Classification theorem of topological vector bundles via presheaves
My issue is mainly set theoretical:
In "Dale Husemoller: Fiber Bundles" it says on page 34 that there is a cofunctor $Vekt_{k}: Prightarrow ens$, where $P$ is the category of paracompact spaces with homotopy classes of maps and ens the category of sets and functions, assigning to each paracompact space $B$ the set of $B$-isomorphism classes of vector bundles over $B$.
Unfortunately,I don't have any profound knowledge of set theory and I do not understand why this actually forms a set and not a proper class. Can anyone help me with this?
algebraic-topology category-theory set-theory vector-bundles
asked Nov 20 at 14:47
Joo
185
add a comment |
My issue is mainly set theoretical:
In "Dale Husemoller: Fiber Bundles" it says on page 34 that there is a cofunctor $Vekt_{k}: Prightarrow ens$, where $P$ is the category of paracompact spaces with homotopy classes of maps and ens the category of sets and functions, assigning to each paracompact space $B$ the set of $B$-isomorphism classes of vector bundles over $B$.
Unfortunately,I don't have any profound knowledge of set theory and I do not understand why this actually forms a set and not a proper class. Can anyone help me with this?
algebraic-topology category-theory set-theory vector-bundles
asked Nov 20 at 14:47
Joo
185
add a comment |
My issue is mainly set theoretical:
In "Dale Husemoller: Fiber Bundles" it says on page 34 that there is a cofunctor $Vekt_{k}: Prightarrow ens$, where $P$ is the category of paracompact spaces with homotopy classes of maps and ens the category of sets and functions, assigning to each paracompact space $B$ the set of $B$-isomorphism classes of vector bundles over $B$.
Unfortunately,I don't have any profound knowledge of set theory and I do not understand why this actually forms a set and not a proper class. Can anyone help me with this?
algebraic-topology category-theory set-theory vector-bundles
asked Nov 20 at 14:47
Joo
185
My issue is mainly set theoretical:
In "Dale Husemoller: Fiber Bundles" it says on page 34 that there is a cofunctor $Vekt_{k}: Prightarrow ens$, where $P$ is the category of paracompact spaces with homotopy classes of maps and ens the category of sets and functions, assigning to each paracompact space $B$ the set of $B$-isomorphism classes of vector bundles over $B$.
Unfortunately,I don't have any profound knowledge of set theory and I do not understand why this actually forms a set and not a proper class. Can anyone help me with this?
algebraic-topology category-theory set-theory vector-bundles
algebraic-topology category-theory set-theory vector-bundles
asked Nov 20 at 14:47
Joo
185
asked Nov 20 at 14:47
Joo
185
asked Nov 20 at 14:47
Joo
185
asked Nov 20 at 14:47
Joo
185
asked Nov 20 at 14:47
Joo
185
185
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I assumed your vector bundles are finite-dimensional, because the isomorphism classes of infinite-dimensional vector bundles never form a set-not even over a point! Given that, the key point is that (real or complex) vector bundles over $B$ have bounded cardinality: their underlying sets cannot be larger than the maximum of the cardinality of $mathbb{R}$ and the cardinality of the base space. There are only a set worth of isomorphism classes of triples $(E,T,f)$ where $E$ is a set of bounded cardinality, $T$ is a topology on $E$, and $f$ is a continuous function $Eto B$. Indeed, we could assume that $E$ is always a subset of some fixed set by applying an appropriate isomorphism. The set of isomorphism classes of vector bundles is a subset of this set.
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
add a comment |
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1 Answer
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I assumed your vector bundles are finite-dimensional, because the isomorphism classes of infinite-dimensional vector bundles never form a set-not even over a point! Given that, the key point is that (real or complex) vector bundles over $B$ have bounded cardinality: their underlying sets cannot be larger than the maximum of the cardinality of $mathbb{R}$ and the cardinality of the base space. There are only a set worth of isomorphism classes of triples $(E,T,f)$ where $E$ is a set of bounded cardinality, $T$ is a topology on $E$, and $f$ is a continuous function $Eto B$. Indeed, we could assume that $E$ is always a subset of some fixed set by applying an appropriate isomorphism. The set of isomorphism classes of vector bundles is a subset of this set.
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
add a comment |
I assumed your vector bundles are finite-dimensional, because the isomorphism classes of infinite-dimensional vector bundles never form a set-not even over a point! Given that, the key point is that (real or complex) vector bundles over $B$ have bounded cardinality: their underlying sets cannot be larger than the maximum of the cardinality of $mathbb{R}$ and the cardinality of the base space. There are only a set worth of isomorphism classes of triples $(E,T,f)$ where $E$ is a set of bounded cardinality, $T$ is a topology on $E$, and $f$ is a continuous function $Eto B$. Indeed, we could assume that $E$ is always a subset of some fixed set by applying an appropriate isomorphism. The set of isomorphism classes of vector bundles is a subset of this set.
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
add a comment |
I assumed your vector bundles are finite-dimensional, because the isomorphism classes of infinite-dimensional vector bundles never form a set-not even over a point! Given that, the key point is that (real or complex) vector bundles over $B$ have bounded cardinality: their underlying sets cannot be larger than the maximum of the cardinality of $mathbb{R}$ and the cardinality of the base space. There are only a set worth of isomorphism classes of triples $(E,T,f)$ where $E$ is a set of bounded cardinality, $T$ is a topology on $E$, and $f$ is a continuous function $Eto B$. Indeed, we could assume that $E$ is always a subset of some fixed set by applying an appropriate isomorphism. The set of isomorphism classes of vector bundles is a subset of this set.
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
I assumed your vector bundles are finite-dimensional, because the isomorphism classes of infinite-dimensional vector bundles never form a set-not even over a point! Given that, the key point is that (real or complex) vector bundles over $B$ have bounded cardinality: their underlying sets cannot be larger than the maximum of the cardinality of $mathbb{R}$ and the cardinality of the base space. There are only a set worth of isomorphism classes of triples $(E,T,f)$ where $E$ is a set of bounded cardinality, $T$ is a topology on $E$, and $f$ is a continuous function $Eto B$. Indeed, we could assume that $E$ is always a subset of some fixed set by applying an appropriate isomorphism. The set of isomorphism classes of vector bundles is a subset of this set.
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
answered Nov 20 at 17:20
Kevin Carlson
32.5k23271
32.5k23271
add a comment |
add a comment |
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