Let $B = \{x \in \mathbb{R}^3 : ||x|| \le 1 \}$. I have seen it claimed that $\mathbb{R}P^3$ is homeomorphic (indeed, diffeomorphic) to $B$ where for $x \in \partial B$, $x$ is identified with $-x$.
I understand how this would work if $B$ were replaced by $S^2$. You would map a line in $\mathbb{R}P^3$ to its normalized direction vector. However, this map fails to be a homeomorphism for (the quotient of)$B$, because it is not surjective. How can we get a homeomorphism between $\mathbb{R}P^3$ and $B$?