$ by the quotient topology [F9]. Thus $q$ is an open map.

The orbit map is a covering with deck group the translations. The space $T=\mathbb{C}\mathrm{/}\mathrm{\Lambda }$ is the orbit space of the action of step 3.1 and $q$ is its orbit map [F9]. By steps 3.1 and 1.2 the orbit-map theorem [F6] applies: $q:\mathbb{C}\to T$ is a covering map, and since $\mathbb{C}$ is path connected its deck group consists exactly of the transformations supplied by $\mathrm{\Lambda }$, that is, $\mathrm{Deck}(q)=\{{\tau }_{\lambda }:\lambda \in \mathrm{\Lambda }\}$. Since $\mathbb{C}$ is simply connected (step 1.2), $q$ is a universal covering space [F7].

Small discs give charts. Fix $z\in \mathbb{C}$ and put $D_z:=D(z,c\mathrm{/}3)$ and $U_z:=q(D_z)$; the set $U_z$ is open in $T$ by step 3.2. If $q(w)=q(w^{\mathrm{\prime }})$ with $w,w^{\mathrm{\prime }}\in D_z$, then $w^{\mathrm{\prime }}=\lambda +w$ for some $\lambda \in \mathrm{\Lambda }$, because the classes of the orbit map are the orbits [F5, F9]; then $w-w^{\mathrm{\prime }}\in \mathrm{\Lambda }$ and $\mathrm{\mid }w-w^{\mathrm{\prime }}\mathrm{\mid }<2c\mathrm{/}3<c$, so $w=w^{\mathrm{\prime }}$ by step 1.1. Hence $q_z:=q{\mathrm{\mid }}_{D_z}$ is a bijection $D_z\to U_z$, and it is an open continuous map: for open $A\subseteq D_z$ the set $q(A)$ is open in $T$ by step 3.2, hence open in $U_z$. Therefore its inverse ${\phi }_z:=q_z^{-1}:U_z\to D_z$ is a homeomorphism onto the open set $D_z\subseteq \mathbb{C}$, that is, a chart [F5, F11]. The sets $U_z$ cover $T$, because $q$ is onto and $z\in D_z$ for every $z$.

The quotient is Hausdorff. Let $[z]\mathrm{\ne }[z^{\mathrm{\prime }}]$ in $T$ and put $P:=z-z^{\mathrm{\prime }}\mathrm{\notin }\mathrm{\Lambda }$. With $R:=2\mathrm{\mid }P\mathrm{\mid }+1$, the set $S:=\mathrm{\Lambda }\cap \bar{D}(0,R)$ is finite: by step 1.1 every $\lambda =m{\omega }_1+n{\omega }_2\in S$ has $\max (\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n\mathrm{\mid })\le R\mathrm{/}c$, and only finitely many integer pairs satisfy this. Since $0\in S$, the number $\delta :=\mathrm{dist}(P,S)=\min \{\mathrm{\mid }P-\lambda \mathrm{\mid }:\lambda \in S\}$ is positive and $\delta \le \mathrm{\mid }P\mathrm{\mid }$; and for $\lambda \in \mathrm{\Lambda }\setminus S$ one has $\mathrm{\mid }P-\lambda \mathrm{\mid }\ge \mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }P\mathrm{\mid }>R-\mathrm{\mid }P\mathrm{\mid }=\mathrm{\mid }P\mathrm{\mid }+1>\delta$ by [F3]. Hence $\mathrm{dist}(P,\mathrm{\Lambda })=\delta >0$. The open sets $q(D(z,\delta \mathrm{/}2))$ and $q(D(z^{\mathrm{\prime }},\delta \mathrm{/}2))$ are then disjoint: a common class would give $u\in D(z,\delta \mathrm{/}2)$ and $u^{\mathrm{\prime }}\in D(z^{\mathrm{\prime }},\delta \mathrm{/}2)$ with $u-u^{\mathrm{\prime }}\in \mathrm{\Lambda }$, whence $\mathrm{\mid }P-(u-u^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\mid }z-u\mathrm{\mid }+\mathrm{\mid }{u}^{\mathrm{\prime }}-z^{\mathrm{\prime }}\mathrm{\mid }<\delta$, contradicting $\mathrm{dist}(P,\mathrm{\Lambda })=\delta$; both sets are open by step 3.2. Therefore $T$ is Hausdorff.

