I'm having some difficulty with the finer points of the construction of the topology of the Mobius ring from Tu's manifolds. Given the square $ R = \{(x,y) \in \mathbb{R}^2 | \, 0 \leq x \leq 1, -1 < y < 1\}$, $ R /\sim $ is formed with the relation $ (0,y) \sim (1,-y) $. I understand that the interior points of $ R $ are only related only to themselves (and are mapped into their own equivalence classes by the projection $ \pi: R \rightarrow R / \sim $), and the edges are glued together into equivalence classes of two elements, $ \pi\big( (0,y) \big) = \pi\big( (1,-y) \big) = \{ (0,y),(1,-y) \}, \, \forall y \in \, ]-1,1[ $.
What I don't understand are the open sets of $ R / \sim $. I know that a set $ U \in R/\sim $ is open if $ \pi^{-1}(U) $ is open in $ R $, but I'm trying to consider an open set in $ R/\sim $ that contains the "glued-together" region. This hypothetical region $ V $ contains some of the points in the glued-together equivalence classes as well as points on both "sides" of the interior. When mapped back to $ R $, this region contains parts on both sides of the square, as well as points on both boundary regions, so how can such a region be open in $ R $? I know that the overlap region of $ R/\sim $ shouldn't be seen any differently from any other region $ R/\sim $, but such regions seem to map back to non-open sets in $ R $. Isn't it true that a set containing boundary points of $ R $ (i.e. at $ x = 0 $ and $ x = 1 $) isn't open in $ R $?
The topology of R is the one inherited in R as a subset of $\mathbb{R}^2$. This means that an open set in R can contain parts of the edge of the square, as long as this open set is the intersection of a regular open ball in $\mathbb{R}^2$ with R.
For clarity, picture a ball of small radius and center one of the vertices.