Why isn't $P^2$ the real projective plane homeomorphic to $\mathbb{R}^2$

Question

The real projective plane $P^2$ is the set of linear subspaces in $\mathbb{R}^3$, with the quotient topology. So this means a set of linear subspaces $U \in P^2$ is open if and only if the points on the union of points in $U$ (regarded as points in $\mathbb{R}^3$)

So if I define the following map:

$f: P^2 \rightarrow \mathbb{R}^2 \times 1$ (Where $\mathbb{R}^2 \times 1$ is the subset of $\mathbb{R}^3$ with z coordinate = $1$.

by $f([x] = x$ where $x$ is the point in the linear subspace $[x]$ with $z$ coordinate as $1$.

This map is well defined, and is continuous because for any open set $V$ in the subspace topology of $\mathbb{R}^2$, $f^{-1}(V)$ is the set of linear subspace that defined by a point in $V$ and the origin. and $f^{-1}(V)$ is open if the union of those these points in the union of the linear subspace is open in $R^3$. Which shouldn't be too hard to show by showing this on the basis for the subspace topology of $\mathbb{R}^2$.

Here I already have the fraudulent result that says there's a surjective continuous function from $P^2$ to $\mathbb{R}^2$. And I have no idea what I am doing wrong here.

Can someone point out why the reasoning above is false?

Answer 1

Your $f$ function is a projection onto $\mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.

What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $\mathbb{R}^n$. But a sphere itself is not homeomoprhic to $\mathbb{R}^n$.

This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:

$$q:S^n\to P^n$$

under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $\mathbb{R}^m$ can't even be an image of $P^n$.