I have a question as follows: Let $C$ be a closed curve in the Mobius band without self intersections. Prove that if $C$ is of non-zero homology, i.e., $C$ does not bound any face, then $C$ winds only once around the Mobius band.
The picture is clear, however I do not know how to prove it. Can someone help me?
Thanks a lot!
Let $\pi \colon M \to S^1$ denote the usual projection and let $C$ denote a non-zero homologous simple closed curve with no self intersections, i.e. an injective map $c$ from $S^1$ to $M$, which induces something non-zero on homology.
We can isotop $c$ s.th. $\pi \circ c$ is a covering of $S^1$ (This is not that hard, you just make the map locally injective by pushing half-circles in).
We think of the Möbius band as a quotient of the square and assume further that $c$ is constant along a neighborhood of $\{1\} \times I$. The number $n$ of preimages of $1$ under $\pi \circ c$ is the absolute value of the degree in this case. Let's distinguish two cases: Firstly if $n$ is odd, we can isotop $c$ s.th the middle point in the preimage is exactly $([1],[1/2])$, but this gives us a contradiction if $n \geq 2$, because then it follows that the image is not connected, because there is no arc from the other points in the preimage to $([1],[1/2])$.
If $n$ is even we can isotop $c$ s.th. the outer two points in the preimages differ exactly by $1/2$, and this gives a contradiction if $n \neq 2$ regarding connectedness in a similiar way as the other case, because then there is no arc connecting the outer two points to the other point in the preimage.