I'm reading Theorem 1.1 in this lecture notes.
Theorem 1.1 (Chain Rule for Manifolds). Suppose $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are smooth maps of manifolds. Then:
$$
\mathrm{d}(g \circ f)_{x}=(\mathrm{d} g)_{f(x)} \circ(\mathrm{d} f)_{x}
$$
Proof. If $\varphi$ is a local parameterisation of $x, \psi$ is a local parameterisation of $y=f(x)$, and $\eta$ is a local parameterisation of $z=g(f(x))$, then this is evident from:
$$
g \circ f=\left[\eta \circ\left(\eta^{-1} \circ g \circ \psi\right) \circ \psi^{-1}\right] \circ\left[\psi \circ\left(\psi^{-1} \circ f \circ \varphi\right) \circ \varphi^{-1}\right]
$$
and the usual chain rule from multivariate analysis (as we can differentiate the RHS as before using the usual chain rule).
In the proof, $g \circ f = \Psi \circ \Phi$ with \begin{align}
\Psi &:= \eta \circ\left(\eta^{-1} \circ g \circ \psi\right) \circ \psi^{-1} \\
\Phi &:= \psi \circ\left(\psi^{-1} \circ f \circ \varphi\right) \circ \varphi^{-1}.
\end{align}
The composition of smooth maps is also smooth, so $\Psi, \Phi$ are smooth. However, $\operatorname{dom} (\Psi) = \operatorname{dom} (\psi^{-1})$ which is open in $Y$ but not necessarily open in its ambient Euclidean space. The same situation holds for $\Phi$. So $\Psi, \Phi$ are not differentiable in the usual sense, so we can not apply the chain rule on $\Psi \circ \Phi$. However, the author said
"…we can differentiate the RHS as before using the usual chain rule…".
Could you please elaborate on my confusion?
Best Answer
In definition 1.8 of the lecture notes, it is said that if $f\colon X \to Y$ is smooth and if $\varphi \colon U \to X$ and $\psi \colon V \to Y$ are local parametrisations in neighbourhoods of $x$ and $f(x)$, then
$$ df_x = d\psi_{\psi^{-1}(f(x))} \circ d(\psi^{-1}\circ f \circ \varphi)_{\varphi^{-1}(x)} \circ \big[(d\varphi_{\varphi^{-1}(x)})^{-1} \big] $$ where $d\varphi_{\varphi^{-1}(x)}$ is considered as an isomorphism from $\Bbb R^n \to T_xX$, that is, forgeting about the ambiant euclidean space $\Bbb R^N$ around $X$.
Same thing with $g$ at the point $f(x)$ gives $$ dg_{f(x)} = d\eta_{\eta^{-1}(g(f(x)))} \circ d(\eta^{-1}\circ g \circ \psi)_{\psi^{-1}(f(x))} \circ \big[(d\psi_{\psi^{-1}(f(x))})^{-1} \big] $$ where $d\psi_{\psi^{-1}(f(x))}$ is considered as an isomiorphism from $\Bbb R^{\tilde{n}} \to T_{f(x)}Y$, that is, forgeting about the ambiant euclidean space $\Bbb R^{\tilde{N}}$ around $Y$.
For $g\circ f$, it yields $$ d(g\circ f)_{g(f(x))} = d\eta_{\eta^{-1}(g(f(x)))} \circ d(\eta^{-1}\circ (g\circ f) \circ \varphi)_{\varphi^{-1}(x)} \circ \big[(d\varphi_{\varphi^{-1}(x)})^{-1} \big] $$
You are thus lead to show that $$ d(\eta^{-1}\circ (g\circ f) \circ \varphi)_{\varphi^{-1}(x)}= \big(d(\eta^{-1}\circ g \circ \psi)_{\psi^{-1}(f(x))}\big)\circ\big( d(\psi^{-1}\circ f \circ \varphi)_{\varphi^{-1}(x)}\big) $$
This is just the chain-rule applied to the composition of
$$ \psi^{-1}\circ f \circ \varphi \colon U \subset \Bbb R^n \to V \subset \Bbb R^{\tilde{n}} $$ and $$ \eta^{-1}\circ g\circ \psi \colon V \subset \Bbb R^{\tilde{n}} \to W \subset \Bbb R^{\tilde{\tilde{n}}} $$ which are smooth functions defined on and with range in open subsets of euclidean spaces.
By the way: there is an obvious typo in the author's sketch: the arrow of at bottom right corner should be $\eta^{-1}\circ g \circ \psi$, not $\eta^{-1}\circ f \circ \psi$.