I want to prove that every closed orientable surface is a Riemann surface i.e. every closed orientable surface admits a complex structure. Several proofs are available which make use of classification theorem for closed surfaces. How can one prove this without assuming the classification theorem?
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Although this question asks only about the non-compact case, the accepted answer explicitly covers the compact case as well. – Lee Mosher May 29 '20 at 18:12
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See also here: First, put a Riemannian metric on the surface, then use isothermal coordinates to define a holomorphic atlas. – Moishe Kohan May 29 '20 at 19:33