Is the Jordan Curve Theorem True for Curves on a Toroid I'm making a presentation on the Jordan Curve Theorem, and I want a counter-example. I think that the toroid is a great example, but I want to make sure that I'm correct.  
Since the toroid has a hole in it, we could technically draw a curve that encloses this hole. Would the resulting curve satisfy the JCT? I don't think that it would, since you wouldn't end up with two distinct regions (like you would on, say, the 2-sphere). It seems to me like the hole allows you to get around the idea of dividing something into two components. Does that seem reasonable?
 A: You are correct that the Jordan Curve theorem needs modifications if you were to implement it on (closed) surfaces other than the sphere. The meridian or longitude curves are examples of curves that do not split the torus into two connected components.
Here's a formulation that works for the torus (including Schoenflies-like statements on the torus): 
Given a closed simple curve in the torus, if the curve bounds an embedded disc, then it separates the torus into two connected components; the non-disc component is homeomorphic to a topological space $X$ not depending on the chosen embedded curve. If it does not bound a disc, its complement is connected, and indeed homeomorphic to the open cylinder (or annulus, if you like) $S^1 \times (-1,1)$.
You can see a proof of this in chapter 2 of Rolfsen's book "Knots and Links". It takes some work, and assumes you already know the Schoenflies theorem for the sphere.
A sort of general JCT might be stated as follows: given a simple closed curve $K$ in a surface $\Sigma$, $K$ separates $\Sigma$ into at most two pieces, and it separates $\Sigma$ into two pieces if and only if $K$ is null-homotopic. In this case, $K$ actually bounds an embedded disc. (This still applied for the sphere - all curves in the sphere are null-homotopic.)
