Intuition behind equivalence of two identifications obtained from Möbius band.

Question

Consider the real projective plane $\Bbb RP^2$ which can be realized as a quotient space obtained from a square by identifying points on each pair of it's opposite edges in reverse order. It has been claimed (without any explicit reasoning) in the lecture note I am following that this space is homeomorphic to the Möbius band with it's edge circle collapsed to a point. But what I can see from the identification that I am familiar with that it is obtained from the Möbius band by identifying the points on it's edge circle in a specific way. Namely after getting a Möbius band by identifying points on one of the pairs of opposite edges of a square in reverse order we need to now identify the points on the unidentified opposite edges (which form the edge circle of the Möbius band) in reverse order. But I don't understand how does this space homeomorphic to the quotient space obtained from the Möbius band by collapsing all the points of it's edge circle to a point. Would anybody please shed some light on it?

Thanks in advance.

Answer 1

Most of these surfaces have multiple different ways to obtain them, and one may might not be as good for one purpose as another way.

In the case of the Möbius band, the way that helps for this problem is to imagine the square with a hole cut out of it near the center. Now, as you say, identify points on each pair of opposite sides of the square in reverse order. Keeping in mind that hole in the middle, the space you get by these identfications will be homeomorphic to the Möbius band (you can perhaps to convince yourself of that by making some additional cuts and pastes).

Observe also that when you identify the entire boundary circle of the hole to a single point, you get the square itself.

So, in a few steps you get the desired homeomorphism: start with:

1. The Möbius band with its circle identified to a point;

that is homeomorphic to

2. The square with a hole, and with the boundary circle of that hole identified to a point, and with its opposite sides identified in reverse order;

and that is homeomorphic to

3. the square (no hole) with its opposite sides identified in reverse order;

and that is the projective plane.

Answer 2

So you start with a square and want to paste together its opposing sides in reverse directions. The corners $A$ and $A'$ are pasted together, as are the corners $B$ and $B'$.

Suppose we were to try actually doing this physically. Our square is made of the famous Topological goo, that stretches as much as you want, but never tears. Pick up the corners $A$ and $A'$ and pinch them together at the top. Let the four sides hang down from the combined $A$. Pushing the center of the square down a little to get it out of the way, we can also bring $B$ and $B'$ together below $A$.

So we have four curves hanging down from $A$ separately until they meet again at $B$. The surface runs up between two curves on the one side and the other two curves on the other side, but the regions between them are still open. Take two of the curves that are diametrically opposite and bring them together in the middle, pasting them together. Finally we need to paste the other two curves. Unfortunately the path through the middle is now blocked by the surface formed by the first pasting. So we push the two curves a little into the $4^{th}$ dimension to get them around the blockage. There we can past them together to finish our projective plane.

The result is commonly called a Crosscap, because it somewhat resembles the shape of a Bishop's miter cap. From our view, it looks pinched with $A$ and $B$ being special points different from the rest. But remember this is just a projection into 3d space. In reality the pinch points extend into a fourth dimension in some directions, undoing the pinch. The projective plane is actually homogenous: the neighborhoods of all points look the same.

Now poke a hole in the cross cap: choose some arbitrary point and remove a small hole about it. If we take the boundary of this hole and collapse all its points back to a single point, we end up back where we started: the crosscap.

Since all the points on the projective plane are the same, it doesn't matter where we create the hole. So let's create it on the intersecting line on our 3D projection of the crosscap. It will extend on one of the crossing surfaces. Expand the hole until it includes $A$ and $B$ on its boundary. We no longer need the 4th dimension to avoid intersection. One surface goes through the hole to connect to its other side. The intersecting surface is removed to make the hole.

We are now back exactly where we were when we pasted together the first two curves to make the crosscap. That is, going back to the square, we've pasted together two opposite sides of the square in reversed directions, but have not yet pasted the other pair of opposite sides. That is, we have made a Möbius strip.