Let $E:y^2=x(x-1)(x-\lambda)$ be the Legendre form of an elliptic curve $E$ defined over $\mathbb{C}$. The ramified covering $E\to \mathbb{P}_{1}$ defined so that $(x,y)\mapsto x$ has two branches and ramification points $0,1,\lambda,\infty$.

The two branches $y=\pm \sqrt{x(x-1)(x-\lambda)}$ are therefore glued at the ramification points. $E$ would turn out to be hence the glueing of two spheres $\mathbb{P}_{1}$ at the 4 distinct points $0,1,\lambda,\infty$.

Such an object has not genus 1, therefore there has to be a mistake.

In J. Silverman's AEC at the very beginning of chapter 6 it is explained that what one does is actually to glue the spheres along the two segments $\overline{\infty 0}$ and $\overline{1\lambda}$. This gives a torus, but still I don't understand why all the points of the segments has to be removed by the two branches? And why not to choose another couple of disjoint segments like $\overline{01}$ and $\overline{\infty\lambda}$? All integrals of $dx/y$ over paths in the complement of the ramification points are perfectly defined in the same branch once the sign of the square root is chosen, see for example figure 6.5 at page 159, where a closed path $\alpha$ is taken across both segments.

So my question is: why do we remove the entire segments if the junction points of the two branches are only 4?