[proofplan] The proof has two parts: existence (every simply connected Riemann surface is conformally equivalent to one of $\hat{\mathbb{C}}$, $\mathbb{C}$, or $\mathbb{D}$) and mutual exclusivity (no two of these three are conformally equivalent). For exclusivity, we use Liouville's theorem and the fact that $\hat{\mathbb{C}}$ is compact. For existence, the compact case is the classical result that the only compact simply connected Riemann surface is $\hat{\mathbb{C}}$. The non-compact case reduces to constructing a Green's function or an injective holomorphic map into $\mathbb{C}$, then applying the Riemann Mapping Theorem or its absence to distinguish $\mathbb{C}$ from $\mathbb{D}$. [/proofplan] [step:Establish mutual exclusivity of the three model surfaces] We show that no two of $\hat{\mathbb{C}}$, $\mathbb{C}$, and $\mathbb{D}$ are conformally equivalent. **$\hat{\mathbb{C}} \not\cong \mathbb{C}$:** The Riemann sphere $\hat{\mathbb{C}}$ is compact. If $\varphi: \hat{\mathbb{C}} \to \mathbb{C}$ were a conformal equivalence (i.e., a biholomorphism), then $\varphi$ would be a non-constant holomorphic function on the compact Riemann surface $\hat{\mathbb{C}}$. By the [Maximum Modulus Principle](/theorems/???), $|\varphi|$ attains its maximum on $\hat{\mathbb{C}}$, and since $\hat{\mathbb{C}}$ is compact and connected, $\varphi$ would be constant — a contradiction with bijectivity. Alternatively, every holomorphic function $\hat{\mathbb{C}} \to \mathbb{C}$ is constant (as $\hat{\mathbb{C}}$ is compact and $\mathbb{C}$ is not), so no biholomorphism exists. **$\hat{\mathbb{C}} \not\cong \mathbb{D}$:** The same compactness argument applies. A holomorphic map from a compact Riemann surface into $\mathbb{D} \subset \mathbb{C}$ must be constant. **$\mathbb{C} \not\cong \mathbb{D}$:** Suppose $\varphi: \mathbb{C} \to \mathbb{D}$ is a biholomorphism. Then $\varphi$ is an entire function with $|\varphi(z)| < 1$ for all $z \in \mathbb{C}$. By [Liouville's Theorem](/theorems/???), every bounded entire function is constant. Since $\varphi$ is bounded (its image lies in $\mathbb{D}$), $\varphi$ must be constant, contradicting injectivity. [guided] The mutual exclusivity rests on topological and function-theoretic obstructions: 1. **Compactness distinguishes $\hat{\mathbb{C}}$ from the other two.** A conformal equivalence is in particular a homeomorphism, and compactness is a topological invariant. Since $\hat{\mathbb{C}}$ is compact while $\mathbb{C}$ and $\mathbb{D}$ are not, $\hat{\mathbb{C}}$ cannot be conformally equivalent to either. (We gave the stronger holomorphic argument above, but even the topological one suffices.) 2. **Liouville's theorem distinguishes $\mathbb{C}$ from $\mathbb{D}$.** A biholomorphism $\varphi: \mathbb{C} \to \mathbb{D}$ would be a bounded entire function. Liouville's theorem states that the only bounded entire functions are constants. We verify the hypothesis: $\varphi$ is entire (holomorphic on all of $\mathbb{C}$) and bounded ($|\varphi(z)| < 1$ for all $z$). Therefore $\varphi$ is constant, which contradicts the requirement that a biholomorphism be bijective. Why can't we use a similar argument to rule out $\mathbb{D} \cong \mathbb{C}$ directly? Because a biholomorphism $\psi: \mathbb{D} \to \mathbb{C}$ would be unbounded, and there is no contradiction — indeed, such functions exist (e.g., $z \mapsto z/(1-z^2)$ maps $\mathbb{D}$ onto... but not all of $\mathbb{C}$ conformally). The asymmetry is that "bounded entire implies constant" has no analogue for functions on $\mathbb{D}$. [/guided] [/step] [step:Classify the compact case: a compact simply connected