12
$\begingroup$

Let $G:\mathbb R \times [-1,1]\to \mathbb R \times [-1,1]$ be a map defined by $G(x,y)=(x+1,-y)$

This space $Q=\mathbb R\times [-1,1]/\sim$, where $(x_1,y_1)\sim (x_2,y_2)$ if and only if there is $n\in \mathbb Z$ such that $G^n(x_1,y_1)=(x_2,y_2)$ is homeomorphic to the Möbius strip? I'm trying to see this intuitively without success. Anyone has an idea why these spaces are "equal"?

Thanks

EDIT

Following the commentaries, The only map I can imagine from the Möbius strip is the one which send a point $(x,y)$ in the Möbius strip onto $(x,y)$ in $Q$. See the picture below:enter image description here

  • 1
    How do you define the Mobius strip?2012-12-28
  • 0
    @ThomasAndrews in the standard way: $[0,1]\times [-1,1]/\sim$, where $(0,y)\sim (1,-y)$2012-12-28
  • 1
    Can you see a way to map one of these spaces to the other?2012-12-28
  • 0
    No, that's why I'm asking :) I don't get used yet with such spaces2012-12-28
  • 0
    @RafaelChavez: What's the simplest (non-constant) map from $[0,1]\times[-1,1]$ to $\mathbb R\times[-1,1]$ you can imagine? Does it preserve the $\sim$'s?2012-12-28
  • 2
    By the way, your $\mathbb R\times[0,1]$ in the beginning of the second paragraph of the question must be a typo; things only make sense if it is to be $\mathbb R\times[-1,1]$.2012-12-28
  • 0
    @HenningMakholm yes, I've already edited myself, thank you2012-12-28
  • 0
    @HenningMakholm It actually works for $\mathbb R\times[0,1]$, too, but much less obviously. Basically, $\mathbb R\times[0,1]$ under the induced equivalence relationship still works...2012-12-28
  • 0
    @ThomasAndrews: Hm, yes. We get a cylinder of perimeter 2 with the bottom edge glued into a crosscap.2012-12-28
  • 0
    @HenningMakholm Please see the new edit.2012-12-28
  • 0
    @ThomasAndrews The map in the picture I added is correct?2012-12-28
  • 0
    @RafaelChavez: Good. Then you only need to prove that this map induces a bijection of equivalence classes and is a homeomorphism.2012-12-28
  • 0
    @HenningMakholm thank you very much, I will try to do it, and after that I will edit my question again, if I have any troubles.2012-12-28

2 Answers 2

23

I'll try to answer the intuition part. You know how a Möbius strip works with a piece of paper, right?

Visualize $\mathbb{R}\times [-1,1]$. This is the whole real line in the $x$-direction, and the interval $[-1,1]$.

Strip 1.

Now, what parts of this does $\sim$ say are the same? When $y=0$, every $x$ is the same as every $x+n$ for any $n\in \mathbb{N}$. Now, move up a little bit in the $y$-direction, to some $a\in (0,1)$. Now any $(x,a)$ is identified with $(x+1,-a)$. So, if you just look at $A_1:=[-1,0]\times [-1,1]$ and $A_2:=[0,1]\times [-1,1]$, you see that the top of $A_1$ is identified with the bottom of $A_2$.

Identify.

In fact, every $(x,y)$ is identified with $(x+1,-y)$, $(x+2,y)$, $(x+3,-y)$, etc. The top of each interval is identified with the bottom of its successor's. We can move down a path like this through points which will later be identified with one another - in the quotient, we would just be going through the same one interval long path over and over.

Strip 2.

Before we take the quotient, the path going through the tops of the even intervals and the path going through the top of the odd intervals are different.

Strip 3.

Here we can see how the paper Möbius strip corresponds to the mathematical one; before you twist and glue, it starts off with two sides. (EDIT: Don't be fooled by the ends of my sticky notes; this strip is supposed to be one interval long.)

Physical Strip Side 1. Physical Strip Side 2.

Once we quotient out by $\sim$, we pull these identified points together according to the orientation given by this equivalence relation. The red and green paths show how the orientation 'flips' at the end of every interval. We are left with the Möbius strip.
Physical strip 2. Physical strip 3.

  • 0
    Thank you very much. Great answer!!!2012-12-30
  • 0
    Awesome helpful diagrams!2015-05-02
2

First, prove by induction that $G^n(x,y)=(x+n,(-1)^ny)$ for $n\geq 0$. The same follows for $n<0$ pretty directly.

Let's distinguish $\sim_1$ as the equivalence on $\mathbb R\times[-1,1]$ defined above, and $\sim_2$ as the equivalence relation on $[0,1]\times[-1,1]$.

Now, obviously, $[0,1]\times[-1,1]\subset \mathbb R\times [-1,1]$. So there is a natural inclusion map from $i:[0,1]\times[-1,1]\to\mathbb R\times[-1,1]$.

First prove that $i(x,y)\sim_1 i(x',y')$ if and only if $(x,y)\sim_2(x',y')$ for all $(x,y),(x',y')\in[0,1]\times [-1,1]$.

Prove that induces a continous map: $$f:[0,1]\times[-1,1]/\sim_2\to \mathbb R\times [-1,1]\sim_1$$

The "only if" part of the "if and only if" proves that $f$ is $1-1$. Show also that $f$ is onto.

This all follows directly from definitions.

The tricky part is proving the inverse is continuous.

  • 0
    Can I say that f is defined as $f([x]_2)=[x]_1$? Thank you for you answer!2012-12-30
  • 1
    Yes, that's fine. The problem is proving the inverse is continuous.2012-12-30
  • 0
    Sorry but how we can prove that a map from a quotient space to another quotient space is continuous?2012-12-30
  • 0
    That's why it is hard. One way follows directly, the inverse takes care, and I have nt even tried to outline the proof here.2012-12-30
  • 0
    yes, but why one way follows directly? I didn't understand why the continuity of this map is obvious.2012-12-30
  • 1
    If $X\subseteq Y$ and $\sim$ is an equivalence relation on $Y$, then there is a natural and continuous map: $X/\sim\to Y/\sim$. You can prove this fairly directly by the definition of quotient topology. Since my $i$ above is an injection and we found the relationship $x\sim_2 y\iff i(x)\sim_1 i(y)$, we have essentially the same prove of continuity2012-12-30