I'm struggling to prove the chain rule for differentiable manifolds. I've found a few other similar questions, but they use different formulations that I haven't been able to relate to my own.
What I want to prove is the following claim, with (hopefully) all of the relevant definitions:
Let $X,Y,Z$ be differentiable manifolds, and $f:X\rightarrow Y,g:Y\rightarrow Z$ be differentiable maps. Then at any point $p \in X$,
\begin{align}
D_p(g\circ f)=D_{f(p)}g \circ D_pf.
\end{align}
The differential $D_pf$ of the map $f$ at $p\in X$ is defined as
\begin{aligned}
D_pf:T_pM &\rightarrow T_f(p)N\newline
[\gamma]=\left.\frac{d}{dt}\right|_{t=0}\gamma(t)&\mapsto \left.\frac{d}{dt}\right|_{t=0}(f\circ\gamma)(t)=[f \circ \gamma],
\end{aligned}
and vectors $[\gamma]$ are the (equivalence classes of) derivatives of curves $\gamma:\mathbb{R}\rightarrow X$ such that $\gamma(0)=p$.
I take $(U,\phi)$, $(V,\psi)$, and $(W,\chi)$ to be charts of $X$, $Y$, and $Z$, respectively.
My first try was to insert charts as follows:
\begin{aligned}
D_p(g\circ f)([\gamma])&= \left.\frac{d}{dt}\right|_{t=0}(g\circ f\circ\gamma)(t)\newline
&=\left.\frac{d}{dt}\right|_{t=0}(\chi^{-1} \circ (\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t).
\end{aligned}
Both the bracketed terms are the maps on $\mathbb{R}^n$ that we can use for the usual chain rule, so if the factor of $\chi^{-1}$ wasn't there, it would be
\begin{aligned}
&\left.\frac{d}{dt}\right|_{t=0}((\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t)\newline
=&D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ\left.\frac{d}{dt}\right|_{t=0}(\psi \circ f\circ\gamma)(t)\newline
=&D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ D_{f(p)}\psi([f\circ \gamma]),
\end{aligned}
where
\begin{aligned}
D_p\phi:T_pX&\rightarrow \mathbb{R}^n\newline
[\gamma]&\mapsto [\phi \circ \gamma]
\end{aligned}
is a vector space isomorphism between the tangent space $T_pX$ and $\mathbb{R}^n$.
So since I needed to 'get rid of' this factor of $\chi^{-1}$, my next try was to use the vector space isomorphism (which I can invert, since it is an isomorphism) to do exactly that. This looks like
\begin{aligned}
(D_{g(f(p))}\chi)^{-1}\circ(D_{g(f(p))}\chi)\left(D_p(g\circ f)([\gamma])\right)&=(D_{g(f(p))}\chi)^{-1}\circ(D_{g(f(p))}\chi)\left([g \circ f \circ \gamma]\right)\newline
&=(D_{g(f(p))}\chi)^{-1}\circ\left([\chi \circ g \circ f \circ \gamma]\right)\newline
&=(D_{g(f(p))}\chi)^{-1}\circ\left(\left.\frac{d}{dt}\right|_{t=0}(\chi \circ \chi^{-1} \circ (\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t)\right)\newline
&=(D_{g(f(p))}\chi)^{-1}\circ\left(D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ D_{f(p)}\psi([f\circ \gamma])\right).
\end{aligned}
At this point, I don't know what to do. The result I want is
\begin{aligned}
D_{f(p)}g\circ D_pf([\gamma])&=D_{f(p)}g\circ [f \circ \gamma],
\end{aligned}
and it looks similar up to the charts I inserted.
How do I get to the end? Or have I gone in a circle and need to take another direction?
Best Answer
If you’re using the equivalence classes of curves approach this is almost obvious (the only perhaps non-obvious thing is showing that the derivative map $[\gamma]\mapsto [f\circ \gamma]$ is well-defined). There’s no need to go down to charts anymore. The value of both sides of the chain rule expression on $[\gamma]$ is obviously equal to $[g\circ f\circ \gamma]$.
In more detail, \begin{align} D_p(g\circ f)\,([\gamma])&:=[(g\circ f)\circ\gamma]\\ &=[g\circ(f\circ\gamma)]\\ &:=D_{f(p)}g\,([f\circ\gamma])\\ &:=D_{f(p)}g\,\left(D_pf\,([\gamma])\right)\\ &=\left(D_{f(p)}g\circ Df_p\right)([\gamma]), \end{align} and thus $D_p(g\circ f)=D_{f(p)}g\circ D_pf$.