title: Uniformization of Riemann surfaces revisited
authors: Cipriana Anghel Rares Stan
text: 
  
  
  Institute of Mathematics of the Romanian Academy
  
  Calea Griviţei 21
  
  010702 Bucharest
   
  Romania
  
  
  
  We give an elementary and self-contained proof of the uniformization theorem for non-compact simply-connected Riemann surfaces.
  
  Uniformization of Riemann surfaces revisited
      Rareş Stan
      December 30, 2023
  ============================================
  
  
  
  
  
  § INTRODUCTION
   
  
  Paul Koebe and shortly thereafter Henri Poincaré are credited with having proved in 1907 the famous uniformization theorem for Riemann surfaces, arguably the single most important result in the whole theory of analytic functions of one complex variable. This theorem generated connections between different areas and lead to the development of new fields of mathematics.
  After Koebe, many proofs of the uniformization theorem were proposed, all of them relying on a large body of topological and analytical prerequisites. Modern authors <cit.>, <cit.> use sheaf cohomology, the Runge approximation theorem, elliptic regularity for the Laplacian, and rather strong results about the vanishing of the first cohomology group of noncompact surfaces. A more recent proof with analytic flavour appears in Donaldson <cit.>, again relying on many strong results, including the Riemann-Roch theorem, the topological classification of compact surfaces, Dolbeault cohomology and the Hodge decomposition. In fact, one can hardly find in the literature a self-contained proof of the uniformization theorem of reasonable length and complexity. Our goal here is to give such a minimalistic proof. 
  
  Recall that
  a Riemann surface is a connected complex manifold of dimension 1, i.e., a connected 
  Hausdorff topological space locally homeomorphic to , endowed with a holomorphic atlas.
  
  [Koebe <cit.>, Poincaré <cit.>]
  Any simply-connected Riemann surface is biholomorphic to either
  the complex plane ℂ, the open unit disk 𝔻, or the Riemann sphere ℂ̂.
  
  A weaker result for domains in ℂ appeared for the first time in Riemann's thesis <cit.>: 
  
  A simply-connected domain strictly contained in ℂ is biholomorphic  to the unit disk.
  
  Riemann's proof for his mapping theorem was notoriously imprecise, as none of the necessary tools were yet fully developed at the time; Osgood <cit.> is credited with completing the argument, building on the work of many other mathematicians. 
  
  What we show below is the following statement:
  
  Every non-compact simply-connected Riemann surface is biholomorphic to a domain in ℂ.
  
  Combined with the Riemann mapping theorem,  
  Theorem <ref> is clearly equivalent to the non-compact case of the uniformization theorem.
  
  The initial proofs of the uniformization theorem implicitly assumed the surface to have a countable basis of topology, but this hypothesis was shown to be unnecessary some 20 years later <cit.>. Most known proofs start from Riemann's idea of searching for so-called Green functions, i.e., harmonic functions with a prescribed logarithmic singularity on sub-surfaces with boundary of a given Riemann surface. Such a Green function will turn out to be the log-norm of a holomorphic map. By exhausting the surface with compact sub-surfaces with boundary, one can find in principle a sequence of embeddings, and compactness results should yield a global embedding into ℂ. Technical complications arise while carrying out this program, and our task is to solve them with the minimum amount of machinery. 
  
  Our paper starts with a self-contained treatment of a few necessary results from Perron's theory <cit.>
  of subharmonic functions, deducing from there Radó's theorem on the second countability of Riemann surfaces. We follow closely the line of proof of <cit.> of constructing an exhaustion by compact surfaces with boundary by means of Sard's lemma. Hubbard <cit.> proves the a priori stronger statement that every non-compact Riemann surface with trivial first Betti number is biholomorphic to either ℂ or the disk. 
  One technical issue in <cit.> is constructing the harmonic dual of the Green function and showing its continuity to the boundary, a particular case of the Osgood-Carathéodory theorem <cit.>, which he uses without proof. He must also rely on Koebe's 1/4 theorem, further complicating the argument. 
  
  The novelty in our approach is to go around these problems by yet another application of Sard's lemma, together with the standard Riemann mapping theorem. We also give a direct proof of the removal of singularities property for bounded harmonic functions, as we were unable to find a clear reference in the literature. Besides these new arguments, we also give full proofs of the necessary auxiliary results in the most economical way possible. 
  
  Our prerequisites are exclusively contained in the standard undergraduate curriculum:
  
  
    * the Schwarz lemma, Montel's theorem on normal families of holomorphic functions, and Riemann's mapping theorem for simply-connected plane domains; we refer to <cit.> or <cit.>;
  
    * the Poisson integral formula on the unit disk (see <cit.>);
  
    * partitions of unity associated to open covers of second-countable manifolds, integration of 1-forms along a smooth path, and Stokes formula in the plane; see <cit.>;
  
    * Sard's lemma; see for instance <cit.>;
  
    * the Tietze extension theorem, the universal cover and the fundamental group π_1; see <cit.>.
  
