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Given smooth manifold $M$ how do you prove that the projection map $\pi : TM\to M$, $(p,v)\mapsto p$ is smooth?

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You can assume $M=\mathbb R^n$. (The notation $(p,v)$ might be misleading.) –  Pierre-Yves Gaillard Mar 26 '12 at 12:34
    
You have to choose an adequate chart for TM! –  checkmath Mar 26 '12 at 12:36
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For 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, 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. –  Willie Wong Mar 26 '12 at 12:53
    
I 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? –  09867 Mar 26 '12 at 12:56
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Let 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?" –  Willie Wong Mar 26 '12 at 13:09
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1 Answer 1

There are nice charts for $M$ and $TM$, given a chart $x$ for $M$: $$ x:U\rightarrow V \text{ and the induced } Tx:TU\rightarrow U \times \mathbb R^n $$ One way to show that a map is smooth, is to express it in charts, which typically is written $y\circ f \circ x^{-1}$ or something like that. And it is enough to show smoothness for one pair of charts around $p$ and $f(p)$ for every point $p$. For $\pi$ and the nice charts this is just $$ x \circ \pi \circ (Tx)^{-1} $$ which is the projection $U \times \mathbb R^n \rightarrow U$

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