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Let $X$ be a compact Riemann surface, and suppose that there exists a meromorphic function $f$ on $X$ with one simple pole and no other poles. Show that $f$ is an isomorphism between $X$ and the Riemann sphere. Using this, prove that the any Riemann surface of genus zero is isomorphic to the Riemann sphere.

Let $(f) = (P) - (Q)$. Since the degree of $f$ is zero, it must be that $P \neq Q$. I am not sure how I can prove that $f$ is surjective, other than the general instruction that I can use the Riemann-Roch theorem. I know that the sum of the divisor of any meromorphic function is equal to zero, and that for any divisor $K$ of $X$, the degree of $K$ is equal to $2g - 2$ where $g$ is the genus.

EDIT: Thanks to @DJ Dowd, I proved the second part: let $Q$ be any point in $X$. Then $\dim L((Q)) \geq 2$ by Riemann Roch, so that there exists a nonconstant function $f \in L((Q))$ and it can be shown that $f$ can have only one simple zero and no other poles by using the fact that the degree of $f$ is zero. It remains to prove the first part. Because the function $f - c$ has a simple zero at $c = f(R)$ for arbitrary $R \in X$, we have injectivity. Thus it remains to prove surjectivity.

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The fact that $(f)$ is of the form $(P) - (Q)$ with $P \neq Q$ follows from the fact that a principal divisor has degree $0$. If a degree $0$ divisor has exactly one pole, it must be of this form: $-1 + 1$ is the only way to add positive integers to $-1$ to get $0$. (You don't need Riemann-Roch for this.)

If $D$ is a divisor, you should interpret the Riemann-Roch space $\mathscr{L}(D)$ as the set of rational functions on $X$ with "poles no worse than $D$," or more precisely that $D + (f)$ is an effective divisor. If $X$ is genus $0$, take $D = (Q)$ to be a point. You should be able to use Riemann-Roch to show that $\mathscr{L}(D)$ contains something other than a constant function, which must be of the form you want.

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  • $\begingroup$ Thank yo so much @CJ Dowd. I got the second part. I was wondering if you could hlep me for the first part? $\endgroup$ Commented Apr 29, 2023 at 7:32
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    $\begingroup$ @CuteBrownie You might be overthinking it: if a function has one and only one pole, and that pole is simple, it must have exactly zero somewhere else in order to have $\deg (f)$ be $0$. No zeros means that $\deg (f) = -1$, and more than two zeros means that $\deg (f) \geq 1$. The zero has to occur at a different point than the pole, of course. $\endgroup$
    – CJ Dowd
    Commented Apr 29, 2023 at 7:39
  • $\begingroup$ Thanks @CJ Dowd. I meant that I wasn't sure how to solve that any $f$ with one simple zero and one simple pole must be an isomorphism. Indeed, we have that if $P,Q$ are zero and pole, respecively, then in any neighborhood that doesn't contain $Q$, $f$ is locally an injection. But how can we prove that $f$ is globally an isomorphism? This is tricky to prove that $f$ is surjective. $\endgroup$ Commented Apr 29, 2023 at 7:43
  • $\begingroup$ @CuteBrownie Maybe one way to see this is that you know the degree of a map between Riemann surfaces is well-defined, and if the degree is $1$ then the map is an isomorphism. But you can check the degree on any fiber of the map if you count with multiplicity. In particular, you can check over the fiber of $0$, which you know is $P$ with multiplicity $1$, since the only zero of $f$ is a simple zero at $P$. Hence $f$ has degree $1$, so it is an isomorphism. $\endgroup$
    – CJ Dowd
    Commented Apr 29, 2023 at 7:51
  • $\begingroup$ Thank you @CJ Dowd for your input. I am wondering if you could further explain your answer in detail as part of your answer? Are you using the Riemann-Hurwitz formula? $\endgroup$ Commented Apr 29, 2023 at 7:55

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