  
  What we do not use here includes: the Riemann-Roch theorem, the ∂∂ lemma, sheaves and their cohomology, singular and de Rham cohomology and their isomorphism, Runge theory, Weierstrass' results on the existence of meromorphic functions, Koebe-Bieberbach's 1/4 theorem, the existence of triangulations on surfaces, or the topological classification of compact surfaces. We hope that in this way, the uniformization theorem becomes accessible to a general audience, including not only students but also working mathematicians wishing to gain access to an elementary, short, and self-contained proof. 
  
  The proof of the uniformization theorem is useful for the study of moduli spaces of hyperbolic metrics of families of non-compact, non simply-connected Riemann surfaces with prescribed geometry near infinity <cit.>, <cit.>. We believe that our approach can be similarly applied to the Schottky uniformization of compact Riemann surfaces, this will be carried out elsewhere.
  
  The reader is referred to  <cit.> for a comprehensive overview of problems in Riemann surface theory leading to the uniformization theorem, and to the collective work <cit.> for a historical account of the theorem and an outline of the proof.
  
  
  
   §.§ Acknowledgements
  
  We would like to thank Sergiu Moroianu for useful discussions. The authors were
  partially supported from the project PN-III-P4-ID-PCE-2020-0794 and from a Bitdefender Junior Scholarship.
  
  
  
  
  § LOCAL BEHAVIOUR OF HOLOMORPHIC MAPS
  
  We will repeatedly use the local description of holomorphic maps as monomials in suitable charts.
  Let f:X ⟶ Y be a non-constant 
  holomorphic map between Riemann surfaces.
  Let x_0∈ X. Choose charts (η,U), (ψ,V) 
  centered at x_0 and y_0=f(x_0) respectively. Then
  ψ∘ f ∘η^-1(z)=z^k u(z) for some k∈ℕ and some holomorphic function u, with u(0)=c≠ 0. Choose z_0≠ 0. Let v(z)=
  e^1/k∫_z_0^zdu/u. Then v(z)^k=u(z)/u(z_0). In the new chart on X defined by
  φ(p)=η(p)v(η(p)), 
  we have 
  ψ∘ f ∘φ^-1(z)=cz^k. In particular, it follows that non-constant holomorphic maps are open: they map open sets in X 
  to open sets in Y.
  
  
  
  § PERRON'S METHOD
   
  
  A  disk centered at x in a Riemann surface X is the preimage of the unit disk through a holomorphic chart φ:U⟶ mapping x to 0, with ⊂φ(U).
  
  
  Let X be a Riemann surface. A continuous function u:X⟶ is called
  subharmonic, respectively harmonic, if for every disk in X, the real number:
  
      1/2π∫_0^2πu(e^it)dt -u(0)
  
  is non-negative, respectively zero.
  
  
  
  Every continuous function f:S^1⟶ℝ admits a unique continuous extension u:𝔻⟶ℝ which is  harmonic on . Moreover, on 𝔻, u is
  given by the formula:
  
      u(z)= 1/2π( ∫_0^2π f( e^iθ ) 
      e^iθ+z/e^iθ-z  dθ) =1/2π∫_0^2π f( e^iθ ) 1-|z|^2/|e^iθ-z|^2 dθ.
   
  
  The proof is elementary, see for instance <cit.>. Since e^iθ+z/e^iθ-z is holomorphic in z, every harmonic function is the real part of a holomorphic function, hence it is real analytic and satisfies the Laplace equation Δ u=0. Conversely,  by Cauchy's integral formula, the real part of any holomorphic function is harmonic, so in particular log |ξ|=( logξ) is harmonic.
  
   
  Let u be a continuous function on a Riemann surface X. The following
  definitions are equivalent:
  
  
    D1:  u is subharmonic;
  
    D2:  (The maximum principle) For any harmonic function h on any open connected subset U⊂ X, either u+h is constant on U, or it does not attain its supremum;
  
    D3: For every disk D⊂ X one has u ≤ u^(D), where u^(D) is the unique continuous function
  defined by modifying u inside the disk by using the Poisson integral formula:
  
      u^(D)(x)= 
      
      u(x)      x ∈ X∖ D;
      harmonic    x∈ D.
  
  
  
  
  We leave to the reader the (immediate) proof of this proposition. Remark that by the maximum principle, a non-constant subharmonic function cannot attain its supremum. Thus, the restriction of a subharmonic function to a compact set attains its maximum on the boundary.
  
  
  Definition D2 is local: a function
  which is subharmonic in the sense of D2 on a neighbourhood of every point is globally subharmonic.
  We deduce that
  the function u^(D) constructed in (<ref>) is subharmonic on X. 
  Indeed, u^(D) is clearly subharmonic on D and on X ∖D.
  Let p be a point in ∂ D. For every disk centered at p we get:
  
      u^(D)(p)=u(p)≤1/2π∫_0^2π u(e^iθ)dθ≤1/2π∫_0^2π u^(D)(e^iθ)dθ.
  
  Hence u^(D) is subharmonic in a neighbourhood of p as well, therefore  subharmonic on X.
  
  
  
  Let X be a Riemann surface. A set ℱ of subharmonic functions on X is
  called a Perron family if the following two properties are verified:
  
  
    * If f,g ∈ℱ, then max(f,g) ∈ℱ.
  
    * If f ∈ℱ, D is a disk, and
  f^(D) is the subharmonic function defined by (<ref>), then f^(D)∈ℱ.  
  
