Why aren't the “higher twist” Möbius bands distinct line-bundles over $S^1$?

It is well known (using for instance sheaf cohomology) that there exist only two possible one-$$\mathbb R$$-dimensional vector bundles over $$S^1$$: the trivial bundle and the Möbius bundle. But what about the Möbius bands with not one, but $$n\in \mathbb N$$ half-twists? It is not obvious to me why these are not distinct line bundles. I am also not entirely comfortable with sheaf cohomology arguments.

This is also physically motivated; if you make a Möbius band, it's easy to see that two half-twists don't "cancel" to make a trivial bundle; so why do they "cancel" topologically?

• If I cut my Möbius band along a line, along a fiber, I can unwind two half-twists and then reglue my Möbius band back together. Because nothing changes along the cutting line, this is a homeomorphism of these two spaces. – Santana Afton Apr 25 at 3:38
• The point is that your picture of the trivial bundle and the two half twist depends on an embedding in $\mathbb{R}^3$. Concretely, let $X_0=\mathbb{S^1}\times\mathbb{R}$ be the trivial bundle and $i_0:X_0\to\mathbb{R}^3$ an embedding as a cylinder. Let $i_1:X_1\to\mathbb{R}^3$ an embedding of the two half twist bundle. Then there is no topological space $X$ with maps $p:X\to[0,1]$ and $i:X\to\mathbb{R}^3$ such that for all $t$, $X_t:=p^{-1}(t)$ is homeomorphic to the total space of a line bundle, and $i_t=i|_{X_t}:X_t\to\mathbb{R}^3$ is an embedding. But such a thing exists in $\mathbb{R}^4$ ! – Roland Apr 25 at 8:03
• The key word here is isotopy : en.wikipedia.org/wiki/Homotopy#Isotopy – Roland Apr 25 at 8:05

As Roland explained in his comments, you have to distinguish between a line bundle $$L$$ over $$S^1$$ and embeddings of $$L$$ into $$\mathbb R ^3$$. In fact, any subspace $$L' \subset \mathbb R ^3$$ which is homeomorphic to $$L$$ is a geometric instantiation of $$L$$, and it may be impossible to deform (inside $$\mathbb R ^3$$) one instantiation $$L'_1$$ into another instantiation $$L'_2$$ without rupturing. The mathematical concept of an "admissible" deformation process inside an ambient space is isotopy.
Take for example the trivial line bundle $$L = S^1 \times \mathbb R$$. Since $$S^1 \subset \mathbb R ^2$$, it is a genuine (unbounded) subset of $$\mathbb R ^3$$. However, there are many other instantiations of $$L$$, for example $$L' = S^1 \times (-\frac{1}{4},\frac{1}{4})$$ or twisted versions $$L'_n$$ of $$L'$$ (with $$n$$ full twists). All these are homeomorphic copies of $$L$$, but no two of them can be deformed into each other.
As an analogue you may also consider the circle $$S^1$$. A homeomorphic copy of $$S^1$$ in $$\mathbb R ^3$$ is a knot, and knot theory is a highly non-trivial field devoted to the classification of knots.