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I'm trying to show that the Möbius strip with boundary circle identified to a point is homeomorphic to $P^2$ (real projective space). I get geometrically why this is so, but, how does one generally construct maps between these spaces to show that they're homeomorphic?

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If you take as your definition that a Moebius strip is the total-space of a non-orientable $I$-bundle over $S^1$, and $\mathbb RP^2$ as $S^2$ modulo antipodal identification, the homeomorphism comes from observing that when you remove the "poles" (one point) from $\mathbb RP^2$ the resulting manifold is an $I$-bundle over the equatorial lines, then check it's non-orientable. – Ryan Budney Oct 11 '10 at 20:35

Maybe you're going to tell me that if all you've got is a hammer, all problems look like nails, but here goes my proposal -following the same strategy as in my answers to this question.

First of all, take as a model for the projective plane the disk

$$ D^2 = \left\{ (x,y) \in \mathbb{R}^2\ \vert \ x^2 + y^2 \leq 1 \right\} $$

and quotient by the equivalence relation among antipodal points on $S^1$: $(x,y) \sim -(x,y) $ if $x^2 + y^2 = 1$. So

$$ \mathbb{RP}^2 = D^2/\sim \ . $$

Put the Moebius strip inside $\mathbb{RP}^2$ as the vertical strip

$$ M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ -\frac{1}{2} \leq x \leq \frac{1}{2} \right\} \ . $$

Let's denote by $M'$ the same set of points inside $D^2$, without any quotient at all. That is, ordinary points of $\mathbb{R}^2$. Thus

$$ M = M'/\sim \ . $$

Now, you want to prove that, when you quotient out the boundary of the Moebius band

$$ \partial M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ x = -\frac{1}{2} \ \quad \text{or} \quad x = \frac{1}{2} \right\} $$

you get the projective plane:

$$ M/\partial M \ \cong \ \mathbb{RP}^2 \ . $$

Ok, the homeomorphism is easily described in some kind of "eliptical" coordinates: for every point $(x,y) \in D^2$ there is one and only one elipse of the form

$$ x^2 + \frac{y^2}{b^2} = 1 \qquad \text{for} \qquad -1 \leq b \leq 1 \ . $$

All these elipses have a common major axis, namely the segment $-1\leq x \leq 1, y = 0$. We make the abuse of considering $b=0$ too as the degenerate "elipse" $y=0$ and in order to distingish the "positive" $y\geq 0$ and "negative" $y\leq 0$ branches of the elipse, we put a sign in $b$: the same as $y$. If all this doesn't sound too rigorous to you, it's ok: look at it as just a motivation. Wait for the resulting formulae.

So a point inside the disk is determined once you know its $x$-coordinate and the elipse to which it belongs; that is, its $b$-coordinate.

Our homeomorphism is going to do the following: points inside the Moebius strip are going to "travel" along the elipse to which they belong. How far? Well, we want all the points in $\partial M$, those with $x=-1/2$ or $x= 1/2$, to reach points $(-1,0)$ or $(1,0)$, depending on the sign of $x$. Hence, we try something like

$$ \widetilde{\varphi }: M \longrightarrow \mathbb{RP}^2 \ , \qquad \widetilde{\varphi} \widetilde{(x,b)} = \widetilde{(2x, b)} \ . $$

So, let's write a decent $\varphi$ in Cartesian coordinates. We want our $(2x, b)$ to belong to the $b$-elipse all the time, hence we have

$$ (2x)^2 + \frac{y^2}{b^2} = 1 \qquad \Longrightarrow \qquad b = \frac{y}{\sqrt{1-(2x)^2}} \ . $$

Hence we begin with a map

$$ \varphi : M' \longrightarrow D^2 $$

defined as

$$ \begin{cases} \varphi (x,y) = \left( 2x , \frac{y}{\sqrt{1-(2x)^2}}\right) & \text{if} \ x \neq \frac{\pm 1}{2} \\ \varphi (\pm 1/2, y) = (\pm 1,0) & {} \end{cases} $$

Exercise. Check that $\varphi$ is continuous, surjective and compatible with the antipodal equivalence relation.

Hence, $\varphi$ induces a well-defined map

$$ M \longrightarrow \mathbb{RP}^2 \ , $$

which is continuous, because of the universal property of the quotient topology, and surjective. It is "almost" injective too, except for all the points in the boundary of the Moebius band $x=-1/2$ and $x=1/2$ that go to $\widetilde{(-1,0)} = \widetilde{(1,0)}$. So, if we quotient out them, we get a homeomorphism

$$ \widetilde{\varphi} : M/\partial M \longrightarrow \mathbb{RP}^2 \ , $$

whit the help, as usual, of the GTET (see the link above, taking into account that, despite being a quotient, $\mathbb{RP}^2$ is a Hausdorff space -see Rotman, "An introduction to Algebraic Topology", theorem 8.4, Hausdorff).

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I would construct a homeomorphism from $A/\sim$ to the Möbius strip where $A$ is the region on the sphere bounded by two antipodal lines of latitude (let's think of them as the Arctic and Antarctic circles) and $\sim$ is the equivalence relation identifying antipodal points. Now deforming the Arctic/Antarctic regions into the north and south poles, induces a homeomorphism from the Möbius strip with edge squeezed to a point and ${S^2/\sim}\ = P^2$.

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Starting from a flat rectangular Möbius strip $ABA'B'$ with the identification $AB'\cong A'B$. The side $AB$ is to be shrunk to a point and similarly the side $A'B'$ to obtain the identification of the boundary circle to a point.

To construct a transformation, we take the side $AB$ is along the x axis and $BA'$ along the y-axis of a Cartesian system. We take the origin of the system at the center of the rectangle and the side lengths 2a (along the x-axis) and 2b along the (y-axis). ($AB$ is along the line $y=-b$).

Then the transformation:

$$ \begin{align*} r &= \sqrt {a^2b^2-(a^2-x^2)(b^2-y^2)} \\ \theta &= \arcsin{\frac{y}{b}} \end{align*} $$

($r$ and $\theta$ are polar coordinates) maps the Möbius strip to a disc of radius $ab$ with the boundary antipodal points identified (which is homeomorphic to $RP^2$. The identified lines $AB$ $A'B'$ are mapped to south and the north poles respectively. The lines $BA'$ and $B'A$ are mapped to the right and left boundary semicircles.

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