  
  
  
  Let X be a (possibly not second countable) Riemann surface.
  
  
    * (Perron's principle)
  If ℱ is a non-empty, locally bounded above Perron family 
  on X, then sup_f ∈ℱ f is harmonic. 
  
    * (Dirichlet's principle) Let Y ⊂ X be a sub-manifold
  with (possibly non-compact) boundary and m,M∈. If f:∂ Y ⟶ [m,M] 
  is a bounded continuous
  function, then there exists a continuous function f̃:Y ⟶ [m,M] which
  extends f, and is harmonic on ∘Y.  
  
  
  
  We prove this theorem in Appendix <ref> and <ref>. As a corollary we get:
   
  Every Riemann surface X is second countable.
  
  
  Let D,D' be two disjoint
  closed disks in X. Consider Y=X ∖(
  D∪D'). By Dirichlet's principle 
  (which does not require second countability), 
  there exists a continuous 
  function g:Y→ℝ, harmonic on ∘Y with g_|_∂ D≡ 0, and 
  g_|_∂ D'≡ 1.
  Since g is harmonic, the 1-form ∂ g=1/2(
  ∂ g/∂ x-i∂ g/∂ y)dz 
  is holomorphic. Fix y_0 ∈∘Y. For every path γ:[0,1] ⟶∘Y
  with γ(0)=y_0, let G(γ)=∫_γ∂ g. Then G is a well-defined, non-constant, holomorphic
  function on the universal cover of ∘Y, which by Appendix <ref> must be second countable. This implies
  immediately that both ∘Y and X are second countable.
  
  
  
  
  § GREEN FUNCTIONS
  
  
  
  
  Every harmonic function f:𝔻^*⟶ℝ bounded
  near 0 extends continuously to a harmonic function on
  𝔻.
  
  
  Recall that from the Poisson integral formula f is smooth, so we can define the 1-form β=∂ f/∂ xdy-∂ f/∂ ydx. Since f is harmonic, it
  follows that β is closed. Choose z_0∈𝔻^*. 
  The multi-valued map 
  
      g: 𝔻^* ⟶ℝ;    g(z)=∫_z_0^zβ,
    
  defined by integration along a path from z_0 to z, depends
  only on the homotopy class of the chosen path. 
  Let γ be the circle centered at 0 passing through z_0. It is a generator of π_1(𝔻^*), so
  the map g is well-defined modulo aℤ, where a=∫_γβ. Hence f+ig is well-defined modulo ia ℤ. We claim
  that f+ig satisfies the Cauchy-Riemann conditions. Indeed,
  
      ∂ g/∂ y(z)= . d/dt|_t=0∫_z^z+it( ∂ f/∂ xdy-∂ f/∂ ydx  )
      = . d/dt|_t=0∫_0^t∂ f/∂
      x (z+i θ) dθ= ∂ f/∂ x (z).
  
  The other condition is verified similarly. If a=0, e^f+ig is a
  well-defined holomorphic function on 𝔻^*. Moreover, e^f+ig is bounded near 0, thus its singularity in 0 is removable. Since 𝔻 is simply-connected and | e^f+ig|=e^f >0, we deduce that f=(log e^f+ig) extends
  harmonically on . Otherwise, if a≠ 0, we apply the same argument to the function e^2 π/a(f+ig).
  
  
  
  Let K⊂ X be a (not necesarily compact) sub-surface with smooth boundary and fix x_0∈∘K. Then there exists a continuous function G:K ∖{ x_0 }⟶ [0,∞) which is harmonic on the
  interior of K ∖{ x_0 }, vanishes on ∂ K, and
  has a logarithmic pole in x_0 (i.e., G+ log|ξ| is harmonic on a
  neighbourhood of 0, where ξ is a local holomorphic coordinate centered at x_0).
  
  
  Consider a disk D centered at x_0 with coordinate ξ, and let D_1/2={|ξ|<1/2}⊂ D. From Dirichlet's principle we can define a continuous map h_1:K ∖ D_1/2⟶ [0,1] by:
  
      h_1(ξ)= 
              1      |ξ|=1/2
      
              0    ξ∈∂ K 
      harmonic,
  
  By the maximum principle, a= sup_∂ D h_1 ∈ (0,1) so we can take A,B > 0 such that: 
  
      Ba < A < B- log 2.
  
  From these inequalities, it follows that the map:
  
      h:D∖ D_1/2⟶ℝ,       h(ξ)=max(-Bh_1(ξ),log|ξ|-A)
  
  can be continuously extended to a subharmonic function on K ∖{ x_0}, by setting it to be -Bh_1 on K ∖ D, and log|ξ|-A on D_1/2.
  
  Let ℱ be the Perron family of those continuous functions g:K
  ∖{ x_0 }⟶ [0,∞) which are subharmonic on the interior,
  restrict to 0 on ∂ K, and satisfy g+h ≤ 0. The family ℱ is bounded above (by -h) and contains the map α: D ∖{ 0 }⟶ℝ,
  α(ξ)=-log|ξ| extended with 0 outside the unit disk. By Perron's principle, the map G:K∖{x_0 }⟶ [0,∞), G=sup_g∈ℱ g is continuous, vanishes on ∂ K and is harmonic on ∘K∖{x_0}. Moreover α≤ G ≤ -h, so -log|ξ| ≤ G ≤ A-log|ξ| on D_1/2 and therefore G+log|ξ| is bounded near 0. Lemma <ref> shows that G+ log|ξ| is actually harmonic near 0, ending the proof.
  
