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Maps between equivalence classes of vector bundles with and without orientation fixed at a point
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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?
vector-bundles orientation
asked 2 days ago
yankyl
62
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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?
vector-bundles orientation
asked 2 days ago
yankyl
62
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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?
vector-bundles orientation
asked 2 days ago
yankyl
62
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?
vector-bundles orientation
asked 2 days ago
yankyl
62
asked 2 days ago
yankyl
62
asked 2 days ago
yankyl
62
asked 2 days ago
yankyl
62
62
add a comment |Â
add a comment |Â
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