Is $RP^2$ homeomorphic to Mobius Band quotient its boundary circle?

Question

If $M = I^2/[(0,x)\sim(1,1-x)]$ is Mobius Band and $C$ = it's boundary circle, then it can be easily shown (by taking CW-structure) that $M/C$ is homotopy-equivalent to $RP^2$ = $S^2/(\text{andipodal points identified})$. Are these homeomorpic? Usual invariants (like compactness, connectedness which are not invariants under homotopy) don't help. Any hint or idea is appreciated.

Answer 1

Yes, they are.

$RP^2$ is a Moebius Band with attached a disc along the boundary, and you can "easily" see that collapsing the boundary is equivalent to attach a disc.

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As usual, in topology, things that are "easy to see" requires long proofs. Here is one.

The main tools for dealing with this kind of problems are:

A) Quotient map theorem: if you have a map $f:X\to Y$, an equivalence relation on $X$ such that $x\sim x'\Rightarrow f(x)=f(x')$, then it induces a map $F:X/\sim\to Y$. If $X,Y$ are topogical spaces and $X/\sim$ is endowed with the quotient topology, and if $f$ is continuous, then the quotient map is continuous. (See also here)

B) The compact Haustorff Theroem: if $f:X\to Y$ is a continuous bijection between a compact space $X$ and a Hausdorff space $Y$, then $f$ is a homeomorphism. (See also here)

C) If $X/\sim$ is Hausdorf anf $Y\subseteq X$ is compact and intersects all equivalence classes, then $X/\sim$ is homeo to $Y/\sim$. This is just Just by A)+B).

Proof of that a moebius band with boundary collapsed is $RP^2$:

Step 1) $RP^2$ is homeomorphic to $S^2$ with antipodal points indentifyied: Since $S^2$ sits in $R^3$, and the relation that defines the projective plane restricts to the identification of antipodals, then, by C) we get that $RP^2$, which is $R^3\setminus\{0\}/\sim$, is homeo to $S^2/\sim$.

Step 2) Let $S^1=\{x^2+y^2=1\}=\{(\cos\theta,\sin\theta): \theta\in [0,2\pi]\}$ be identified with $[0,2\pi]$ with $0=2\pi$. The moebius band is homeomorphic to the quotient of $X=S^1\times [-1,1]$ by the equivalence relation that identifies $(x,y)$ with $(x+\pi,-y)$ (where the sum $x+\pi$ is taken modulo $2\pi$: The set $Y=[0,\pi]\times [-1,1]$ is compact and intersect all equivalence classes, so by $C)$ we have $S^1\times [-1,1]$ is homeo to $Y/\sim$. Now $Y$ is homeo to $I^2$ and the equivalence relation is the one that give the moebius band.

Step 3) Let $X$ be the space of step 2). Identifying $S^1$ as the unit circle in $R^2$, we have that $X$ is a subset of $R^3$. The map $X\to R^3$ given by $$f(x,y,t)=(x\cos(\pi t),y\cos(\pi t),\sin(\pi t))$$ has image in $S^2$, and if we compose with the antipodal relation of $S^2$, provides a map $F:X\to RP^2$, which is continuous. Moreover, by construction the map is constant on equivalence classes, so by A) this gives a map $F:X/\sim\to RP^2$ which is continuous. Finally, it is readly checked that that map is bijective. Since $X$ is compat, then $X/\sim$ is compact. So B) gives that our map is a homeo. Since $X/\sim$ is homeomorphic to the moebius band, than you have the desired claim.

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Generalisation: In general, you can prove by means of A) and B) that

Theorem: if you have a surface $M$, you remove the interior of an embedded closed disc, and you quotient the result by collapsing the boundary to a point, you obtain a space which is homeo to $M$ itself.

So if you accept the fact that removing the interior of an embedded closed disc from $RP^2$ you have the moubius strip, then you can apply that theorem.

In fact, Steps 1), 2) provides exactly that $RP^2$ minus a disc gives the moebius band, while step 3) is the one that can be generealised to the theorem.

Answer 2

Just want to give a very clean proof to this rather old post.

We prove first the following . The closed disc with antipodal boundary points identified is homeomorphic to $\Bbb{RP}^{2}$ .

To do this we identify $D^{2}$ , the closed unit disc with the upper hemisphere with the equatorial circle. So as to not confuse, let us denote this "$D^{2}$" as $N$ (N stands for the northern hemisphere).

Now $\Bbb{RP}^{2}$ is $S^{2}$ with antipodal points identified. So let $q$ denote the quotient map from $S^{2}\to S^{2}/\sim$ where $\sim$ denotes the antipodal identification. And henceforth , by $\Bbb{RP}^{2}$ we will mean $S^{2}/\sim$

We now restrict this quotient map $q$ to $N$ and call $q|_{N}=Q$ .

So $Q:N\to \Bbb{RP}^{2} $ is a continuous map. Now $\{Q^{-1}[x]\}=\{x\}$ if $x$ lies on the upper hemisphere but not on the equatorial circle and $\{Q^{-1}[x]\}=\{x,-x\}$ if $x$ is on the equatorial circle.

Thus , if $p$ denotes the quotient map from $N$ to $N/\sim$ such that $p$ identifies the antipodal points on the equatorial circle, then we have that $Q$ descends to a unique bijection $\overline{Q}$ from the quotient $N/\sim\,\,\to \Bbb{RP}^{2}$ by property of quotient topology.

And thus $\overline{Q}$ is directly a homemorphism as it is a continuous bijection from a compact space to Hausdorff space.

Now a Mobius Band with boundary identified is just a closed disc with antipodal points identified and hence we are done. To see this , we have the following picture. And thus we are done .