Fiber bundle homeomorphism with the fiber

cianfa72 said:

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##.

mathwonk said:

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 said:

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 said:

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

cianfa72 said:

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

cianfa72 said:

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

jbergman said:

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 said:

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 said:

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

cianfa72 said:

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.

jbergman said:

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 said:

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

jbergman said:

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

cianfa72 said:

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## ?

cianfa72 said:

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#

cianfa72 said:

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 said:

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 said:

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 said:

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 said:

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##) ?

cianfa72 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.

jbergman said:

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

cianfa72 said:

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##.

cianfa72 said:

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.

martinbn said:

This cannot happen.

cianfa72 said:

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

cianfa72 said:

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

jbergman said:

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

cianfa72 said:

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 said:

Isomorphisms of the bundles not ##M##.

cianfa72 said:

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.