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I am trying to show $P^2 (\mathbf R) $ is homeomorphic to $D^2/\sim$ where $D^2$ is the disc and it is quotiented by the relation $(r,\phi) \sim (r', \phi')$ if and only if either $(r,\phi) = (r', \phi')$ or $r=r'=1$ and $\phi = \phi' + \pi (\mod 2 \pi)$.

The $P^2 (\mathbf R) $ is the space $(\mathbf R^3 \setminus \{0\}) /\sim$ where $(x,y,z) \sim (x',y',z')$ if and only if $(x',y',z') = \lambda (x,y,z)$.

Let $q_1 : \mathbf R^3 \setminus \{0\} \to (\mathbf R^3 \setminus \{0\}) /\sim = P^2 (\mathbf R)$ and $q_2 : D^2 \to D^2/\sim$ be the quotient maps. To show $P^2 (\mathbf R) \cong D^2 \sim$ my idea was to show that $f = q_1 \circ q_2^{-1}$ is a homeomorphism. But I fail to show any of the properties. For example, I want to show $f$ is continuous. But I can't show $q_2^{-1}$ is continuous: It is not clear that it is true: quotient map in general is not open.

How to show the spaces are homeomorphic? Is it not possible to prove $f$ is homeomorphism?

and:

What is $D^2$? It means: is it correct to write in proof: "for $q_2 : D^2 \to D^2/\sim$ assume that domain $D^2$ lies in $\mathbf R^3$ on upper half of the sphere $S^2$ centered at origin" ? Or is such assumption not general enough and $D^2$ is abstract space?

Thank you.

goobie
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Let $S= \mathbb{S}^2 \cap \{(x,y,z) \in \mathbb{R}^3 : z \geq 0 \}$ and $R$ the smallest equivalence relation on $S$ such that $(x,y,0) R (x',y',0)$ if $x'=-x$ and $y'=-y$.

Then $p : P^2\mathbb{R} \to S/R$ defined by $p([x:y:z])=(a,b,c)$ with $a^2+b^2+c^2=1$ and $c \geq 0$ is an homeomorphism.

Finally, just compose with the projection $\pi : S/R \to D^2/\sim$ defined by $\pi(x,y,z)=(x,y)$.

Seirios
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  • Is it not possible to prove with quotient maps $q_1 \circ q_2^{-1}$? – goobie Feb 26 '13 at 15:00
  • @goobie: First of all, $q_2^{-1}$ is not well defined ($q_2$ is not one-to-one). – Seirios Feb 26 '13 at 15:32
  • I understand $q_2^{-1}$ is not well-defined but $q_1 \circ q_2^{-1}$ is. I don't yet understand your answer: why does $R$ have to be the smallest relation? I am sorry, this is all new to me. – goobie Feb 28 '13 at 14:15
  • I don't understand the map $p$. It maps $x,y,z$ to $a,b,c$ but how? – goobie Feb 28 '13 at 14:30
  • @goobie: I agree that $f=q_1 \circ q_2^{-1}$ can be well-defined, however $f$ is not surjective, the image of $f$ in contained in a plane. For $R$, it only identificates antipodal points on the equator of the demi-sphere $S$. For $p$, any line of $P^2\mathbb{R}$ intersects $S$ in exactly one point or in two antipodal points on the equator of $S$ then collapsing in one point by $R$; so $p([x:y:z])$ is the only point in the intersection between the line associated to $[x:y:z]$ and the quotiented demi-sphere $S/R$. – Seirios Feb 28 '13 at 14:49
  • Maybe I understand! Do you say you pick without loss of generality the representative $x:y:z$ of the equivalence class $[x:y:z]$ to be such that $| (x:y:z) | = 1$ and $z \ge 0$ and therefore $p$ is the identity map? – goobie Feb 28 '13 at 15:05
  • @goobie: Exactly, but there is a problem for $z=0$, because there are two such representatives; it is why $R$ is introduced. – Seirios Feb 28 '13 at 15:19
  • I get error about unregistered users. Last bounty was awarded automatically to accepted answer. So we just wait. – goobie Feb 28 '13 at 21:02