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:

Solution 1:

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.

Solution 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.