What is my mistake in the following proof? I want to prove antipodal map $a:\mathbb{S}^n \rightarrow \mathbb{S}^n$ is orientation-preserving if and only if $n$ is odd. I have a idea to prove this as follows: 
Since $a$ is a diffeomorphism, it is sufficient to check matrix $a_*$ has positive determinant. I can use a chart for $\mathbb{S}^n$ such that $a_*$ has local representation $-1I_{n\times n}$ so det $a_* = (-1)^n$ and so this is wrong. What is my mistake?
 A: Note that $a_*:T_pM\to T_{a(p)}M$ is a map between different tangent spaces. Without specifying charts, the statement that the local representative of $a_*$ at some point $p$ is $-I_{n}$ is vacuous: with suitable choices of local coordinates around $p$ and $a(p)$, you could make the local representative of $a_*$ whatever you wanted.
To make the orientation of local representatives meaningful, you would need to choose consistently oriented charts around $p$ and $a(p)$. If you do, you will find that the local representatives have appropriately signed determinant.
Alternately, a common approach for this problem is to work with the standard embedding $S^n\subset\mathbb{R}^{n+1}$ since both the antipodal map $a$ and the standard orientation form on $S^n$ are easy to work with in this setting. Since $\mathbb{R}^{n+1}$ is covered by a single chart (which, of course, is consistently oriented with itself), we can work in coordinates and the determinant of local representatives of $a$ are meaningful. It will take a bit more work to restrict the results to $S^n$, of course.
A: This answer invokes some facts which are possibly not yet available to the OP. Perhaps it provides motivation for further study.
Perhaps the best way to see it is to use the fact that $S^n$ is a smooth submanifold of $\mathbb R^{n+1}$. The tangent space $T_x S^n$ can thus be identified naturally with its Euclidean tangent space $T'_xS^n$ which can be understood as the set of all derivatives $u'(0)$ of all smooth curces $u : (-r,r) \to \mathbb R^{n+1}$ such that $u(0) = x$ and $u((-r,r)) \subset S^n$.
It is easy to see $T'_xS^n$ is the orthogonal complement of $x$. Thus $T'_xS^n = T'_{-x}S^n = T'_{a(x)}S^n = V$. The differential $T_x a : T_xS^n \to T_{a(x)}S^n$ can then be identified with the restriction of the ordinary Euclidean derivative $Da(x) : \mathbb R^{n+1} \to \mathbb R^{n+1}$ to $T'_xS^n$. We have $Da(x) = a$ because $a$ is linear. Therefore $T_x a$ is identified with $a : V \to V$.
How does $T'_xS^n$ get its orientation? One can show that it inherits one from $\mathbb R^{n+1}$ as follows: Call an ordered basis $(b_1,\ldots,b_n)$ of $T'_xS^n$ positively oriented if $(x,b_1,\ldots,b_n)$ is a positively oriented basis of $\mathbb R^{n+1}$. The canonical orientation $\omega_x$ of $T'_xS^n$ is the equivalence class of positively oriented bases of $T'_xS^n$ . This shows the oriented Euclidean tangent spaces $T'_xS^n = (V,\omega_x)$ and $T'_{a(x)}S^n = (V,\omega_{a(x)})$ have opposite orientations (although they agree as unoriented vector spaces).
See also https://math.stackexchange.com/q/10248. 
Now consider $a : (V,\omega_x) \to (V,\omega_{a(x)})$. It splits as the composition of $a : (V,\omega_x) \to (V,\omega_x)$ and $id : (V,\omega_x) \to (V,\omega_{a(x)})$. The first map is orientation preserving iff $n$ is even, the second is always orientation reserving. Thus $a : (V,\omega_x) \to (V,\omega_{a(x)})$ is orientation preserving iff $n$ is odd.