  
  
  
  
  
  § A HOLOMORPHIC EMBEDDING IN THE UNIT DISK
   
  
  
  Let X be a Riemann surface and K⊂ X a compact simply-connected sub-surface with precisely one (smooth) boundary component. Then there exists a biholomorphism φ:∘K⟶.
  
  
  Let G be the Green function from Proposition <ref> and consider the 1-form given in holomorphic coordinates by ω=∂ G/∂ xdy-∂ G/∂ ydx ∈Ω^1(∘K). One can easily check that it is invariant under holomorphic changes of variables (in fact, this form is just JdG, where J is the almost complex structure).  
  Fix z_0 in ∘K∖{ x_0 } and
  consider the multi-valued map 
  
      F:∘K∖{ x_0 }⟶ℝ; 
         F(z)= ∫_z_0^zω
  
  defined by integrating along any smooth path from z_0 to z. As above,
  the value of F(z) depends only on the homotopy class of the chosen path. We claim that F(z) is well-defined modulo 2π.
  
  One could try to prove that π_1(∘K∖{ x_0 }) is cyclic as an application of Van Kampen's theorem, since
  ∘K∖{ x_0 } is obtained from ∘K by removing a disk. In reality, all we can say is that π_1(∘K∖{ x_0 }) equals the normalizer of a cyclic subgroup. We will avoid this problem by using the Sard lemma.
  
  
  Take γ:S^1⟶∘K∖{x_0} a smooth loop with γ(1)=z_0. Since ∘K is simply-connected there exists h:⟶∘K, a smooth extension of γ to the closed unit disk. 
  
  We can assume x_0 to be a regular value for h. If not, consider a disk D centered at x_0 which does not intersect γ. By Sard's lemma, there exists a regular value y_0 for h inside D. Pick ψ:∘K⟶∘K a diffeomorphism which maps y_0 to x_0, and acts as the identity outside D. Then ψ∘ h is a new homotopy for which x_0 is a regular value. 
  
  Hence h^-1(x_0) is a (possibly empty) finite set in the disk . By pulling back and using Stokes' theorem, we get:
  
      ∫_γω =∫_S^1h^*ω=∑_q∈ h^-1(x_0)∫_𝒞_qh^*ω=∑_q∈ h^-1(x_0)∫_h(𝒞_q)ω,
  
  where the 𝒞_q's are disjoint circles centered at q, for each q∈ h^-1(x_0), chosen such that their image through h lies inside a punctured disk D∖{x_0}. Every circle 𝒞_q is homotopic in D∖{x_0} to the circle |ξ|=r, where ξ is a local coordinate in a disk centered at x_0 and r<1 is a positive number. Therefore the right-hand side is an integer multiple of ∫_|ξ|=rω. 
  From Proposition
  <ref>, we know that H:=G+ log |ξ| is harmonic on
  { |ξ| < 1 }, hence, using Stokes' theorem:
  
      ∫_ |ξ|=r∂ G/∂ xdy-∂ G/∂
      ydx    =∫_|ξ|≤ r( ∂^2 H/∂
      x^2+∂^2H/∂ y^2) dx∧ dy -∫_|ξ|=rxdy-ydx/x^2+y^2=-2π.
  
  This proves the claim that F is well-defined modulo 2π, therefore
  φ=e^-G-iF:∘K∖{x_0}⟶ is holomorphic and well-defined. The singularity in 0 of such a function is removable because |e^-G|=|xi|e^H is bounded, so we can extend φ: ∘K⟶𝔻 with φ(x_0)=0. Clearly,
  φ^-1(0)={x_0}.
  
  Let us prove that φ is injective. Let A
  be the set of those points x ∈∘K for which dφ_x ≠ 0 and the preimage φ^-1(φ(x)) only contains x. First we prove
  that A is open. Let x ∈ A. By continuity and using Section <ref> we find a disk
  D centered at x where dφ≠ 0 and on which φ is injective. Suppose there
  existed
  sequences (x_j)_j ∈ℕ, (y_j)_j ∈ℕ⊂∘K with x_j ≠ y_j, x_j
  → x and
  φ(x_j)=φ(y_j). Since φ is injective on
  D, there exists a rank j_0 such that for any j ≥ j_0,
  x_j ∈ D and y_j ∉ D. Since
  (y_j)_j ≥ j_0⊂ K ∖ D,
  there exist y ∈ K∖ D and a subsequence (y_j_k)_k ∈ℕ such that lim_k→∞ y_j_k=y. If y∈∘K then we have φ(x)=φ(y), contradicting x∈ A. Otherwise, if y∈∂ K, we have |φ(y_j_k)|→ e^-G(y)=1 while |φ(x_j_k)|→ |φ(x)|<1, contradiction.
  
