Fiber bundle homeomorphism with the fiber

[cianfa72]

TL;DR Summary

About the Fiber bundle homeomorphism between the preimage of a point in the base space under the continuous map $\pi$ and the fiber

Hi, in the definition of fiber bundle there is a continuous onto map $\pi$ from the total space $E$ into the base space $B$. Then there are local trivialization maps $\varphi: \pi^{-1}(U) \rightarrow U \times F$ where the open set $U$ in the base space is the trivializing neighborhood. These maps by definition are homeomorphisms using the subspace topology from $E$ for $\pi^{-1}(U)$ and the product topology for $U \times F$.

Then there is a claim that each $\pi^{-1}(\{p\})$ is homeomorphic with the fiber $F$ (that is a topological space).

My question: is this homeomorphism done using the subspace topology from $\pi^{-1}(U), p \in U$ for the set $\pi^{-1}(\{p\})$ ? Thanks.

Edit: since $\pi$ is continuous then the subspace topology for $\pi^{-1}(\{p\})$ from $\pi^{-1}(U)$ is the same from $E$.

 

[jbergman]

TL;DR Summary: About the Fiber bundle homeomorphism between the preimage of a point in the base space under the continuous map $\pi$ and the fiber

Hi, in the definition of fiber bundle there is a continuous onto map $\pi$ from the total space $E$ into the base space $B$. Then there are local trivialization maps $\varphi: \pi^{-1}(U) \rightarrow U \times F$ where the open set $U$ in the base space is the trivializing neighborhood. These maps by definition are homeomorphisms using the subspace topology from $E$ for $\pi^{-1}(U)$ and the product topology for $U \times F$.

Then there is a claim that each $\pi^{-1}(\{p\})$ is homeomorphic with the fiber $F$ (that is a topological space).

My question: is this homeomorphism done using the subspace topology from $\pi^{-1}(U), p \in U$ for the set $\pi^{-1}(\{p\})$ ? Thanks.

Edit: since $\pi$ is continuous then the subspace topology for $\pi^{-1}(\{p\})$ from $\pi^{-1}(U)$ is the same from $E$.

It seems you have answered your own question. For fiber bundles you have three ingredients.

1. A model topological space $F$.
2. A base space $B$ with it's own topology.
3. The total space $E$ which also has a topology and all the maps you mentioned.

When talking about restrictions of the various maps on the total space we are always talking about the subspace topology.

 

[mathwonk]

you have omitted the crucial fact that these maps fit into a commutative diagram, so that π corresponds under phi to projection on U, which forces π^-1(p) to map homeomorphically to {p}xF, (with subspace topologies).

 

[cianfa72]

you have omitted the crucial fact that these maps fit into a commutative diagram, so that π corresponds under phi to projection on U, which forces π^-1(p) to map homeomorphically to {p}xF, (with subspace topologies).

You mean that since these maps fit into a commutative diagram, we get $\pi^{-1}(\{p \})$ via $\varphi^{-1} \circ \text{proj}^{-1}(\{ p\})$ and the first is by definition $\{ p \} \times F$. Then $\{ p \} \times F$ with subspace topology maps homeorphically to $F$, hence by composition of homeomorphisms $\pi^{-1}(\{p \})$ is homeomorphic to $F$.

 

[cianfa72]

Another point: the fiber bundle definition requires that for any $p \in B$ there is an open neighborhood in $B$ such that there exists the homeomorphism $\varphi$ as in post #1.

My question: is the above true for any open neighborhood of point $p \in B$ ?

 

[martinbn]

Another point: the fiber bundle definition requires that for any $p \in B$ there is an open neighborhood in $B$ such that there exists the homeomorphism $\varphi$ as in post #1.

My question: is the above true for any open neighborhood of point $p \in B$ ?

No, because you can take the whole $B$ as a neighborhood, and it will not be true unless the bundle is trivial.

 

[cianfa72]

No, because you can take the whole $B$ as a neighborhood, and it will not be true unless the bundle is trivial.

Ok, the same applies in general to open sets in the base space $B$ chart's atlas that contain the point $p$ ?

 

[jbergman]

Ok, the same applies in general to open sets in the base space $B$ chart's atlas that contain the point $p$ ?

Interesting question. My guess is no but I don't know the answer.

I think of a fiber bundle as line a product $B \times F$, but with twisting. That twisting can result in open sets in $E$ associated with local trivializations that are smaller or larger than the inverse image of open sets in the base space in a chart, or at least that is my guess.

 

[jbergman]

Ok, the same applies in general to open sets in the base space $B$ chart's atlas that contain the point $p$ ?

Actually, this question, https://math.stackexchange.com/questions/186145/a-fiber-bundle-over-euclidean-space-is-trivial, makes me think you might be correct, since every open set corresponding to a chart is homeomorphic to euclidean space. Hence there should be an open trivialization of the fiber bundle over it.

