Why isn't $\mathbb{RP}^2$ orientable? How to prove that  $\mathbb{RP}^2$ isn't orientable? My book (do Carmo "Riemannian Geometry") gives a hint: "Show that it has a open subset diffeomorphic to the Möbius band", but I don't know even who is the "open subset".
 A: $\mathbb{RP}^2$ is the 2-sphere, modulo the relation identifying opposite points. The open subset referred to is a neighborhood of the equator on the sphere. Consider the tangent vector at the equator pointing upwards, and rotate the sphere horizontally by $\pi$. Using the quotient map from the sphere to $\mathbb{RP}^2$, you will see that your tangent vector is in the same fiber as your original tangent vector, but which way will the tangent vector point now?
A: Here are some general facts which will help us to solve the problem on a more general level.

*

*The Antipodal map $a:S^n\longrightarrow S^n$ on the n-sphere is defined by $$a(x^1,\dots,x^{n+1})=(-x^1,\dots,-x^{n+1})$$ is orientation preserving map iff n is odd.


*Suppose $f(x^1,\dots,x^{n+1})$ is a $\mathbb{C}^{\infty}$ function on $\mathbb{R}^{n+1}$ with 0 as  a regular value. Then the zero set of $f$ is an orientable submanifold of $\mathbb{R}^{n+1}$, In particular $S^n$ is orientable in $\mathbb{R}^{n+1}$ So use these facts and you will get any odd dimensional projective space is orientable.
A: For geometric understanding of the real projective plane, I prefer to think of it as gotten from a closed circular disk by identifying opposite points on the boundary. If you accept this, then a Möbius band subset is easily found: take any diameter of your original circle, and widen it to a strip.
