My question is pretty much summed up in the title. We have a tangent vector $X\in T_pM$ at a point $p\in M$ for a smooth manifold $M$, a diffeomorphism $\varphi:M\to N$ for another smooth manifold $N$ and a function $f\in\mathcal{F}(N)$.
Why do we get that $X(\varphi^*f)=(\varphi_*X)f$ and what happens to the point $p$ in the transformation? Where $\varphi_*,\,\varphi^*$ are the pushforward and the pullback respectively. I came upon this question, when reading this answer.
Best Answer
Consider a function $f\colon N\to \Bbb R$, a tangent vector $X\in T_pM$ and a diffeomorphism $\varphi\colon M \to N$. Note that $X\in T_pM$ and $\varphi_*X = d_p\varphi(X) \in T_{\varphi(p)}N$. Then one has \begin{align} X(\varphi^* f) &= \left(d_p(\varphi^*f)\right) X & \text{by definition of the action of $X\in T_pM$},\\ &= d_p(f\circ \varphi) X & \text{by definition of $\varphi^*f$},\\ &= (d_{\varphi(p)}f \circ d_p\varphi)X & \text{by the chain rule},\\ &= d_{\varphi(p)}f(d_p\varphi X) &\text{by associativity},\\ &= d_{\varphi(p)}f(\varphi_*X) & \text{by definition of $\varphi_*X$},\\ &= (\varphi_* X)f & \text{by definition of the action of $\varphi_*X\in T_{\varphi(p)}N$}. \end{align} Of course, this proof relies on the chain rule, so using this to prove the chain rule would be a circular argument. So let's just pretend that the chain rule has been shown, for instance in coordinates.