In Chapter 6 of Silverman's "Arithmetic of Elliptic Curves," he introduces elliptic functions by first attempting to integrate the invariant differential of an elliptic curve over $\mathbb{C}$ $$\omega=dx/y=dx/\sqrt{x(x-1)(x-\lambda)}.$$ But we have path-independence issues, so we need to make some branch cuts. If we cut out a line from 0 to $\infty$ and a line from 1 to $\lambda$, then this square root is single-valued and integrating the differential on this subset is path-independent. We form the Riemann surface of this function by taking two copies of $\mathbb{P}^1$ with the indicated slits and gluing them together. This surface is readily seen to be a torus.
So finally we arrive to my question in the title. If we follow the same procedure with the function $\sqrt{z}$, we make the Riemann surface by cutting a line $0$ to $\infty$ in our two copies of $\mathbb{P}^1$ and then gluing them together. This surface is readily seen to be just another copy of $\mathbb{P}^1$.
But the Riemann surface of $\sqrt{z}$ is not $\mathbb{P}^1$. So what is going wrong with the above construction? The best I can surmise is that I'm constructing the Riemann surface for the function $\sqrt{z}$ whose domain is $\mathbb{P}^1$, rather than the function $\sqrt{z}$ whose domain is $\mathbb{C}$.
Any explanation would be appreciated :)