Show that function gives a bijection from curve to projective line but $\textrm{div}(f) = [P] - [Q] \implies P = Q$?

Question

Washington's book on Elliptic Curves Chapter 11 on Divisors, page 380, question 11.3 says:

Suppose $f$ is a function on an algebraic curve $C$ such that $\textrm{div}(f) = [P] - [Q]$ for points $P$ and $Q$. Show that $f$ gives a bijection of $C$ with $\mathbb{P}^1$.

Doesn't this mean that $f$ is a constant function?

$$f \in \mathcal{L}([Q]) = \bar{K}$$

So how can it give a bijection to the projective line $\mathbb{P}^1$? Wouldn't this imply then $P = Q$ and $\textrm{div}(f) = 0$?

Answer 1

A non-constant morphism of curves is finite surjective (ref). $(f)_0$ and $(f)_\infty$ are the preimages (with multiplicity) of $0$ and $\infty$ under the map $f:C\to\Bbb P^1$, so if they're a single point, then $f$ must be finite of degree one, or an isomorphism.

Answer 2

This is only true when the genus $g > 1$, in which case $\ell([Q]) \geq 2 - g$. So if $P \neq Q$, then $g > 1$.

To show bijection, we first show injective property. Since $\textrm{ker} f = \{ P \}$, this means $f(S) = f(T) \iff S = T$. Secondly for surjectivity, $f(Q) = \infty$, and for the other points, let $p \in \mathbb{P}^1$. We want to find $R$ with $f(R) = p$. $f$ has the same divisor $f(X) = (X - P)$ in affine space which is the equation of a line.

(leaving this unanswered in case someone has a better comment)