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

Question

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?

Answer 1

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 $C/\sim \to S/\sim$.

Answer 2

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.