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For a smooth n-manifold $M$ we have a tangent bundle $TM$, which is a vector bundle of dimension 2n. Now let $p_M: TM \rightarrow M$ be the canonical projection map $(x, v_x) \mapsto x$. I am looking at an exercise that requires me to show that the derivative of $p_M$, $D p_M$ gives a smooth map $T(T(M)) \rightarrow TM$.

My first difficulty is interpreting what is meant by derivative here? Also, could you provide me with a hint on how to go about the question itself?

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The derivative is the same as it is for any map of manifolds: at every point it gives you a map of tangent spaces. For the question itself, since the tangent bundle is locally trivial, the question (essentially) reduces to looking at the derivative of a projection $U \times \mathbb{R}^n \to U$, which itself reduces to looking at the derivative of a projection $\mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^m$ (why?).

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  • $\begingroup$ So if $S: X \rightarrow Y$ is a smooth map of manifolds then $DS: X \rightarrow \{(T_x X \rightarrow T_{S(x)} Y): x\in X\}$ assigns to $x\in X$ a map $T_x X \rightarrow T_{S(x)} Y$ that maps $v_x \in T_x X$ to $f \mapsto v_x(S \circ f), f \in C^{\infty}(Y)$. Is my understanding correct? $\endgroup$ – gen May 22 at 17:19

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