Orientation twisted on the $2$-sphere If one is asked to verify that the canonical orientation on the $2$-sphere $S^2$ is twisted by the antipodal map, is it right to say that this is so simply because the Jacobian matrix of the map $A\colon\mathbb{R}^3\to\mathbb{R}^3$ defined by 
$$(x,y,z)\mapsto (-x,-y,-z)$$
has determinant $-1$?
 A: Actually, it's not that simple. Consider a similar example: Let $M$ be the $xy$-plane in $\mathbb R^3$ -- that is, $M$ is the set of all points of the form $(x,y,0)$ for $(x,y)\in \mathbb R^2$.  Even though the antipodal map of $\mathbb R^3$ has Jacobian determinant $-1$ as you noted, and it takes $M$ to itself, in turns out that its restriction to $M$ is actually orientation-preserving. 
The thing about $S^2$ that's different is that the antipodal map also takes the outward unit normal $N$ of $S^2$ to itself.  The area form $\omega$ of $S^2$ is given by 
$$
\omega = i_N (dx\wedge dy\wedge dz)|_{TS^2},
$$
where $i_N$ represents interior multiplication by $N$. It follows that 
\begin{align*}
(A|_{S^2})^*\omega 
&= i_{A^{-1}_*N} A^*(dx\wedge dy\wedge dz)|_{TS^2}\\
&= i_N ( -dx\wedge dy\wedge dz)|_{TS^2}\\
&= -\omega.
\end{align*}
(The reason this calculation doesn't apply to the $xy$-plane is that $A$ reverses the direction of its normal.)
A: Just wondering, is it right to write down the antipodal map in local coordinates as
$$(\theta,\phi)\mapsto(\pi-\theta,\phi+\pi)$$
so that 
$$ dA_{(\theta,\phi)}=\begin{bmatrix}-1 & 0 \\0 & 1\end{bmatrix}$$
and its determinant is everywhere negative so that the orientation is twisted?
