# The projective plane and its lines

I have a problem about the projective plane. They said that the only thing that I must know is the definition. In the problem, the "line" of a homogeneous polynomial is defined. It's the set $$L_{a,b,c} = \{ {\left[ {x,y,z} \right] \in P_\mathbf{R}^2 :ax + by + cz = 0} \}.$$ This exercise has two parts. I want to understand it )=. First I must show that $$P_\mathbf{R}^2 - L \simeq \mathbf{R}^2$$ for every line $L$. I've tried to define an embedding of $\mathbf{R}^2$ first, and I think that the natural way to proceed is try to define a parametrization of the sphere, and then compose this function with the quotient map $q\colon \mathbf{R}^3 - \{0\} \to P_\mathbf{R}^2$. And this function I suppose will be useful, but I don't see how to proceed )= sorry for be so stupid <.< , but it´s wear to me visualize this things like the projective plane, i need more practice

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I manage to write this always wrong in Spanish, in English and in French: projective is with a J. :) –  Mariano Suárez-Alvarez Oct 1 '11 at 2:36
I would try to first give a bijection between $\mathbf{R}^2$ and $\mathbf{P}^2 \setminus L_{1, 0, 0}$; the latter space consists of those points which can be written in the form $[1, y, z]$ for some $y, z \in \mathbf{R}$. If you can figure this out (writing down a map, showing that it's well-defined, a homeomorphism, and so on) then the only thing left to do, more or less, is change coordinates. –  Dylan Moreland Oct 1 '11 at 2:36
Dear Daniel: I tried to fix up the spelling, TeX and such without touching the more personal sentences. I hope I haven't distorted your meaning. –  Dylan Moreland Oct 1 '11 at 2:46

If you see $P^2$ as a quotient space of $S^2$ by identifying antipodal points, then a line as defined above makes sense in that sphere too, only with $|(x, y, z)|=1$ instead of them being homogeneous. You will then have that the curve divides the sphere into two parts.
As you transition into $P^2$, these two parts will go together to be that one subspace $P^2 - L$
To show that what is left is homeomorphic to $R^2$ is easy, really. You just have to show that the transition of $S^2$ to $P^2$ is homeomorphic on a single open hemisphere, and that an open hemisphere is homeomorphic to $R^2$.