  Now we prove that A is closed in ∘K. Let (x_j)_j ∈ℕ⊂ A, with x_j → p ∈∘K. If
  φ'(p)=0, then again by Section <ref>, in a suitable chart U near p, the map φ is given
  by z↦ cz^k, for some positive integer k≥ 2. Let ζ≠ 1 be a k-th
  root of unity. Then, for j large enough, we have that φ(x_j)=φ(ζ x_j), which is a contradiction. Otherwise,
  suppose there existed another point q∈∘K with
  φ(p)=φ(q). Since φ is an open map, we can find two
  disjoint neighborhoods U∋ p and V∋ q with φ(U)=φ(V).
  But this means that, for j large enough, φ^-1(x_j)∩ V ≠∅, which is again a contradiction. Hence A is closed in
  ∘K.
  
  To conclude that A=∘K, we are left to prove that A≠∅. Note that near x_0 we have
  |φ(ξ)|=|e^log|ξ|-H(ξ)|=|ξ||e^-H(ξ)|.
  Thus φ'(x_0)≠ 0, so x_0∈ A.
  
  Since non-constant holomorphic maps are open, the injection φ is a biholomorphism onto its image φ(∘K)⊂. 
  
  The Riemann mapping theorem allows us to conclude that ∘K is biholomorphic to . 
  
  
  
  
  
  
  
  § EXHAUSTION BY COMPACT SIMPLY-CONNECTED SUBMANIFOLDS
  
  Let X be a simply-connected, non-compact Riemann surface and x_0 ∈ X a
  fixed point.
  
  There exists a compact exhaustion of X
  
      x_0 ∈ K_0 ⊂ K_1 ⊂ ... ⊂ X,      K_n⊂∘K_n+1,      ⋃_n≥ 0K_n=X
  
  with simply-connected
  surfaces with non-empty connected smooth boundary.
  
  
  By Radó's theorem we can choose a countable locally finite cover of X by relatively compact sets
  (U_m)_m∈ℕ
  and a subordinated partition of unity (ϕ_m)_m∈ℕ. Define a smooth function
  
      g:X⟶ [0,∞),       
      g(x)=∑_m∈ℕ m ϕ_m (x).
  
  For every integer m, g^-1([0,m]) ⊂ϕ_1 ∪...∪ϕ_m is compact, hence
  g is proper and unbounded above.
  By Sard's lemma, there exists an increasing sequence of regular values
  (a_n)_n ∈ℕ with a_n→∞ and we may suppose that
  a_0>g(x_0).
  Consider Y_n'={ x ∈ X | g(x)≤ a_n } and let Y_n be the
  connected component of Y_n' containing x_0. Obviously Y_n' ⊂
  Y_n+1', so Y_n ⊂ Y_n+1.
  Since X is a connected manifold, it is path-connected 
  so, for every x∈ X, there exists a path γ:
  [0,1]⟶ X with γ(0)=x_0 and γ(1)=x. The set
  g(γ[0,1]) is compact, so 
  g(γ[0,1]) ⊂ [0,a_k] for some k ∈ℕ. 
  Then γ([0,1]) ⊂ Y_k' and, by connectedness,
  γ([0,1]) ⊂ Y_k, hence x∈ Y_k so ⋃_n Y_n=X. 
  
  The compact sub-manifolds Y_n are not exactly what we are looking for because they might have 'holes', like an annulus in  (they are not holomorphically convex). The natural way to fill these holes is to add those connected components of X ∖ Y_n with compact closure. Since any two components have distinct boundaries and the boundaries are subsets of g^-1(a_n) (which is a compact 1-manifold) we can define the compact sub-manifold K_n as the finite union of Y_n with these connected components. We can easily convince ourselves that ⋃_n ∈ℕ K_n=X, K_n ⊂∘K_n+1, and K_n is connected (see picture).
  
  
  
      < g r a p h i c s >
  
  
   
  To finish the proof we have to show that π_1(K_n,x_0)=0. The idea is to find a retraction from X to K_n. Let Z be the closure of a connected component of
  X ∖ K_n. First, we need to show that ∂ Z is connected.
  Assume by contradiction that there existed two components γ_1 and γ_2 of ∂ Z. We take a
  point on each and connect them by smooth simple paths δ_1 in Z and δ_2
  in K_n, both transversal to γ_1 and γ_2 (the existence of such paths linking any two points on a
  Riemann surface can be easily proved: the set
  of points which can be connected by a simple curve to a given point is both
  open and closed). Moreover we can assume that δ=δ_1 ∪δ_2 is smooth. Then δ is
  diffeomorphic to S^1 and we can consider a tubular neighbourhood S^1× (-ϵ,ϵ) ⊂ X with
  coordinates (x,y), where x is the coordinate on the circle. Choose η :
  ℝ⟶ℝ a test function such that η⊂ (-ϵ,ϵ) and ∫_-∞^∞η(y) dy=1. Define a closed 1-form ω∈Ω ^1(X) by ω = η (y) dy, extended by 0 outside the tubular neighbourhood. From the homotopy invariance of the integral, we can assume that the intersection of γ_1 with the tubular neighbourhood is a segment of type {x=constant}, hence ∫_γ_1ω=± 1. However X
  is simply-connected, therefore the integral of any closed 1-form on a loop is 0, yielding the desired contradiction. Thus
  ∂ Z is connected, and we parametrize it by γ :[0,1] ⟶∂ Z ⊂∂ K_n, where γ (0)=γ (1)=p. 
  
