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In another question I asked for help to proving the two different definitions of projective planes are homeomorphic. I tried to write the details in proof then realised I can't finish the proof. I need help for the last step, please. What I have done:

The definitions are: Let $P^2 ( \mathbf R) = (\mathbf R^3 \setminus > \{0\}) / \sim$ where $(x,y,z) \sim (x', y', z')$ if and only if $(x,y,z) = \lambda (x', y', z')$ and let $D^2 / \sim $ be the closed disk quotient by the equivalence relation $(r, \varphi) \sim (r', > \varphi ' )$ if and only if $[(r, \varphi) = (r', \varphi ' )] > \text{or} [r=r'=1 \text{ and } \varphi = \varphi ' \pm \pi] $ where $\varphi, \varphi', \varphi' \pm \pi \in [0, 2 \pi)$.

I proved that $D^2$ is homeomorphic to the closed upper half of a sphere $S^{2+}$. Then I wrote the following proof:

Define $\varphi : P^2 ( \mathbf R) \to S^{2+} / \sim$ as follows: For $[(x,y,z)] \in P^2 ( \mathbf R)$, without loss of generality, pick the representative $(x,y,z)$ such that $\|(x,y,z)\| = 1$ and $z \ge 0$. Then $(x,y,z) \in S^{2+}$. Define $\varphi ( [(x,y,z)]) = [(x,y,z)]$. Then $\varphi$ is a homeomorphism:

It is well-defined since if $ [(x,y,z)] = [(x',y',z')]$ and both representatives are in $S^{2+}$ then it follows that $\lambda $ must be $1$ and hence either $(x,y,z) = (x', y', z')$ or if $z=0$ then $x = -x'$ and $y=-y'$. If the former, $\varphi ( [(x,y,z)] ) = \varphi ( [(x',y',z')] )$. If the latter the same holds since in $S^{2+}/ \sim$ the two points are identified.

It follows from the definition that $\varphi$ is injective and surjective.

$\varphi$ is continuous: Let $O$ be open in $S^{2+} / \sim$. Then $\pi^{-1} O$ is open in $S^{2+}$ and hence in $\mathbf R^3 \setminus \{0\}$. The points in $\pi_{S^{2+}}^{-1} O$ are the representatives of the points in $\varphi^{-1}O$. Hence $\varphi^{-1}O = \pi_{P^2 ( \mathbf R)} \circ \pi^{-1}_{S^{2+}} O$ is open.

How to show $\varphi$ is open? Thank you for help.

goobie
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    do you know that (if not try to prove it) bijective continuous maps $f: X \to Y$ are automatically homeomorphisms provided that $X$ is quasicompact and $Y$ is a Hausdorff-space? A hint is - try to prove that $f$ is closed instead of open. – mland Mar 03 '13 at 16:32
  • @mland Thank you for comment! but closed map need not be open. Does in this case map is closed imply map is open? – goobie Mar 03 '13 at 16:38
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    If a map is bijective then being closed and being open are equivalent. This is just as you an define a map to be continuous using closed or open sets. – mland Mar 03 '13 at 18:48
  • @mland If I want to use the theorem you suggest I have to prove $P^2 ( \mathbf R)$ is compact. It seems to be a consequence of what I am trying to show. Is there a simple way to show projective plane is compact? – goobie Mar 04 '13 at 13:47
  • well, it indeed is strongly related to what you want to show. you could start with an open cover of $P^2(R)$ and pull it back to $R^3$. Then you see that this is equivalent to ask that $S^2$ is compact which follows from the Heine-Borel property for example. – mland Mar 04 '13 at 16:34

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We can also calculate the inverse of $\varphi$, as the following composition of continuous maps: $$\psi_0:= S^{2+}\hookrightarrow \Bbb R^3\setminus\{0\}\to P^2(\Bbb R)$$ then $\psi_0(v)=\psi_0(w)$ whenever $v\sim w$, so this factor through (by a continuous map) $S^{2+}/\sim$.

Berci
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  • Did you mean $$ \psi_0 := S^{2+}/\sim \hookrightarrow \mathbb R^3 \setminus {0} \to P^2 (\mathbb R)$$? $\varphi$ is map $P^2 (\mathbb R) \to S^{2+}/\sim$. – goobie Mar 04 '13 at 08:00
  • No, I meant $\psi_0$ starts from $S^{2+}$ (the upper semisphere with boundary). Then, it determines a continuos function $\psi:S^{2+}/\sim\to P^2(\Bbb R)$. – Berci Mar 04 '13 at 10:40