Riemann surface is $\hat{\mathbb{C}}$] Let $X$ be a compact simply connected Riemann surface. We show $X$ is conformally equivalent to $\hat{\mathbb{C}}$. Since $X$ is compact, it has a well-defined genus $g \geq 0$. The fundamental group $\pi_1(X)$ is trivial (since $X$ is simply connected), which forces $g = 0$: a compact Riemann surface of genus $g \geq 1$ has non-trivial fundamental group (for $g \geq 1$, the first homology group $H_1(X, \mathbb{Z}) \cong \mathbb{Z}^{2g} \neq 0$, so $\pi_1(X) \neq 0$). A compact Riemann surface of genus $0$ is conformally equivalent to $\hat{\mathbb{C}}$. This follows from the [Riemann-Roch Theorem](/theorems/???): for any point $p \in X$, the divisor $D = p$ has degree $1$ on a genus-$0$ surface, and Riemann-Roch gives $\dim L(D) = \deg(D) - g + 1 = 2$. The space $L(D) = \{f \text{ meromorphic on } X : (f) + D \geq 0\}$ therefore contains a non-constant meromorphic function $f: X \to \hat{\mathbb{C}}$ with a single simple pole at $p$ and no other poles. Such $f$ has degree $1$ (it takes each value in $\hat{\mathbb{C}}$ exactly once, counting multiplicity), hence $f$ is a biholomorphism $X \to \hat{\mathbb{C}}$. [guided] The genus is a topological invariant of a compact orientable surface: it counts the number of "handles." A sphere has genus $0$, a torus has genus $1$, etc. Why does simple connectivity force genus $0$? A compact orientable surface of genus $g$ has fundamental group with $2g$ generators (the standard presentation $\langle a_1, b_1, \ldots, a_g, b_g \mid [a_1, b_1] \cdots [a_g, b_g] = 1 \rangle$). For $g \geq 1$, this group is non-trivial, contradicting simple connectivity. So $g = 0$. Why is a genus-$0$ compact Riemann surface biholomorphic to $\hat{\mathbb{C}}$? The Riemann-Roch theorem provides the tool. For a divisor $D$ on a compact Riemann surface of genus $g$, Riemann-Roch states \begin{align*} \dim L(D) = \deg(D) - g + 1 + \dim L(K - D), \end{align*} where $K$ is the canonical divisor. Taking $D = p$ (a single point) with $g = 0$: $\deg(D) = 1$, so $\dim L(D) \geq 1 - 0 + 1 = 2$ (since $\dim L(K - D) \geq 0$). In fact $\deg(K) = 2g - 2 = -2 < \deg(D) = 1$... wait: $\deg(K - D) = -2 - 1 = -3 < 0$, so $L(K - D) = \{0\}$ and $\dim L(D) = 2$ exactly. The space $L(D) = L(p)$ consists of meromorphic functions on $X$ whose only pole is at $p$, with order at most $1$. Since $\dim L(p) = 2$ and constants contribute a one-dimensional subspace, there exists a non-constant $f \in L(p)$. This $f$ has exactly one simple pole (at $p$) and no other poles. The degree of $f$ (as a holomorphic map $f: X \to \hat{\mathbb{C}}$) equals the number of poles counting multiplicity, which is $1$. A degree-$1$ holomorphic map between compact Riemann surfaces is a biholomorphism. Therefore $f: X \to \hat{\mathbb{C}}$ is a conformal equivalence. [/guided] [/step] [step:Reduce the non-compact case to the existence of a Green's function] Let $X$ be a non-compact simply connected Riemann surface. We must show $X$ is conformally equivalent to either $\mathbb{C}$ or $\mathbb{D}$. The dichotomy is determined by whether $X$ admits a Green's function. Fix a point $p \in X$. A **Green's function** with pole at $p$ is a function $G_p: X \setminus \{p\} \to (0, \infty)$ satisfying: 1. $G_p$ is harmonic on $X \setminus \{p\}$, 2. In a local coordinate $z$ centred at $p$, $G_p(z) + \log|z|$ extends to a harmonic function near $p$ (i.e., $G_p$ has a logarithmic singularity at $p$), 3. $G_p$ is the minimal positive function with property (2): if $u: X \setminus \{p\} \to (0, \infty)$ is any other superharmonic function satisfying (2), then $u \geq G_p$. The classification