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Can the tangent bundle $TM$ be considered as the product manifold of a smooth manifold $M$ and its tangent planes $T_pM$ with $p \in M$?.
If this is the case, then can we conclude that it is a smooth manifold because it is a product of smooth manifolds and we can find a differentiable structure that comes from the maps of $M$ and the maps of $T_pM$?

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    $\begingroup$ No, in general $TM$ may not even be diffeomorphic to $\Bbb R^n\times M$. The manifolds such that $TM$ is isomorphic as a bundle to $M\times \Bbb R^n$ are called parallelizable. $\endgroup$ – user228113 Oct 20 '17 at 8:52
  • $\begingroup$ @G.Sassatelli But, can't we define a map $\phi$ that takes a point $p$ from $R^n$ to $M$(using a map $\psi$ of $M$) and also takes a vector from $T_p R^n$ to $T_{\psi (p)} M$? Isn't this also the kind of maps that we use for product manifolds? (i.e. mapping a point from one manifold with its map and mapping a point of another manifold using its map). My problem is that I can't quite understand geometrically the reason of why it can't be considered as a product manifold $\endgroup$ – TheQuantumMan Oct 20 '17 at 11:39

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