Understanding the pullback of the differential of the projection map of a cotangent bundle

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Let $X$ be a smooth manifold. Then any point $p \in T^*X$ is of the form $p = (x, \xi)$ where $x \in X$ and $\xi \in T^*_xX$. We have the projection map $$\pi: T^*X \rightarrow X \\ (x, \xi) \mapsto x$$ and the pushforward $$d\pi_p: T_p\big(T^*X\big) \rightarrow T_xX.$$

Recalling that a one-form on a manifold $M$ is a function defined on $T_pM$ for $p \in M$, I would expect that $$(d\pi_p)^*: T^*\big(T_xX) \rightarrow T^*\Big(T_p\big(T^*X\big)\Big) \tag{1}$$ because a one-form on $T_p\big(T^*X)$ takes as an input a vector in the tangent space to $T_p\big(T^*X\big)$.

However, the definition of the tautological 1-form is instead given by $$(d\pi_p)^*: T_x^*X \rightarrow T_p^*\big(T^*X\big). \tag{2}$$

Can anyone help explain why the definition in (2) makes sense as opposed to (1)?