Show that the cylinder is not ambient isotopic to the Mobius band.

Question

Here is my definition for ambient isotopy:

We say if there is an orientation preserving piecewise linear homeomorphism $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ (or replace $\mathbb{R}^3$ with $S^3$) for two knots, say, $K_1$ and $K_2$ then the two knots are $\textit{equivalent}$, or $f$ is an $\textit{ambient isotopy}$, such that $f(K_1)=K_2$.

Obviously it doesn't have to be in terms of knots specifically. I want to show why the cylinder (which is a closed ribbon with an odd number of twists including $0$) is not ambient isotopic to the Mobius band (which is a closed ribbon with an odd number of twists).

Answer 1

Hint: Consider what would happen to the boundary.