Quotient of an elliptic curve by involution biholomorphic to $S^2$ I stumbled upon the following exercise in my lecture notes and I don't manage to come up with a solution:
Let $\Lambda$ be a fixed lattice in $\mathbb{C}$ and $\sigma(z) = -z$ the involution on $\mathbb{C}$. Show that the quotient of $\mathbb{C}/\Lambda$ induced by the action of $\{id, \sigma\}$ is biholomorphic to the Riemannian sphere. 
First, I tried to apply my knowledge on covering spaces, but since the action has a fixed point, namely the origin, the canonical projection is not necessarily a covering map. Then, I tried to determine the genus of the quotient since it is still a compact Riemann surface. However, I could not conclude that the genus is zero.
EDIT:
I also know that if $f: \mathbb{C}P^2 \to S$ is a non-constant holomorphic map between Riemann surfaces, then $S$ is the Riemannian sphere. Let's denote by $S$ the quotient in question. In this case, we have the projections $\pi_1: \mathbb{C} \to \mathbb{C}/\Lambda$ and $\pi_2: \mathbb{C}/\Lambda \to S$ which are non-constant and holomorphic. If we denote the composition $\eta = \pi_2 \circ \pi_1: \mathbb{C} \to S$, it suffices to show that $\eta$ passes to the quotient. However, I have a hard time to see, why it should be constant on the equivalence classes of the projection $p: \mathbb{C} \to \mathbb{C}P^2$
 A: It's true that the origin is a fixed point, but beware that there are actually a total of four fixed points: these are the four 2-torsion points of the elliptic curve, one of which is the origin.  A 2-torsion point is a point $x \in \mathbb{C}/\Lambda$ such that $2x=0$.  In other words $x=-x$, which says it's fixed under the involution.  I'm not sure, perhaps it was this issue preventing the Riemann-Hurwitz computation from working for you.  
Your involution along with the fixed points can actually be seen quite intuitively.  Imagine lying the torus flat on the table, and rotating by 180 degree about a horizontal line.  Imagine this line skewering straight through the torus such that it punctures it four times; those are your four fixed points.  And by thinking about the fundamental domain, it's clear the quotient is topologically a sphere.  
To make this intuition rigorous, you do the Riemann-Hurwitz computation.  You have a degree 2 map $\pi: E \to C$ from an elliptic curve $E$ to a genus $g$ curve $C$.  The formula say that:
$$2(1) -2 = 2(2g-2) + \sum_{p \in E} (e_{p}-1)$$
where $e_{p}=1$ for all points in $E$ except the four fixed points discussed above, in which case $e_{p}=2$.  It follows easily that $g=0$.      
