In one of the questions of my homework, tangent bundles are defined as in the picture. So I was wondering how to prove the openness of $\pi^{-1}(U)$, since the question only states that $\pi^{-1}(U)$ with the subspace topology is hemeomorphic to $V\times \mathbb{R}^n$. Actually, I do not understand how the topology is defined at all. (I do know how the tangent bundle is usually defined).

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Well you see that $T\varphi$ is a bijection, so since you know the product topology on $V \times \mathbb{R}^n $, that automatically give you a topology on $\pi^{-1}(U)$. If you want, you give $\pi^{-1}(U)$ the topology which makes the map $T\varphi$ a homeomorphism.

This is the analogue of when you define the quotient topology, that is you give the target space the topology which makes the surjection continous.

  • $\begingroup$ You mean that $\pi^{-1}(U)$ is open by definition? $\endgroup$ – user150248 Oct 8 '18 at 0:29
  • $\begingroup$ Yes, but also any subset of $\pi^{-1}(U)$ is open iff its image uder $T\varphi$ is open. $\endgroup$ – Baol Oct 8 '18 at 13:43

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