Let P be the real projective plane obtained by identifying antipodal points on the unit sphere of $R^3$.
How to prove that P is nonorientable in a rigorous and elementary way? I do not want mere intuition.
My idea is to consider the closed curve $a(t)=(\cos t,\sin t, 0)$ , $0 \leq t \leq \pi$
Then the tangent vectors $a'(0)$ and $a'(\pi)$ are identical.
However, for a vector field V on $a(t)$ defined by $V(a(t))=(0,0,1)$, the tangent vectors at $a(0)$ and $a(\pi)$ differ by a sign. Therefore the normal vector cannot be continuously defined.
Is the constant vector field V continuous on the curve? Are my arguments right?