I have some trouble solving Hartshorne chapter Ex.IV.4.5.
For an elliptic curve $X$, a point $P_0$ of $X$, and a morphism $f:X \rightarrow X$ of degree 2, let $\pi : X \rightarrow \mathbb{P}^1$ be the morphism defined by the divisor $2P_0$.
(a) states that there exist a morphism $\pi' : X \rightarrow \mathbb{P}^1$ and $g : \mathbb{P}^1 \rightarrow \mathbb{P}^1$ both of degree 2 such that $\pi f = g \pi'.$
Here we have $(\pi f)^*\mathcal{O}_{\mathbb{P}^1}(1) = (g \pi')^*\mathcal{O}_{\mathbb{P}^1}(1)$, so if we let $D$ be the divisor on $X$ corresponding to $\pi'$, we have $2D \sim 2f^*(P_0)$.
If we let $D = f^*(P_0)$, this solves (a). However (observing other solutions), later in (c) I felt that it is supposed to take $D = 2f(P_0)$, but then I cannot solve (a). How does this go?