nonorientability of the projective plane 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?
 A: Preliminary Result
If $M$ is an orientable manifold and if $(U,\phi),(V,\psi)$ are two charts with $U,V$ connected, then the jacobian determinant of the change of coordinates $\psi\circ \phi^{-1}$ has constant sign (even if $U\cap V$ is not connected)
Application
Let us apply this to projective space and to its two charts $$\phi : U=\{x\neq 0\}\to \mathbb R^2:[x:y:z]\mapsto (y/x,z/x)$$  
$$\psi : V=\{y\neq 0\}\to \mathbb R^2:[x:y:z]\mapsto (x/y,z/y)$$
The change of coordinates $\psi\circ \phi^{-1}$ is the morphism $$\mathbb R^2\setminus \{u=0\}\to \mathbb R^2\setminus \{u=0\}:(u,v)\mapsto (\frac 1u,\frac vu)$$
whose jacobian determinant is $\frac {-1}{u^3}$.
Since this jacobian determinant changes sign on $\mathbb R^2\setminus \{x=0\}$, $\mathbb P^2$ is non-orientable as a consequence of the Preliminary Result.  
Remark
The Preliminary Result seems not to be often mentioned in books on differential geometry.
It is a pity since as a consequence non orientability of manifolds is often treated in a hand-waving fashion.
