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.
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.
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.