Yes, in that case (topological manifolds alone) the isomorphism is actually an homeomorphism.WWGD said:Maybe the manifolds themselves only have a topological structure, not a differentiable one?
The cylinder and the mobius strip are two different vector bundles over the base manifold of the circle. The cylinder is a trivial bundle and the Mobius strip is not. They are not isomorphic. I would work through Lee's chapter on vector bundles because it goes more in depth.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.
You may want to look up some more into Stieffel-Whitney classes.cianfa72 said:Yes, in that case (topological manifolds alone) the isomorphism is actually an homeomorphism.
Btw up to dimension 3, every topological manifold admits a differentiable structure and all these structures are equivalent.
Yes, the initial topology will contain whole fibers. The bundle topology also has open subsets of fibers.cianfa72 said:As far as I can tell, the topology assigned on the tangent bundle is not the same as the initial topology from the canonical projection map ##\pi## on the first factor (see Tu's book section 12.1).
Yes, open sets in initial topology contain whole fibers, however each fiber ##\pi^{-1} \{p \}## will be open in the initial topology iff the base space B has the discrete topology.martinbn said:Why!
No, i meant why do you make these statements/questions?cianfa72 said:Yes, open sets in initial topology contain whole fibers, however each fiber ##\pi^{-1} \{p \}## will be open in the initial topology iff the base space B has the discrete topology.
Just to double check my understanding (also sometimes I've problem with reading english).martinbn said:No, i meant why do you make these statements/questions?