Given smooth manifold $M$ how do you prove that the projection map $\pi : TM\to M$, $(p,v)\mapsto p$ is smooth?
Projection map from TM to M is smooth
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differential-geometry
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1You can assume $M=\mathbb R^n$. (The notation $(p,v)$ might be misleading.) – 2012-03-26
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0You have to choose an adequate chart for TM! – 2012-03-26
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4For fundamental issues like this, it may partially depend on how you define a manifold and how you define its tangent bundle. For example, if you define smooth manifolds via [local rings](http://en.wikipedia.org/wiki/Smooth_manifold#Structure_sheaf), the proof would be much different from if you define it via charts and atlases, and it would be still different if you define smooth manifolds by their embeddings in Euclidean space. So in this sense, please specify how a smooth manifold and a tangent bundle are defined for you. – 2012-03-26
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0I still don't understand. I have as charts for $TM$ the maps $\tau_i: U_i\times\mathbb{R}^n\to \bigsqcup_{p\in U_i}T_pM$ which send $(x,v)\mapsto (\sigma_i(x),[\sigma_i,v]_p)$ where $\sigma_i:U_i\to U_i'\subseteq M$ is a chart for $M$. Where do I go from here? – 2012-03-26
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0@WillieWong, I define it using charts and atlases – 2012-03-26
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1Let me go back to @Pierre-Yves' comment: once you have chosen an atlas on $TM$ and another on $M$ such that in local coordinates $\pi:(p,v)\mapsto p$, then you are done, as obviously coordinate projections $\mathbb{R}^{2k} \to \mathbb{R}^k$ are smooth maps. So if you defined the smooth structure on $TM$ using the charts you gave above, there is pretty much nothing left to prove. What you need to think about is "why am I guaranteed to be able to find local coordinates such that $\pi$ is given as the canonical projection?" – 2012-03-26