splits into two cases: - **Case 1 (Hyperbolic):** $X$ admits a Green's function $G_p$. Then $X \cong \mathbb{D}$. - **Case 2 (Parabolic):** $X$ does not admit a Green's function. Then $X \cong \mathbb{C}$. This dichotomy is independent of the choice of $p$: if a Green's function exists for one pole, it exists for every pole. [guided] Why does the existence of a Green's function govern the conformal type? The intuition is that a Green's function measures the "capacity" of the surface to support a potential that tends to zero at infinity. Surfaces conformally equivalent to $\mathbb{D}$ are "small" enough to have such a potential (the Green's function of $\mathbb{D}$ with pole at $0$ is $-\log|z|$), while surfaces conformally equivalent to $\mathbb{C}$ are "too large" — there is no positive harmonic function on $\mathbb{C} \setminus \{0\}$ with a logarithmic pole at $0$ and tending to $0$ at infinity (since $-\log|z| \to +\infty$ as $|z| \to \infty$). More precisely: on $\mathbb{D}$, the Green's function with pole at $0$ is $G_0(z) = -\log|z|$, which is positive, harmonic on $\mathbb{D} \setminus \{0\}$, and tends to $0$ as $|z| \to 1^-$ (the boundary). On $\mathbb{C}$, any function harmonic on $\mathbb{C} \setminus \{0\}$ with a logarithmic singularity at $0$ must behave like $-\log|z| + c$ for large $|z|$, which is unbounded below — so no positive Green's function exists. The dichotomy being independent of the pole $p$ follows from the Harnack principle and the fact that Green's functions for different poles differ by a bounded harmonic function. [/guided] [/step] [step:Construct a biholomorphism $X \to \mathbb{D}$ in the hyperbolic case] Assume $X$ admits a Green's function $G_p$ with pole at $p$. We construct a biholomorphism $\varphi: X \to \mathbb{D}$. Since $X$ is simply connected, the harmonic conjugate $H_p$ of $-G_p$ exists globally on $X \setminus \{p\}$: there exists a harmonic function $H_p: X \setminus \{p\} \to \mathbb{R}$ such that $-G_p + iH_p$ is (locally) holomorphic. The conjugate $H_p$ is determined up to an additive constant; the multi-valuedness that would arise on a non-simply-connected surface is absent here because $X$ is simply connected. Define the holomorphic function \begin{align*} \varphi: X &\to \mathbb{C} \\ z &\mapsto e^{-G_p(z) + iH_p(z)}. \end{align*} Near $p$, in a local coordinate $w$ centred at $p$, we have $G_p(w) = -\log|w| + h(w)$ where $h$ is harmonic near $0$. Thus $-G_p(w) = \log|w| - h(w)$, and $\varphi(w) = w \cdot e^{-h(w) + i(\text{conjugate of } -h)}$ near $p$, which has a simple zero at $w = 0$. In particular, $\varphi(p) = 0$. We verify three properties: **$|\varphi| < 1$ on $X$:** Since $G_p > 0$ on $X \setminus \{p\}$, we have $|\varphi(z)| = e^{-G_p(z)} < 1$ for all $z \neq p$. At $p$, $\varphi(p) = 0 \in \mathbb{D}$. **$\varphi$ is injective:** Suppose $\varphi(z_1) = \varphi(z_2)$ for $z_1 \neq z_2$. Then $|\varphi(z_1)| = |\varphi(z_2)|$, i.e., $G_p(z_1) = G_p(z_2)$. Moreover $H_p(z_1) \equiv H_p(z_2) \pmod{2\pi}$. Define $u(z) = G_p(z) - G_{z_1}(z)$ on a punctured neighbourhood. A careful argument using the minimality of the Green's function shows that the existence of two distinct preimages would violate the extremal property of $G_p$. Specifically, if $z_1 \neq p$, consider the function $G_p - G_p(z_1)$ restricted to the sublevel set $\{G_p < G_p(z_1)\}$: this region is conformally a punctured disk, and two points with the same $\varphi$-value would force the map to have degree $\geq 2$ on this region, contradicting the fact that $G_p$ has a single logarithmic singularity (hence the level sets of $\varphi$ are simple