# Vector space structure on fibre of line bundle

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.

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 $$(2/3,5/6)$$.