 

[cianfa72]

Actually, this question, https://math.stackexchange.com/questions/186145/a-fiber-bundle-over-euclidean-space-is-trivial, makes me think you might be correct, since every open set corresponding to a chart is homeomorphic to euclidean space. Hence there should be an open trivialization of the fiber bundle over it.

A chart is homeomorphism to an open set in Euclidean space. Take for example a disjoint union of open sets in Euclidean space. There is a (global) chart from it to an open set (the homeomorphism is just the Identity map) however it is not homeomorphic to the entire Euclidean space.

 

[jbergman]

A chart is homeomorphism to an open set in Euclidean space. Take for example a disjoint union of open sets in Euclidean space. There is a (global) chart from it to an open set (the homeomorphism is just the Identity map) however it is not homeomorphic to the entire Euclidean space.

That's wrong. https://en.m.wikipedia.org/wiki/Topological_manifold

 

[jbergman]

That's wrong. https://en.m.wikipedia.org/wiki/Topological_manifold

Also see, https://en.m.wikipedia.org/wiki/Topological_manifold

 

[cianfa72]

I was thinking about the atlas with only a (global) chart for the topological manifold made up of disjoint union of Euclidean open sets. Such a chart is not homeomorphism with the entire Euclidean space (e.g. that manifold is not connected). Nevertheless the maximal atlas for it contains charts locally homeomorphic to the entire Euclidean space.

In other words: not every chart of a topological manifold has to be homeomorphic with the entire Euclidean space.

 

[jbergman]

I was thinking about the atlas with only a (global) chart for the topological manifold made up of disjoint union of Euclidean open sets. Such a chart is not homeomorphism with the entire Euclidean space (e.g. that manifold is not connected). Nevertheless the maximal atlas for it contains charts locally homeomorphic to the entire Euclidean space.

I don't think that is a chart. A chart has to be homeomorphic to an open ball of some sort, or at least that is the definition that I know. That is the whole point of describing something as locally euclidean.

 

[cianfa72]

A chart has to be homeomorphic to an open ball of some sort, or at least that is the definition that I know. That is the whole point of describing something as locally euclidean.

See for instance here Atlas chart. It talks about generic open sets in Euclidean space.

 

[jbergman]

See for instance here Atlas chart. It talks about generic open sets in Euclidean space.

Interesting. Ok. Then maybe not because if you have disconnected components then things are more complex.

 

[cianfa72]

Interesting. Ok. Then maybe not because if you have disconnected components then things are more complex.

Do you refer to my post#7 ? Hence even for base space's chart open sets $U_i$ their preimage under $\pi$ might not be homeomorphic to $U_i \times F$ ?

 

[jbergman]

Do you refer to my post#7 ? Hence even for base space's chart open sets $U_i$ their preimage under $\pi$ might not be homeomorphic to $U_i \times F$ ?

Yes. But, it's just a guess at this point.

 

[cianfa72]

A very basic doubt about topological manifold. What if the target/codomain of the atlas's charts homeomorphisms would not open in $\mathbb R^n$ ?

The transition maps using the subspace topologies in the intersections from the target spaces (each target space endowed with the subspace topology from $\mathbb R^n$) would result in homeomorphisms as well.

So what is the reason a such condition is actually required in the definition of topological manifold ?

 

[jbergman]

Do you refer to my post#7 ? Hence even for base space's chart open sets $U_i$ their preimage under $\pi$ might not be homeomorphic to ##U_i \times F#

A very basic doubt about topological manifold. What if the target/codomain of the atlas's charts homeomorphisms would not open in $\mathbb R^n$ ?

The transition maps using the subspace topologies in the intersections from the target spaces (each target space endowed with the subspace topology from $\mathbb R^n$) would result in homeomorphisms as well.

So what is the reason a such condition is actually required in the definition of topological manifold ?

Not sure I fully understand. If the domain of a chart is open in $M$ and the codomain is not open in $\mathbb R^n$, how can you expect the manifold to be locally homeomorphic to $\mathbb R^n$?

 

[cianfa72]

If the domain of a chart is open in $M$ and the codomain is not open in $\mathbb R^n$, how can you expect the manifold to be locally homeomorphic to $\mathbb R^n$?

From a general point of view I see no reason why an open set in a topological space (endowed with the subspace topology) might not be homeomorphic to a non open set in $\mathbb R^n$ with the subspace topology.

 

[jbergman]

From a general point of view I see no reason why an open set in a topological space (endowed with the subspace topology) might not be homeomorphic to a non open set in $\mathbb R^n$ with the subspace topology.

