How to write the real projective plane as a pushout of a disk and the mobius strip?

Question

I heard in topology class that the real projective plane is obtained by gluing a disk along the boundary of the mobius strip. I was wondering - how can I write this as a pushout?

Also, how can I write a mobius strip itself as a pushout?

Answer 1

We have two copies of $S^1$ here: the $S^1$ that is the boundary of $D^2$, and the $S^1$ that forms the boundary of the Möbius band.

$$\require{AMScd} \begin{CD}S^1 @>>> D^2 \\ @VVV @VVV \\ M @>>> \mathbb{RP^2}\end{CD}$$

Actually showing that these are the same might take a little more work, depending on how familiar you are with manipulating these spaces! Perhaps the easiest way to see it is to do the gluing and then shrink $M$ onto the middle of the band - after you've done this you see you've identified antipodal points on the boundary of $D^2$, which is $\mathbb{RP}^2$ as desired.

I don't know how you'd write a Möbius strip as a pushout - I can't see how you can express it as gluing two separate topological spaces together, which is what a pushout does. (Normally you obtain the Möbius strip as a quotient space of the unit square $I^2$.)

Answer 2

The Möbius strip can be built like this:

$$\require{AMScd} \begin{CD}S^0 \times [0,1] @>>> [0,1] \times [0,1] \\ @VVV @VVV \\ [0,1] \times [0,1] @>>> M\end{CD}$$

The map across the top send one of the $[0,1]$'s to the left edge of the square and the other to the right edge. The map going down does the exact same to one of the $[0,1]$'s, and flips the other to go in the opposite direction.