Expressing $\mathbb{R} P^3$ as a fibre bundle This question came up in office hours with my differential topology prof and since then I've almost settled on an answer.
The question was whether we could write $\mathbb{R} P^3$ as a fiber bundle with base space $\mathbb{R} P^2$. We spent a few moments thinking about it before deciding it wasn't relevant to the conversation at hand and moved on. Since then I've done some reading and become aware of a result that says if $M, N$ are compact and connected smooth manifolds and $f:M\rightarrow N$ is a submersion, then the fibers of $f$ are all diffeomorphic to a manifold $F$, and this gives rise to a fibre bundle with total space $M$, fiber $F$ and base $N$. Following that, I wrote a function
$$f: \mathbb{R}P^3 \rightarrow  \mathbb{R}P^2$$
defined on homogeneous coordinates by
$$(x,y,z,w) \mapsto (x,y,z).$$
$f$ is well-defined, smooth, and unless I've made a silly mistake in calculating Jacobians it's also a submersion. Now, I'm having trouble understanding what the preimage $f^{-1}(p)$ for $p \in \mathbb{R}P^2$ is. Is it simply $\mathbb{R}$ or is it something more complicated than that?
 A: Let $f:S^3 \to S^2$ be the Hopf map, and $\pi:S^3 \to \mathbf{RP}^{2}$ the composition with the projection $S^2 \to \mathbf{RP}^{2}$. The map $\pi$ sends antipodes of $S^{3}$ to the same place (since the Hopf map collepses equatorial circles), so $\pi$ factors through the projection $S^{3} \to \mathbf{RP}^{3}$. It's "clear" this expresses $\mathbf{RP}^{3}$ as a bundle over $\mathbf{RP}^{2}$ whose fibres are pairs of circles.
Added in edit: I couldn't get @Sanath Devalapurkar's idea (using $SO(3)$) to work, probably because the fibres aren't subgroups, but it did suggest an alternative, and arguably more geometric/elementary construction. View $\mathbf{RP}^{3} \simeq SO(3)$ as the total space $US^{2}$ of the unit circle bundle over $S^2$. (The first column $A_{1}$ of a matrix $A$ in $SO(3)$ is a unit vector, a.k.a., point of $S^2$, and the second column $A_{2}$ is an orthogonal unit vector, a.k.a., a unit tangent vector at $A_{1}$;  since $\det A = 1$, these columns uniquely determine $A_{3} = A_{1} \times A_{2}$.) The fibration $US^{2} \to \mathbf{RP}^{2}$ in question is the projection $US^{2} \to S^2$ followed by the quotient $S^2 \to \mathbf{RP}^{2}$; the fibre over a point $[p]$ in $\mathbf{RP}^{2}$ is the pair of unit circles over the antipodal points $\{p, -p\}$ of $S^2$.
