I am interested in a proof of the following fact :


Suppose that $X$ is a Riemann surface homeomorphic to the Riemann sphere. Then $X$ is *conformally equivalent* to the Riemann sphere.


Of course, this follows from the uniformization theorem which states that every simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.


However, I was wondering if it was possible to prove this without using the full classification given by the uniformization theorem.


Any relevant reference is welcome.


Thank you and best regards,
Malik 



A special case of the uniformization theorem
An alternative is to simply use Riemann-Roch, which does not depend on the Uniformization Theorem. 


R-R says that $\ell(D)-\ell(K-D) = \mathrm{deg}(D)-g+1$, where $\ell(D)$ is the dimension of the space of meromorphic functions $f$ on your surface $C$ such that $(f)+D\ge0$. 


If $C$ is homeomorphic to the sphere, then, for topological reasons, $g=0$, so when $D=(p)$ for any $p\in C$, the formula simplifies to $\ell(D)-\ell(K-D) = 2$. In particular, $\ell(D)\ge2$, which implies that there is at least one nonconstant meromorphic function $f$ that has a simple pole at $p$ (and no other poles). This then defines a holomorphic mapping $f:C\to\mathbb{CP}^1$ that is $1$-to-$1$ (because it only assumes the value $\infty$ once), and this gives the conformal equivalence. 


