TITLE: Orientability of $\mathbb{RP}^3$
QUESTION [11 upvotes]: I was wondering if there is a nice way to see that $\mathbb{RP}^{3}$ is orientable without using  tools of algebraic topology, like homology.
The only think I could think of was to argue that $\mathbb{RP}^{3}=\mathbb{R}^3 \cup \mathbb{RP}^{2}$ and perhaps you could argue that to get back to any starting position you have to cross the $\mathbb{RP}^{2}$ boundary but I'm pretty sure that what I'm thinking is nonsense.
This was a question on the homework for one of my topics courses and I plan on asking the professor about it tomorrow, but I was curious to see if anyone had any interesting ways of thinking about or picturing this space.

REPLY [5 votes]: $RP^{2k-1}$ is a quotient of a codimension 1 submanifold with induced orientation ($S^{2k-1}$)in $R^{2k}$, by an orientation-preserving transformation of $R^{2k}$.  Because there is a consistent notion of normal vector (i.e., of "inside" and "outside") on the submanifold, preserving orientation of the surrounding manifold also preserves orientation on the submanifold.  Transporting the orientation along a path between two points of the submanifold that are identified in the quotient can be viewed as transporting the orientation in the surrounding manifold, and this has to be consistent (have positive determinant on the local frames) because of the orientability of the surrounding manifold.