# Chain rule for smooth functions between smooth manifolds

Following this thread, I'm trying to prove in detail the chain rule to unveil the subtle machinery. Could you have a check on my proof?

Let $$X \subseteq \mathbb R^M$$, $$Y \subseteq \mathbb R^N$$, and $$Z \subseteq \mathbb R^P$$ be $$m$$-, $$n$$-, and $$p$$-dimensional smooth manifolds respectively. Let $$f:X \to Y$$ and $$g:Y \to Z$$ be smooth. Fix $$x \in X$$. Let $$\varphi:U \to V$$, $$\psi:A \to B$$, and $$\eta:E \to F$$ be local parameterizations around $$x$$, $$f(x)$$, and $$g \circ f(x)$$ respectively.

• There exists $$A'$$ open in $$A$$ such that $$\psi^{-1} \circ f (x) \in A'$$ and $$t := \eta^{-1} \circ g \circ \psi_{\restriction A'}$$ is well-defined, i.e., the domains and images of functions in the composition are compatible.

• Similarly, there exists $$U'$$ open in $$U$$ such that $$\varphi^{-1} (x) \in U'$$ and $$h := \psi^{-1} \circ f \circ \varphi_{\restriction U'}$$ is well-defined.

• Also, there exists $$U''$$ open in $$U$$ such that $$\varphi^{-1} (x) \in U''$$ and $$k := \eta^{-1} \circ (g \circ f) \circ \varphi_{\restriction U''}$$ is well-defined.

• Composition of smooth maps is smooth, so $$t,h,k$$ are smooth. Wlog, we assume $$U' =U''$$ such that $$\operatorname{im} h \subseteq A'$$.

The derivative of $$f$$ at $$x$$ is $$\mathrm d f_x := \mathrm d \psi_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1}.$$

The derivative of $$g$$ at $$f(x)$$ is $$\mathrm d g_{f (x)} := \mathrm d \eta_{\eta^{-1} \circ g \circ f (x)} \circ \mathrm d t_{\psi^{-1} \circ f (x)} \circ (\mathrm d \psi_{\psi^{-1} \circ f (x)})^{-1}.$$

The derivative of $$g \circ f$$ at $$x$$ is $$\mathrm d (g \circ f)_{x} := \mathrm d \eta_{\eta^{-1} \circ g \circ f (x)} \circ \mathrm d k_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1}.$$

It follows that \begin{align} & \mathrm d g_{f (x)} \circ \mathrm d f_x \\ ={} & \left ( \mathrm d \eta_{\eta^{-1} \circ g \circ f (x)} \circ \mathrm d t_{\psi^{-1} \circ f (x)} \circ (\mathrm d \psi_{\psi^{-1} \circ f (x)})^{-1} \right ) \circ \left ( \mathrm d \psi_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1} \right ) \\ ={} & \mathrm d \eta_{\eta^{-1} \circ g \circ f (x)} \circ \mathrm d t_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1}. \end{align}

Notice that $$t$$ and $$h$$ are smooth functions whose domain is open in a Euclidean space, so we can apply the chain rule. We have \begin{align} & \mathrm d k_{\varphi^{-1} (x)} \\ ={} & \mathrm d \big (\eta^{-1} \circ (g \circ f) \circ \varphi_{\restriction U'} \big)_{\varphi^{-1} (x)} \\ ={} & \mathrm d \big (\eta^{-1} \circ g \circ \psi_{\restriction A'} \circ \psi^{-1} \circ f \circ \varphi_{\restriction U'} \big)_{\varphi^{-1} (x)} \quad \text{because} \quad \operatorname{im} h \subseteq A'\\ ={} & \mathrm d (t \circ h)_{\varphi^{-1} (x)} \\ ={} & \mathrm d t_{h \circ \varphi^{-1} (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \quad \text{by chain rule} \\ ={} & \mathrm d t_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)}. \end{align}

Hence $$\mathrm d g_{f (x)} \circ \mathrm d f_x = \mathrm d \eta_{\eta^{-1} \circ g \circ f (x)} \circ \mathrm d k_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1} = \mathrm d (g \circ f)_{x}.$$

This completes the proof.