The deck group is ${\mathbb{Z}}^2$. By step 4.1 the map $\lambda \mapsto {\tau }_{\lambda }$ is a bijection $\mathrm{\Lambda }\to \mathrm{Deck}(q)$ and ${\tau }_{\lambda }\circ {\tau }_{\mu }={\tau }_{\lambda +\mu }$, ${\tau }_0={\mathrm{id}}_{\mathbb{C}}$, so it is a group isomorphism [F15]. The map ${\mathbb{Z}}^2\to \mathrm{\Lambda }$, $(m,n)\mapsto m{\omega }_1+n{\omega }_2$, is surjective by the definition of $\mathrm{\Lambda }$ and injective because $m{\omega }_1+n{\omega }_2=0$ forces $m=n=0$, and it is additive; hence it too is an isomorphism [F15]. Therefore $\mathrm{Deck}(q)\cong \mathrm{\Lambda }\cong {\mathbb{Z}}^2$.

Transitions are translations. Let $z,z^{\mathrm{\prime }}\in \mathbb{C}$ with $W:=U_z\cap U_{z^{\mathrm{\prime }}}\mathrm{\ne }\mathrm{\varnothing }$. For $u\in {\phi }_z(W)\subseteq D_z$ the point ${\phi }_{z^{\mathrm{\prime }}}(q(u))$ lies in $D_{z^{\mathrm{\prime }}}$ and satisfies $q({\phi }_{z^{\mathrm{\prime }}}(q(u)))=q(u)$, since $q(u)\in W\subseteq U_{z^{\mathrm{\prime }}}$ and ${\phi }_{z^{\mathrm{\prime }}}$ inverts $q$ on $D_{z^{\mathrm{\prime }}}$ (step 4.2); hence $\lambda (u):={\phi }_{z^{\mathrm{\prime }}}(q(u))-u\in \mathrm{\Lambda }$ by the orbit description of the classes. The map $u\mapsto \lambda (u)$ is continuous on the open set ${\phi }_z(W)$ (compositions of continuous maps and subtraction, steps 2.1 and 4.2) and its values are separated: $\mathrm{\mid }\lambda -\mu \mathrm{\mid }\ge c$ for distinct $\lambda ,\mu \in \mathrm{\Lambda }$ (step 1.1). Given $u$ in the domain, continuity gives $\delta >0$ with $\mathrm{\mid }\lambda (u^{\mathrm{\prime }})-\lambda (u)\mathrm{\mid }<c$ whenever $\mathrm{\mid }{u}^{\mathrm{\prime }}-u\mathrm{\mid }<\delta$; two distinct values of $\lambda$ would differ by at least $c$, so $\lambda$ is constant on ${\phi }_z(W)\cap D(u,\delta )$. Thus the transition ${\phi }_{z^{\mathrm{\prime }}}\circ {\phi }_z^{-1}$, which on ${\phi }_z(W)$ is the map $u\mapsto u+\lambda (u)$, agrees near each of its points with a single translation $u\mapsto u+{\lambda }_0$, an entire function [F11]; the same argument with $z$ and $z^{\mathrm{\prime }}$ interchanged shows that the inverse transition ${\phi }_z\circ {\phi }_{z^{\mathrm{\prime }}}^{-1}$ is holomorphic too. Hence the charts ${\phi }_z$ are pairwise compatible.

