Show that the chain rule $(G\circ F)_* = G_* \circ F_*$ holds. Show that the chain rule $(G\circ F)_* = G_* \circ F_*$ holds.
I understand the chain rule from calculus but I am not sure how to show that it holds when also considering the pushforward.
So if we let $F: M \to N$ and $G: N \to P$ be two diffeomorphisms of smooth manifolds.
I know that I could use
$$d(g\circ f) = \sum_{i=1}^n \frac{\partial(g\circ f)}{\partial x^i}dx^i$$
and then apply the chain rule to that but I'm not sure how to do that or how it helps me.
Let $F:M \to N$ be a diffeomorphism of smooth manifolds. Then the pushforward $F_*$ of F, is defined by
$$F_* : X(M) \to X(N),     X \mapsto F_*X = dF\circ X \circ F^{-1}$$
map pushes forward vector fields on M to vector fields on N.
I am not sure how to use this knowledge to show the chain rule.
I understand that I could write
$$F_*X = dF\circ X \circ F^{-1}$$
$$G_*X = dG\circ X \circ G^{-1}$$
Therefore
$$G_*X \circ F_*X = dG\circ X \circ G^{-1} \circ dF\circ X \circ F^{-1}$$
However, I'm not sure how to manipulate this such that $(G\circ F)_* = G_* \circ F_*$.
 A: Careful! The expression $G_* X \circ F_* X$ does not make sense. A vector field takes in a point on a manifold and outputs a tangent vector (formally known as a section of the tangent bundle). But, if $p \in M$, then $G_* X$ would be a vector field "evaluated" on a tangent vector $F_* X$, which of course is non-sense. Moreover if $M \neq N$ then the expression $G_* X$ does not make sense either, since $G_*$ takes in vector fields on $N$.
What $G_* \circ F_*$ really does to vector fields $X$ on $M$ is the following:
\begin{align*}
(G_* \circ F_*)(X) &= G_* (dF \circ X \circ F^{-1}) = dG \circ dF \circ X \circ F^{-1} \circ G^{-1} \\&= d(G \circ F) \circ X \circ (G \circ F)^{-1} = (G \circ F)_* (X)
\end{align*}
Note $G_* (dF \circ X \circ F^{-1})$ is an evaluation of $G_*$ on $F_* X = dF \circ X \circ F^{-1}$, not a composition, and $dF$ sends a tangent vector of $p$ to a tangent vector of $F(p)$ (the intuition here is that the derivative of a function at a point is a linear approximation, so the derivative of a function is just a pairing of points and these linear maps).
