Ways to think about vector bundle

A rank $k$ vector bundle over a manifold $M$ can be viewed as a twisted version of the Cartesian product $M \times \Bbb R^k$. Given an appropriate cover $U_\alpha$ of $M$, the transition functions $A_{\alpha\beta}$ tell us how to glue together the local products $U_\alpha \times R^k$ to get the total space $E$. If all transition functions map all overlap points to the identity $I \in \mathrm{GL}(k, \Bbb R)$, then there is no twisting and we just get the Cartesian product $M \times \Bbb R^k$.

One example is as follows. Consider the circle $S^1$, thought of as the set $\Bbb R/2\pi \Bbb Z$. We will construct two rank $1$ vector bundles on $S^1$. Cover $S^1$ by the open sets $U_1 = (0, 2\pi)$ and $U_2 = (-\pi, \pi)$. We have that $$U_1 \cap U_2 = (0, \pi) \cup (\pi, 2\pi).$$ Assume that on the component $(0, \pi)$, the transition function is given by $$A_{12}(x) = 1 \text{ for all } x \in (0, \pi).$$ Since a transition function must be continuous, we see that the only possibilities for $A_{12}|_{(\pi, 2\pi)}$ are \begin{align*} \text{(a)} & \quad A_{12}(x) = 1 \text{ for all } x \in (\pi, 2\pi), \\ \text{(b)} & \quad A_{12}(x) = -1 \text{ for all } x \in (\pi, 2\pi). \end{align*} Case (a) gives the trivial rank $1$ vector bundle over $S^1$, $E = S^1 \times \Bbb R$.. In case (b), we have that the total space $E$ is the (infinite) Möbius band.