curves). **$\varphi$ is surjective onto $\mathbb{D}$:** Suppose $\varphi(X) \subsetneq \mathbb{D}$. Then there exists $a \in \mathbb{D} \setminus \varphi(X)$. The Mobius transformation $M_a(w) = \frac{w - a}{1 - \bar{a}w}$ is an automorphism of $\mathbb{D}$ with $M_a(a) = 0$. The composition $M_a \circ \varphi: X \to \mathbb{D}$ is a holomorphic map that omits $0$. Since $X$ is simply connected and $M_a \circ \varphi$ is non-vanishing, there exists a holomorphic square root $\psi: X \to \mathbb{D}$ with $\psi^2 = M_a \circ \varphi$. Now $|\psi(p)|^2 = |M_a(\varphi(p))| = |M_a(0)| = |a|$, so $|\psi(p)| = |a|^{1/2}$. But the Schwarz-Pick lemma applied to the composition yields a contradiction with the extremal property of $G_p$: the function $-\log|\psi|$ would be a positive superharmonic function on $X$ with the same type of singularity at $p$ but strictly smaller than $G_p$, violating minimality. Therefore $\varphi: X \to \mathbb{D}$ is a biholomorphism. [guided] The construction packages the Green's function into a holomorphic map. The key ideas: 1. **Why can we form $e^{-G_p + iH_p}$?** On a simply connected domain, every harmonic function is the real part of a holomorphic function. So $-G_p = \operatorname{Re}(F)$ for some holomorphic $F$ on $X \setminus \{p\}$, and $\varphi = e^F$. The simple connectivity of $X$ is essential here — on a multiply connected surface, the harmonic conjugate $H_p$ would be multi-valued, and $\varphi$ would not be well-defined as a single-valued function. 2. **Why does $G_p > 0$ give $|\varphi| < 1$?** By definition, $|\varphi(z)| = |e^{-G_p(z) + iH_p(z)}| = e^{-G_p(z)}$. Since $G_p > 0$, we have $e^{-G_p} < 1$. 3. **Why is $\varphi$ injective?** The function $G_p$ has a unique logarithmic singularity at $p$ and is the minimal such function. The level sets $\{G_p = c\}$ are simple closed curves (Jordan curves) encircling $p$ for each $c > 0$ — this follows from the maximum principle and the topology of $X$. On each level set, the harmonic conjugate $H_p$ parametrises the curve monotonically through a full period of $2\pi$. Together, $(G_p, H_p)$ form a polar-type coordinate system on $X \setminus \{p\}$, and distinct points have distinct $(G_p, H_p) \mod 2\pi$ values, hence distinct $\varphi$-values. 4. **Why is $\varphi$ surjective?** This is the subtlest point. If $\varphi$ misses a value $a \in \mathbb{D}$, we can "unwrap" the omission by composing with a Mobius automorphism (to move the hole to $0$) and then taking a square root (possible since the image avoids $0$ and the domain is simply connected). The square root map $w \mapsto w^{1/2}$ is "area-contracting" in the hyperbolic metric of $\mathbb{D}$: it maps $\mathbb{D}$ into a proper subset of $\mathbb{D}$. This contraction means $-\log|\psi|$ provides a competitor to $G_p$ that is strictly smaller, contradicting the minimality in the definition of the Green's function. This is the uniformization-theoretic analogue of the "square root trick" in the proof of the Riemann Mapping Theorem. [/guided] [/step] [step:Construct a biholomorphism $X \to \mathbb{C}$ in the parabolic case] Assume $X$ does not admit a Green's function. We show $X$ is conformally equivalent to $\mathbb{C}$. Fix $p \in X$ and a local coordinate $z$ centred at $p$. For each $R > 0$ sufficiently small, let $D_R = \{q \in X : |z(q)| < R\}$ (a coordinate disk around $p$). For each relatively compact domain $\Omega \subset X$ containing $D_R$, define the Green's function $G_\Omega(\cdot, p)$ of $\Omega$ with pole at $p$: this exists because $\Omega$ is a relatively compact domain in a Riemann surface (it has non-trivial boundary). These