# Topology of Mobius Ring

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.