Probing a manifold with closed curves Since fedja's excellent comment on Joseph's question on probing a manifold with geodesics remained uncommented (especially by topologists), I'd like to make a question out of it:

Conjecture: Given an orientable 2-dimensional manifold and two
  closed curves on it which intersect
  transversally in exactly one
  point. Then the two curves cannot be
  homotopic.

(An immediate consequence of this would be that living on a surface with two such curves, one would know, that it is not homeomorphic to the sphere.)

How to proof this conjecture (if it's true)? 

 A: I wanted to give a somewhat "low-tech" answer to this, here's the best I came up with:
Call the curves $C_1$ and $C_2$ and the surface that they lie in $\Sigma$.  Take a small open neighborhood $N$ of $C_1 \cup C_2$.  What does $N$ look like?  Well, it's a neighborhood of $C_1$ stuck to a neighborhood of $C_2$ in an operation called a "plumbing" (which is the same operation as sticking together two strips of paper to make a cross).
The neighborhood of $C_i$ is a cylinder that has been plumbed to itself a few times (one plumbing for each self-intersection of $C_i$).
Now if $C_1$ can be homotoped to coincide with $C_2$ inside $\Sigma$ then it can be homotoped to $C_2$ inside the space $S$ that I make by crushing $\Sigma \setminus N$ to a point.
The idea is to convince yourself that $S$ is a torus with a finite number of points identified (one point for each boundary circle of $N$), and that $C_1$ wraps once around one direction of the torus while $C_2$ wraps once around the transverse direction.
A: The conjecture is false.  For example, in a Moebius band, the central circle self-intersects once with many transverse perturbations of itself. 
If you want a conjecture like yours to be true, you'll have to assume that a tubular neighbourhood of one of the curves is trivial.  That ensures a regular neighbourhood of the union is a punctured torus or Klein bottle. 
A: There are exactly two one plane bundles over the circle, the open annulus, and open Moebius band, where we take projection to be onto the central circle. We can now ask what the mod 2 self intersection of the zero section is. In the case of an annulus it is zero, in the case of the Moebius band it is one.  
The tubular neighborhood theorem says that any smooth submanifold of a smooth manifold has a neighborhood homeomorphic to a k-plane bundle over the submanifold, where k is it's codimension.
You can't embed a Moebius band in an orientable surface, so the self intersection of any simple closed curve in an orientable surface is zero.  The self intersection number is invariant under homotopy, so if gamma and gamma' are homotopic simple closed curves in a surface that intersect each other transversely then they intersect in an even number of points.
