TITLE: Probing a manifold with closed curves
QUESTION [5 upvotes]: 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)?

REPLY [7 votes]: 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.