Exercise IV.4.5 of Hartshorne, Algebraic geometry This is point $(c)$ in Exercise IV.4.5 of Hartshorne, Algebraic geometry. I have solved the first two points, but I am stuck with this one.
It goes:
let $X$ be an elliptic curve, $P_0\in X$ a point and $f:X\to X$ a degree two morphism. If $\pi:X\to\mathbb{P}^1$ is the morphism associated to the linear system $|2P_0|$ and $\pi':X\to\mathbb{P}^1$ the one associated to $|2f(P_0)|$, then there exists a morphism $g:\mathbb{P}^1\to\mathbb{P}^1$ such that $\pi\circ f=g\circ\pi'$. Moreover $g$ can be chosen to be the morphism $x\mapsto x^2$ (this is the content of the first two points of the exercise). Show that $g$ is branched over two of the four branch points of $\pi$ and that $g^{-1}$ of these points consists of the four branch points of $\pi'$. 
First of all, since $\deg f=\deg g=2$ by Hurwitz formula I know that $f$ and $g$ are ramified exactly in four points with index $1$ and similarly $\pi$ and $\pi'$ are ramified in exactly four points with index one again. In particular this means that the sets of branch points have, respectively, exactly two and four points (the images of the ramification points, sorry for the sloppiness here). Moreover, since $g$ has that shape, I know that the ramification points are $0$ and $\infty$ and coincides with the branch points.
By construction $\pi(P_0)=\infty$ and similarly $\pi'(f(P_0))=\infty$ are brunch points for $\pi$ and $\pi'$. This implies that $g$ is branched at least over one of the branch points of $\pi$. I guess the other one is $0$ which can be reach by a change of coordinate in $\mathbb{P}^1$. But I am not sure why $0$ is a branch point of $\pi$.
If $r$ is a branch point of $\pi$, then $\pi^{-1}(r)$ consists of a single point $s\in X$ by definition. The preimage $f^{-1}(s)$ is then composed by two points $t_1,t_2\in X$ (possibly coincident if $s$ is a branch point for $f$). By commutativity of the square, $g^{-1}(r)$ consists of the points $\pi'(t_1),\pi'(t_2)\in\mathbb{P}^1$. Hence I should prove that $t_i$ is a ramification point of $\pi'$ for $i=1,2$. 
When $r$ is the point at the infinity, by construction it should follow that $t_1=t_2=P_0$, which means that $P_0$ is a ramification point of $f$. But $f$ was chosen at the beginning as a general morphism which has nothing to do with the point $P_0$, so the last sentence above sounds quite wrong... and probably also the road I have taken is not the right one.
Can anyone point me out the several mistakes I am making and give a clue toward the right way?
Thank you very much!
 A: I realize this is a bit old and you may not be around anymore, but I've been thinking about this question lately and felt I should leave an answer for posterity.
The error in the question above is that the ramification of $f$ and $g$ has been mis-identified. Let's draw the diagram first:
$$\require{AMScd}
\begin{CD}
X @>{f}>> X\\
@V{\pi'}VV @VV{\pi}V \\
\Bbb P^1 @>{g}>> \Bbb P^1
\end{CD}$$
Riemann-Hurwitz for a map $m:A\to B$ says $2g(A)-2=(\deg m)\cdot(2g(B)-2)+\deg R$ where $R$ is the ramification divisor.

*

*Applied to the case of $f:X\to X$, we have $g(X)=1$ so $2g(X)-2=0$, so $\deg R=0$, and $f$ is unramified.

*Applied to $g:\Bbb P^1\to\Bbb P^1$, we have $g(\Bbb P^1)=0$ and $\deg g=2$, so $\deg R=2$ and therefore $g$ is ramified over two points. The content of (b) is showing that you can choose those points to be $0,\infty$ on both copies.

*Applied to $\pi':X\to\Bbb P^1$ and $\pi:X\to\Bbb P^1$, both of degree two, we have that $\deg R=4$, so they are each ramified over four points.

Now we compare the two ways to go around the diagram from the top left $X$ to the bottom right $\Bbb P^1$. Travelling along $f$ and $\pi$, we see that the four branch points of $\pi$ on $\Bbb P^1$ have two distinct preimages in the top left copy of $X$, while every point on the lower right $\Bbb P^1$ which is not a branch point of $\pi$ has four distinct preimages. Travelling along $\pi'$ and $g$, we see that $0$ and $\infty$ each have a single preimage under $g$, and therefore must have two preimages under $\pi'$. On the other hand, the other two branch points of $\pi$ must have two preimages each under $g$, and each of those four points must have a single preimage under $\pi'$ in order that the non-zero non-infinity branch points of $\pi$ have exactly two preimages in the top left $X$. This means we've found our four ramification points of $\pi'$ on $X$!
