Tangent Bundle of Product is diffeomorphic to Product of Tangent Bundles

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SUMMARY

The tangent bundle of the product manifold T(MxN) is diffeomorphic to the product of the tangent bundles TM x TN. This conclusion is established by constructing projection maps from MxN to M and N, which facilitates the formation of the diffeomorphism. The proof involves recognizing that the sets are identical and requires a proper differentiable structure, which is achieved by using contractible neighborhoods and defining the topology as the product topology. The local diffeomorphisms TU and TV are combined to demonstrate that T(UxV) is equivalent to U x V x R^(n+m).

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Amateur659
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My apologies if this question is trivial. I have searched the forum and haven't found an existing answer to this question.

I've been working through differential geometry problem sets I found online (associated with MATH 481 at UIUC) and am struggling to show that T(MxN) is diffeomorphic to TM x TN.

My intuition is that I could set up the product of projection maps from MxN to M and MxN to N to form the diffeomorphism. However, I seem to be stuck on the implementation (which probably indicates gaps in my understanding of previous material).

Thanks for your assistance.
 
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This is the kind of question that requires (only) a precise application of all definitions. The sets are clearly identical:
$$
T(M\times N)= \bigcup_{(p,q)\in M\times N}\{(p,q)\} \times T_{(p,q)}(M\times N) = \bigcup_{p\in M} \{p\}\times T_{p}(M)\times \bigcup_{q\in N} \{q\}\times T_{q}(N)=T(M) \times T(N)
$$
Next, we need a differentiable structure on both. Differentiability is a local property. So we can pick a point ##(p,q)\in U\times V## from a contractible neighborhood. By defining it as ##U\times V## we already fixed the topology, namely the product topology. Now gather all local diffeomorphisms ##TU=U\times \mathbb{R}^n\, , \,TV=V\times \mathbb{R}^m## and combine them to a diffeomorphism ##T(U\times V)= U\times V \times \mathbb{R}^{n+m}.##
 
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