Basic definition of orientation/preservation question 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.
 A: 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 
