# This quotient space is homeomorphic to the Möbius strip?

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:

• How do you define the Mobius strip? – Thomas Andrews Dec 28 '12 at 22:09
• @ThomasAndrews in the standard way: $[0,1]\times [-1,1]/\sim$, where $(0,y)\sim (1,-y)$ – user42912 Dec 28 '12 at 22:13
• Can you see a way to map one of these spaces to the other? – Thomas Andrews Dec 28 '12 at 22:22
• No, that's why I'm asking :) I don't get used yet with such spaces – user42912 Dec 28 '12 at 22:37
• 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 $\mathbb R\times[-1,1]$. – hmakholm left over Monica Dec 28 '12 at 22:41

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 $[-1,1]$.

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 $A_2$.

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.

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.

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.)

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.

• Thank you very much. Great answer!!! – user42912 Dec 30 '12 at 1:34
• Awesome helpful diagrams! – Y-dog May 2 '15 at 15:03

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.

• Can I say that f is defined as $f([x]_2)=[x]_1$? Thank you for you answer! – user42912 Dec 30 '12 at 1:44
• Yes, that's fine. The problem is proving the inverse is continuous. – Thomas Andrews Dec 30 '12 at 1:52
• Sorry but how we can prove that a map from a quotient space to another quotient space is continuous? – user42912 Dec 30 '12 at 2:33
• 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. – Thomas Andrews Dec 30 '12 at 3:27
• 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 continuity – Thomas Andrews Dec 30 '12 at 17:27