Let $X$ be a one-dimensional, compact, complex manifold, let $x_0\in X$, and define $\varphi:X\to\text{Pic}^0(X)$ by $\varphi(x)=[x-x_0]$.
I am showing that $X$ is isomorphic to $\mathbb P^1$ if and only if $\varphi$ is not injective.
Suppose that $\varphi$ is not injective. Then there are $x_1,x_2\in X$ such that $x_1\neq x_2$ and $[x_1-x_0]=[x_2-x_0]$. Moreover, $L=[x_i-x_0]$ admits two holomorphic sections $s_1,s_2\in H^0(X,L)$ such that $Z(s_i)=x_i$. This implies that $\psi:X\to\mathbb P^1$ defined by $\psi(x)=[s_1(x),s_2(x)]$ is defined everywhere.
I am stuck here, but I was told that $\psi$ is injective in a neighborhood of $x_i$ and thus injective everywhere. I fail to see both of these assertions.
Since $Z(s_1)=x_1$, the only zero of $\psi$ is a simple zero at $x_1$ (note that here we use the fact that $x_2\neq x_1$, so that $s_2(x_1)\neq 0$ and so $\psi$ really does have a zero at $x_1$; if $x_1$ and $x_2$ were the same then $\psi$ would have no zeroes and thus would be constant). So $\psi$ is a meromorphic function on $X$ with only a single simple zero. A meromorphic function on a compact Riemann surface takes every value the same number of times (counted with multiplicity), so $\psi$ takes every value exactly once, and thus is injective.