  Let δ be a simple path in Z starting at p escaping from every compact subset of Z. We cut Z along δ to obtain a topological surface Z̃
  with connected boundary ∂Z̃ = δ' ∪ [p',p”] ∪δ” (see picture). 
  
  
  
      < g r a p h i c s >
  
  
   
  Let
  μ: ∂Z̃⟶ [0,1] be the continuous function 
  
      μ(x)= {[          0      x ∈δ';    γ^-1(x) x ∈γ∖{ p };          1     x ∈δ”. ].
  
  By the Tietze extension theorem, we can extend μ to a continuous function
  μ̃:Z̃⟶ [0,1]. Note that the composition γ∘μ̃ takes the same value (namely p) on δ' and δ”, therefore
  it descends to a well defined map Ψ_Z :Z ⟶γ([0,1])⊂ X:
  
  
      3pcZ̃[d] [r]^μ̃   [0,1][r]^γ    X
      
      Z @–>[urr]_Ψ_Z
  
  
  Now let
  ρ_n:X⟶ K_n be the retraction which acts as the identity on K_n and
  equals Ψ_Z on every connected component Z of X ∖
  K_n. Since the composition
   
      K_n↪ X  K_n
   
   is the identity map, it induces the identity on the fundamental groups. But
  π_1(X)=0, thus K_n is simply-connected. 
  
  
  
  
  § END OF THE PROOF OF THEOREM <REF>
  
  For n ≥ 0, let ϕ_n:∘K_n⟶ r_n be the unique
   biholomorphism given by Theorem <ref> with ϕ_n(x_0)=0 and ϕ_n'(x_0)=1, where r_n is a positive number determined by K_n,
   and the derivative is computed in a fixed coordinate near x_0.
  
  From Schwarz's Lemma for the function ϕ_n+1∘ϕ_n^-1(z):D_r_n⟶ D_r_n+1, it
  follows that r_n ≤ r_n+1. Let R be the limit of this sequence. 
  We claim that there exists a subsequence (ϕ_n_k)_k ∈ℕ of
  (ϕ_n)_n ∈ℕ such that for any m ∈ℕ, (ϕ_n_k|_X_m)_k ≥ m  converges on
  ∘X_m.
  
  
  For every Riemann surface S, let 𝒪(S) be the topological vector space of holomorphic functions from S to  with the topology defined by uniform convergence on compact subsets of S. 
  
  Let (f_n:⟶ℂ)_n ∈ℕ be a sequence of injective holomorphic functions with f_n(0)=0 and f_n'(0)=1. Then there exists a convergent subsequence f_n_k→ f∈𝒪().
  
  We defer the proof of this lemma to Appendix <ref>.
  Consider the holomorphic map g_n: 𝔻⟶ℂ, g_n(z)=1/r_0ϕ_n ∘ϕ_0^-1(r_0z). Clearly g_n is injective, g_n(0)=0, and g_n'(0)=1. 
  By Lemma
  <ref>, there exists a convergent subsequence (g_n_0k)_k ∈ℕ. 
  Therefore (ϕ_n_0k)_k ∈ℕ converges on ∘K_0. Similarly, from the
  sequence ( 1/r_1ϕ_n_0k∘ϕ_1^-1(r_1z)
  )_k∈ℕ, extract a subsequence ( ϕ_n_1k)_k ∈ℕ which converges on ∘K_1. 
  Repeating this
  process, we obtain the sequences (ϕ_n_mk)_k ∈ℕ which
  converges on ∘K_m. Set ϕ_n_k=ϕ_n_kk. Now (ϕ_n_k)_k ∈ℕ converges on X to a holomorphic function ϕ:X ⟶ℂ, uniformly on every compact of X. By continuity ϕ'(0)=1, so ϕ is non-constant. Let us show that ϕ is injective: suppose there existed z_1≠ z_2 ∈ X with ϕ(z_1)=ϕ(z_2). Since ϕ is non-constant, there exists a circle 𝒞 centered at z_2 whose interior does not contain z_1, and such that the difference ϕ-ϕ(z_2) does not vanish on 𝒞. Using the dominated convergence theorem, we get
  
      lim_n→∞1/2 π i∫_𝒞( ϕ_n(z)-ϕ_n(z_1) )'(z)/( ϕ_n(z)-ϕ_n(z_1) )(z)dz= 1/2 π i∫_𝒞(ϕ-ϕ(z_1)) '(z)/( ϕ-ϕ(z_1) ) (z)dz,
  
  The first integral is zero, while the second one counts the number of zeros of the function ϕ-ϕ(z_1) in the disk bounded by 𝒞, which is at least 1, yielding a contradiction. Thus, ϕ is a biholomorphism onto
  its image, ending the proof of Theorem <ref>.
  
