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I have read that thanks to Riemann Roch theorem, if get $\Sigma$ a compact Riemann Surface of genus $g$ there exists a conformal branch covering $\phi: \Sigma \rightarrow S^2$ of degree less than $g+1$. Unfortunately I have found only very abstract references which not clearly implies this fact. does any one can explain this to me? Ideally with a basic reference.

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When a reference is abstract is probably dependent on who you are, but the result you're mentioning is Corollary 16.12 of Forster's book "Lectures on Riemann Surfaces" which is fairly concrete. –  fuglede Jun 12 at 12:03

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Let $D$ be a finite collection of $d$ points on $\Sigma$. The RR theorem shows that the vector space of meromorphic functions on $\Sigma$ with at worst a simple pole at each $D$ has dimension $\geq d + 1 - g$, which is $> 1$ if $d > g$. Thus (choosing any $D$ with $d > g$) this space contains a non-constant meromorphic function $f$. (The constant functions contribute just one dimension.) Since $f$ has at most $d$ simple poles, it induces a degree $\leq d$ branched covering $\Sigma \to \mathbb C P^1$.

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