# Derivative of a tangent bundle projection

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?

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?).
• 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? – gen May 22 at 17:19