# Topology of $\;\mathbb{R}^3$ without origin and antipodal points identified.

Setting: Let $$\sim \subseteq {\Bbb R}^3 \times {\Bbb R}^3$$ be the relation identifying antipodal points, i.e. $$\vec{x} \sim \vec{y}$$ iff $$\vec{x} = \vec{y}$$ or $$\vec{x} = - \vec{y}$$.

Question: What is the quotient topology of $$({\Bbb R}^3 \setminus \{ \vec{0} \})/\sim$$ like?

Background: I understand that $${\Bbb S}^2 / \sim$$ is the real projective plane (with $${\Bbb S}^2 \subseteq {\Bbb R}^3$$ having the subspace topology) and I understand how I get the real projective plane from $${\Bbb R}^3\setminus \{ \vec{0} \}$$ by identifying $$\vec{x}$$ with $$\lambda \cdot \vec{x}$$ for non-zero $$\lambda$$, and I understand how to get the $${\Bbb S}^2$$ by identifying $$\vec{x}$$ with $$\lambda\cdot \vec{x}$$ for positive $$\lambda$$. My goal is to fill in all parts of the below diagram and I am missing the ?? part. I also would be interested what we can say about the diagonals (which cannot be drawn in MathJax afaik, the main diagonal direction is obvious, but can we say something about the other diagonal).

$$\require{AMScd}$$ $$\begin{CD} {\Bbb R}^3\setminus\{ \vec{0} \} @>a>> ??\\ @V b V V e,f @VV c V\\ {\Bbb S}^2 @>>d> P{\Bbb R}^2 \end{CD}$$

$$\newcommand{\Reals}{\mathbf{R}}$$Note that $$\Reals^{3}\setminus\{0\}$$ is diffeomorphic to $$S^{2} \times \Reals$$ via $$x \leftrightarrow \biggl(\frac{x}{|x|}, \log|x|\biggr),$$ and the antipodal map acts as the antipodal map on $$S^{2}$$ but as the identity on $$\Reals$$. The quotient is therefore $$P\Reals^{2} \times \Reals$$, the total space of the trivial real line bundle over the real projective plane. The anti-diagonal map is the double covering of the zero section.
This quotient space, denoted ?? in your diagram, can be identified with $$P\mathbb R^2 \times (0,\infty)$$. You could also use the logarithm function to identify it with $$P\mathbb R^2 \times \mathbb R$$, but $$P\mathbb R^2 \times (0,\infty)$$ is maybe better in this context.