Let $C_1$ and $C_2$ be two simple closed curves on an orientable compact surface $S$, such that:
- They are homotopic to each other.
- They are set-theoretically disjoint.
Is $S\setminus(C_1 \cup C_2)$ is disconnected?
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5$\begingroup$ More than that, they bound an embedded cylinder -- a 2-manifold in $S$ diffeomorphic to $S^1 \times [0,1]$. $\endgroup$– Ryan BudneyCommented Apr 6, 2012 at 8:24
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$\begingroup$ 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. $\endgroup$– Ryan BudneyCommented Apr 6, 2012 at 8:26
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1$\begingroup$ 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. $\endgroup$– Tony HuynhCommented Apr 6, 2012 at 9:33
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1$\begingroup$ 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 $\endgroup$– George LowtherCommented Apr 6, 2012 at 13:51
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