# The derivative of a smooth function between smooth manifolds is independent of local parameterizations

Following this thread, I'm trying to prove that the definition of such a derivative is well-defined. Could you have a check on my justification?

Let $$X \subseteq \mathbb R^M$$ and $$Y \subseteq \mathbb R^N$$ be $$m$$- and $$n$$-dimensional smooth manifolds respectively. Let $$f:X \to Y$$ be smooth. Fix $$x \in X$$.

• Let $$\varphi:U \to V$$ and $$\psi:A \to B$$ be local parameterizations around $$x$$ and $$f(x)$$ respectively.

• Let $$\hat \varphi:\hat U \to \hat V$$ and $$\hat \psi:\hat A \to \hat B$$ be another local parameterizations around $$x$$ and $$f(x)$$ respectively.

• Wlog, we assume $$V = \hat V$$ and $$B = \hat B$$.

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

Clearly, $$\lambda := \varphi^{-1} \circ \hat \varphi : \hat U \to U$$ and $$\gamma := \psi^{-1} \circ \hat \psi : \hat A \to A$$ are diffeomorphisms. Also, $$\hat \varphi = \varphi \circ \lambda$$ and $$\hat \psi = \psi \circ \gamma$$. It follows that $$\hat h = \gamma^{-1} \circ h \circ \lambda.$$

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

Next we show $$\mathrm d f_x$$ is independent of the local parameterizations of $$x$$ and $$f(x)$$. Indeed, we have \begin{align} & \mathrm d \hat \psi_{\hat \psi^{-1} \circ f (x)} \circ \mathrm d \hat h_{\hat \varphi^{-1} (x)} \circ (\mathrm d \hat \varphi_{\hat \varphi^{-1} (x)})^{-1} \\ ={} & \underbrace{\mathrm d (\psi \circ \gamma)_{\gamma^{-1} \circ \psi^{-1} \circ f (x)}}_{A} \circ \underbrace{\mathrm d (\gamma^{-1} \circ h \circ \lambda)_{ \lambda^{-1} \circ \varphi^{-1} (x)}}_{B} \circ \underbrace{\left (\mathrm d (\varphi \circ \lambda)_{\lambda^{-1} \circ \varphi^{-1} (x)} \right )^{-1}}_{C}. \end{align}

Notice that $$\psi, \gamma, \gamma^{-1}, h, \lambda$$ are smooth maps whose domain is open in some Euclidean space, so we can apply the chain rule on their compositions. First, \begin{align} A &= \mathrm d \psi_{\psi^{-1} \circ f (x)} \circ \mathrm d \gamma_{\gamma^{-1} \circ \psi^{-1} \circ f (x)}. \end{align}

Notice that $$\gamma$$ is a diffeomorphism, so $$\mathrm d \gamma_{\gamma^{-1} (a)} \circ \mathrm d \gamma^{-1}_a = \operatorname{id}_{\mathbb R^m}$$ and thus $$\mathrm d \gamma^{-1}_a = \left ( \mathrm d \gamma_{\gamma^{-1} (a)} \right )^{-1}$$ Hence, \begin{align} B &= \mathrm d \gamma^{-1}_{h \circ \varphi^{-1} (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ \mathrm d \lambda_{ \lambda^{-1} \circ \varphi^{-1} (x)} \\ &= \mathrm d \gamma^{-1}_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ \mathrm d \lambda_{ \lambda^{-1} \circ \varphi^{-1} (x)} \\ &= \left (\mathrm d \gamma_{\gamma^{-1} \circ \psi^{-1} \circ f (x)} \right )^{-1} \circ \mathrm d h_{\varphi^{-1} (x)} \circ \mathrm d \lambda_{ \lambda^{-1} \circ \varphi^{-1} (x)}. \end{align}

Also, \begin{align} C &= \left ( \mathrm d \varphi_{\varphi^{-1} (x)} \circ \mathrm d \lambda_{\lambda^{-1} \circ \varphi^{-1} (x)} \right )^{-1} \\ &= \left ( \mathrm d \lambda_{\lambda^{-1} \circ \varphi^{-1} (x)} \right )^{-1} \circ \left ( \mathrm d \varphi_{\varphi^{-1} (x)} \right )^{-1}. \end{align}

Finally, we multiple $$A,B,C$$ together and see that $$\mathrm d \hat \psi_{\hat \psi^{-1} \circ f (x)} \circ \mathrm d \hat h_{\hat \varphi^{-1} (x)} \circ (\mathrm d \hat \varphi_{\hat \varphi^{-1} (x)})^{-1} = \mathrm d \psi_{\psi^{-1} \circ f (x)} \circ \mathrm d h_{\varphi^{-1} (x)} \circ (\mathrm d \varphi_{\varphi^{-1} (x)})^{-1}.$$

This completes the justification.