= Solution
The degree is the sum of divisor coefficients; principal divisors are $(f)$ for nonzero rational <functions>. $\operatorname{Cl}(C)=\operatorname{Div}(C)/\operatorname{Prin}(C)$ and $\operatorname{Cl}^0$ is its degree-zero <subgroup>. On $\mathbb P^1$, $K\sim-2P$; on a hyperelliptic genus-$g$ curve with degree-two fibre $H$, $K\sim(g-1)H$. If $\operatorname{Cl}^0=0$, any point divisor has $\ell(P)=2$ by Riemann-Roch, giving a degree-one map to $\mathbb P^1$; the converse follows because every degree-zero divisor on $\mathbb P^1$ is principal. For $g\ge1$, $[p-p_0]=[q-p_0]$ would give a <function> with divisor $p-q$, hence a degree-one map unless $p=q$, proving injectivity. If $p+q\sim r+s$ and the pairs differ, the resulting degree-two pencil makes $C$ hyperelliptic; therefore on a nonhyperelliptic curve the unordered pairs coincide.
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