general-topologymobius-bandprojective-space
Let $\mathbb{S}^2$ be the unit sphere in $\mathbb{R}^3$. Define the quotient space
$$\mathbb{R}P^2:=\mathbb{S}^2/(x\sim -x).$$
How can we show that $\mathbb{R}P^2$ contains an embedding of the Möbius strip
$$ M:=([0,1]\times (-1,1))/((0,y)\sim (1,1-y))$$
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.
We can construct the quotient directly by identifying $$(z,-1)\sim \left(\frac{1}{z},1\right)$$ We can see that this works correctly by noting that if we map $[0,1]\times [0,1]\to \mathbb{S}^1\times [-1,1]$ via the function $$(x,y)\mapsto (e^{2\pi i (x-\frac{1}{2})},2(y-\frac{1}{2}))$$ then the identification $(x,0)\sim (1-x,1)$ is translated to $$(e^{2\pi i (x-\frac{1}{2})},-1)\sim (e^{2\pi i (\frac{1}{2}-x)},1)$$ and this coincides with the definition $(z,-1)\sim \left(\frac{1}{z},1\right)$.
EDIT: Note that there is no way to obtain a Mobius strip from $\mathbb{S}^1\times [-1,1]$ by identifying opposite edges. The space constructed here is homeomorphic to the Klein bottle.
We obtain the Mobius strip by identifying $(z,x)\sim (-z,-x)$. To see this, map $[-1,1]\times [-1,1]\to \mathbb{S}^1\times [-1,1]$ by sending $(x,y)\mapsto (e^{\pi i x},y)$. When we identify $(z,x)\sim (-z,-x)$, we are identifying $(e^{\pi i x},y)\sim (e^{\pi i (x+1)},-y)$. Thus if $x<0$ we have $(x,y)\sim (x+1,-y)$. Map the quotient space to the Mobius strip, considered as the quotient of $[0,1]\times [0,1]$ by identifying $(0,y)\sim (1,1-y)$, by sending $[(x,y)]\mapsto [(x+1,\frac{1}{2}(1+y))]$ for $x\leq 0$. Verify that this is a homeomorphism between the quotient space and the Mobius strip.
Best Answer
The natural way to show this would be to explicitly write down the embedding.
For example, the Möbius strip can map to a thin belt around the equator of the sphere. Then the relation between this belt (in spherical coordinates!) and your defining parameterization of the Möbius strip is a simple affine transformation of each coordinate separately. Remember that $[0,1]$ should map to only one hemisphere of longitude due to the identification of antipodes.
Then there remains some slightly tedious fiddlework proving that your map respects the two relations you quotient out, and you're set.
But actually there's no reason to make it a thin belt. If instead you make the belt extend all the way to the poles, you'll have proved the stronger result that the real projective plane minus one point is homeomorphic to the Möbius strip without boundary.