Tangent Bundles, T(MxN) is Diffeomorphic to TM x TN

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The discussion centers on proving that the tangent bundle T(MxN) is diffeomorphic to TM x TN for smooth manifolds M and N. Participants analyze the structure of tangent vectors at points in the product manifold, specifically addressing the representation of vectors in T(MxN) and their relation to tangent spaces TM and TN. There is confusion regarding the assumptions made about vector fields and the need for careful mapping between the tangent spaces. One participant suggests finding a map that accurately translates vectors from T(MxN) to the product of the tangent spaces. The conversation highlights the complexities involved in establishing the diffeomorphism and the necessity for precise definitions in the proof.
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Homework Statement



If M and N are smooth manifolds, then T(MxN) is diffeomorphic to TM x TN

Homework Equations





The Attempt at a Solution



So I'm here

let ((p,q),v) \in T(MxN)

then p \in M and q \in N and v \in T(p,q)(MxN).

so T(p,q)(MxN) v = \sum_{i=1}^{m+n} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)}

= \sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{p} + \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{q}

not sure if this is it?
 
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You kind of assume that \frac{\partial}{\partial x^i} are vector fields on both M\times N and N. I think you should be a little more careful than this...
 
Hi Micro, I'm not sure I understand what you mean, am am assuming however that

\sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{p} and \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{q}

are bases for TM and TN respectively and I'm not sure if that's good enough.
 
Well, you assume \frac{\partial}{\partial x_i}\vert_p is both in T_pM and T_pN. You can't do that.
 
v = \sum_{i=1}^{m} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)} + \sum_{i=}^{m+1} v_{i}\frac{\partial}{∂x_{i}}|_{(p,q)}

so i need to find a map that takes v \rightarrow (w,y)

where w\inTpM and y\inTpN

and (w,y) \in TpM x TqN

does this sound better? I was thinking about this problem a lot. I'm not sure if that's a diffeo right off the get.
 
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