  
  
  § PROOF OF PERRON'S PRINCIPLE
   
  
  Since the family ℱ is locally bounded above, we can
  define
  u:X ⟶ℝ, u=sup_f∈ℱ f.
  Our aim is to show that u is harmonic. Since harmonicity is a local property, it suffices to prove that u is harmonic on a
  disk D⊂ X. Let A={ z_0, z_1,... }⊂ D be a dense subset. For each j, there exists a sequence (v_jk)_k ∈ℕ⊂ℱ such that u(z_j)=lim_k→∞ v_jk(z_j).
  Since ℱ is a Perron family, the map h_1=v_11^(D) belongs to ℱ, is harmonic on D, and h_1 ≥ v_11. Suppose we constructed
  the functions {h_1,...,h_n}⊂ℱ harmonic on D such that
  h_k ≥ h_k-1 and h_k ≥ v_ij for all k∈{1,...,n}, i,j∈{1,...,k}. Pick h_n+1=max (v_11,v_12,...,h_n)^(D)∈ℱ harmonic on D, with h_n+1≥ h_n and h_n+1≥ v_ij for any i,j ∈{1,...,n+1}. Then
  h_n(z_j) ≥ v_jk(z_j) for all n ≥ k ≥ j. Letting k→∞
  in the previous inequality, we get: 
  
      lim_n→∞h_n(z_j)=u(z_j).
       
  Since (h_n)_n ∈ℕ is an increasing sequence, we can consider
  the Lebesgue measurable function
  
      w:D⟶ℝ,        w=lim_n→∞ h_n≤sup_g∈ℱg_|_D=u_|_D.
   
  By the dominated convergence theorem, for every disk D' centered at a point x∈ D we have
  
      w(x)=lim_n→∞ h_n(x)=lim_n→∞1/π∫_D' h_n(z)dz=1/π∫_D'lim_n→∞
      h_n(z)dz=1/π∫_D' w(z)dz.
  
  Here we used the mean property on disks for harmonic functions, which is an immediate consequence of the definition.
  This implies easily that w is C^0. Again by dominated convergence,
  
      1/2π∫_0^2π w(e^it) dt=1/2π∫_0^2πlim_n→∞ h_n(e^it) dt= lim_n→∞1/2π∫_0^2π h_n(e^it) dt=lim_n→∞
      h_n(x)=w(x)
  
  thus showing that w is harmonic. We
  claim that w=u. 
  
  Since h_n ∈ℱ for any n ∈ℕ, then h_n ≤ u, so w
  ≤ u. But w(z_j)=u(z_j) for all j ∈ℕ, hence w ≥ v on the 
  dense subset A ⊂ D for every v ∈ℱ. Since ℱ is a family of
  continuous functions, it follows that w ≥ v for any v ∈ℱ,
  therefore w ≥ u, which ends the proof. 
  
  
  
  
  § PROOF OF DIRICHLET'S PRINCIPLE
   
  
  If f is constant, the conclusion is clear, otherwise let m<M be the infimum, respectively the supremum of f on ∂ Y.
  Consider the family ℱ of continuous functions g:Y ⟶ [m,M]
  which are subharmonic on the interior of Y with g_|_∂ Y≤ f. 
  The first condition for ℱ to be a Perron family is evident.
  Let g ∈ℱ and D' a disk in Y. Using Remark <ref>, g^(D') is subharmonic. By the maximum principle, g^(D') still takes values in [m,M]. Thus g^(D')∈ℱ. 
  Now ℱ≠∅ since it contains the constant function
  m. Let F=sup_g ∈ℱ g. By Perron's principle,
  F is harmonic on ∘Y. 
  
  
  
  
  We must prove that for every x∈∂ Y, lim_ξ→ xF(ξ)=f(x). Let us first show
  that lim inf_ξ→ x
  F(ξ)≥ f(x). If f(x)=m, there is nothing to prove, hence let us suppose that f(x)>m. We work in a disk D⊂ X centered at x with local coordinate ξ.
  
  Fix ϵ>0 such that f(0)-ϵ>m. There exists 1>δ>0 such that for every ξ∈∂ Y
  with |ξ|<δ we have f(0)-ϵ <f(ξ). Consider the exterior normal to ∂ Y at x=0. Take a circle centered at a point p on this normal, of radius r small enough such that it touches 
  Y only in 0. Choose t>1 such that log1/t-1+f(0)-ϵ<m and set R=rt. By decreasing r if needed, we can assume that R<δ. Clearly, |p|≤ |ξ-p| for every ξ∈ Y∩ D.
  Define
  
      u:D→ [m, ∞),      
      u(ξ)=max(m,log|p||ξ-p|
      +f(0)-ϵ).
  
  Notice that u is the maximum of two harmonic functions, hence subharmonic. When |ξ| is close to 0,  
  u(ξ)=log|p||ξ-p|+f(0)-ϵ, while for |ξ|≥ R, u(ξ)=m. Hence u can be continuously extended (by the constant value m) to Y∪ D, and is subharmonic on ∘Y. It is straightforward to check that u_|_Y belongs to the family ℱ. Thus, for small |ξ|, we get:
  
      F(ξ) ≥ u(ξ)=log|p||ξ-p|+f(0)-ϵ.
  
