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Definition: Let $M\subseteq \mathbb{R}^n$ be a smooth k-manifold and fix a point $p\in M$. Let $C_p$ denote the collection of $C^1$ maps of the form $\gamma:(-1,1)\to M$ such that $\gamma(0)=p$. Fix a regular embedding $\phi :U\to v$ such that $p\in V$ and let $\phi^{-1}:V\to U$ denote its inverse. Endow $C_p$ with and equivalence relation such that $\gamma_1 \sim \gamma_2$ if $(\phi^{-1}\circ \gamma_1)^\prime(0)=(\phi^{-1}\circ \gamma_2)^\prime(0)$.

Show that this definition doesn't depend on the choice of regular embedding.

So previous to this I had to show that for 2 regular embeddings $\phi_1,\phi_2$ that $\phi_2^{-1}\circ\phi_1$ is a diffeomorphism.

So let $\phi_1,\phi_2$ be 2 regular embeddings for $M$. Suppose $\gamma_1\sim\gamma_2$.

then $(\phi^{-1}\circ \gamma_1)^\prime(0)=(\phi^{-1}\circ \gamma_2)^\prime(0)$

then $$(\phi_2^{-1}\circ\gamma_1)^\prime(0)=(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$=(\phi_2^{-1}\circ\phi_1)^\prime\circ\phi_1^{-1}\circ\gamma_1)(\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$=(\phi_2^{-1}\circ\phi_1)^\prime\circ\phi_1^{-1}\circ\gamma_1)(\phi_1^{-1}\circ\gamma_2)^\prime(0)$$

Really not sure where to go from here.

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The chain rule states that

$$D(g\circ f)(x)=D(g)(f(x))\cdot D(f)(x)$$

Since $\gamma_1(0)=\gamma_2(0)=p$ your computation goes like

$$(\phi_2^{-1}\circ\gamma_1)^\prime(0) =(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$ =(\phi_2^{-1}\circ\phi_1)^\prime(\phi_1^{-1}(p))\cdot(\phi_1^{-1}\circ\gamma_1)^\prime(0)$$ $$=(\phi_2^{-1}\circ\phi_1)^\prime(\phi_1^{-1}(p))\cdot(\phi_1^{-1}\circ\gamma_2)^\prime(0)$$

$$ =(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_2)^\prime(0)=(\phi_2^{-1}\circ\gamma_2)^\prime(0)$$

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