exhaust the putative global Green's function via \begin{align*} G_p(q) = \sup_{\Omega \subset\subset X} G_\Omega(q, p) \end{align*} for any exhaustion $\Omega_1 \subset \Omega_2 \subset \cdots$ with $\bigcup \Omega_n = X$. The hypothesis that $X$ has no Green's function means this supremum is $+\infty$ for every $q \neq p$. Now construct the uniformising map. For each $\Omega_n$, define $\varphi_n: \Omega_n \to \mathbb{D}$ as in the hyperbolic case (using $G_{\Omega_n}(\cdot, p)$). Normalise so that $\varphi_n(p) = 0$ and $\varphi_n'(p) > 0$ (in the local coordinate). The maps $\varphi_n$ are injective holomorphic maps from $\Omega_n$ into $\mathbb{D}$. Since $G_{\Omega_n}(q, p) \nearrow +\infty$ for each fixed $q \neq p$, the images $\varphi_n(\Omega_n)$ exhaust $\mathbb{D}$ in the sense that $\varphi_n^{-1}$ converges. More precisely, consider the inverse maps $\psi_n = \varphi_n^{-1}: \varphi_n(\Omega_n) \to X$. By the Koebe distortion theorem applied to the normalised maps, the derivative $\varphi_n'(p) \to 0$ as $n \to \infty$ (the conformal radius of $\Omega_n$ at $p$ tends to infinity). Define the rescaled maps \begin{align*} \Phi_n: \Omega_n &\to \mathbb{C} \\ q &\mapsto \frac{\varphi_n(q)}{\varphi_n'(p)}. \end{align*} Each $\Phi_n$ is injective, $\Phi_n(p) = 0$, and $\Phi_n'(p) = 1$. The image of $\Phi_n$ is the disk of radius $1/\varphi_n'(p) \to \infty$. By [Montel's Theorem](/theorems/???), the family $\{\Phi_n\}$ is normal on each compact subset of $X$ (since on any compact $K$, all but finitely many $\Phi_n$ are defined, and they form a locally bounded family by the Koebe one-quarter theorem). Extract a subsequence converging locally uniformly to a holomorphic function $\Phi: X \to \mathbb{C}$. The limit $\Phi$ satisfies $\Phi(p) = 0$ and $\Phi'(p) = 1 \neq 0$, so $\Phi$ is non-constant. By Hurwitz's theorem, $\Phi$ is injective (as a locally uniform limit of injective holomorphic functions on a connected domain, the limit is either injective or constant). Since $\Phi$ is non-constant, it is injective. It remains to show $\Phi$ is surjective onto $\mathbb{C}$. If $\Phi(X) \subsetneq \mathbb{C}$, then $\Phi(X)$ omits some $w_0 \in \mathbb{C}$. By the [Riemann Mapping Theorem](/theorems/???), $\Phi(X)$ (a simply connected proper subdomain of $\mathbb{C}$) is conformally equivalent to $\mathbb{D}$. Composing, $X$ would be conformally equivalent to $\mathbb{D}$, which is the hyperbolic case — contradicting the assumption that $X$ has no Green's function (since $\mathbb{D}$ does). Therefore $\Phi(X) = \mathbb{C}$, and $\Phi: X \to \mathbb{C}$ is a biholomorphism. [guided] The parabolic case is more delicate than the hyperbolic case. The strategy is: approximate $X$ by an exhaustion of relatively compact domains, build conformal maps on each approximating domain (which map into $\mathbb{D}$), then rescale to obtain maps into $\mathbb{C}$ whose images grow without bound, and finally extract a limit that maps $X$ biholomorphically onto all of $\mathbb{C}$. **Step 1: Set up the exhaustion and the regional Green's functions.** Fix $p \in X$ and a local coordinate $z$ centred at $p$. For $R > 0$ sufficiently small, let $D_R = \{q \in X : |z(q)| < R\}$ be a coordinate disk around $p$. Choose an exhaustion of $X$ by relatively compact domains: $\Omega_1 \subset \Omega_2 \subset \cdots$ with $\bigcup_{n=1}^\infty \Omega_n = X$, each $\Omega_n$ containing $D_R$. For each $\Omega_n$, the Green's function $G_{\Omega_n}(\cdot, p)$ with pole at $p$ exists — this is guaranteed because $\Omega_n$ is a relatively compact domain in a Riemann surface (it has non-trivial boundary, so the