  Since ϵ was arbitrarily small, we obtain lim inf_ξ→ 0 F(ξ) ≥ f(0).
  We now prove that lim sup_ξ→ 0 F(ξ) ≤ f(0). If f(0)=M, this is clear. Otherwise, take ϵ>0 such that f(0)+ϵ< M. As above, consider t>1 and R=rt such that -log1/t-1+f(0)+ ϵ>M, and f(ξ)<f(0)+ ϵ for |ξ| ≤ R. Define 
  
      U:D→ (-∞,M],      
      U(ξ)=min(M,-log|p||ξ-p| + f(0)+ϵ).
  
  Notice that U is the minimum of two harmonic functions, thus -U is subharmonic. Also U(ξ)=M for
  |ξ|>R, U(ξ)=-log|p||ξ-p| + f(0)+ϵ for small |ξ|, U ≥ m, and
  U(ξ)≥ f(ξ) on ∂ Y. For every g∈ℱ, the
  continuous function
  g-U is non-positive on the boundary of the compact domain {|ξ|≤ R }∩
  Y and subharmonic in the interior.
  By the maximum principle it follows that on this compact set g≤ U, hence
  F≤ U. Thus for small |ξ|, 
  
      F(ξ)≤ -log|p||ξ-p|+f(0)+ϵ.
  
  Therefore lim sup_ξ→ 0 F(ξ) ≤ f(0), showing that lim_ξ→ 0 F(ξ) exists and equals f(0).
  
  
  
  § SECOND COUNTABILITY IN THE PRESENCE OF A HOLOMORPHIC FUNCTION
  
  Let us prove that if there exists a non-constant holomorphic function f:X⟶ℂ, then the Riemann surface X is second countable.
  
  Let ℬ be a countable basis of topology for ℂ. Let 𝒜 be the set of those connected components of each f^-1(U), U ∈ℬ,
  which are second countable. We claim that A is a basis of topology for X.
  Indeed, let D ⊂ X be an open set and x ∈ D. By the identity theorem for holomorphic
  functions, f^-1 (
  f(x) ) is discrete, hence there exists an open  neighbourhood
  W of x, relatively compact in D, such that W ∩ f^-1( f(x) )= { x  }. We have that f ( ∂W) is compact in ℂ. Since f(x) ∉
  f ( ∂W), there exists U ∈ℬ which contains
  f(x) such that U ∩ f ( ∂W) =∅. Let V be the connected component of f^-1(U) which contains x. Since V ∩∂W=∅, we get that V ⊂ W, hence V is second countable.
  We found x ∈ V ⊂ D, with V ∈𝒜, therefore the claim is
  proved. We next prove that 𝒜 is countable.
  
  Each V ∈𝒜 intersects countably many other
  elements in 𝒜. Otherwise, there would exist U ∈ℬ such that
  V intersects uncountably many connected components of f^-1(U). It would follow that V contains uncountably many disjoint open subsets, which is a
  contradiction.
  
  Consider V_0 ∈𝒜. There exists a countable number of open sets from
  𝒜 which intersect V_0, denote their union with V_0 by V_1. For k ≥ 1, define V_k+1 as the union of V_k with those open sets in 𝒜 which intersect V_k. By induction, V_k is countable. It is clear that ∪_k ≥ 1 V_k is both open and closed, hence ∪_k ≥ 1 V_k=X. Therefore all the open sets from 𝒜 appear in this process, so 𝒜 is a countable union of countable sets, hence countable. 
  
  
  
  
  § COMPACTNESS OF FAMILIES OF INJECTIVE HOLOMORPHIC FUNCTIONS
   
  
  
  Denote by r_n=sup{ r ∈ℝ: r⊂ f_n(𝔻) }. Let a_n ∈∂ (r_n ) ∖ f_n(𝔻). From the Schwarz lemma for the
  function f_n^-1_|_r_n, it
  follows that |a_n|=r_n ≤ 1. By extracting a subsequence, we can assume that (a_n)_n ∈ is convergent. Consider the
  function g_n: 𝔻⟶,
  g_n=a_n^-1 f_n. It is easy to see that 𝔻⊂
  g_n(𝔻) and also 1 ∉ g_n(𝔻). Since g_n is holomorphic
  and injective, it follows that g_n(𝔻) is simply-connected, hence we can construct the square root
  
      ψ_n:g_n(𝔻)⟶ℂ^*   ψ_n(z)=ie^1/2∫_0^z dw/w-1.
  
  Then ψ_n(0)=i and ψ_n^2(z)=z-1 for z ∈ g_n(𝔻).
  Since ψ_n^2 is injective, the image of ψ_n cannot contain pairs of the form (w,-w). Thus, there exists a disk D centered at -i disjoint from the image of ψ_n for every n.
  Define h_n=ψ_n ∘ g_n:
  𝔻⟶∖ D. There exists 0<r<∞ so that
  the homography α(z)=1/z+i maps ∖ D into r. Using
  Montel's Theorem and relabeling the sequence, we can assume that α∘ h_n converges to a
  holomorphic map h:𝔻⟶ r. If
  h is non-constant, it is open by Section <ref>, hence it maps 𝔻 to r.
  Thus
  we can compose it with the inverse homography α^-1. Otherwise,
  h equals the constant 1/2i. In both cases, h_n
  converges to α^-1∘ h, thus f_n=a_n( 1+ h_n^2 ) converges in 𝒪 ().
  
  
  
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