Why is this a degree 2 morphism? Let $X$ be a smooth projective curve of genus 1, let $P_0\in X$ and consider the linear system $|2P_0|$. By Riemman-Roch $l(2P_0)=2$. I understand why this linear system is base-point free, and it defines a morphism 
\begin{align*}
f:X &\to\mathbb{P}^1\\
P &\mapsto [f_0(P):f_1(P)],\end{align*} where $\{f_0,f_1\}$ is any basis of the $K$-vector space $L(2P_0)$. Hartshorne [IV, §4] says it is a degree $2$  morphism, why is that? I would appreciate an answer as elementary as possible.
 A: Choose $x \in L(2P_0)$ such that $\{1,x\}$ is a basis for $ L(2P_0)$, and consider the map $f: X \longrightarrow \mathbb{P}^1$, with  $P\mapsto [1,x(P)]$.  We will  use the following  fact:
If $f:\mathcal{C}_1 \longrightarrow \mathcal{C}_2$
 is a nonconstant map of smooth curves, then for all but finitely many points $Q\in  \mathcal{C}_2$
 $$\deg f=\# f^{-1}(Q). $$
Now, let $Q=[1:\alpha] \in \mathbb{P}^1$ be a generic point.    If $P_1,P_2,\cdots,P_n \in X$ are such that $x(P_1)=x(P_2)=\cdots=x(P_n)=\alpha $, then $P_1,P_2,\cdots,P_n$ are zeros of $(x-\alpha)$. However,
the fact that $P_0$ is the only pole (a double pole) of $x$ gives that  $P_0$ will be the only pole (a double pole) of $x-\alpha $. This implies that $x-\alpha$ has only two zeros. Generically the two zeros  will be distinct, and  so  $\deg f=\# f^{-1}(Q)=2$.
A: We dicuss this more scheme-theorically.
Let $L=\mathscr{O}(2P_0)$. Actually we get a basis $\{x,1\}\in\Gamma(X,L)$. So if we let $f_0$ defines $P_0$ locally (over $V\ni P_0$), then $x|_{V}=f_0^{-2}$, as $x$ has pole of order 2 only at $P_0$.
This defines $f:X\to\mathbb{P}^1$ (classically, $P\mapsto[x(P):1]$) as follows:
Let cordinates of $\mathbb{P}^1$ is $x_0,x_1$. Let $U_0,U_1$ are non-vanising sets of $x$ and $1$.
First $k[x_1/x_0]\to\Gamma(U_0,\mathscr{O}_{U_0})$ is $x_1/x_0\mapsto 1/x$. Second $k[x_0/x_1]\to\Gamma(U_1,\mathscr{O}_{U_1})$ is $x_0/x_1\mapsto x$.
Next we need to find the image of $P_0$ in $\mathbb{P}^1$.
(a) Actually $x$ not vanish for all but $P_0$, as near $P_0$, $x=f_0^{-2}$. Using the fact that $\mathfrak{m}_{P_0}=(f_0)$. So if $x_{P_0}\in\mathfrak{m}_{P_0}L_{P_0}$, then $x_{P_0}=uf_0^k\cdot f_0^{-2}$ where $u$ be a unit and $k>0$. But this is impossible! So $x_{P_0}\notin\mathfrak{m}_{P_0}L_{P_0}$. So $U_0=X$!
(b) Now as (a), we get $1\in\mathfrak{m}_{P_0}L_{P_0}$! So $U_1=X-\{P_0\}$!
SO there is only $P_0$ maps to $Q=[1:0]\in\mathbb{P}^1$.
Now using the proposition II.6.9, we get $\deg(f)=v_{P_0}(t_Q)$. Now $t_Q=x_1/x_0$ mapping to the $x$ in $\mathscr{O}_{X,P_0}$ with $t_{P_0}=f_0$! As $x$ locally is $f_0^{-2}$, we get $v_{P_0}(t_Q)=2$, hence $\deg(f)=2$.
