In reading these notes (elliptic curves starting from elliptic integrals) I came across a couple claims about the topology of some complex surfaces.
On page 4, they discuss the integral $\phi(x) = \int_0 ^x \frac{dt}{\sqrt{1 - t^2}}$ In order to define it on all of $\mathbb C$, you have to use a branch cut; they glue two copies of $\mathbb C$ together along $[-1,1]$, in the same crossing-over manner as you do when dealing with $\sqrt z$ (at least, I think). They then claim that this surface $C$ is homeomorphic to a cylinder. However, I'm having trouble seeing a way to explicitly bend $C$ into a cylinder. I think I might be missing some intuition on what a complex cylinder looks like.
I think I understand why $C$ would be homotopically equivalent (not sure if that's the best term) to a cylinder, because there's one set of loops from going around the branch cut, and if you avoid integrating across the branch cut the other ones are all null-homotopic. But why glue two copies of $\mathbb C$ together at all if you're going to avoid integration across the branch cut?
I also don't quite get they say that $C$ can also be defined as $\{ (x,y) \in \mathbb C ^2 : x^2 + y^2 = 1 \}$; is it just that we can integrate $dx/y$ on $C$, because that differential on $C$ looks like the differential in $\phi$?
They make similar claims a page later about $ \psi (x) = \int_0^x \frac{dt}{\sqrt{t(t-1)(t-\lambda)}}$ but I think my issues are basically the same.