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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.

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