The square schema realizes the quotient. Let $g:[0,1{]}^2\to T$ be $g(s,t):=q(s{\omega }_1+t{\omega }_2)$, continuous as the composite of the continuous map $T_0$ (step 2.1) with $q$ [F9]. It respects the side pairings: $g(1,t)=q({\omega }_1+t{\omega }_2)=q(t{\omega }_2)=g(0,t)$ and $g(s,1)=q(s{\omega }_1+{\omega }_2)=q(s{\omega }_1)=g(s,0)$, since ${\omega }_1,{\omega }_2\in \mathrm{\Lambda }$ and the classes of $q$ are the orbits [F5]. By the characteristic property of the quotient $Y$ of the square by these pairings [F9, F12], $g$ induces a continuous map $\bar{g}:Y\to T$ with $\bar{g}\circ \pi =g$, where $\pi$ is the quotient map of the schema. The map $\bar{g}$ is surjective: given $[z]\in T$ write $z=T_0(s,t)$ (step 1.1), decompose $s=m+s^{\mathrm{\prime }}$, $t=n+t^{\mathrm{\prime }}$ with $m,n\in \mathbb{Z}$ and $s^{\mathrm{\prime }},t^{\mathrm{\prime }}\in [0,1)$, and use $T_0(s,t)-T_0(s^{\mathrm{\prime }},t^{\mathrm{\prime }})=T_0(m,n)\in \mathrm{\Lambda }$ to get $[z]=g(s^{\mathrm{\prime }},t^{\mathrm{\prime }})$. It is injective: if $g(s,t)=g(s^{\mathrm{\prime }},t^{\mathrm{\prime }})$ then $T_0(s-s^{\mathrm{\prime }},t-t^{\mathrm{\prime }})\in \mathrm{\Lambda }=T_0({\mathbb{Z}}^2)$, so $(s-s^{\mathrm{\prime }},t-t^{\mathrm{\prime }})\in {\mathbb{Z}}^2$ by the injectivity of $T_0$ (step 1.1); since all four coordinates lie in $[0,1]$, the differences $s-s^{\mathrm{\prime }}$ and $t-t^{\mathrm{\prime }}$ lie in $\{-1,0,1\}$ and each is nonzero exactly when the two points lie on a paired pair of sides, so $(s,t)$ and $(s^{\mathrm{\prime }},t^{\mathrm{\prime }})$ have the same image under $\pi$. Hence $\bar{g}$ is a continuous bijection; since $Y$ is compact [F12] and $T$ is Hausdorff (step 4.3), $\bar{g}$ is a homeomorphism [F10].

The quotient is a compact Riemann surface. The charts $\{{\phi }_z{\}}_{z\in \mathbb{C}}$ cover $T$ and have holomorphic transitions in both directions (steps 4.2 and 5.2), so they form a holomorphic atlas; the space $T$ is nonempty, connected as the image of the connected space $\mathbb{C}$ under the continuous surjection $q$ [F8, step 1.2], Hausdorff by step 4.3, and second countable because it is homeomorphic to $Y$ (step 5.3) and $Y$ is second countable [F12]. Therefore $T$ is a Riemann surface by [F11], and it is compact because $Y$ is compact [F12] and homeomorphic to $T$ (step 5.3). The map $q$ is holomorphic for this atlas: on $D_z$ the chart expression ${\phi }_z\circ q$ is the identity, because ${\phi }_z$ inverts $q{\mathrm{\mid }}_{D_z}$ (step 4.2).

The genus is one. By [F12] the same square schema realizes the torus $T^2$, so $Y\cong T^2$, and with step 5.3 this gives $T\cong T^2={\mathrm{\#}}_1{T}^2$ [F13]. The topological classification of compact Riemann surfaces [F13] supplies exactly one $g\ge 0$ with $T\cong {\mathrm{\#}}_g{T}^2$; since $g=1$ has this property, the genus of the compact Riemann surface $T$ is $1$.

The type is parabolic. By steps 6.1 and 7.1 the space $T$ is a compact Riemann surface of genus $1$, so the compact-genus corollary [F14] gives $T$ parabolic universal-covering type under the Axiom of Choice [F1]. Moreover the exhibited covering $q:\mathbb{C}\to T$ is holomorphic (step 6.1) with simply connected total space $\mathbb{C}$ (step 1.2), hence is a holomorphic universal cover of $T$ whose model is $\mathbb{C}$, in agreement with the definition of parabolic type [F14]. This proves all three assertions of the Example.