Vector space structure on fibre of line bundle

Question

Let $I=[0,1]$ and $E=I \times \mathbb{R}/\sim,$ where $\sim$ identifies $(0,t) \sim (1,-t)$. The projection $I \times \mathbb{R} \to I$ induces $p:E \to S^1$. This is supposed to be a $1$-dimensional vector bundle, but I don't see what structure the fibres are given.

If I am not mistaken, the fibres of $[v] \in B$ are $p^{-1}([v])=\{[(v,w)] \in E \vert w \in \mathbb{R}\}$ for every $v$, not only $v \notin \{0,1\}$, where $B$ is $I/\sim$ with $0 \sim 1$. I also see that there is a bijection $p^{-1}([v]) \to \mathbb{R}$ for $v \neq 0,1$ which can induce a vector space structure, but I don't see how this is possible for $v=0,1$. Perhaps I made a mistake or the vector space structure is even more obvious and I just can't see it, but I am stuck here.

Answer 1

Since this is a quotient construction, we can look at the vector properties through local trivializations. Take a covering of the base by, say $B_1$ the image of $(1/6, 5/6)$, and $B_2$ the image of $[0,1/3)\cup(2/3,1]$. The pullback of each $B_j$ is $B_j\times\mathbb{R}$. The map $B_1\times\mathbb{R}\to E$ sends the fibers right side up (can I be informal?) and the map $B_2\times\mathbb{R}\to E$ sends the fibers over $[0,1/3)$ right side up and those over $(2/3,1]$ upside down. The fiber isomorphisms over $B_1\cap B_2$ are then multiplication by one over $(1/6,1/3)$ and by $-1$ over $(2/3,5/6)$.