Space homeomorphic to Mobius Strip? I have a space $G$ of distinct pairs of points that are not ordered on  $S^1$ with metric:
$D = min(d(a,b) + d(a', b') + d(a, b') + d(a', b))$ 
Is $G$ homeomorphic to a mobius strip?
This has been a problem that I have been stuck on for the past week, would love an explanation/proof of this if possible. Thanks. :) 
 A: Here's a proof in pictures without the pictures.
First, the space of ordered pairs on $S^1 \times S^1$ which is the torus. Let's model the torus as $[0,1] \times [0,1]$ with opposite faces identified: $(x,0)\sim (x,1)$ and $(0,y) \sim (1,y)$.
Now we want to get rid of equal pairs $(x,x)$ points in $S^1 \times S^1$, afterwards we'll get rid of the ordering. But, we'll carry this all out on $[0,1] \times [0,1]$. 
We'll remove from the torus its diagonal line $y=x$, and we'll also remove the two vertices $(0,1)$ and $(1,0)$ since those get identified with $(0,0)$ and $(1,1)$ on the diagonal line. So, what we are left with after this removal are two triangles each with some of its boundary data "deleted": the "lower-right" triangle $(0,0),(1,0),(1,1)$ with all three vertices deleted and its long side deleted; and the "upper-left" triangle $(0,0), (0,1), (1,1)$ also with all three vertices deleted and with its long side deleted.
To review what we have so far, these union of these two "deleted" triangles is identified with your space $G$ in a many-to-one fashion. In particular: each point $(a,b)$ in the lower right triangle is identified in $G$ with the point $(b,a)$ in the upper left triangle. So, we can simply remove one of those two triangles from our scenario. Let's remove the upper left triangle and keep the lower right triangle.
What we have now is a gluing diagram for $G$: in the lower right "deleted" triangle $(0,0)$, $(1,0)$, $(1,1)$, its lower side $(0,1) \times 0$ is identified with its right side $1 \times (0,1)$. If we can figure out exactly how this identifiction is carried out, we'll be able to see $G$ as a Möbius band.
Consider a point $(x,0)$ on the lower side of the lower right triangle. That point is at first identified with $(x,1)$ on the upper side of the upper-left triangle, and then identified with $(1,x)$ on right side of the lower right triangle. So the gluing of those lower side and the right side is specified by putting an arrow on the lower side that points rightwards, and an arrow on the right side that points upwards. 
You may already be able to see this gluing diagram as producing a Möbius band, but I'll go one more step to be sure. Keep in mind: this is an open Möbius band, i.e. a Möbius band missing its boundary circle.
So, the lower right triangle is missing its long diagonal side from $(0,0)$ to $(1,1)$ and it is missing its as well as its lower right vertex $(1,0)$. The missing point $(1,0)$ on the lower right triangle can be stretttttccchhhhhhed out, so that the triangle becomes a rectangle missing two opposite sides; perhaps you can visualize this as the rectangle with vertices $(0,0)$, $(1,1)$, $(3/2,1/2)$, $(1/2,-1/2)$, in which its two "positive slope" sides have been deleted, and its two "negative slope" sides have been retained. When this stretching happens, the rightward pointing arrow on the bottom side of the triangle becomes a down-right pointing arrow from $(0,0)$ to $(1/2,-1/2)$. Also, the upper arrow on the right side of the triangle becomes an up-left pointing arrow from $(3/2,1/2)$ to $(1,1)$. Now it should be pretty clear: the quotient is a Möbius band.
