The existence of a square!

Question

My question is the following:

There exists an homeomorphism from a Moebius band to a subset of $\mathbb{R}^3$ such that $\partial M$ is mapped in a continuous closed simple curve of $\mathbb{R}^3$ contained in a plane?

I've done this question because I would to prove the following interesting property:

I think that this problem is well known, but I don't understand how to prove it. Let $\gamma=(\gamma_1,\gamma_2)$ be a continuous closed simple curve of $\mathbb{R}^2$, so $\gamma:[0,1]\to \mathbb{R}^2$ is injective in $(0,1)$ , $\gamma(0)=\gamma(1)$ and it's a continuous function.

The question is if there exists $4$ points $A,B,C,D$ on this curve for which the polygon $ABCD$ results to be a square.

I guess is possible to prove a weaker thesis in a simple way, substituting the existence of a square with the existence of a rectangle.

We know that $ABCD$ would be a rectangle if and only if $AB\cap CD=\{M\}$ where $M$ is the middle point of $AB$ and $CD$, and $AB\cong CD$. Now we can interpret this property in this way:

We define the following map $F:[0,1]\times [0,1]\to \mathbb{R}^3$ that maps each couple $(a,b)$ to

$F(a,b):=\left(\frac{\gamma_1(a)+\gamma_1(b)}{2},\frac{\gamma_2(a)+\gamma_2(b)}{2}, ||\gamma(a)\gamma(b)||\right)$

This map associate to each couple of points $A$ and $B$ on the curve $\gamma$, their middle point $M$ and the distance between them.

We can observe that this map is continuous and it holds the following property:

1. $F(0,b)=F(1,b)$;
2. $F(a,0)=F(a,1)$;
3. $F(a,b)=F(b,a)$;
4. $F(a,a)=(\gamma(a),0)$;

We define the following equivalence relation $\sim$ on $[0,1]\times [0,1]$:

for each $(a,b), (a',b')$ we say $a\sim b \iff (a,b)=(a',b')$ or $b=b'$, $a=0,a'=1$ or $a=a'$, $b=0,b'=1$ or $a=b'$,$b=a'$

It's easy to check that $[0,1]\times[0,1]/\sim$ is the Moebius band $M$ and it's boundary is the image with respect the projection map of the diagonal $\Delta:=\{(a,a): a\in [0,1]\}$.

Now we observe that the map $F$ with that $4$ property induces a natural map $F^\sim : M\to \mathbb{R}^3$ defined in the following way:

$F^\sim([(a,b)]):=F(a,b)$

Is clear that $F^\sim$ is a continuous map.

Moreover we have $F^\sim([(a,a)])=(\gamma(a),0)\in \gamma([0,1])\times \{0\}$, thus the boundary of the Moebius band $M$ is mapped by $F^\sim$ in a simple closed continuous curve of $\mathbb{R}^3$ contained in the plane $\{z=0\}$.

We observe that the existence of the rectangle $ABCD$ on the curve $\gamma$ is equivalent to say that the map $F^\sim$ is not injective.

Arguing by contradiction, if $F^\sim$ would be injective, then $F^\sim: M\to F^\sim(M)\subseteq \mathbb{R}^3$ would be an homeomorphism because $M$ is compact and $\mathbb{R}^3$ is Housdorff, that means $F^\sim$ is also a closed map.

Now the question is:

There exists an homeomorphism from $M$ to a subset of $\mathbb{R}^3$ such that $\partial M$ is mapped in a continuous closed simple curve of $\mathbb{R}^3$ contained in a plane?

I think that the answer is no, but I don't understand how to prove it.

Answer 1

The answer to your question is yes. There even exists a diffeomorphism from the Möbius strip $M$ to $\mathbb R^3$ such that $\partial M$ is the unit circle of the $xy$-plane (or any other unknot for that matter). Abstractly, this is an immediate consequence of the Smooth Isotopy Extension Theorem, which in particular implies that if any a smooth isotopy $f\colon S^1\times[0,1]\to\mathbb R^3$ can be extended to a smooth isotopy $F\colon\mathbb R^3\times[0,1]\to\mathbb R^3$.

One can perform this construction explicitly, which results in the Sudanese Möbius strip.