# Which space do we obtain if we take a Möbius strip and identify its boundary circle to a point?

I know that the boundary circle of a Möbius strip is actually formed by the horizontal sides of $$[0 ,1] \times [0,1]$$.If we identify all the points of the 1st horizontal side to a single point and do the same for the second horizontal side we get a disc with the antipodal points identified to itself(which is $$\mathbb{RP}^2$$).

However I am not sure of how to write it down rigorously.

$$\require{AMScd}$$ $$\begin{CD} Mobius strip @>i(x)>> [0,1] \times[0,1] @>g(x)>> D @>f(x)>> D/\sim_1\\ @VV\pi(x) = cl(x)V \\ Mobius Strip / \sim \\ \end{CD}$$ where $$\sim$$ denotes identifying the boundary cirlce of a Mobius strip and $$\sim_1$$ denotes identifying the opposite points of the disc.

I am not sure of what $$g(x)$$ is ? or the fact that mobius strip will be compact or the composition of $$f \circ g \circ i(x)$$ will be onto?

Any other method will be appreciated but I wanted to know whether the mobius strip is compact?

One approach is to lift everything to double covers in Cartesian three-space. Here's a detailed sketch:

• Let $$S = \{(x, y, z) : x^{2}+ y^{2} + z^{2} = 1\}$$ be the unit sphere, and $$S_{0}$$ the complement of the north and south poles $$\pm(0, 0, 1)$$.
• Let $$C = \{(x, y, z) : x^{2}+ y^{2} = 1, |z| \leq 1\}$$ be the closed cylinder circumscribed about $$S$$, and $$C_{0}$$ the open cylinder with the boundary removed, i.e., with $$|z| < 1$$.

Radial projection toward the $$z$$-axis maps $$C_{0}$$ to $$S_{0}$$, and maps the respective boundary circles of $$C$$ to the north pole and south pole of $$S$$.

The antipodal map $$(x, y, z) \mapsto (-x, -y, -z)$$ of three-space defines an involution on each surface; the quotient of $$C_{0}$$ (and of $$S_{0}$$, if it matters) is an open Möbius strip (non-compact), the quotient $$C/\sim$$ is a closed Möbius strip (compact as a continuous image of $$C$$, which is compact), and the quotient $$S/\sim$$ is the real projective plane.

Radial projection toward the $$z$$-axis induces the desired mapping $$C/\sim \to S/\sim$$.

• D.Hwang thank you for the answer .Do you think I can continue from where I left? Jul 27, 2021 at 12:54
• You mean via some more-or-less explicit homeomorphism from the square (perhaps with two opposite sides identified to points) and the disk? Without passing to the double covers or using the double cover as motivation? Something else...? <> Whether or no, writing down an explicit mapping between the square and a disk looks unnecessarily complicated, but is possible. Jul 27, 2021 at 17:00

I think there is mistake in your specified definition of $$\mathbb{R}\mathbb{P}^2$$. Here, $$\mathbb{R}\mathbb{P}^2=\frac{S^2}{Antipodals}$$ ($$S^2$$ is a sphere in $$\mathbb{R}^3$$). Now, I am just trying to clarify your approach, though I am not sure if this huge composition of maps will work or not,

Here, you are using the following theorem, $$\require{AMScd}$$ $$\begin{CD} X @>f>> Y \\ @V\pi VV \\ X / \sim \end{CD}$$ If $$f$$ is an identification then $$Y$$ is homeomorphic with $$X/\sim$$ (where $$\pi(x) = cl(x)$$). Here, $$\sim$$ is an eq. relation using the partition made by $$f$$.

For this problem I may draw the picture like, $$\require{AMScd}$$ $$\begin{CD} X =I\times I @>f_1(x)=cl(x)>> S^1\times S^1@>f_2(x)=cl(x)>> S^1\times S^1/\sim_3=S^2 @>f_3(x)=cl(x)>> S^2/\sim\ =\mathbb{R}\mathbb{P}^2\\ @V\pi_1(x) = cl(x)VV \\ I\times I / \sim_1 \\ @V\pi_2(x) = cl(x)VV \\ Möbius Strip / \sim_2 \end{CD}$$ where,
$$X =I\times I = [0,1]\times [0,1]$$.
$$\sim$$ = Identifying the antipodal points.
$$\sim_1$$ = Usual eq. relation to make Möbius.
$$\sim_2$$ = Identifying the points on the boundary circle of the Möbius strip.
$$\sim_3$$ is an eq. relation such that, $$S^1\times S^1/\sim_3 \ \cong \ S^1\times S^1/ S^1\vee S^1$$
$$f=f_1\circ f_2\circ f_3$$ is an identification from $$I\times I$$ to $$\mathbb{R}\mathbb{P}^2$$ and
$$\pi=\pi_1\circ \pi_2$$ is an identification from $$I\times I$$ to $$Möbius Strip / \sim_2$$.

But here to prove $$Möbius Strip / \sim_2 \ \cong \ \mathbb{R}\mathbb{P}^2$$, we need to show $$\pi$$ is the identification map using the partition made by $$f$$. I think this will not hold for this maps, I am not sure. This maps may clarify your approach that you are asking for, but may not serve the purpose.