I'm reading Hatcher's book on K-Theory, and on pages 25-26, he talks about real vector bundles over $S^k$. He defines the object $\text{Vect}_0^n(S^k)$ as
the n dimensional vector bundles over $S^k$ with an orientation specified in the fiber over one point $x_0 \in S^{k−1}$, with the equivalence relation of isomorphism preserving the orientation of the fiber over $x_0$.
I'm interpreting this to mean that an element of $\text{Vect}_0^n(S^k)$ is a vector bundle plus a specified orientation of the fiber over some arbitrary point $x_0$ (chosen for convenience to lie on the "equator" $S^{k-1}$ of $S^k$). Another way to say this is, the trivializing functions $\phi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times\mathbb{R}^n$ must send the oriented basis of $\pi^{-1}(x_0)$ to the standard basis of $\mathbb{R}^n$ for any $\alpha$ with $x_0\in U_\alpha$. And we consider any two vector bundles to be equivalent if they are isomorphic AND have the same specified orientation on $x_0$.
However, I'm worried this interpretation is incorrect, or there is something else I'm missing, because later he says:
The map $\text{Vect}^n_0(S^k)\rightarrow\text{Vect}^n(S^k)$ that forgets the orientation over $x_0$ is a surjection that is two-to-one except on vector bundles that have an automorphism (an isomorphism from the bundle to itself) reversing the orientation of the fiber over $x_0$, where it is one-to-one.
According to my interpretation, this is exactly backwards. I would assume that if a vector bundle has an automorphism that reverses the orientation of $x_0$, there would be TWO elements of $\text{Vect}^n_0(S^k)$ that map to it, one for each orientation of $x_0$, while if a vector bundle does NOT have an automorphism that reverses the orientation of $x_0$, then there would only be one element in $\text{Vect}^n_0(S^k)$ that maps to it.
Can someone point out where I am thinking about this incorrectly, or just provide more details on the object $\text{Vect}^n_0(S^k)$?
The elements of $\text{Vect}^n(S^k)$ are equivalence classes $[E]$ of $n$-dimensional vector bundles $E$ over $S^k$, where $E, E'$ are equivalent iff there exists a bundle isomorphism $\phi : E \to E'$. The elements of $\text{Vect}_0^n(S^k)$ are equivalence classes $[E,\omega]$ of pairs $(E,\omega)$ consisting of an $n$-dimensional vector bundle $E$ over $S^k$ and an orientation $\omega$ of the fiber over $x_0$ (which has two orientations $\omega^\pm$), where $(E,\omega), (E',\omega')$ are equivalent iff there exists a bundle isomorphism $\phi : E \to E'$ such that $\phi(\omega) = \omega'$.
The map $$P : \text{Vect}_0^n(S^k) \to \text{Vect}^n(S^k),\quad P:[E,\omega] \mapsto [E]$$ is obviously a surjection. We certainly have $$P^{-1}([E]) = \{ [E,\omega^+], [E,\omega^-] \} .$$ But $[E,\omega^+] = [E,\omega^-]$ iff there exists a bundle isomorphism $\phi : E \to E$ such that $\phi(\omega^+) = \omega^-$, i.e. a bundle automorphism on $E$ which reverses the orientation of the fiber over $x_0$.
Thus Hatcher is right.