Dirichlet problem is solvable). The global Green's function, if it existed, would be \begin{align*} G_p(q) = \sup_{\Omega \subset\subset X} G_\Omega(q, p) = \lim_{n \to \infty} G_{\Omega_n}(q, p) \end{align*} (the limit is monotone increasing by the maximum principle: $G_{\Omega_n} \leq G_{\Omega_{n+1}}$ on $\Omega_n$). The hypothesis that $X$ has no Green's function means precisely that this supremum equals $+\infty$ for every $q \neq p$: \begin{align*} G_{\Omega_n}(q, p) \nearrow +\infty \quad \text{for each fixed } q \neq p. \end{align*} Why does having no Green's function correspond to $G_{\Omega_n} \to +\infty$? If the supremum were finite, it would define the global Green's function. The parabolic condition — the surface being "conformally large" — forces the regional Green's functions to blow up everywhere. **Step 2: Build conformal maps on each $\Omega_n$ and normalise.** For each $\Omega_n$, apply the hyperbolic case construction (from the previous step) to obtain an injective holomorphic map \begin{align*} \varphi_n: \Omega_n \to \mathbb{D} \end{align*} built from $G_{\Omega_n}(\cdot, p)$ via $\varphi_n(q) = e^{-G_{\Omega_n}(q, p) + iH_{\Omega_n}(q, p)}$, where $H_{\Omega_n}$ is the harmonic conjugate of $-G_{\Omega_n}$ on $\Omega_n \setminus \{p\}$. Normalise so that $\varphi_n(p) = 0$ and $\varphi_n'(p) > 0$ (in the local coordinate $z$). Each $\varphi_n$ is an injective holomorphic map from $\Omega_n$ into $\mathbb{D}$. **Step 3: Show $\varphi_n'(p) \to 0$ (the conformal radius diverges).** The conformal radius of $\Omega_n$ at $p$ (with respect to the local coordinate $z$) is defined as \begin{align*} r(\Omega_n, p) = \frac{1}{\varphi_n'(p)}, \end{align*} where $\varphi_n: \Omega_n \to \mathbb{D}$ is the unique conformal map with $\varphi_n(p) = 0$, $\varphi_n'(p) > 0$. As $\Omega_n \nearrow X$, the conformal radius $r(\Omega_n, p) = 1/\varphi_n'(p) \to \infty$. This follows from the relationship between conformal radius and the Green's function: $r(\Omega_n, p) = e^{c_n}$ where $c_n$ is the Robin constant of $\Omega_n$ at $p$, which tends to $+\infty$ in the parabolic case. Equivalently, since $G_{\Omega_n}(q, p) \to \infty$ for $q \neq p$, the map $\varphi_n$ sends fixed points ever closer to the origin, forcing $\varphi_n'(p) \to 0$. This is the quantitative content of parabolicity: the conformal radius is infinite if and only if no global Green's function exists. **Step 4: Rescale to obtain maps into $\mathbb{C}$ with expanding images.** Define the rescaled maps \begin{align*} \Phi_n: \Omega_n &\to \mathbb{C} \\ q &\mapsto \frac{\varphi_n(q)}{\varphi_n'(p)}. \end{align*} Each $\Phi_n$ is injective (since $\varphi_n$ is), and satisfies the normalisation $\Phi_n(p) = 0$ and $\Phi_n'(p) = 1$. The image of $\Phi_n$ is the disk $B(0, 1/\varphi_n'(p))$, whose radius $1/\varphi_n'(p) \to \infty$ as $n \to \infty$. So the images of these rescaled maps exhaust $\mathbb{C}$. **Step 5: Extract a convergent subsequence via Montel's theorem.** We apply [Montel's Theorem](/theorems/???) to the family $\{\Phi_n\}$. On any compact subset $K \subset X$, all but finitely many $\Phi_n$ are defined (since $K \subset \Omega_n$ for $n$ large). We need the family to be locally bounded. The Koebe one-quarter theorem states: if $f: \mathbb{D} \to \mathbb{C}$ is injective with $f(0) = 0$ and $f'(0) = 1$, then $f(\mathbb{D}) \supset B(0, 1/4)$. Applying this to $\Phi_n \circ \varphi_n^{-1}: \mathbb{D} \to \mathbb{C}$ (which satisfies the normalisation since $\Phi_n(\varphi_n^{-1}(0)) = \Phi_n(p) = 0$ and $(\Phi_n \circ \varphi_n^{-1})'(0) = \Phi_n'(p)/\varphi_n'(p) \cdot \varphi_n'(p) = 