Mixing
Must every manifold bundle be a fiber bundle?
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I have been learning differential geometry from this lecture series by Frederic Schuller's. Here he defines a bundle (of topological manifolds) as
A triple, $(E, pi, M),$ where $E$ and $M$ are topological manifolds and $pi:E longrightarrow M$ is a surjective and continuous function.
Here he defines a fiber bundle as
A bundle $(E, pi, M)$ such that $forall p in M: text{preim}_pi big({p}big) cong F$ for some manifold $F$.
Using these definitions (which I hope are standard), does there exist some bundle of topological manifolds that is not a fiber bundle (i.e., has fibers that are not homeomorphic to each other)?
Sidenote, Schuller provides examples here and here of bundles over a set that are not fiber bundles, but the base set is not a manifold in either case, as he clarifies in this later video.
EDIT: I'll clarify that the $F$ in the definition of fiber bundle is intended to be the same $F$ for each point $p$, i.e., a fiber bundle is a bundle whose fibers are all homeomorphic to each other.
differential-geometry manifolds fiber-bundles
asked Jan 24 at 17:12
WillGWillG
488310
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|
show 3 more comments
$begingroup$
I have been learning differential geometry from this lecture series by Frederic Schuller's. Here he defines a bundle (of topological manifolds) as
A triple, $(E, pi, M),$ where $E$ and $M$ are topological manifolds and $pi:E longrightarrow M$ is a surjective and continuous function.
Here he defines a fiber bundle as
A bundle $(E, pi, M)$ such that $forall p in M: text{preim}_pi big({p}big) cong F$ for some manifold $F$.
Using these definitions (which I hope are standard), does there exist some bundle of topological manifolds that is not a fiber bundle (i.e., has fibers that are not homeomorphic to each other)?
Sidenote, Schuller provides examples here and here of bundles over a set that are not fiber bundles, but the base set is not a manifold in either case, as he clarifies in this later video.
EDIT: I'll clarify that the $F$ in the definition of fiber bundle is intended to be the same $F$ for each point $p$, i.e., a fiber bundle is a bundle whose fibers are all homeomorphic to each other.
differential-geometry manifolds fiber-bundles
asked Jan 24 at 17:12
WillGWillG
488310
$endgroup$
7
$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
1
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
2
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@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
2
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@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
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– Moishe Kohan
Jan 25 at 16:31
2
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37
|
show 3 more comments
$begingroup$
I have been learning differential geometry from this lecture series by Frederic Schuller's. Here he defines a bundle (of topological manifolds) as
A triple, $(E, pi, M),$ where $E$ and $M$ are topological manifolds and $pi:E longrightarrow M$ is a surjective and continuous function.
Here he defines a fiber bundle as
A bundle $(E, pi, M)$ such that $forall p in M: text{preim}_pi big({p}big) cong F$ for some manifold $F$.
Using these definitions (which I hope are standard), does there exist some bundle of topological manifolds that is not a fiber bundle (i.e., has fibers that are not homeomorphic to each other)?
Sidenote, Schuller provides examples here and here of bundles over a set that are not fiber bundles, but the base set is not a manifold in either case, as he clarifies in this later video.
EDIT: I'll clarify that the $F$ in the definition of fiber bundle is intended to be the same $F$ for each point $p$, i.e., a fiber bundle is a bundle whose fibers are all homeomorphic to each other.
differential-geometry manifolds fiber-bundles
asked Jan 24 at 17:12
WillGWillG
488310
$endgroup$
I have been learning differential geometry from this lecture series by Frederic Schuller's. Here he defines a bundle (of topological manifolds) as
A triple, $(E, pi, M),$ where $E$ and $M$ are topological manifolds and $pi:E longrightarrow M$ is a surjective and continuous function.
Here he defines a fiber bundle as
A bundle $(E, pi, M)$ such that $forall p in M: text{preim}_pi big({p}big) cong F$ for some manifold $F$.
Using these definitions (which I hope are standard), does there exist some bundle of topological manifolds that is not a fiber bundle (i.e., has fibers that are not homeomorphic to each other)?
Sidenote, Schuller provides examples here and here of bundles over a set that are not fiber bundles, but the base set is not a manifold in either case, as he clarifies in this later video.
EDIT: I'll clarify that the $F$ in the definition of fiber bundle is intended to be the same $F$ for each point $p$, i.e., a fiber bundle is a bundle whose fibers are all homeomorphic to each other.
differential-geometry manifolds fiber-bundles
differential-geometry manifolds fiber-bundles
asked Jan 24 at 17:12
WillGWillG
488310
asked Jan 24 at 17:12
WillGWillG
488310
edited Jan 24 at 23:44
WillG
asked Jan 24 at 17:12
WillGWillG
488310
asked Jan 24 at 17:12
WillGWillG
488310
asked Jan 24 at 17:12
WillGWillG
488310
488310
7
$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
1
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
2
$begingroup$
@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
2
$begingroup$
@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
$endgroup$
– Moishe Kohan
Jan 25 at 16:31
2
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37
|
show 3 more comments
7
$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
1
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
2
$begingroup$
@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
2
$begingroup$
@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
$endgroup$
– Moishe Kohan
Jan 25 at 16:31
2
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37
7
7
$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
1
1
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
2
2
$begingroup$
@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
$begingroup$
@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
2
2
$begingroup$
@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
$endgroup$
– Moishe Kohan
Jan 25 at 16:31
$begingroup$
@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
$endgroup$
– Moishe Kohan
Jan 25 at 16:31
2
2
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37
|
show 3 more comments
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$begingroup$
This is an offensively bad definition, sure to cause confusion. Most surjective continuous maps are not fiber bundles. And the definition of fiber bundle is not even correct.
$endgroup$
– user98602
Jan 24 at 17:38
1
$begingroup$
@WillG Take the projection $[0,1]to S^1=[0,1]/sim$, where here $sim$ identifies $0$ and $1$. Then the fiber above the equivalence class of $0$ has two points in it but the fiber above any other point has only one point
$endgroup$
– leibnewtz
Jan 25 at 0:41
2
$begingroup$
@WillG The example has essentially nothing to do with manifolds with boundary. Some other options: collapse an arc in $S^1$ to a point; then you have a continuous map from the circle itself in which all fibers but one are points, and the remaining fiber is an arc. Or consider the exponential map $Bbb R to S^1$ restricted to $(-epsilon, 2pi+epsilon)$, which mostly mostly has 1-point preimage but has a few 2-point preimages. You can rig up self-maps of the circle so that some fibers are Cantor sets!
$endgroup$
– user98602
Jan 25 at 11:08
2
$begingroup$
@leibnewtz: Unfortunately, yes, see math.stackexchange.com/questions/2176044/…
$endgroup$
– Moishe Kohan
Jan 25 at 16:31
2
$begingroup$
I suggest, you avoid using either definition that he gives. They are good for the intuition but you cannot prove much with these definitions; the actual definition (that is of a locally trivial fiber bundle or, more generally, a fibration in the sense of Hurewicz or Serre) is much more restrictive. Better, first read the wikipedia article: en.wikipedia.org/wiki/Fiber_bundle
$endgroup$
– Moishe Kohan
Jan 25 at 16:37