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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?

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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.

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  • $\begingroup$ Shouldn't the top left corner be $[0,1]\amalg [0,1]$? $\endgroup$
    – Arrow
    May 26 '16 at 13:35
  • $\begingroup$ @Arrow Yes, and it is. $S^0$ is the discrete space with two points, so $S^0 \times [0,1] \cong [0,1] \amalg [0,1]$. $\endgroup$
    – wckronholm
    May 26 '16 at 20:36
  • $\begingroup$ I understand they're homeo as spaces, I just thought writing $[0,1]\amalg [0,1]$ would make the maps easier to explain by the universal properties :) $\endgroup$
    – Arrow
    May 26 '16 at 20:40
  • $\begingroup$ @Arrow Feel free to edit the answer if you think it makes it better. $\endgroup$
    – wckronholm
    May 26 '16 at 20:49
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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 $I^2$.)

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  • $\begingroup$ The Möbius strip is obtained by gluing together two antiparallel edges of a square. You can express this as a pushout fairly straightforwardly. $\endgroup$
    – Zhen Lin
    May 5 '16 at 18:21
  • $\begingroup$ Could you show me how this is done? (i.e. what spaces are at the three other corners of the pushout square?) $\endgroup$
    – Josh Hunt
    May 5 '16 at 21:33
  • $\begingroup$ It appears someone else has posted the solution below. $\endgroup$
    – Zhen Lin
    May 6 '16 at 6:20
  • $\begingroup$ @ZhenLin shouldn't the top left corner of this solution be $[0,1]\amalg [0,1]$? $\endgroup$
    – Arrow
    May 26 '16 at 13:34
  • $\begingroup$ That is homeomorphic to $S^0 \times [0, 1]$. $\endgroup$
    – Zhen Lin
    May 26 '16 at 13:44

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