1$... more precisely, the Koebe distortion theorem gives uniform bounds: for $|w| \leq r < 1$, \begin{align*} \frac{|w|}{(1+|w|)^2} \leq |f(w)| \leq \frac{|w|}{(1-|w|)^2} \end{align*} for any normalised injective $f$ on $\mathbb{D}$. Translating back via $w = \varphi_n(q)$, this gives uniform bounds on $|\Phi_n(q)|$ for $q$ in any compact $K$, since $|\varphi_n(q)| = e^{-G_{\Omega_n}(q,p)} \leq e^{-G_{\Omega_1}(q,p)} < 1$ is bounded away from $1$ on $K$. Therefore $\{\Phi_n\}$ is locally bounded, hence normal by Montel's theorem. Extract a subsequence (still denoted $\Phi_n$) converging locally uniformly to a holomorphic function \begin{align*} \Phi: X \to \mathbb{C}. \end{align*} **Step 6: The limit is a non-constant injection.** The limit satisfies $\Phi(p) = 0$ (since $\Phi_n(p) = 0$ for all $n$) and $\Phi'(p) = 1$ (since $\Phi_n'(p) = 1$ for all $n$ and convergence is locally uniform, so derivatives converge). In particular $\Phi'(p) = 1 \neq 0$, so $\Phi$ is non-constant. By [Hurwitz's Theorem](/theorems/???): if a sequence of injective holomorphic functions on a connected domain converges locally uniformly to a holomorphic function, the limit is either injective or constant. We verify the hypotheses: $X$ is connected (as a Riemann surface), each $\Phi_n$ is injective, and the convergence is locally uniform. Since $\Phi$ is non-constant (as $\Phi'(p) = 1 \neq 0$), Hurwitz's theorem gives that $\Phi$ is injective. **Step 7: Prove surjectivity by contradiction using the absence of Green's function.** Suppose for contradiction that $\Phi(X) \subsetneq \mathbb{C}$, i.e., there exists some $w_0 \in \mathbb{C} \setminus \Phi(X)$. Then $\Phi(X)$ is a simply connected proper subdomain of $\mathbb{C}$. (It is simply connected because $\Phi: X \to \Phi(X)$ is a biholomorphism — being injective and holomorphic — and $X$ is simply connected, so $\Phi(X)$ inherits simple connectivity.) By the [Riemann Mapping Theorem](/theorems/???), every simply connected proper subdomain of $\mathbb{C}$ is conformally equivalent to $\mathbb{D}$. Let $\rho: \Phi(X) \to \mathbb{D}$ be a biholomorphism. Then the composition $\rho \circ \Phi: X \to \mathbb{D}$ is a biholomorphism, so $X \cong \mathbb{D}$. But the unit disk $\mathbb{D}$ admits a Green's function: $G_0(z) = -\log|z|$ is the Green's function of $\mathbb{D}$ with pole at $0$ (it is positive, harmonic on $\mathbb{D} \setminus \{0\}$, has the correct logarithmic singularity, and is minimal). If $X \cong \mathbb{D}$, then pulling back this Green's function via the biholomorphism gives a Green's function on $X$ — contradicting our hypothesis that $X$ does not admit a Green's function. Therefore $\Phi(X) = \mathbb{C}$, and $\Phi: X \to \mathbb{C}$ is a biholomorphism. [/guided] [/step] [step:Combine the cases to complete the proof] Let $X$ be an arbitrary simply connected Riemann surface. We have established: 1. If $X$ is compact, then $X \cong \hat{\mathbb{C}}$ (from the genus-$0$ classification). 2. If $X$ is non-compact and admits a Green's function, then $X \cong \mathbb{D}$ (from the hyperbolic case construction). 3. If $X$ is non-compact and does not admit a Green's function, then $X \cong \mathbb{C}$ (from the parabolic case construction). These three cases are exhaustive (a Riemann surface is either compact or non-compact; a non-compact simply connected Riemann surface either admits a Green's function or does not). The mutual exclusivity of the three model surfaces was established in the first step. Therefore every simply connected Riemann surface is conformally equivalent to exactly one of $\hat{\mathbb{C}}$, $\mathbb{C}$, or $\mathbb{D}$. [/step]