Yes, but that's not the point of a topological manifold. An n-dimensionql topological manifold is something that is locally like $\mathbb R^n$. For example with your definition, you could define a 3-dimensional manifold with chart maps from it to $\mathbb R^3$ that are all 1-d closed subsets of $\mathbb R^3$, in other words a 1-dimensional manifold.

 

[cianfa72]

For example with your definition, you could define a 3-dimensional manifold with chart maps from it to $\mathbb R^3$ that are all 1-d closed subsets of $\mathbb R^3$, in other words a 1-dimensional manifold.

Ah ok, so for example we could end up with an open set of $\mathbb R^3$ homeomorphic to a 2-d plane in $\mathbb R^3$ (the latter understood as topological space of its own with the subspace topology from $\mathbb R^3$) ?

Edit: I think the above example doesn't make sense thanks to Invariance of domain theorem. Perhaps it could be true using topologies other than Euclidean topology in $\mathbb R ^3$ as 'naked' set.

 

[jbergman]

Ah ok, so for example we could end up with an open set of $\mathbb R^3$ homeomorphic to a 2-d plane in $\mathbb R^3$ (the latter understood as topological space of its own with the subspace topology from $\mathbb R^3$) ?

Yes, or a line like I said.

Edit: I think the above example doesn't make sense thanks to Invariance of domain theorem. Perhaps it could be true using topologies other than Euclidean topology in $\mathbb R ^3$ as 'naked' set.

Invariance of domain doesn't apply because a 2-d plane is a closed subset. I think I have given you enough information that this should be perfectly clear at this point. This is the most basic concept about topological manifolds that you must understand before understanding anything else. To repeat, a topological manifold is locally like an open ball in $\mathbb{R}^n$.

 

[cianfa72]

Invariance of domain doesn't apply because a 2-d plane is a closed subset.

Sorry, I believe it does apply the other way. Suppose the open set $A$ of $\mathbb R ^3$ was homeomorphic via $\varphi$ to a 2-plane in $\mathbb R^3$ with the subspace topology. Then there would be an injective continuous map from $A$ to $\mathbb R ^3$. The conditions of Invariance of domain theorem apply hence the image $\varphi(A)$ would be open in $\mathbb R^3$.

 

[jbergman]

Sorry, I believe it does matter. Suppose the open set $A$ of $\mathbb R ^3$ was homeomorphic via $\varphi$ to a 2-plane in $\mathbb R^3$ with the subspace topology. Then there would be an injective continuous map from $A$ to $\mathbb R ^3$. The conditions of Invariance of domain theorem apply hence the image $\varphi(A)$ would be open in $\mathbb R^3$.

Well, yes your example was silly. An example is the 2-plane being homeomorphic to a closed set of $\mathbb R^3$ like you wanted to allow. Anyways, I think this discussion has gone on long enough.

 

[martinbn]

Suppose the open set $A$ of $\mathbb R ^3$ was homeomorphic via $\varphi$ to a 2-plane in $\mathbb R^3$ with the subspace topology.

This cannot happen.

 

[cianfa72]

This cannot happen.

Right, we can prove it by contraddiction using the Invariance of domain theorem.

 

[cianfa72]

A general question on tangent and cotangent bundle. Are they always trivial bundle regardless of the base space topological manifold ?

 

[jbergman]

A general question on tangent and cotangent bundle. Are they always trivial bundle regardless of the base space topological manifold ?

Definitely not. Look at the tangent space of the Mobius strip. If tangent bundles were always trivial then you wouldn't need connections as you could identify tangent vectors at different points globally like in $\mathbb{R}^n$. Tangent bundles are trivial if and only if the manifold admits a smooth global frame.

You seem to have a lot pf confusion about some of the basic geometric notions here. Have you systematically worked through a book like Tu's Introduction to Manifolds?

 

[WWGD]

A general question on tangent and cotangent bundle. Are they always trivial bundle regardless of the base space topological manifold ?

They're only guanteed to be locally , only at times globally trivial. IIRC, the latter is true only for product spaces. The term " Trivial Bundle" wouldn't have been coined.

 

[cianfa72]

Have you systematically worked through a book like Tu's Introduction to Manifolds?

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

 

[martinbn]

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

Isomorphisms of the bundles not $M$.

 

[cianfa72]

Isomorphisms of the bundles not $M$.

However, in general, bundles do not have a vector space structure.

Edit: eventually the term isomorphism is not restricted to just vector spaces (e.g. in the context of differentiable manifolds a diffeomorphism is actually an isomorphism).

 

[WWGD]

Yes, I'm keeping study Tu's book.

In particular at the end of section 12.3 he talks of isomorphism between vector bundles over the same manifold $\mathbb M$. But the manifold $\mathbb M$ in general may not have a vector space structure, so why he talks of isomorphisms and non just diffeomorphism ? Thanks.

Maybe the manifolds themselves only have a topological structure, not a differentiable one?