# Do homotopic non-intersecting simple closed curves separate the surface?

Let $C_1$ and $C_2$ be two simple closed curves on an orientable compact surface $S$, such that:

1. They are homotopic to each other.
2. They are set-theoretically disjoint.

Is $S\setminus(C_1 \cup C_2)$ is disconnected?

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More than that, they bound an embedded cylinder -- a 2-manifold in $S$ diffeomorphic to $S^1 \times [0,1]$. – Ryan Budney Apr 6 '12 at 8:24
The idea is that if $S \setminus C$ is not disconnected, then you can write $S$ as a lower genus surface connect-sum a torus, and $C$ is an essential curve in that torus. – Ryan Budney Apr 6 '12 at 8:26
If you are considering surfaces with boundary, then the only case when you get something connected is if $S$ is a cylinder and $C_1$ and $C_2$ are the two boundary curves. – Tony Huynh Apr 6 '12 at 9:33
As the other comments mention, you can say rather more than what you ask. But, an easy way to get a positive answer to your question is as follows. Choose a neighbourhood of a point in $C_1$ homeomorphic to the plane, disjoint from $C_2$ and in which $C_1$ coincides with the x-axis. Then $C_1$ divides this nhood into two components. Any curve joining these two components in the surface and not intersecting your curves can be extended to a closed curve intersecting $C_1$ once, so intersects $C_2$ once, giving a contradiction. So the two components of the nhood lie in distinct components – George Lowther Apr 6 '12 at 13:51