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Let's consider two manifolds M and N.

I am trying to figure out if it is true that the tangent bundle of the product $M\times N$ is isomorphic to the cartesian product of the pullbacks of the tangent bundle through the projections.

In formulae:

$T(M\times N)\cong \pi_M^*(TM)\times\pi^*_N(TN)$

I think that is the case but I'm quite struggling proving it.

popoolmica
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    Have you managed to define a map in either direction? – Mariano Suárez-Álvarez Nov 27 '15 at 18:28
  • No I didn't actually. – popoolmica Nov 27 '15 at 19:03
  • There is an obvious map $M\times N\to M$, whose differential is a map $T(M\times N)\to TM$. Can you construct from that a map $T(M\times N)\to \pi_M^*(TM)$? – Mariano Suárez-Álvarez Nov 27 '15 at 19:08
  • Is that map $\pi_M$ itself? Is it correct to think that given $x\in M$, $\forall y\in N$ the fiber of $\pi_M^*(TM)$ on $(x,y)$ is $T_xM$ itself? – popoolmica Nov 27 '15 at 19:26
  • You can't really do that, for the the fibers over (x, y) and over (x, y') would be the same set! You probably have an explicit construction for the pullback: use it. – Mariano Suárez-Álvarez Nov 27 '15 at 23:18
  • Sorry I must be missing something. What I wrote in my previous comment came from my understanding of the definition of the pull-back of a vector bundle. – popoolmica Nov 28 '15 at 09:50
  • The fibers of $\pi_M^(TM)$ over different points of $M\times N$ have* to be disjoint (this follows simply the definition of fibers!) It is true, though, that the fiber of that pullback over $(x,y)$ is in bijection with the fiber of $TM$ over $x$ and, in some cirtumstances, you can even think of them as being the same. In this particular case, where you need the details of things, you can't. – Mariano Suárez-Álvarez Nov 28 '15 at 17:42

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