I'm reading through do Carmo's book on Riemannian Geometry and reviewing a bit about orientation. He uses a definition of orientation preserving which I find difficult to verify in practice, but perhaps I'm missing something. He says a manifold is orientable if every change of coordinates function has a positive (determinant) Jacobian, and a choice of such a differentiable structure is a choice of orientation.
1) First, I wanted to verify that if $\varphi: M_1 \to M_2$ is a diffeomorphism, then $M_2$ gets an orientation from $M_1$. My argument was: if $y_\beta^{-1} \circ y_\alpha$ is a change of coordinate map on $M_2$, then $y_\beta^{-1} \circ y_\alpha = (x_\beta \varphi^{-1})^{-1} \circ (x_\alpha \varphi^{-1}) = \varphi (x_\beta^{-1} \circ x_\alpha) \varphi^{-1}$ for suitable coordinate maps $x_\alpha, x_\beta$ on $M_1$. Hence the determinant of the Jacobian for coordinate maps on $M_2$ are positive because the ones on $M_1$ are. Does this work?
2) Now how do I actually compare orientations under a map? For example, I want to do the classic exercise of showing the antipodal map $A: S^n \to S^n$ is orientation preserving iff $n$ is odd. I see that the determinant of the Jacobian of $A$ is $(-1)^{n+1}$ (I think Jacobians behave well under restriction?), but I don't see how to use this to compare the induced orientation on $A(S^n)$ using the above formulation.
I would prefer to try to solve this using this level of information, and not bringing anything fancier like differential forms into the picture yet. Thanks.
Actually, let's even forget about volume forms.
Start with the manifold ${\bf R}^{N}$, regard it as a vector space (in the future, as a tangent space at a point of a manifold), and choose a basis: $$ {\bf E} = \{{\bf e}^{k} \; : \; k = 1, 2, \ldots, N\} $$ (in the future, a frame on the manifold), with the qualification that the basis is ordered (say, by the indices $k$). (For $N$, the order ${\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$ corresponds to what we know colloquially as the counterclockwise orientation; to help the visual, traverse the points points: ${\bf 0} = (0,0), {\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$, in that order).
A linear transformation $A$ which transforms one basis (frame) to another will, therefore, be invertible, hence has a nonzero determinant. The transformation with a positive determinant is said to $\textit{preserve orientation}$ with a negative determinant, to $\textit{change orientation}$. (Test this on the 2-D example I gave above.)
Two frames have the same orientation if one is transformed into the other by a transformation with a positive determinant. This establishes an equivalence relation on the set of frames, with two equivalence classes. These classes are called orientations.
Hope this sheds light on it.
A source: Mathematical Analysis II, V.A. Zorich