When we create a Mobius loop from a square with side identifications, are we treating the square as having 2 faces? I am considering a Mobius loop as the quotient space of $([0,1] \times [0,1])/R$, where $(0,y)R(1,1-y)$ is the equivalence relation. My question is whether we are treating the initial (before side identifications) square (or 4-gon) as having both faces (i.e. if we imagine looking at it from behind the page), or if we only consider the front face.
To me it seems that if we treat the square as only having one face, then the quotient space that we create is in some sense half of a Mobius loop. That is, if we shaded the face of the square, only half of the Mobius loop would be shaded.
Specifically I am thinking about the inclusion map from $S^1$ the unit circle in $\Bbb R^2$ into the Mobius strip, given by $i(e^{(2 \pi i x)}) = (x, \frac 12)$, as part of a homotopy retract. The retraction map in this case is $r: M \to i(S^1)$, $r((x,y)) = (x, \frac 12)$.
To me it seems like this does only consider one side of the Mobius loop.
Mobius loop side identifications
 A: It's one-sided. The space $[0, 1] \times [0, 1]$ is a subset of the plane $\Bbb{R}^2$, and has only one "side". You can't have, for example, two distinct points at $[0.1, 0.2]$, one on one "side" and the other on the other.
In this way, despite the prima facie similarity in construction, the standard identification you're describing differs a little from the paper and glue construction using a paper strip. A strip of paper is really more like two rectangles pressed together, one for each side. It could be more accurately modelled by two rectangles, say $[0, 1] \times [0, 1] \times \{0\}$ and $[0, 1] \times [0, 1] \times \{1\}$. That way, the two "sides" can indeed contain independent elements.
How is the Mobius strip constructed in this way? We need to make two identifications now, taking each parallel edge of each side, and identifying it with an edge of the other side. If you want a visual aid, I suggest cutting a strip of paper, labelling the sides $0$ and $1$, then labelling each corner on each side with one of the four points $(0, 0), (0, 1), (1, 1), (1, 0)$. How you label these corners will determine the exact identifications you're using, so there are multiple possibilities.
Let's say that you're labelling each corner the same as on the other side of the paper (e.g. the physical corner of the paper that says $(0, 0)$ on side $0$ will also say $(0, 0)$ on side $1$), and we're twisting the edge with $(1, 0)$ and $(1, 1)$ upside down. At this point, I really recommend making and labelling a paper model!
In this case, we are identifying $(1, 1, 0)$ with $(0, 0, 1)$ and $(1, 0, 0)$ with $(0, 1, 1)$. Simultaneously, on the other side, $(1, 1, 1)$ maps to $(0, 0, 0)$ and $(1, 0, 1)$ maps to $(0, 1, 0)$.
This produces two pairs of identified lines: we identify the point $(1, y, 0)$ with $(0, 1 - y, 1)$ for all $y \in [0, 1]$, as well as $(0, y, 0)$ with $(1, 1 - y, 1)$ for all $y \in [0, 1]$. Performing these identifications, we get more of the "two-sided" rectangle construction that we get with paper and glue.
Does it produce the same space? Absolutely. Just think of the two "sides" adjoined along one of the edges (perhaps with one "side" mentally turned upside down so that the two sides next to each other look nice and "flat"). What do we get? We get a wider flat rectangle, which is, of course, homeomorphic to $[0, 1] \times [0, 1]$. Performing the final identification corresponds to the standard identification on $[0, 1] \times [0, 1]$, and produces the same space up to homeomorphism.
A: The short answer is: A square doesn’t have faces.I’ll explain that but at first it may seem I’m going off on a tangent, so you’ll have to bear with me.
It is usual to say that a square has four sides. In this case side means edge. It is usual to say that a cube has six sides. In this case we do not mean edges. We mean faces or surfaces. These surfaces are square and they have two sides – an inside and an outside. In this case side means adjacent area. I use the word side in this sense when I say there is a tree on the west side of my house. There is an area adjacent to each square surface that is inside the cube and an adjacent area that is outside the cube. A square surface is a face of a cube. It is a face and it does not have faces. The areas adjacent to the squares are not faces. If a face had faces, then those faces would have faces and a square would have an infinite number of faces.
Let me approach this differently. You place a square book on a table. The book is in contact with the table; there is nothing between the book and the table (at least in theory). This nothing is something. It is a two dimensional space that has length and width and no thickness. It is an actual square. This square had two sides. It has an adjacent area were the book is and an adjacent area were the table is. If you are going to shade one side of this square, you are going to shade either the book or the table. If you want to shade the square itself there is nothing to shade. There is nothing you can draw on, or paint on. Remember this square has no thickness; you can’t divide it through its thickness into two squares; so even if you could somehow shade a square, you can’t do it selectively, it’s all or nothing.
Now let’s consider a Mobius strip. To do that, we’ll make one. Start with two strips of paper, identical in size but of different colors. Place one on top of the other. Between them is a two dimensional space. Keep them together and give them a half twist and join the ends. The space between the strips becomes an actual Mobius strip and the two strips join together to become one long strip that surrounds the Mobius strip. The Mobius strip has only one side because it has only one adjacent area. That area is occupied by the two-colored strip. Like a square a Mobius strip is a two dimensional surface. Unlike a square it cannot be the surface or face of anything in three dimensional space. It can, however, be a surface or face in four dimensional space.
The unit square you make a Mobius strip from does not have sides in the sense of faces. It has sides in the sense of edges and it has sides in the sense of adjacent areas. If you keep that in mind, your problem goes away.
