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.
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.