There is no immersion of the Möbius band in the plane.
There is no immersion of the Möbius band in the plane.
I believe we have to work with the tangent bundle of the Möbius band, but I'm not getting no useful result.
Solution 1:
If $f:M\to\Bbb R^2$ is a differentiable map from the Möbius band to the plane, then for the standard volume form $\omega=dx\wedge dy$ on $R^2$, its pull back via $f$, denoted by $f^*\omega$, induces a $2$-form on $M$. Since $M$ is not orientable, $f^*\omega=0$ at some $p\in M$, so the tangent map of $f$ is degenerate at $p$. That is to say, $f$ cannot be an immersion.