# How to see the real projective plane is a Möbius band glued to a disk?

I am seeking an easily comprehended, convincing explanation that $${RP}^2$$ is topologically the same as gluing the circle boundary of a disk to the edge of a Möbius band.

Let $$D$$ be the closed unit disk in $$\mathbb{R}^2$$, and $$D/\sim$$ the disk with antipodal points on the boundary identified, which is homeomorphic to $$\mathbb{RP}^2$$.

Now decompose $$D$$ into an annulus $$A$$ and a smaller disk, so that attaching a disk to $$A$$ along the inner circle gives you $$D$$.

So, attaching a disk to $$A/ \sim$$ along the inner circle will give you $$(D/\sim) \cong \mathbb{RP}^2$$. If we can show that $$A/\sim$$ is homeomorphic to a Möbius band, we're done.

Here's how we do that.

(The image is from the Oxford Part A Topology lecture notes)

• Thanks! May I ask: Why does "reflect top" maintain $\cong$? Oct 20 '18 at 16:20
• @JosephO'Rourke You're still identifying the same points (e.g. the new left edge of the top, which is the old right edge of the top, still gets glued to the left edge of the bottom). So there's no actual change at all. Oct 20 '18 at 16:41

It's really the same thing as Daniel's answer, but I find this easier to visualize:

Consider $$\mathbb{RP}^2$$ as a quotient of $$S^2$$. The image of the disk $$\{(x,y,z)\in S^2\mid z\geq1/2\}$$ is the disk, and the image of the strip $$\{(x,y,z)\in S^2\mid y\geq0,\,|z|\leq1/2\}$$ is the Möbius strip.

• Very nice! ${}$ Jun 6 at 7:04