What exactly is a vector bundle isomorphism Recall that a vector bundle (of rank $n$) is a family of vector spaces $V_x$ of dimension $n$ that is parameterized by a topological space $X$. In addition there is a continuous surjective map $\pi: E \to X$ where $E$ is also a topological space. On top of this it is required that for every $x \in X$ there be an open set $x \in U$, some $k$ and a homeomorphism $\varphi: U \times \mathbb R^k \to \pi^{-1}(U)$. (local triviality condition)
Two vector bundles are isomorphic if there is a homeomorphism on the total spaces $\phi : E \to E'$ with the property that $\phi|_{\pi^{-1}(x)}: \pi^{-1}(x) \to \pi'^{-1}(x)$ is a linear isomorphism.
Consider the example $X = S^1, E = \mathbb R^2$, $k=1$. From reading about vector bundles I know that there exist only two non-isomorphic vector bundles: the Moebius band and the annulus. But why this is so I don't understand: it's not clear to me why these two can't be isomorphic given the definition above and I also don't see why there are no others. To illustrate why I don't understand why there can't be others consider the vector bundle that "twists around" twice. (If the Moebius band is "twisting around" once and the annulus zero times). It's not clear to me why this is isomorphic to either the annulus or the Moebius band. Also why is it a twist that distinguishes isomorphic bundles? Slightly warping the fibres still leaves the bundles isomorphic: how does it work?
 A: The annulus and the Möbius band can't be isomorphic because the total spaces are not homeomorphic - the annulus is orientable, but the Möbius band is not. An alternative proof is that the annulus admits nowhere vanishing sections, while the Möbius band doesn't (essentially by the intermediate value theorem).
It's not so easy (to me) to quickly justify why these are the only possibilities, but I claim that for any two line bundles $E$ and $E'$ on the the map $E\to E'$ given by mapping each $(x,v)$ in $E$ to the point $(x,v)'$ in $E'$ with the same coordinates* is a homeomorphism exactly when the "number of twists" has the same parity for both $E$ and $E'$.
*I'm not sure this explanation is very clear - what I mean is that once you have identified each fibre with $\mathbb{R}$ via the local trivialization, the points in both $E$ and $E'$ look like pairs $(x,v)$ for $x\in S^1$ and $v\in\mathbb{R}$, but this means something different depending on which bundle you are in.
A: The doubly-twisted annulus is isomorphic to the annulus (in general, twisted annuli are isomorphic if their twist numbers are equivalent mod 2) because it is allowed to "pass through itself" on the way - in other words, the particular embedding of these objects in $\mathbb{R}^3$ is irrelevant. The motion that turns a doubly-twisted annulus into an annulus seems counterintuitive only because, if we were (unnecessarily) restricting ourselves to 3-space, it would have to pass through itself.
Lastly, the reason why a twist is something that changes the isomorphism class of the bundle is that it is a global change. In fact, the very idea of "bundle" means that two bundles of the same rank on the same space will locally be isomorphic - the only distinguishing features of bundles must be things that take into account what the entire bundle does at once.
