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I have no one else to correct my work since I am not going to school therefore I'd be very grateful if someone check my work. I was trying to show that the projective real plane is a manifold:

On $D^2 = \{(r,\varphi) \mid r \le 1\}$ define the equivalence relation $(r,\varphi) \sim (r', \varphi')$ if and only if $[(r,\varphi) = (r', \varphi')]$ or $[r=r'=1 \land \varphi = \varphi' + \pi]$. Let $q: D^2 \to D^2 / \sim$ denote the quotient map.

$D^2 / \sim$ is Hausdorff and second countable since $q$ is perfect: That $q$ is surjective and continuous follows from $q$ being a quotient map. The closures of the fibres $\overline{q^{-1}(y)} = q^{-1}(y)$ each consist of at most two points and are therefore compact. That $q$ is closed can be verified as follows: Let $C$ be closed in $D^2$. Define $D = q^{-1}(q(C))$ so that $q(C)$ is closed if and only if $D$ is. Let $f: D^2 \to D^2$ be the map ${r, \varphi} \mapsto (r, \varphi + \pi)$. Then $f$ is a homeomorphism. Since $\partial D^2$ is closed in $D^2$, the intersection $C \cap \partial D^2$ is closed in $D^2$. Therefore $D$ is closed in $D^2$ since $D= C \cup f(C \cap \partial D^2)$. Hence $D^2/\sim$ is Hausdorff and second countable.

$D^2 / \sim$ is locally Euclidean: Let $x \in D^2 / \sim$. If $x \in \mathrm{int}(D^2)$ then there is $\varepsilon > 0$ such that $B(x, \varepsilon) \subseteq \mathrm{int}(D^2)$ and the inclusion map $i: B(x, \varepsilon) \hookrightarrow B(x, \varepsilon) \subseteq \mathbf R^2$ provides a local homeomorphism.

If $x \in \partial D^2$ let $f$ be rotation by $\pi$ around the origin (imagine $D^2$ lying centered at $(0,0)$) and let $N = B(x, \varepsilon) \cap D^2 \sqcup B(f(x), \varepsilon) \cap D^2$. Then $N$ is an open set in $D^2$ containing $x$. We know that $D^2 \cong [0,1] \times [0,1]$. Let $\varphi$ be a homeomorphism $D^2 \to [0,1] \times [0,1]$. Pick $\varphi$ so that two of the sides of the square are parallel to the tangents at $D^2$ at $x$ and $f(x)$. Then $\varphi$ maps $N$ to two halves of a ball $B(x, \varepsilon)$ whose straight boundary is the tangent at $x$ and $f(x)$ respectively. Let $P$ be the line through the origin perpendicular to the tangents. Let $f_1$ and $f_{-1}$ be the maps that translate points in the direction of $P$ by $1$ and $-1$ respectively. Let $f_P$ be reflection at $P$. Then the map $f = f_{-1} \sqcup (f_1 \circ f_P)$ defines a local homeomorphism from $N$ to $B((0,0), \varepsilon)$. Hence $D^2 / \sim$ is locally Euclidean.

I am sorry for posting long text. Many thank you for help me!

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Your proof looks really long. Wouldn't you like to prove that $\mathbb RP^2$ is a manifold by studying the action of $\mathbb R$ on $\mathbb R^3$ minus the zero vector, and show that the quotient is a nice manifold ? –  Damien L Feb 12 '13 at 17:42
    
It seems correct, anyway. –  Berci Feb 12 '13 at 18:14
    
@Berci Many thank you for helping me! –  goobie Feb 12 '13 at 20:06

1 Answer 1

You can use the result that spaces that result from some types of group actions (proper discontinuous actions) are manifolds; you can see projective space as the quotient space $ S^2 / Z_2 $. This result then tells you $\mathbb RP^2 $ is a manifold. In your case, you have $Z_2$ , a finite group acting freely on $S^2$ , so the quotient is a manifold. For an actual proof, see John M. Lee's Smooth Manifolds book on the chapter on Lie groups ( different pages for different editions; I have page 153 in my 2nd ed.)

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