For two unit non-oriented bivectors $A,B\in \mathbb{R}P^2\subset \Lambda^2\mathbb{R}^3$ is the mapping $\phi:(A,B)\rightarrow AB$ bijective?

Question

For two non-oriented unit bivectors $A,B\in \mathbb{R}P^2\subset \Lambda^2\mathbb{R}^3$ is the mapping $\phi:\mathbb{R}P^2\times \mathbb{R}P^2/\mathbf{D} \rightarrow S^3$, where $\mathbf{D}$ is the diagonal $\{(x,x):x\in \mathbb{R}P^2\}$, of the form $\phi:(A,B)\rightarrow AB$ bijective? That is, is the mapping from two non-oriented ($A\sim -A,B\sim -B$) unit bivectors to their geometric product bijective, as long as the two bivectors are not the same? Intuitively this makes sense, but I'm having a hard time justifying it algebraically.

Answer 1

HINT:

An equivalent question is: what are the fibers of the map $$\mathbb{R}^3\times \mathbb{R}^3\to\mathbb{R}^3, (v,w)\mapsto v\times w$$