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I'm reading Hatcher's "Vector Bundles and K-Theory" and he says something that seems to be false.

Here is a screenshot of the part of the book in question.

He says that if we compare $Vect^n_0(S^k)$ (equivalence classes of vector bundles over $S^k$ with orientation fixed at some point $x_0 \in S^{k-1} \subset S^k$) and $Vect^n(S^k)$ then the map from $Vect^n_0(S^k)$ to $Vect^n(S^k)$ is two to one unless we have an orientation reversing isomorphism in which case it's one to one.

Shouldn't it be just the opposite? Suppose I had some $E \in Vect^n(S^k)$ such that there was no orientation reversing isomorphism from $E$ to itself then the two objects in $Vect^n_0(S^k)$ map to two different objects in $Vect^n(S^k)$ because there was never an isomorphism between them anyways so the map is 1-1. Conversely, if there was an orientation reversing isomorphism from $E$ to itself its only then that I get two different objects in $Vect^n_0(S^k)$ which both correspond to the same $E \in Vect^n(S^k)$ so it is in this case that the map is 2-1.

Am I missing something?

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